## LOG#035. Doppler effect and SR (I).

The Doppler effect is a very important phenomenon both in classical wave motion and relativistic physics. For instance, nowadays it is used to the detect exoplanets and it has lots of applications in Astrophysics and Cosmology.

Firstly, we remember the main definitions we are going to need here today.

$\omega =\dfrac{2\pi}{T}=2\pi\nu=2\pi f$

Sometimes, we will be using the symbol $f$ for the frequency $\nu$. We also have:

$\mathbf{n}=\dfrac{\mathbf{k}}{\vert \mathbf{k}\vert}$

$k=\vert \mathbf{k}\vert=\dfrac{2\pi}{\lambda}=\dfrac{\omega}{c}$

A plane (sometimes electromagnetic) wave is defined by the oscillation:

$A(\mathbf{r},t)=A_0\exp\left(iKX\right)$

where

$KX=\mathbb{K}\cdot\mathbb{X}=\mbox{PHASE}=\mathbf{k}\cdot \mathbf{r}-\omega t$

If $K^2=0$, then the wave number is lightlike (null or isotropic) and then $(k^0)^2=\dfrac{\omega^2}{c^2}$

Using the Lorentz transformations for the wave number spacetime vector, we get:

$k'^0=\gamma \left(k^0-\beta k^1\right)$

i.e., if the angle with the direction of motion is $\alpha$ so $k^1=\dfrac{\omega}{c}\cos\alpha$

$\dfrac{\omega '}{c}=\gamma \left(\dfrac{\omega}{c}-\beta\left(\dfrac{\omega}{c}\cos \alpha\right)\right)$

we deduce that

$\boxed{\omega=\omega_0\dfrac{\sqrt{1-\beta^2}}{1-\beta\cos\alpha}}$

or for the normalized frequency shift

$\boxed{D=\dfrac{\Delta \omega}{\omega_0}=\dfrac{\sqrt{1-\beta^2}}{1-\beta\cos\alpha}-1}$

This is the usual formula for the relativistic Doppler effect when we define $\omega'=\omega_0$, and thus, the angular frequency (also the frecuency itself, since there is only a factor 2 times the number pi of difference) changes with the motion of the source. When the velocity is “low”, i.e., $\beta<<1$, we obtain the classical Doppler shift formula:

$\omega \approx (1+\beta\cos\alpha)\omega_0$

We then calculate the normalized frequency shift from $\Delta \omega=\omega-\omega_0$

$\boxed{D=\dfrac{\Delta \omega}{\omega_0}=\dfrac{V\cos\alpha}{c}}$

The classical Doppler shift states that when the source approaches the receiver ($\cos\alpha>0$), then the frequency increases, and when the source moves away from the receiver ($\cos\alpha<0$), the frequency decreases. Interestingly, in the relativistic case, we also get a transversal Doppler shift which is absent in classical physics. That is, in the relativistic Doppler shift, for $\alpha=\pi/2$, we obtain

$\boxed{\omega=\dfrac{\omega_0}{\gamma_V}=\omega_0\sqrt{1-\beta^2}}$

and the difference in frequency would become

$\boxed{D=\dfrac{\Delta \omega}{\omega_0}=\sqrt{1-\beta^2}-1}$

There is an alternative deduction of these formulae. The time that an electromagnetic wave uses to run a distance equal to the wavelength, in a certain inertial frame S’ moving with relative speed V to another inertial frame S at rest, is equal to:

$t=\dfrac{\lambda}{c-V}=\dfrac{c}{(c-V)f_s}=\dfrac{1}{(1-\beta_V)f_s}$

where $f_s$ is the frequency of the source. Due to the time dilation of special relativity

$t=t_0\gamma$ and thus

$f_0=\dfrac{1}{t_0}=\gamma(1-\beta_V)f_s=\sqrt{\dfrac{1-\beta_V}{1+\beta_V}}f_s$

so we get

$\boxed{\dfrac{f_s}{f_o}=\dfrac{f_{source}}{f_{obs}}=\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}}$

The redshift (or Doppler displacement) is generally defined as:

$\boxed{z=-D=\dfrac{\lambda_o-\lambda_s}{\lambda_s}=\dfrac{f_s-f_o}{f_o}=\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}-1}$

When $\beta\rightarrow 0$, i.e., $V\rightarrow 0,V<, we get the classical result

$z\approx\beta=\dfrac{v}{c}$

The generalization for a general non-parallel motion of the source/observer is given by

$\boxed{f_0=\dfrac{f_s}{\gamma\left(1+\dfrac{v_s\cos\theta_o}{c}\right)}}$

If we use the stellar aberration formula:

$\boxed{\cos\theta_o=\dfrac{\cos\theta_s-\dfrac{v_s}{c}}{1-\dfrac{v_s}{c}\cos\theta_s}}$

the last equation can be recasted in terms of $\theta_s$ instead of $\theta_o$ as follows:

$\boxed{f_0=\gamma\left(1-\dfrac{v_s}{c}\cos\theta_s\right)}$

and then

$\boxed{z=\dfrac{f_s-f_0}{f_0}=-D=\dfrac{\Delta f}{f_0}=\dfrac{1}{\gamma\left(1-\dfrac{v_s\cos\theta_s}{c}\right)}-1}$

Again, for a transversal motion, $\theta_o=\pi/2$, we get a transversal Doppler effect:

$\boxed{f_0=\dfrac{f_s}{\gamma}=\sqrt{1-\beta^2}f_s} \leftrightarrow \boxed{z_T=-D=\dfrac{f_o-f_s}{f_s}=\sqrt{1-\beta_s^2}-1}$

Remark: Remember that the Doppler shift formulae are only valid if the relative motion (of both source and observer/receiver) is slower than the speed of the (electromagnetic) wave, i.e., if $v\leq c$.

In the final part of this entry, we are going to derive the most general formula for Doppler effect, given an arbitrary motion of source and observer, in both classical and relativistic Physics. Recall that the Doppler shift in Classical Physics for an arbitrary observer is given by a nice equation:

$\boxed{f'=f\left[\dfrac{v-v_o\cos\theta_o}{v-v_s\cos\theta_s}\right]}$

Here, $v$ is the velocity of the (electromagnetic) wave in certain medium, $v_o$ is the velocity of the observer in certain direction forming an angle $\theta_o$ with the “line of sight”, while $v_s$ is the velocity of the source forming an angle $\theta_s$ with the line of sight/observation. If we write

$V_o=-v_o\cos\theta_o$ $V_s=-v_s\cos\theta_s$ $V_{so}=V_s-V_o$

we can rewrite this last Doppler formula in the following way ( for $v=c$):

$f'=\left(\dfrac{c+V_o}{c+V_s}\right)f$ or with $f'=f_o$ and $f=f_s$

$\boxed{f_o=f_s\left(\dfrac{c+V_s}{c+V_o}\right)=\left(1+\dfrac{V_s-V_o}{c+V_o}\right)f_s=\left(1+\dfrac{V_{so}}{c+V_o}\right)f_s}$

and

$\boxed{z=-D=\dfrac{f_0-f_s}{f_0}=\left(\dfrac{c+V_0}{c+V_s}\right)-1=\dfrac{V_o-V_s}{c+V_s}=\dfrac{-V_{so}}{c+V_s}}$

The most general Doppler shift formula, in the relativistic case, reads:

$\boxed{\dfrac{f_o}{f_s}=\dfrac{1-\dfrac{\vert\vert \mathbf{v}_o\vert\vert}{\vert\vert \mathbf{c}\vert\vert}\cos\theta_{co}}{1-\dfrac{\vert\vert \mathbf{v}_s\vert\vert}{\vert\vert \mathbf{c}\vert\vert}\cos\theta_{cs}}\sqrt{\dfrac{1-\dfrac{v_s^2}{c^2}}{1-\dfrac{v_o^2}{c^2}}}}$

or equivalently

$\boxed{\dfrac{f_o}{f_s}=\dfrac{1-\vert\vert \beta_o \vert\vert \cos\theta_{co}}{1-\vert\vert \beta_s\vert\vert\cos\theta_{cs}}\sqrt{\dfrac{1-\beta_s^2}{1-\beta_o^2}}}$

and where $\mathbf{v}_s,\mathbf{v}_o$ are the velocities of the source and the observer at the time of emission and reception, respectively, $\beta_s,\beta_o$ are the corresponding beta boost parameters, $\mathbf{c}$ is the “light” or “wave” velocity vector and we have defined the angles $\theta_{cs},\theta_{co}$ to be the angles formed at the time of emission and the time of reception/observation between the source velocity and the “wave” velocity, respectively, between the source and the wave and the observer and the wave. Two simple cases of this formula:

1st. Parallel motion with $\mathbf{c}\parallel\mathbf{v}_s\rightarrow \theta_{cs}=0\textdegree\rightarrow \cos\theta_{cs}=1\rightarrow f_o>f_s$.

2nd. Antiparallel motion with $\mathbf{c}$ going in the contrary sense than that of $\mathbf{v}_s$. Then, $\theta_{cs}=180\textdegree\rightarrow \cos\theta_{cs}=-1\rightarrow f_o.

The deduction of this general Doppler shift formula can be sketched in a simple fashion. For a signal using some propagating wave, we deduce:

$\vert \mathbf{r}_o-\mathbf{r}_s\vert ^2=\vert\mathbf{C}\vert^2\left(t_o-t_s\right)^2$

with

$\mathbf{C}=\dfrac{\mathbf{r}_o-\mathbf{r}_s}{t_o-t_s}$

Differentiating with respect to $t_o$ carefully, it provides

$\mathbf{C}\cdot\left[\mathbf{v}_o-\mathbf{v}_s\dfrac{dt_s}{dt_o}\right]=\vert \mathbf{C}\vert^2\left(1-\dfrac{dt_o}{dt_s}\right)$

Solving for $\dfrac{dt_o}{dt_s}$ we get

$\dfrac{dt_o}{dt_s}=\dfrac{\vert \mathbf{C}\vert^2-\mathbf{C}\cdot\mathbf{v}_o }{\vert \mathbf{C}\vert^2-\mathbf{C}\cdot\mathbf{v}_s}=\dfrac{\mathbf{C}\cdot\left(\mathbf{C}-\mathbf{v}_o\right)}{\mathbf{C}\cdot\left(\mathbf{C}-\mathbf{v}_s\right)}=\dfrac{1-\dfrac{\mathbf{C}\cdot\mathbf{v}_o}{\vert\mathbf{C}\vert^2}}{1-\dfrac{\mathbf{C}\cdot\mathbf{v}_s}{\vert\mathbf{C}\vert^2}}$

Using the known formula $\vert \mathbf{r}\cdot\mathbf{s}\vert=\vert\mathbf{r}\vert\vert\mathbf{s}\vert\cos\theta_{rs}$, we obtain in a simple way:

$\dfrac{dt_o}{dt_s}=\dfrac{1-\dfrac{\vert\mathbf{v}_o\vert}{\vert\mathbf{C}\vert}\cos\theta_{\mathbf{c},\mathbf{v}_o}}{1-\dfrac{\vert\mathbf{v}_s\vert}{\vert\mathbf{C}\vert}\cos\theta_{\mathbf{c},\mathbf{v}_s}}=\dfrac{\vert\mathbf{C}-\mathbf{v}_o\vert\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_o}}{\vert\mathbf{C}-\mathbf{v}_s\vert\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_s}}$

Finally, using the fact that $\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_s}=\cos\theta_{\mathbf{C},\mathbf{v}_s}$, the similar result $\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_o}=\cos\theta_{\mathbf{C},\mathbf{v}_o}$, and that the proper time induces an extra gamma factor due to time dilation

$dt=d\tau\gamma$

and we calculate for the frequency:

$\dfrac{n_o}{n_s}=\dfrac{\nu_o}{\nu_s}=\dfrac{f_o}{f_s}=(\mbox{PREFACTOR})\dfrac{\gamma_s dt_s}{\gamma_odt_o}$

where the PREFACTOR denotes the previously calculated ratio between differential times. Finally, elementary algebra let us derive the expression:

$\dfrac{f_o}{f_s}=\dfrac{d\tau_s}{d\tau_o}=\dfrac{1-\vert\vert \beta_o \vert\vert \cos\theta_{co}}{1-\vert\vert \beta_s\vert\vert\cos\theta_{cs}}\sqrt{\dfrac{1-\beta_s^2}{1-\beta_o^2}}$

Q.E.D.

An interesting remark about the Doppler effect in relativity: the Doppler effect allows us to “derive” the Planck’s relation for quanta of light. Suppose that a photon in the S’-frame has an energy $E'$ and momentum $(p'_x,0,0)=(-E',0,0)$ is being emitted along the negative x’-axis toward the origin of the S-frame. The inverse Lorentz transformation provides:

$E=\gamma (E'+vp'_x)=\gamma (E'-\beta E')=\dfrac{1-\beta}{\sqrt{1-\beta^2}}E'$, i.e.,

$E=\sqrt{\dfrac{1-\beta}{1+\beta}}E'$

By the other hand, by the relativistic Doppler effect we have seen that the frequency f’ in the S’-frame is transformed into the frequency f in the S-frame if we use the following equation:

$f=f'\sqrt{\dfrac{1-\beta}{1+\beta}}$

If we divide the last two equations we get:

$\dfrac{E}{f}=\dfrac{E'}{f'}=constant \equiv h$

Then, if we write $h=6.63\cdot 10^{-34}J\cdot s$, and $E=hf$ and $E'=hf'$.

AN ALTERNATIVE HEURISTIC DEDUCTION OF THE RELATIVISTIC DOPPLER EFFECT

In certain rest frame S, there is an observer receiving light beams/signals. The moving frame is the S’-frame and it is the emitter of light. The source of light approaches at velocity V, and it sends pulses with frequency $f_0=\nu_0$. What is the frequency that the observer at S observes? Due to time dilation, the observer at S observes a longer period

$T=T_0\dfrac{1}{\sqrt{1-\beta_V^2}}$

with

$\beta_V=\dfrac{V}{c}$

The distance between two consecutive light beams seen by the observer at S will be:

$\lambda=cT-VT=(c-V)T=(c-V)\dfrac{T_0}{\sqrt{1-\beta_V^2}}$

Therefore, the observer frequency in the S-frame is:

$f=\nu=\dfrac{c}{\lambda}=\dfrac{c\sqrt{1-\beta_V^2}}{T_0(c-V)}=\nu_0\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}$

i.e.

$\boxed{f=f_0\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}}$

If the source approaches the observer, then $f>f_0$. If the source moves away from the observer, then $f. In the case, the velocity of the source forms certain angle $\alpha$ in the direction of observation, the same argument produces:

$f=\nu=\dfrac{c}{\lambda}=\dfrac{c\sqrt{1-\beta_V^2}}{T_0(c-V\cos\alpha)}=f_0\dfrac{\sqrt{1-\beta_V}}{1-\beta\cos\alpha}$

that is

$\boxed{f=f_0\dfrac{\sqrt{1-\beta_V}}{1-\beta\cos\alpha}}$

In the case of transversal Doppler effect, we get $\alpha=90\textdegree=\pi/2$, and so:

$\boxed{f=f_0\sqrt{1-\beta^2}}$

Q.E.D.

Final remark (I): If $\theta=\alpha=\dfrac{\pi}{2}$ or $\theta=\alpha=\dfrac{3\pi}{2}$ AND $v<, therefore we have that there is no Doppler effect at all.

Final remark (II): If $v\approx c$, there is no Doppler effect in certain observation directions. Those directions can be deduced from the above relativistic Doppler effect formula with the condition $f=f_0$ and solving for $\theta$. This gives the next angular direction in which Doppler effect can not be detected

$\boxed{\cos\theta=\dfrac{\sqrt{1-\beta^2}-1}{\beta}}$

## LOG#034. Stellar aberration.

In this entry, we are going to study a relativistic effect known as “stellar aberration”.

From the known Lorentz transformations of velocities (inverse case), we get:

$v_x=\dfrac{v'_x+V}{1+\dfrac{v'_xV}{c^2}}$

$v_y=\dfrac{v'_y\sqrt{1-\beta^2}}{1+\dfrac{v'xV}{c^2}}$

$v_z=\dfrac{v'_z\sqrt{1-\beta^2}}{1+\dfrac{v'zV}{c^2}}$

The classical result (galilean addition of velocities) is recovered in the limit of low velocities $V\approx 0$ or sending the light speed get the value “infinite” $c\rightarrow \infty$. Then,

$v_x=v'_x+V$ $v_y=v'_y$ $v'_z=v_z$

Let us define

$\theta =\mbox{angle formed by x}\; \mbox{and}\; v_x$

$\theta' =\mbox{angle formed by x'}\; \mbox{and}\; v'_x$

Thus, we get the component decomposition into the xy and x’y’ planes:

$v_x=v\cos\theta$ $v_y=v\sin\theta$

$v'_x=v'\cos\theta'$ $v'_y=v'\sin\theta'$

From this equations, we get

$\tan \theta=\dfrac{v'\sin\theta'\sqrt{1-\beta^2}}{v'\cos\theta'+V}$

If $v=v'=c$

$\boxed{\tan \theta=\dfrac{\sin\theta'\sqrt{1-\beta_V^2}}{\cos\theta'+\beta_V}}$

and then

$\boxed{\cos\theta=\dfrac{\beta_V+\cos\theta'}{1+\beta_V\cos\theta'}}$

$\boxed{\sin\theta=\dfrac{\sin\theta'\sqrt{1-\beta_V^2}}{1+\beta_V\cos\theta'}}$

From the last equation, we get

$\sin\theta'\sqrt{1-\beta_V^2}=\sin\theta\left(1+\beta_V\cos\theta'\right)$

From this equation, if $V<, i.e., if $\beta_V<<1$ and $\theta'=\theta+\Delta\theta$ with $\Delta \theta<<1$, we obtain the result

$\Delta \theta=\theta'-\theta=\beta_V\sin\theta$

By these formaulae, the angle of a light beam propagating in space depends on the velocity of the source respect to the observer. We can observe this relativistic effect every night (supposing a good approximation that Earth’s velocity is non-relativistic, as it shows). The physical interpretation of the above aberration formulae (for the stars we watch during a skynight) is as follows: due to the Earth’s motion, a star in the zenith is seen under an angle $\theta\neq \dfrac{\pi}{2}$.

Other important consequence from the stellar aberration is when we track ultra-relativistic particles ($\beta\approx 1$). Then, $\theta'\rightarrow \pi$ and then, the observer moves close to the source of light. In this case, almost every star (excepting those behind with $\theta=\pi$) are seen “in front of” the observer. If the source moves with almost the speed of light, then the light is “observed” as it were concentrated in a little cone with an aperture $\Delta\theta\sim\sqrt{1-\beta_V^2}$

## LOG#033. Electromagnetism in SR.

The Maxwell’s equations and the electromagnetism phenomena are one of the highest achievements and discoveries of the human kind. Thanks to it, we had radio waves, microwaves, electricity, the telephone, the telegraph, TV, electronics, computers, cell-phones, and internet. Electromagnetic waves are everywhere and everytime (as far as we know, with the permission of the dark matter and dark energy problems of Cosmology). Would you survive without electricity today?

The language used in the formulation of Maxwell equations has changed a lot since Maxwell treatise on Electromagnetis, in which he used the quaternions. You can see the evolution of the Mawell equations “portrait” with the above picture. Today, from the mid 20th centure, we can write Maxwell equations into a two single equations. However, it is less know that Maxwell equations can be written as a single equation $\nabla F=J$ using geometric algebra in Clifford spaces, with $\nabla =\nabla \cdot +\nabla\wedge$, or the so-called Kähler-Dirac-Clifford formalism in an analogue way.

Before entering into the details of electromagnetic fields, let me give some easy notions of tensor calculus. If $x^2=\mbox{invariant}$, how does $x^\mu$ transform under Lorentz transformations? Let me start with the tensor components in this way:

$x^\mu e_\mu=x^{\mu'}e_{\mu'}=\Lambda^{\mu'}_{\;\; \nu}x^\mu e_{\mu'}=\Lambda^{\mu'}_{\;\; \mu}x^\mu e_{\mu'}$

Then:

$e_\mu=\Lambda^{\mu'}_{\;\; \mu} e_{\mu'}\rightarrow e_{\mu'}=\left(\Lambda^{-1}\right)_{\;\; \mu'}^{\mu}e_\mu=\left[\left(\Lambda^{-1}\right)^T\right]^{\;\; \mu}_{\nu}e_\mu$

Note, we have used with caution:

1st. Einstein’s convention: sum over repeated subindices and superindices is understood, unless it is stated some exception.

2nd. Free indices can be labelled to the taste of the user segment.

3rd. Careful matrix type manipulations.

We define a contravariant vector (or tensor (1,0) ) as some object transforming in the next way:

$\boxed{a^{\mu'}=\Lambda^{\mu'}_{\;\; \nu}a^\nu}\leftrightarrow\boxed{a^{\mu'}=\left(\dfrac{\partial x^{\mu'}}{\partial x^\nu}\right)a^\nu}$

where $\left(\dfrac{\partial x^{\mu'}}{\partial x^\nu}\right)$ denotes the Jabobian matrix of the transformation.
In similar way, we can define a covariant vector ( or tensor (0,1) ) with the aid of the following equations

$\boxed{a_{\mu'}=\left[\left(\Lambda^{-1}\right)^{T}\right]_{\mu'}^{\:\;\; \nu}a_\nu}\leftrightarrow\boxed{a_{\mu'}=\left(\dfrac{\partial x^{\nu}}{\partial x^{\mu'}}\right)a_\nu}$

Note: $\left(\dfrac{\partial x^{\nu}}{\partial x^{\mu'}}\right)=\left(\dfrac{\partial x^{\mu'}}{\partial x^{\nu}}\right)^{-1}$

Contravariant tensors of second order ( tensors type (2,0)) are defined with the next equations:

$\boxed{b^{\mu'\nu'}=\Lambda^{\mu'}_{\;\; \lambda}\Lambda^{\nu'}_{\;\; \sigma}b^{\lambda\sigma}=\Lambda^{\mu'}_{\;\; \lambda}b^{\lambda\sigma}\Lambda^{T \;\; \nu'}_{\sigma}\leftrightarrow b^{\mu'\nu'}=\dfrac{\partial x^{\mu'}}{\partial x^\lambda}\dfrac{\partial x^{\nu'}}{\partial x^\sigma}b^{\lambda\sigma}}$

Covariant tensors of second order ( tensors type (0,2)) are defined similarly:

$\boxed{c_{\mu'\nu'}=\left(\left(\Lambda\right)^{-1}\right)^{T \;\;\lambda}_{\mu'}\left(\left(\Lambda\right)^{-1T}\right)^{\;\; \sigma}_{\nu'}c_{\lambda\sigma}=\left(\Lambda^{-1T}\right)^{\;\; \lambda}_{\mu'}c_{\lambda\sigma}\Lambda^{-1 \;\; \nu'}_{\sigma}\leftrightarrow c_{\mu'\nu'}=\dfrac{\partial x^{\lambda}}{\partial x^{\mu'}}\dfrac{\partial x^{\sigma}}{\partial x^{\nu'}}c_{\lambda\sigma}}$

Mixed tensors of second order (tensors type (1,1)) can be also made:

$\boxed{d^{\mu'}_{\;\; \nu'}=\Lambda^{\mu'}_{\;\; \lambda}\left(\left(\Lambda\right)^{-1T}\right)^{\;\;\;\; \sigma}_{\nu'}d^{\lambda}_{\;\;\sigma}=\Lambda^{\mu'}_{\;\; \lambda}d^{\lambda}_{\;\;\sigma}\left(\left(\Lambda\right)^{-1}\right)^{\sigma}_{\;\;\; \nu'}\leftrightarrow d^{\mu'}_{\;\; \nu'}=\dfrac{\partial x^{\mu'}}{\partial x^{\lambda}}\dfrac{\partial x^{\sigma}}{\partial x^{\nu'}}d^{\lambda}_{\;\; \sigma}}$

We can summarize these transformations rules in matrix notation making the transcript from the index notation easily:

1st. Contravariant vectors change of coordinates rule: $X'=\Lambda X$

2nd. Covariant vectors change of coordinates rule: $X'=\Lambda^{-1T} X$

3rd. (2,0)-tensors change of coordinates rule: $B'=\Lambda B \Lambda^T$

4rd. (0,2)-tensors change of coordinates rule: $C'=\Lambda^{-1T}C\Lambda^{-1}$

5th. (1,1)-tensors change of coordinates rule: $D'=\Lambda D \Lambda^{-1}$

Indeed, without taking care with subindices and superindices, and the issue of the inverse and transpose for transformation matrices, a general tensor type (r,s) is defined as follows:

$\boxed{T^{\mu'_1\mu'_2\ldots \mu'_r}_{\nu'_1\nu'_2\ldots \nu'_s}=L^{\nu_s}_{\nu'_s}\cdots L^{\nu_1}_{\nu'_1}L^{\mu'_r}_{\mu_r}\cdots L^{\mu'_1}_{\mu_1}T^{\mu_1\mu_2\ldots\mu_r}_{\nu_1\nu_2\ldots \nu_s}}$

We return to electromagnetism! The easiest examples of electromagnetic wave motion are plane waves:

$x=x_0\exp (iKX)=x_0\exp (ix^\mu p_\mu)$

where $\phi=XK=KX=X\cdot K=x^\mu p_\mu=\mathbf{k}\cdot\mathbf{r}-\omega t$

Indeed, the cuadrivector K can be “guessed” from the phase invariant ($\phi=\phi'$ since the phase is a dot product):

$K=\square \phi$

where $\square$ is the four dimensional nabla vector defined by

$\square=\left(\dfrac{\partial}{c\partial t},\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z}\right)$

and so

$K^\mu=(K^0,K^1,K^2,K^3)=(\omega/c,k_x,k_y,k_z)$

Now, let me discuss different notions of velocity when we are considering electromagnetic fields, beyond the usual notions of particle velocity and observer relative motion, we have the following notions of velocity in relativistic electromagnetism:

1st. The light speed c. It is the ultimate limit in vacuum and SR to the propagation of electromagnetic signals. Therefore, it is sometimes called energy transfer velocity in vacuum or vacuum speed of light.

2nd. Phase velocity $v_{ph}$. It is defined as the velocity of the modulated signal in a plane wave, if $\omega =\omega (k)=\sqrt{c^2\mathbf{k}-K^2}$, we have

$v_{ph}=\dfrac{\omega (\mathbf{k})}{k}$ where k is the modulus of $\mathbf{k}$. It measures how much fast the phase changes with the wavelength vector.

From the definition of cuadrivector wave length, we deduce:

$K^2=\omega^2\left(\dfrac{1}{v_p}-\dfrac{1}{c^2}\right)$

Then, we can rewrite the distinguish three cases according to the sign of the invariant $K^2$:

a) $K^2>0$. The separation is spacelike and we get $v_p.

b)$K^2=0$. The separation is lightlike or isotropic. We obtain $v_p=c$.

c)$K^2<0$. The separation is timelike. We deduce that $v_p>c$. This situation is not contradictory with special relativity since phase oscillations can not transport information.

3rd. Group velocity $v_g$. It is defined like the velocity that a “wave packet” or “pulse” has in its propagation. Therefore,

$v_g=\dfrac{dE}{dp}=\dfrac{d\omega}{dk}$

where we used the Planck relationships for photons $E=\hbar \omega$ and $p=\hbar k$, with $\hbar=\dfrac{h}{2\pi}$

4th. Particle velocity. It is defined in SR by the cuadrivector $U=\gamma (c,\mathbf{v})$

5th. Observer relative velocity, V. It is the velocity (constant) at which two inertial observes move.

There is a nice relationship between the group velocity, the phase velocity and the energy transfer, the lightspeed in vacuum. To see it, look at the invariant:

$K^2=\mathbf{k}^2-\omega^2/c^2$

Deriving this expression, we get $v_g=d\omega/dk=kc^2/\omega=c^2/v_{ph}$

so we have the very important equation

$\boxed{v_gv_{ph}=c^2}$

Other important concept in electromagnetism is “light intensity”. Light intensity can be thought like the “flux of light”, and you can imagine it both in the wave or particle (photon corpuscles) theory in a similar fashion. Mathematically speaking:

$\mbox{Light intensity=Flux of light}=\dfrac{\mbox{POWER}}{\mbox{Area}}\rightarrow I=\dfrac{\mathcal{P}}{A}=\dfrac{E/V}{tA/V}=\dfrac{uV}{tA}=uc$

so $I=uc$ where u is the energy density of the electromagnetic field and c is the light speed in vacuum. By Lorentz transformations, it can be showed that for electromagnetic waves, energy, wavelength, energy density and intensity change in the following way:

$E'=\sqrt{\dfrac{1+\beta}{1-\beta}}E$

$\lambda'=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda$

$u'= \dfrac{E'}{\lambda' NA}=\dfrac{1-\beta}{1+\beta}\dfrac{E}{N\lambda A}$

$I'=\dfrac{1-\beta}{1+\beta}I$

The relativistic momentum can be related to the wavelength cuadrivector using the Planck relation $P^\mu=\hbar K^\mu$. Under a Lorentz transformation, momenergy transforms $P'=\Lambda P$. Assign to the wave number vector $\mathbf{k}$ a direction in the S-frame:

$\vert \mathbf{k}\vert \left( \cos \theta, \sin \theta, 0 \right)=\dfrac{\omega}{c}\left(\cos\theta,\sin\theta,0\right)$

and then

$K^\mu=\dfrac{\omega}{c}\left(1,\cos\theta,\sin\theta,0\right)$

In matrix notation, the whole change is written as:

$\begin{pmatrix}\dfrac{\omega'}{c}\\ k'_x\\ k'_y\\k'_z\end{pmatrix}=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}\dfrac{\omega}{c}\begin{pmatrix}1\\ \cos\theta\\ \sin\theta\\ 0\end{pmatrix}$

so

$K'\begin{cases}\omega'=\gamma \omega(1-\beta\cos\theta)\\ \;\\ k'_x=\gamma\dfrac{\omega}{c}\\ \;\\ k'_y=\dfrac{\omega}{c}\sin\theta\\\;\\ k'_z=0\end{cases}$

Using the first two equations, we get:

$k'_x=\dfrac{\omega'}{c}\dfrac{\cos\theta-\beta}{1-\beta\cos\theta}$

Using the first and the third equation, we obtain:

$k'_y=\dfrac{\omega'}{c}\dfrac{\sin\theta}{\gamma\left(1-\beta\cos\theta\right)}$

Dividing the last two equations, we deduce:

$\dfrac{k'_x}{k'_y}=\dfrac{\sin\theta}{\gamma\left(\cos\theta-\beta\right)}=\dfrac{u'_y}{u'_x}=\tan\theta'$

This formula is the so-called stellar aberration formula, and we will dedicate it a post in the future.

If we write the first equation with the aid of frequency f (and $f_0$) instead of angular frequency,

$f=f_0\dfrac{1}{\gamma(1-\beta\cos\theta)}$

where we wrote the frequency of the source as $\omega'=2\pi f_0$ and the frequency of the receiver as $\omega=2\pi \nu$. This last formula is called the relativistic Doppler shift.

Now, we are going to introduce a very important object in electromagnetism: the electric charge and the electric current. We are going to make an analogy with the momenergy $\mathbb{P}=m\gamma\left(c,\mathbf{v}\right)$. The cuadrivector electric current is something very similar:

$\mathbb{J}=\rho_0\gamma\left(c,\mathbf{u}\right)=\rho\left(c,\mathbf{u}\right)=\left(\rho c,\mathbf{j}\right)$

where $\rho=\gamma \rho_0$ is the electric current density, and $\mathbf{u}$ is the charge velocity. Moreover, $\rho_0=nq$ and where $q$ is the electric charge and $n=N/V$ is the electric charge density number, i.e., the number of “elementary” charges in certain volume. Indeed, we can identify the components of such a cuadrivector:

$\mathbb{J}=\left(J^0,J^1,J^2,J^3\right)=\rho_0\left(c\gamma,\gamma\mathbf{v}\right)=\rho_0\gamma\left(c,\mathbf{u}\right)$. We can make some interesting observations. Suppose certain rest frame S where we have $\rho=\rho_++\rho_-=0$, i.e., a frame with equilibred charges $\rho_+=-\rho_-$, and suppose we move with the relative velocity of the electron (or negative charge) observer. Then $u=v(e)$ and $j_x=\rho_-v$, while the other components are $j_y=j_z=0$. Then, the charge density current transforms as follows:

$\begin{pmatrix}\rho'c\\ j'_x\\ j'_y\\j'_z\end{pmatrix}=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}\begin{pmatrix}0\\ \rho_- v\\ 0\\ 0\end{pmatrix}=\begin{pmatrix}-\gamma \beta \rho_- v\\ \gamma \rho_- v\\ 0\\ 0\end{pmatrix}$

and

$\rho'=-\gamma \beta^2\rho_-=\gamma\beta^2\rho_+$

$j'_x=\gamma\rho_- v=-\gamma \rho_+ v$

We conclude:

1st. Length contraction implies that the charge density increases by a gamma factor, i.e., $\rho_+\rightarrow \rho_+\gamma$.

2nd. The crystal lattice “hole” velocity $-v$ in the primed frame implies the existence in that frame of a current density $j'_x=-\gamma \rho_+ v$.

3rd. The existence of charges in motion when seen from an inertial frame (boosted from a rest reference S) implies that in a moving reference frame electric fields are not alone but with magnetic fields. From this perspective, magnetic fields are associated to the existence of moving charges. That is, electric fields and magnetic fields are intimately connected and they are caused by static and moving charges, as we do know from classical non-relativistic physics.

Remember now the general expression of the FORPOWER tetravector, or Power-Force tetravector, in SR:

$\mathcal{F}=\mathcal{F}^\mu e_\mu=\gamma\left(\dfrac{\mathbf{f}\cdot\mathbf{v}}{c},f_x,f_y,f_z\right)$

and using the metric, with the mainly plus convention, we get the covariant componets for the power-force tetravector:

$\mathcal{F}_\mu=\gamma\left(-\dfrac{\mathbf{f}\cdot\mathbf{v}}{c},f_x,f_y,f_z\right)$

We define the Lorentz force as the sum of the electric and magnetic forces

$\mathbf{f}_L=\mathbf{f}_e+\mathbf{f}_m=q\mathbf{E}+\mathbf{v}\times \mathbf{B}$

Noting that $(\mathbf{v}\times\mathbf{B})\cdot \mathbf{v}=0$, the Power-Force tetravector for the Lorentz electromagnetic force reads:

$\mathcal{F}_L=\mathcal{F}^\mu e_\mu=\gamma q\left(\dfrac{\mathbf{E}\cdot{\mathbf{v}}}{c},\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)$

And now, we realize that we can understand the electromagnetic force in terms of a tensor (1,1), i.e., a matrix, if we write:

$\mathcal{F}=\dfrac{q}{c}\mathbb{F}\mathbb{U}$

so

$\begin{pmatrix}\mathcal{F}^0\\ \mathcal{F}^1\\ \mathcal{F}^2\\ \mathcal{F}^3\end{pmatrix}=\dfrac{q}{c}\begin{pmatrix}0 &E_x & E_y & E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y & -cB_z & 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}\begin{pmatrix}\gamma c\\ \gamma v_x\\ \gamma v_y\\ \gamma v_z\end{pmatrix}$

Therefore, $\mathcal{F}^\mu=\dfrac{q}{c}F^\mu_{\;\; \nu}U^\nu\leftrightarrow \mathcal{F}=\dfrac{q}{c}\mathbb{F}\mathbb{U}$

where the components of the (1,1) tensor can be read:

$\mathbb{F}=\mathbf{F}^\mu _{\;\; \nu}=\begin{pmatrix}0& E_x& E_y& E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y& -cB_z& 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}$

We can lower the indices with the metric $\eta=diag(-1,1,1,1)$ in order to have a more “natural” equation and to read the symmetry of the electromagnetic tensor $F_{\mu\nu}$ (note that we can not study symmetries with indices covariant and contravariant),

$\mathbf{F}_{\mu\nu}=\eta_{\mu \alpha}\mathbf{F}^{\alpha}_{\;\; \nu}$

with

$\mathbf{F}_{\mu\nu}=\begin{pmatrix}0& -E_x& -E_y& -E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y& -cB_z& 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}$

Similarly

$\mathbf{F}^{\mu\nu}=\mathbf{F}^{\alpha}_{\;\; \beta}\eta^{\beta \nu}=\begin{pmatrix}0& E_x& E_y& E_z\\ -E_x & 0 & cB_z& -cB_y\\ -E_y& -cB_z& 0 & cB_x\\ -E_z & cB_y& -cB_x& 0\end{pmatrix}$

Please, note that $F_{\mu\nu}=-F_{\nu\mu}$. Focusing on the components of the electromagnetic tensor as a tensor type (1,1), we have seen that under Lorentz transformations its components change as $F'=LFL^{-1}$ under a boost with $\mathbf{v}=\left(v,0,0\right)$ in such a case. So, we write:

$\boxed{F'=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}\begin{pmatrix}0& E_x& E_y& E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y& -cB_z& 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}\begin{pmatrix}\gamma & \beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}}$

$\boxed{F'=\mathbf{F}^{\mu'}_{\;\; \nu'}=\begin{pmatrix}0& E_x& \gamma (E_y-vB_z)& \gamma (E_z+vB_y)\\ E_x & 0 & c\gamma(B_z-\frac{v}{c^2}E_y) & -c\gamma (B_y+\frac{v}{c^2}E_z)\\ \gamma (E_y-vB_z)& -c\gamma (B_z-\frac{v}{c^2}E_y) & 0 & cB_x\\ \gamma (E_z+vB_y) & c\gamma (B_y+\frac{v}{c^2}E_z)& -cB_x& 0\end{pmatrix}}$

From this equation we deduce that:

$\mbox{EM fields after a boost}\begin{cases}E_{x'}=E_x,\; \; E_{y'}=\gamma \left( E_y-vB_z\right),\;\; E_{z'}=\gamma \left(E_z+vB_y\right)\\ B_{x'}=B_x,\;\; B_{y'}=\gamma \left(B_y+\frac{v}{c^2}E_z\right),\;\;B_{z'}=\gamma \left(B_z-\frac{v}{c^2}E_y\right)\end{cases}$

Example: In the S-frame we have the fields $E=(0,0,0)$ and $B=(0,B_y,0)$. The Coulomb force is $f_C=qE=(0,0,0)$ and the Lorentz force is $f_L=(0,0,qvB_y)$. How are these fields seen from the S’-frame? It is easy using the above transformations. We obtain that

$E'=(0,0,\gamma vB_y)$, $B'=(0,\gamma B_y,0)$, $f'_C=qE'=(0,0,\gamma qvB_y)$, $f'_L=(0,0,0)$

Surprinsingly, or not, the S’-observer sees a boosted electric field (non null!), a boosted magnetic field,  a boosted non-null Coulomb force and a null Lorentz force!

We can generalize the above transformations to the case of a general velocity in 3d-space $\mathbf{v}=(v_x,v_y,v_z)$

$\mathbf{E}_{\parallel'}=\mathbf{E}_\parallel$ $\mathbf{B}_{\parallel'}=\mathbf{B}_{\parallel}$

$\mathbf{E}_{\perp'}=\gamma \left[\mathbf{E}_\perp+(\mathbf{v}\times \mathbf{B})_\perp\right]=\gamma \left[\mathbf{E}_\perp+(\mathbf{v}\times \mathbf{B})\right]$

$\mathbf{B}_{\perp'}=\gamma \left[\mathbf{B}_\perp-\dfrac{1}{c^2}(\mathbf{v}\times \mathbf{E})_\perp\right]=\gamma \left[\mathbf{B}_\perp-\dfrac{1}{c^2}(\mathbf{v}\times \mathbf{E})\right]$

The last equal in the last two equations is due to the orthogonality of the position vector to the velocity in 3d space due to the cross product. From these equations, we easily obtain:

$E_\parallel=\dfrac{(v\cdot E)v}{v^2}=\dfrac{\beta\cdot E}{\beta^2}$

$E_\perp=E-E_\parallel=E-\dfrac{(v\cdot E)v}{v^2}$

and similarly with the magnetic field. The final tranformations we obtain are:

$\boxed{E'=E_{\parallel'}+E_{\perp'}=\dfrac{(v\cdot E)v}{v^2}+\gamma \left[ E-\dfrac{(v\cdot E)v}{v^2}+v\times B\right]}$

$\boxed{B'=B_{\parallel'}+B_{\perp'}=\dfrac{(v\cdot B)v}{v^2}+\gamma \left[ B-\dfrac{(v\cdot B)v}{v^2}-v\times E\right]}$

Equivalently

$\boxed{E'=\gamma \left(E+v\times B\right)-\left(\gamma-1\right)\dfrac{\left(v\cdot E\right) v}{v^2}}$

$\boxed{B'=\gamma \left(B-v\times \dfrac{E}{c^2}\right)-\left(\gamma-1\right)\dfrac{\left(v\cdot B\right) v}{v^2}}$

In the limit where $c\rightarrow \infty$ or $\dfrac{v}{c}\rightarrow 0$, we get that

$E'=E+v\times B$ $B'=B-\dfrac{v\times E}{c^2}$

There are two invariants for electromagnetic fields:

$I_1=\mathbf{E}\cdot\mathbf{B}$ and $I_2=\mathbf{E}^2-c^2\mathbf{B}^2$

It can be checked that

$\mathbf{E}\cdot\mathbf{B}=\mathbf{E}'\cdot\mathbf{B}'=invariant$

and

$\mathbf{E}^2-c^2\mathbf{B}^2=\mathbf{E'}^2-c^2\mathbf{B'}^2=invariant$  under Lorentz transformations. It is obvious since, up to a multiplicative constant,

$I_1=\dfrac{1}{4} F^\star_{\mu\nu}F^{\mu\nu}=\dfrac{1}{8}\epsilon_{\mu\nu\sigma \tau}F^{\sigma \tau}F^{\mu\nu}=\dfrac{1}{2}tr \left(F^{\star T}F\right)$

$I_2=\dfrac{1}{2}F_{\mu\nu}F^{\mu\nu}=tr\left(F^TF\right)$

and where we have defined the dual electromagnetic field as

$\star F=F^\star_{\mu\nu}=\dfrac{1}{2}\epsilon_{\mu\nu \sigma \tau}F^{\sigma \tau}$

or if we write it in components ( duality sends $\mathbf{E}$ to $\mathbf{B}$ and $\mathbf{B}$ to $-\mathbf{E}$)

$\star F=F^\star_{\mu\nu}=\begin{pmatrix}0& -B_x& -B_y& -B_z\\ B_x & 0 & -cE_z& cE_y\\ B_y& cE_z& 0 & -cE_x\\ B_z & -cE_y& cE_x& 0\end{pmatrix}$

We can guess some consequences for the electromagnetic invariants:

1st. If $E\perp B$, then $E\cdot B=0$ and thus $E_\perp B$ in every frame! This fact is important, since it shows that plane waves are orthogonal in any frame in SR. It is also the case of electromagnetic radiation.

2nd. As $E\cdot B=\vert E\vert \vert B\vert \cos \varphi$ can be in the non-orthogonal case either positive or negative. If $E\cdot B$ is positive, then it will be positive in any frame and similarly in the negative case. Morevoer, a transformation into a frame with $E=0$ (null electric field) and/or $B=0$ (null magnetic field) is impossible. That is, if a Lorentz transformation of the electric field or the magnetic field turns it to zero, it means that the electric field and magnetic field are orthogonal.

3rd. If E=cB, i.e., if $E^2-c^2B^2=0$, then it is valid in every frame.

4th. If there is a electric field but there is no magnetic field B in S, a Lorentz transformation to a pure B’ in S’ is impossible and viceversa.

5th. If the electric field is such that $E>cB$ or $E, then they can be turned in a pure electric or magnetic frame only if the electric field and the magnetic field are orthogonal.

6th. There is a trick to remember the two invariants. It is due to Riemann. We can build the hexadimensional vector( six-vector, or sixtor) and complex valued entity

$\boxed{\mathbf{F}=\mathbf{E}+ic\mathbf{B}}$

The two invariants are easily obtained squaring F:

$F^2=\mathbf{E}^2-c^2\mathbf{B}^2+2ic\mathbf{E}\cdot\mathbf{B}=invariant$

We can introduce now a vector potencial tetravector:

$\mathbb{A}=A^\mu e_\mu=\left( A^0,A^1,A^2,A^3\right)=\left(\dfrac{V}{c},\mathbf{A}\right)=\left(\dfrac{V}{c},A_x,A_y,A_z\right)$

This tetravector is also called gauge field. We can write the Maxwell tensor in terms of this field:

$F_{\mu\nu}=\partial_\mu A_\nu-\partial _\nu A_\mu$

It can be easily probed that, up to a multiplicative constant in front of the electric current tetravector, the first set of Maxwell equations are:

$\boxed{\partial_\mu F^{\mu\nu}=j^\nu \leftrightarrow \square \cdot \mathbf{F}=\mathbb{J}}$

The second set of Maxwell equations (sometimes called Bianchi identities) can be written as follows:

$\boxed{\partial_\mu F^{\star \mu \nu}=\dfrac{1}{2}\epsilon^{\nu\mu\alpha\beta}\partial_\mu F_{\alpha\beta}=0}$

The Maxwell equations are invariant under the gauge transformations in spacetime:

$\boxed{A^{\mu'}=A^\mu+e\partial^\mu \Psi}$

where the potential tetravector and the function $\Psi$ are arbitrary functions of the spacetime.

Some elections of gauge are common in the solution of electromagnetic problems:

A) Lorentz gauge: $\square \cdot A=\partial_\mu A^\mu=0$

B) Coulomb gauge: $\nabla \cdot \mathbf{A}=0$

C) Temporal gauge: $A^0=V/c=0$

If we use the Lorentz gauge, and the Maxwell equations without sources, we deduce that the vector potential components satisfy the wave equation, i.e.,

$\boxed{\square^2 A^\mu=0 \leftrightarrow \square^2 \mathbb{A}=0}$

Finally, let me point out an important thing about Maxwell equations. Specifically, about its invariance group. It is known that Maxwell equations are invariant under Lorentz transformations, and it was the guide Einstein used to extend galilean relativity to the case of electromagnetic fields, enlarging the mechanical concepts. But, the larger group leaving invariant the Maxwell equation’s invariant is not the Lorentz group but the conformal group. But it is another story unrelated to this post.

## LOG#032. Invariance and relativity.

Invariance, symmetry and invariant quantities are in the essence, heart and core of Physmatics. Let me begin this post with classical physics. Newton’s fundamental law reads:

$\mathbf{F}=m\mathbf{a}=\begin{cases}m\ddot{x}, \;\; m\ddot{x}=m\dfrac{d^2x}{dt^2}\\\;\\m\ddot{y}, \;\; m\ddot{y}=m\dfrac{d^2y}{dt^2}\\\;\\ m\ddot{z}, \;\; m\ddot{z}=m\dfrac{d^2z}{dt^2}\end{cases}$

Suppose two different frames obtained by a pure translation in space:

$\mathbf{x}'=\mathbf{x}-\mathbf{x}_0$

or

$\mathbf{r}'=\mathbf{r}-\mathbf{r}_0$

We select to make things simpler

$\mathbf{x}_0=\mathbf{r}_0=(x_0,0,0)$

We can easily observe by direct differentiation that Newton’s fundamental is invariant under translations in space, since mere substitution provides:

$\mathbf{F}'=\mathbf{F}$

since

$m\dfrac{d^2\mathbf{r}'}{dt^2}=m\dfrac{d^2\mathbf{r}}{dt^2} \leftrightarrow \mathbf{a}'=\mathbf{a}$

By the other hand, rotations around a fixed axis, say the z-axis, are transformations given by:

$\boxed{\mathbf{r}'=R\mathbf{r}\leftrightarrow \begin{pmatrix}x'\\y'\\z'\end{pmatrix}=R\begin{pmatrix}x\\y\\z\end{pmatrix}\rightarrow\begin{cases}x'=x\cos\theta+y\sin\theta\\y'=-x\sin\theta+y\cos\theta\\z'=z\end{cases}}$

If we multiply by the mass these last equations and we differentiate with respect to time twice, keeping constant $\theta$ and $m$, we easily get

$\mathbf{F}'=R\mathbf{F}\leftrightarrow \begin{cases}F_{x'}=F_x\cos\theta+F_y\cos\theta\\F_{y'}=-F_x\sin\theta+F_y\cos\theta\\F_{z'}=F_z\end{cases}$

or

$\boxed{\begin{pmatrix}F'_x\\F'_y\\F'_z\end{pmatrix}=\begin{pmatrix}\cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0& 1 \end{pmatrix}\begin{pmatrix}F_x\\F_y\\F_z\end{pmatrix}\rightarrow \begin{cases}F'_x=F_x\cos\theta+F_y\sin\theta\\F'_y=-F_x\sin\theta+F_y\cos\theta\\F'_z=F_z\end{cases}}$

Thus, we can say that Newton’s fundamental law is invariant under spatial translations and rotations. Its form is kept constant under those kind of transformations. Generally speaking, we also say that Newton’s law is “covariant”, but nowadays it is an abuse of language since the word covariant means something different in tensor analysis. So, be aware about the word “covariant” (specially in old texts). Today, we talk about “invariant laws”, or about the symmetry of certain equations under certain set of (group) transformations.

Newton’s law use the concept of acceleration:

$\mathbf{a}=\dfrac{d\mathbf{v}}{dt}=\left(\dfrac{d}{dt}\right)\left(\dfrac{d}{dt}\right)\mathbf{r}=\dfrac{d^2\mathbf{r}}{dt^2}$

with

$a_x=\dfrac{dv_x}{dt}=\dfrac{d^2x}{dt^2}$ $a_y=\dfrac{dv_y}{dt}=\dfrac{d^2y}{dt^2}$ $a_z=\dfrac{dv_z}{dt}=\dfrac{d^2z}{dt^2}$

or, in compact form

$a_i=\dfrac{dv_i}{dt},\;\; i=x,y,z$

And then, the following equations are invariant under translations in space and rotations:

$\mathbf{F}=m\mathbf{a}$ or $\mathbf{F}=\dfrac{d\mathbf{p}}{dt}$ with $\mathbf{p}=m\mathbf{v}$

Intrinsic components of the aceleration provide a decomposition

$\mathbf{a}=\mathbf{a}_\parallel+\mathbf{a}_\perp$

where we define

$a_\parallel=\dfrac{dv}{dt}\leftrightarrow \mathbf{a}_\parallel=\dfrac{dv}{dt}\mathbf{u}_\parallel$

where $\mathbf{u}_\parallel$ is a unit vector in the direction of the velocity, and

$\mathbf{a}_\perp=\mathbf{a}-\mathbf{a}_\parallel$

In the case of motion along a general curve, we can approximate the motion in every point of the curve by a circle of radius R, and thus

$s=R\Delta \theta\rightarrow \Delta \theta=\dfrac{s}{R}=\dfrac{v\Delta t}{R}\rightarrow \dfrac{\Delta \theta}{\Delta t}=\omega=\dfrac{v}{R}$

By the other hand,

$\Delta v_\perp=v\Delta \theta\rightarrow a_\perp=v\dfrac{\Delta \theta}{\Delta t}=\dfrac{v^2}{R}=\omega^2R$

and we get the known expression for the centripetal acceleration:

$\mathbf{a}_\perp=a_\perp\mathbf{u}_\perp=\dfrac{v^2}{R}\mathbf{u}_\perp$

More about invariant quantities in Classical Physics: the scalar (sometimes called dot) product of two vectors is invariant, since the length of every vector is constant in euclidean spaces under rotations and translations. For instance,

$\boxed{r^2=x^2+y^2+z^2=\mathbf{r}\cdot\mathbf{r}=\mathbf{r'}\cdot\mathbf{r'}=\mbox{INVARIANT}=\mbox{SQUARED LENGTH}}$

In matrix form,

$r^2=X^TX=\delta _{ij}x^ix^j=\begin{pmatrix}x & y & z\end{pmatrix}\begin{pmatrix}1 &0& 0\\ 0 &1& 0\\ 0& 0& 1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}$

where we have introduced the $\delta_{ij}$ symbol to be the so-called Kronecker delta as “certain object” with components: its components are “1” whenever $i=j$ and “0” otherwise. Of course, the Kronecker delta symbol “is” the identity matrix when the symbol have two indices. However, let me remark that “generalized delta Kronocker” with more indices do exist and it is not always posible to express easily that “tensor” in a matrix way, excepting using some clever tricks.

The scalar (dot) product can be computed with any vector quantity:

$\mathbf{a}\cdot\mathbf{a}=a^2=a_x^2+a_y^2+a_z^2\rightarrow \mathbf{a}\cdot\mathbf{b}=a_xb_x+a_yb_y+a_zb_z$

Moreover, there is a coordinate free definition as well:

$\mathbf{a}\cdot\mathbf{b}=ab\cos\theta,\;\; \theta=\mbox{angle formed by}\; \mathbf{a},\mathbf{b}$

Note that the invariance of the dot product implies the invariance of classical kinetic energy, since:

$K.E.=T=\dfrac{1}{2}m\mathbf{v}\cdot\mathbf{v}=\dfrac{\mathbf{p}\cdot\mathbf{p}}{2m}=\dfrac{1}{2}mv^2=\dfrac{1}{2m}p^2=\mbox{INVARIANT}$

We have also the important invariant quantities:

$\mbox{WORK}=W=\int \mathbf{F}\cdot d\mathbf{r}$

$\mbox{POWER}=P=\dfrac{dW}{dt}=\mathbf{F}\cdot\mathbf{v}$

where the second equality holds if the force is constant along the trajectory. Moreover, in relativistic electromagnetism, you also get the wave-number 4-vector:

$\boxed{\mathbb{K}=(K^0,\mathbf{K})}\leftrightarrow \mbox{WAVE NUMBER SPACETIME VECTOR}$

and the invariant $\mathbb{K}\cdot\mathbb{K}=K^2$, that you can get from the plane wave solution:

$A=A_0\exp (i(K^\mu X_\mu))=A_0\exp (i(\mathbb{K}\cdot \mathbb{X}))$

where the phase invariant reads

$\phi =\mathbb{K}\cdot \mathbb{X}=\mathbf{K}\cdot\mathbf{X}-\omega t$

Therefore, we deduce that

$K^0=\dfrac{\omega}{c}$

and the wave number  vector satisfies the following relation with the wave-length

$\vert \mathbf{K}\vert =K=\dfrac{2\pi}{\lambda}$

There is another important set of transformations or symmetry in classical physics. It is related to inertial frames. Galileo discovered that the laws of motion are the same for every inertial observer, i.e., the laws of Mechanics are invariant for inertial frames! A Galilean transformation is defined by:

$\boxed{\mbox{GALILEAN TRANSFORMATIONS}\begin{cases}\mathbf{x}'=\mathbf{x}-\mathbf{V}t\\t'=t\end{cases}}$

where $\mathbf{V}=constant$. Differentiating with respect to time, we get

$\boxed{\mbox{GALILEAN TRANSFORMATIONS}\begin{cases}\dfrac{d\mathbf{x}'}{dt}=\dfrac{d\mathbf{x}}{dt}-\mathbf{V}\\ \;\\ \dfrac{dt'}{dt}=1\end{cases}}$

and then

$\boxed{\mbox{GALILEAN TRANSFORMATIONS}\begin{cases}\dfrac{d^2\mathbf{x}'}{dt^2}=\dfrac{d^2\mathbf{x}}{dt^2}\\ \;\\\dfrac{d^2t'}{dt^2}=0\end{cases}}$

And thus, the accelerations (and forces) that observe different inertial ( i.e., reference frames moving with constant relative velocity) frames are the same

$\mathbf{a}'=\mathbf{a}$

And now, about symmetry. What are the symmetries of Physics? There are many interesting transformations and space-time symmetries. A non-completely exhaustive list is this one:

1. Translations in space.

2. Translations in time.

3. Rotations around some axis ( and with fixed angle).

4. Uniform velocity in straight line, a.k.a., galilean transformations for inertial observers. This symmetry “becomes” Lorentz boosts in the spacetime analogue of special relativity.

5. Time reversal ( inversion of the direction of time), T.

6. Reflections in space (under “a mirror”). It is also called parity P.

7. Matter-antimatter interchange, or charge conjugation symmetry, C.

8. Interchange of identical atoms/particles.

9. Scale transformations $\mathbb{X}'=\lambda\mathbb{X}$.

10. Conformal transformations (in the complex plane or in complex spaces).

11. Arbitrary coordinate transformations (they are also called general coordinate transformations).

12. Quantum-mechanical (gauge) phase symmetry: $\Psi\rightarrow \Psi'=\Psi \exp (i\theta)$.

Beyon general vectors, in classical physics we also find “axial” vectors (also called “pseudovectors”). Pseudovectors or axial vectors are formed by the 3d “cross”/outer/vector  product:

$\mathbf{C}=\mathbf{A}\times \mathbf{B}=\begin{vmatrix}e_1 & e_2 & e_3\\ A_x & A_y & A_z\\ B_x & B_y & B_z \end{vmatrix}=e_1\begin{vmatrix}A_y & A_z\\ B_y & B_z\end{vmatrix}-e_2\begin{vmatrix}A_x & A_z\\ B_x & B_z\end{vmatrix}+e_3\begin{vmatrix}A_x & A_y\\ B_x & B_y\end{vmatrix}$

Some examples are the angular momentum

$\mathbf{L}=\mathbf{r}\times\mathbf{p}=\mathbf{r}\times m\mathbf{v}$

or the magnetic force

$\mathbf{F}_m=q\mathbf{v}\times \mathbf{B}$

Interestingly, the main difference between axial and polar vectors, i.e., between common vectors and pseudovectors is the fact that under P (parity), pseudovectors are “invariant” while common vectors change their sign, i.e., polar vectors become the opposite vector under reflection, and pseudovectors remain invariant. It can be easily found by inspection in the definition of angular momentum or the magnetic force ( or even the general definition of cross product given above).

Now, we turn our attention to invariants in special relativity. I will introduce a very easy example to give a gross idea of how the generalization of “invariant theory” works as well in special relativity. From the classical definition of power:

$P=\dfrac{dE}{dt}=\mathbf{F}\cdot\mathbf{v}$

Using the relativistic definition of 4-momentum:

$\mathbb{P}=\left(Mc,\dfrac{d\mathbf{p}}{d\tau}\right)=\left(E/c,M\mathbf{v}\right)$

where $M=m\gamma$, we are going to derive a known result, since $E=Mc^2$. Note that, in agreement with classical physics, from this

$\mathbf{F}=\dfrac{d\mathbf{P}}{dt}$

Therefore, inserting the relativistic expressions for energy and force into the power equation, we obtain:

$\dfrac{d(Mc^2)}{dt}=\mathbf{v}\cdot\dfrac{d(M\mathbf{v})}{dt}$

Multiplying by 2M, and using the Leibniz rule for the differentiation of a product of two functions:

$2M\dfrac{d(Mc^2)}{dt}=2M\mathbf{v}\cdot\dfrac{d(M\mathbf{v})}{dt}=2\mathbf{P}\cdot\dfrac{d\mathbf{P}}{dt}$

or equivalently

$2Mc^2\dfrac{dM}{dt}=2\mathbf{P}\cdot\dfrac{d\mathbf{P}}{dt}=\dfrac{d(\mathbf{P})^2}{dt}$

and so

$c^2\dfrac{dM^2}{dt}=\dfrac{d(\mathbf{P})^2}{dt}$

Integrating this, we deduce that

$M^2c^2=M^2v^2+\mbox{constant}$

If we plug $\mbox{constant}=m^2c^2=M^2c^2(v=0)$

then

$M^2c^2=M^2v^2+m^2c^2\rightarrow M^2=m^2\gamma^2,\;\; \gamma^2=\dfrac{1}{1-\frac{v^2}{c^2}}$

and thus, we have rederived the notion of “relativistic mass”

$M=m\gamma=\dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Special relativity generalizes the notion of dot product ( scalar product) to a non-euclidean (pseudoeuclidean to be more precise) geometry. The dot product in special relativity is given by:

$\boxed{\mathbb{A}\cdot\mathbb{B}=A^x B_x+A^y B_y+A^z B_z-A^t B_t}$

The sign of the temporal fourth component is conventional in the sense some people uses a minus sign for the purely spatial components and a positive sign for the temporal component. Using a more advanced notation we can write the new scalar product as follows:

$\boxed{A^\mu B_\mu=A_\mu B^\mu=A^x B_x+A^y B_y+A^z B_z-A^t B_t}$

where the repeated dummy index implies summation over it. This convention of understanding summation over repeated indices is called Einstein’s covention and it is due to Einstein himself. Another main point about notation is that some people prefer the use of a $\mu=0,1,2,3$ while other people use $\mu=1,2,3,4$. We will use the notation with $\mu=0,1,2,3$ unless we find some notational issue. Unlikely to the 3d world, the 4d world of special relativity forces us to use something different to the Kronecker delta in the above scalar product. This new object is generally called pseudoeuclidean “metric”, or Minkovski metric:

$\boxed{A^\mu B_\mu=\eta_{\mu\nu}A^\mu B^\nu=\mathbb{A}\cdot\mathbb{B}=A^xB_x+A^yB_y+A^zB_z-A^tB_t}$

In matrix form,

$\boxed{A^\mu B_\mu=A^\mu\eta_{\mu\nu}B^\mu=A^T\eta B=\begin{pmatrix}A^t & A^x & A^y & A^z\end{pmatrix}\begin{pmatrix}-1 & 0 & 0 & 0\\ 0& 1 & 0 & 0\\ 0 & 0 & 1& 0\\ 0 & 0& 0& 1\end{pmatrix}\begin{pmatrix}B^t \\ B^x \\ B^y \\ B^z\end{pmatrix}}$

Important remarks:

1st. $\eta=\eta_{\mu\nu}=diag(-1,1,1,1)$ in our convention. The opposite convention for the scalar product would give $\eta=diag(1,-1,-1,-1)$.

2nd. The square of a “spacetime” vector is its “lenght” in spacetime. It is given by:

$\boxed{A^2=\mathbb{A}\cdot\mathbb{A}=A^\mu A_\mu=A_x^2+A_y^2+A_z^2-A_t^2=-(\mbox{SQUARED SPACETIME LENGTH})}$

In particular, for the position spacetime vector

$S^2=x^\mu x_\mu=-c^2\tau ^2$

3rd. Unlike the euclidean 3d space, 4d noneuclidean spacetime introduces “objects” non-null that whose “squared lenght” is equal to zero, and even weirder, objects that could provide a negative dot product!

4th. For spacetime events given by a spacetime vector $\mathbb{X}=x^\mu e_\mu=(ct,\mathbf{r})$, and generally for any arbitrary events A and B (or 4-vectors) we can distinguish:

i) Vectors with $A^2=\mathbb{A}\cdot\mathbb{A}>0$ are called space-like vectors.

ii) Vectors with $A^2=\mathbb{A}\cdot\mathbb{A}=0$ are called null-vectors, isotropic vectors or sometimes light-like vectors.

iii) Vectos with $A^2=\mathbb{A}\cdot\mathbb{A}<0$ are called time-like vectors.

Thus, in the case of the spacetime (position) vector, every event can be classified into space-like, light-like (null or isotropic) and time-like types, depending on the sign of $s^2=\mathbb{X}\cdot\mathbb{X}=X^T\eta X$. Moreover, the metric itself allows us to “raise or lower” indices, defining the following rules for components:

$x^\mu=\begin{pmatrix}ct \\ \mathbf{r}\end{pmatrix}\rightarrow x_\mu =\eta _{\mu \nu}x^\nu=x^\nu \eta_{\mu\nu}=X^T\eta=(-ct,\mathbf{r})$

The minkovskian metric has a very cool feature too. Its “square” is the identity matrix. That is,

$\eta^2=\eta^T\eta=\eta\eta^T=\mathbb{I}$

Then, the metric is its own inverse:

$\eta=\eta^{-1}$

In components, it reads

$\eta^{\mu\nu}\eta_{\nu\sigma}=\delta^{\mu}_{\;\;\sigma}$

with

$\eta^{\mu\nu}=(\eta^{-1})_{\mu\nu}$

and where we have introduced the Kronecker delta symbol in four dimensions in the same manner we did in the 3d space. Therefore, the Kronecker delta has only non-null components when $\mu=\sigma$, so that $\delta^0_{\;\;0}=\delta^1_{\;\;1}=\ldots=1$

Subindices are called generally “covariant” components, while superindices are called “contravariant” components. It is evident that euclidean spaces don’t distinguish between covariant and contravariant components. The metric is the gadget we use in non-euclidean metric spaces to raise and lower indices/components from tensor quantities. Tensors are multi-oriented objects. The metric itself is a second order tensor, more precisely, the metric is a second order rank 2 covariant tensor. 4-vectors are contravariant object with a single index. Upwards single-indexed tensors are contravariant vectors, downwards single-indexed tensores are covariant vectors. When a metric is introduced, there is no need to distinguish covariant and contravariant tensors, since the components can be calculated with the aid of the metric, so we speak about n-th rank tensors. Multi-indixed objects can have some features of symmetry. The metric is symmetric itself, for instance, under the interchange of subindices ( columns and rows). So, then

$\eta_{\mu\nu}=\eta_{\nu\mu}\rightarrow \eta =\eta^T$

What kind of general objects can we use in Minkovski spacetime or even more general spaces? Firstly, we have scalar fields or functions, i.e., functions depending only on the spacetime coordinates:

$\psi (x)=\psi' (x') \leftrightarrow \psi (x^\mu) =\psi ' (x'^\mu) \leftrightarrow \psi (x,y,z, ct)= \psi ' (x',y',z',ct')$

Another objetct we have found are “vectors” or “oriented segments”. In 3d space, they transform as $\mathbf{x}'=R\mathbf{x}$. In 4d spacetime, we found $\mathbb{X}'=L\mathbb{X}$.

In 3d space, we also found pseudovectors. They are defined via the cross product, that in components read: $c^i=\epsilon ^{ijk}a_jb_k$, where the new symbol $\epsilon^{ijk}$ is a completely antisymmetric object under the interchange of any pair of indices (with $\epsilon^{123}=+1$) is generally called Levi-Civita symbol. This symbol is the second constant object that we can use in any number of dimensions, like the Kronecker delta

$\delta ^i_{\;\; j}=\begin{cases}+1,\mbox{if}\; i=j\\ 0,\mbox{otherwise}\end{cases}$

The completely antisymmetric Levi-Civita symbol has some interesting identities related with the Kronecker delta. Thus, for instance, in 2d and 3d respectively:

$\epsilon^{ij}\epsilon_{ik}=\delta^{j}_{\;\;k}$

$\epsilon^{ijk}\epsilon_{ilm}=\delta^{j}_{\;\; l}\delta^{k}_{\;\; m}-\delta^{j}_{\;\; m}\delta^{k}_{\;\; l}$

or

$\epsilon^{ijk}a_ib_jc_k=det(\mathbf{a},\mathbf{b},\mathbf{c})$

We have also the useful identity:

$\epsilon^{i_1i_2\ldots i_n}\epsilon_{i_1i_2\ldots i_n}=n!$

The n-dimensional Levi-Civita symbol is defined as:

$\epsilon^{i_1i_2\ldots i_n}a_1a_2\ldots a_n=det(\mathbf{a_{1}},\mathbf{a_2},\ldots,\mathbf{a_{n}})$

and its product in n-dimensions

$\varepsilon_{i_1 i_2 \dots i_n} \varepsilon_{j_1 j_2 \dots j_n} = \begin{vmatrix}\delta_{i_1 j_1} & \delta_{i_1 j_2} & \dots & \delta_{i_1 j_n} \\\delta_{i_2 j_1} & \delta_{i_2 j_2} & \dots & \delta_{i_2 j_n} \\\vdots & \vdots & \ddots & \vdots \\ \delta_{i_n j_1} & \delta_{i_n j_2} & \dots & \delta_{i_n j_n} \\ \end{vmatrix}$

or equivalently, given a nxn matrix $A=(a_{ij})$

$\epsilon^{i_1i_2\ldots i_n}a_{1i_i}a_{2i_2}...a_{ni_n}=det( a_{ij})=\dfrac{1}{n!}\epsilon^{i_1i_2\ldots i_n}\epsilon^{j_1j_2\ldots j_n}a_{i_1 j_1}a_{i_2 j_2}\ldots a_{i_n j_n}$

This last equation provides some new quantity called pseudoscalar, different from the scalar function in the sense it changes its sign under parity in 3d, while a common 3d scalar is invariant under parity! Generally speaking, determinants (pseudoscalars) in even dimensions are parity conserving, while determinants in odd dimensions don’t conserve parity.

Like the Kronecker delta, the epsilon or Levi-Civita can be generalized to 4 dimensions (or even to D-dimensions). In 4 dimensions:

$\epsilon^{\mu\nu\sigma\tau}=\epsilon_{\mu\nu\sigma\tau}=\begin{cases}+1,\mbox{if} (\mu\nu\sigma\tau)\mbox{is an even permutation of 0,1,2,3}\\-1,\mbox{if} (\mu\nu\sigma\tau)\mbox{is an odd permutation of 0,1,2,3}\\ 0,\mbox{otherwise}\end{cases}$

In general, unlike the Kronecker deltas, the Levi-Civita epsilon symbols are not ordinary “tensors” (quantities with subindices and superindices, with some concrete properties under coordinate transformations) but more general entities called “weighted” tensors (sometimes they are also called tensorial densities). Indeed, the generalized Kronecker delta can be defined of order 2p is a type (p,p) tensor that is a completely antisymmetric in its ”p” upper indices, and also in its ”p” lower indices.  This characterization defines it up to a scalar multiplier.

$\delta^{\mu_1 \dots \mu_p }_{\;\;\;\;\;\;\;\;\;\; \nu_1 \dots \nu_p} =\begin{cases}+1 & \quad \text{if } \nu_1 \dots \nu_p \text{ are an even permutation of } \mu_1 \dots \mu_p \\-1 & \quad \text{if } \nu_1 \dots \nu_p \text{ are an odd permutation of } \mu_1 \dots \mu_p \\ \;\;0 & \quad \text{in all other cases}.\end{cases}$

Using an anti-symmetrization procedure:
$\delta^{\mu_1 \dots \mu_p}_{\;\;\;\;\;\;\;\;\;\;\nu_1 \dots \nu_p} = p! \delta^{\mu_1}_{\lbrack \nu_1} \dots \delta^{\mu_p}_{\nu_p \rbrack}$

In terms of an pxp determinant:
$\delta^{\mu_1 \dots \mu_p }_{\;\;\;\;\;\;\;\;\;\;\nu_1 \dots \nu_p} =\begin{vmatrix}\delta^{\mu_1}_{\nu_1} & \cdots & \delta^{\mu_1}_{\nu_p} \\ \vdots & \ddots & \vdots \\ \delta^{\mu_p}_{\nu_1} & \cdots & \delta^{\mu_p}_{\nu_p}\end{vmatrix}$

Equivalently, it could be defined by induction in the following way:

$\delta^{\mu \rho}_{\nu \sigma} = \delta^{\mu}_{\nu} \delta^{\rho}_{\sigma} - \delta^{\mu}_{\sigma} \delta^{\rho}_{\nu}$

$\delta^{\mu \rho_1 \rho_2}_{\nu \sigma_1 \sigma_2} = \delta^{\mu}_{\nu} \delta^{\rho_1 \rho_2}_{\sigma_1 \sigma_2} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2}_{\nu \sigma_2} + \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2}_{\sigma_2 \nu}$
$\delta^{\mu \rho_1 \rho_2 \rho_3}_{\nu \sigma_1 \sigma_2 \sigma_3} = \delta^{\mu}_{\nu} \delta^{\rho_1 \rho_2 \rho_3}_{\sigma_1 \sigma_2 \sigma_3} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2 \rho_3}_{\nu \sigma_2 \sigma_3} + \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2 \rho_3}_{\sigma_2 \nu \sigma_3} - \delta^{\mu}_{\sigma_1} \delta^{\rho_1 \rho_2 \rho_3}_{\sigma_2 \sigma_3 \nu}$
and so on.

In the particular case where  p=n  (the dimension of the vector space), in terms of the Levi-Civita symbol:
$\delta^{\mu_1 \dots \mu_n}_{\nu_1 \dots \nu_n} = \varepsilon^{\mu_1 \dots \mu_n}\varepsilon_{\nu_1 \dots \nu_n}$

Under a Lorentz transformation, we have ( using matrix notation) the next transformations:

$A'=LA \leftrightarrow (A')^T=A^TL^T$

$\eta A'=\eta LA\rightarrow (A')^T\eta A'=A^T(L^T\eta L)A$

$A'^T\eta A'=A^T \eta A$ iff $L^T \eta L=\eta$

so the metric itself is “invariant” under a Lorentz transformation (boost). I would like to remark that the metric can be built from the basis vectors in the following way:

$\eta_{\mu \nu}=e_\mu e_\nu=e_\mu \cdot e_\nu= g\left( e_\mu ,e_\nu\right)=\begin{cases}-1, \mu =\nu =0\\ +1, \mu =\nu =1,2,3\\ 0,\mu \neq \nu \end{cases}$

For Lorentz transformations, we get that

$x^\mu\rightarrow x'^\mu =\Lambda^\mu _{\;\;\; \nu} x^\nu$

with

$\Lambda^\mu _{\;\;\; \nu}=\begin{pmatrix}\Lambda^0_{\;\;\; 0}& \Lambda^0_{\;\;\; 1}& \Lambda^0_{\;\;\; 2}& \Lambda^0_{\;\;\; 3}\\ \Lambda^1_{\;\;\; 0}& \Lambda^1_{\;\;\; 1}& \Lambda^1_{\;\;\; 2}& \Lambda^1_{\;\;\; 3} \\ \Lambda^2_{\;\;\; 0}& \Lambda^2_{\;\;\; 1}&\Lambda^2_{\;\;\; 2}& \Lambda^2_{\;\;\; 3}\\ \Lambda^3_{\;\;\; 0}& \Lambda^3_{\;\;\; 1}& \Lambda^3_{\;\;\; 2}& \Lambda^3_{\;\;\; 3}\end{pmatrix}$

Moreover, the equation $\Lambda^{-1}=\eta^{-1}\Lambda \eta$, i.e., for pseudo-orthogonal Lorentz transformations, taking the determinant, we deduce that

$\det (\Lambda)=\pm 1\leftrightarrow \det (\Lambda)^2=1$

We can fully classify the Lorentz transformations according to the sign of the determinant and the sign of the element $\Lambda^0_{\;\;\; 0}$ as follows:

$\Lambda\begin{cases} \mbox{Proper Lorentz transf.(e.g.,boosts,3d rotations, Id)}: L^\uparrow_+ \det (\Lambda)=+1, \Lambda^0_{\;\;\; 0}\ge 1\\ \mbox{Improper Lorentz transf.:}\begin{cases}L^\downarrow_+ \det (\Lambda)=+1, \; \Lambda^0_{\;\;\; 0}\le -1\\ L^\uparrow_- \det (\Lambda)=-1,\; \Lambda^0_{\;\;\; 0}\ge 1\\ L^\downarrow_- \det (\Lambda)=-1,\; \Lambda^0_{\;\;\; 0}\le 1\end{cases} \end{cases}$

For instance, let us write 5 kind of Lorentz transformations:

1) Orthogonal rotations. They are continuous (proper) Lorentz transformations with a 3×3 submatrix $\Omega +\Omega^T=0$:

$\Lambda=\begin{pmatrix}1 & \mathbf{0}\\ \mathbf{0} & \Omega\end{pmatrix}$

2) Boosts. They are continuous (proper) Lorentz transformations mixing spacelike and timelike coordinates. The matrix has in this case the form:

$\Lambda =\begin{pmatrix}\gamma & -\beta \gamma & \mathbf{0}\\ -\beta \gamma & \gamma & \mathbf{0}\\ \mathbf{0}&\mathbf{0}& \mathbb{I}\end{pmatrix}$

3) PT symmetry. Discrete non-proper Lorentz transformation. It “inverts” space and time coordinates in the sense $\mathbf{r}\rightarrow -\mathbf{r}$ and $t\rightarrow -t$. They belongs to $L_+^\downarrow$. The matrix of this transformation is:

$\Lambda_{PT}=diag(-1,-1,-1,-1)$

4) Parity. Discrete non-proper Lorentz transformation. It inverts only the spacelike components of true vectors ( be aware of pseudovectors!) in the sense $\mathbf{r}\rightarrow -\mathbf{r}$. Sometimes, it is denoted by P, parity, and this transformation belongs to $L_-^\uparrow$. It is defined as follows:

$P=\Lambda_P=diag(1,-1,-1,-1)$

5) Time reversal, T. Discrete non-proper Lorentz transformation. It inverts the direction of time in the sense that $t\rightarrow -t$. $\Lambda_T$ belongs to the set $L_-^\downarrow$

Remark: If $X^2>0$, then $L_+^\uparrow, L_-^\uparrow$ don’t change the sense of time. This is why they are called orthochronous!

## LOG#031. Entropic Gravity (II).

We will generalize the entropic gravity approach to include higher dimensions in this post. The keypoint from this theory of entropic gravity, according to Erik Verlinde, is that gravity does not exist as “fundamental” force and it is a derived concept. Entropy is the fundamental object somehow. And it can be generalized to a a d-dimensional world as follows.

The entropic force is defined as:

$\boxed{F=-\dfrac{\Delta U}{\Delta x}=-T\dfrac{\Delta S}{\Delta x}}$

The entropic force is a force resulting from the tendency of a system to increase its entropy. Since $\Delta S>0$ the sign of the force (whether repulsive or attractive) is determined by how we take the definition of $\Delta x$ as it is related to the system in question.

An arbitrary mass distribution M induces a holographic screen $\Sigma$  at some distance R that has encoded on it gravitational information. Today, we will consider the situation in d spatial dimensions.Using the holographic principle, the screen owns all physical information contained within its volume in bits on the screen whose number N is given by:

$\boxed{N=\dfrac{A_\Sigma (R)}{l_p^{d-1}}}$

This condition implies the quantization of the hyperspherical surface, where the hyperarea (from the hypersphere) is defined as:

$\boxed{A_\Sigma=\dfrac{2\pi^{d/2}}{\Gamma \left(\frac{d}{2}\right)}R^{d-1}}$

By the equipartition principle:

$\boxed{E=Mc^2=\dfrac{N}{2}k_BT}$

Therefore,

$k_BT=\dfrac{2Mc^2l_p^{d-1}}{A_\Sigma}$

The entropy shift due to some displacement is:

$\Delta S=2\pi k_B \dfrac{\Delta x}{\bar{\lambda}}=2\pi k_B mc\dfrac{\Delta x}{\hbar}$

Plugging the expression for the temperature and the entropy into the entropic force equation, we get:

$F=-T\dfrac{\Delta S}{\Delta x}=-\dfrac{2Mc^2}{k_B}\dfrac{2\pi k_B mc}{\hbar}\dfrac{l_p^{d-1}}{A_\Sigma}=-\dfrac{4\pi Mmc^3l_p^{d-1}}{\hbar A_\Sigma}$

and thus we finally get

$F=-\dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}Mmc^3}{\hbar R^{d-1}}=-G_d\dfrac{Mm}{R^{d-1}}$

i.e.,

$\boxed{F=-G_d\dfrac{Mm}{R^{d-1}}}\leftrightarrow \boxed{F=-\dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}Mmc^3}{\hbar R^{d-1}}}$

where we have defined the gravitational constant in d dimensions to be

$\boxed{G_d\equiv \dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}c^3}{\hbar }}$

## LOG#030. Entropic Gravity (I).

In 2010, Erik Verlinde made himself famous once again. Erik Verlinde is a theoretical physicist who has made some contributions to String Theory. In particular, the so-called Verlinde formula. However, this time was not apparently a contribution related to string theory. He guessed a way to derive both the Newton’s second law and the Newton’s law of gravity. He received a prize time later, and some critical voices against his approach were raised.

I will review in this post his deductions.

A. Newton’s second law.

There are some hypothesis to begin with:

1st. Entropic force ansatz. Forces aren’t really fundamental, they are derived from some entropy functional. More precisely, forces are ”entropy fluxes”. Mathematically speaking:

$F\Delta x=T\Delta S$

or

$\boxed{F=T\dfrac{\Delta S}{\Delta x}}$

2nd. Acceleration has a temperature. Equivalently, this is the well known Unruh’s effect from QFT in curved spacetime. Any particle that is accelerated is equivalent to some thermal system. This parallels the Hawking’s effect in black hole physics as well. It stands mysterious for me yet, since indeed, the temperature is relative to some vacuum or rest system.

$T=\dfrac{\hbar a}{2\pi k_B c}$

3rd. Holographic principle.   A variant of the holographic principle is postulated to hold. The idea is that a particle separated certain distance from a “holographic screen” has an entropy shift:

$\Delta S = 2\pi k_B \dfrac{mc}{\hbar} \Delta x$

Then, plugging the holographic entropy and the Unruh’s temperature into the entropic force ansatz, we get easily

$F= 2\pi k_B \dfrac{mc}{\hbar} \dfrac{\hbar a}{2\pi k_B c}$

i.e. we get the Newton’s second law of Dynamics

$F=ma$

B. Newton’s gravity.  In order to get the Newton’s law of gravitation, we have to modify a bit the auxiliary hypothesis but yet we conserve the core approach.

1st. The entropic force ansatz. Again,

$\boxed{F=T\dfrac{\Delta S}{\Delta x}}$

2nd. Holographic principle. Again,

$\Delta S = 2\pi k_B \dfrac{mc}{\hbar} \Delta x$

3rd. Equipartition principle of relativistic energy.   Temperature is obtained at the statistical level when you distribute N quanta in thermal equilibrium, and they equal the relativistic energy formula. Equivalently,

$\dfrac{Nk_BT}{2}=Mc^2$

4th. Microscopical degrees of freedom and minimal length ( or area quantization). The number of allowed microscopical quanta or microstates can not exceed and must match in the extreme case the ration of the area available and the square of Planck’s length ( or some other squared fundamental length). In other words, the number of bits can not overcome the area of a ball in Planck’s units. EQuivalently, the (hyper)area must be quantized (through a number N). Mathematically speaking,

$N=\dfrac{A}{l_{p}^{2}}$

Then, we plug the hypothesis 2 and 3 into 1, to have:

$F=\dfrac{2Mc^2}{k_B N}2\pi k_B \dfrac{mc}{\hbar}$

and now we use 4, in order to get

$F=\dfrac{2Mc^2}{k_B A}l_{p}^{2} 2\pi k_B \dfrac{mc}{\hbar}$

that is

$F=4\pi Mm \dfrac{l_{p}^{2}c^3}{\hbar A}$

And now, recalling that in 3 spatial dimensions, a ball has an area $A=4\pi r^2$ and that the Newton’s constant of gravity is indeed given as function of Planck’s length as $G=\dfrac{l_{p}^{2}c^3}{\hbar}$, we have what we wanted to derive

$F=4\pi Mm \dfrac{G}{4\pi r^2}$

i.e. the Newton’s gravitational law has been derived from the entropic force too

$F=G\dfrac{Mm}{r^2}$

It is done. Is it just a trick or something deeper is behing all this stuff? Nobody knows for sure…People think he is probably wrong, but there are a whole line of research opened from his works. It is quite remarkable his approach is quite general and he suggests that every fundamental force is “entropic” or “emergent”, i.e., also electromagnetic fields, or even Yang-Mills fields could be entropic according to this approach.

Is he right? Time will tell…Some doubts arise from the fact that he has only derived the “temporal” components of Einstein Field Equations for gravity (a.k.a. newtonian gravitation) but, indeed, he and other physicists have been able to derive the remaining components as well. Perhaps, the strongest critics comes from neutron interferometry results. However, theoretical ideas like this one, like extra dimensions, could be saved by some clever argument.

## LOG#029. Interstellar trips in SR.

My final article dedicated to the memory of Neil Armstrong. The idea is to study quantitatively the relativistic rocket motion with numbers, after all we have deduced the important formulae, and we will explain what is happening in the two frames: S’-frame (in motion), S-frame (in rest on Earth). There are many “variations” of this problem, also called “Langevin’s paradox” or “the Langevin’s interstellar trip” problem by some authors. Here, we will follow the approach suggested in the book Gravitation, by Misner, Wheeler and Thorne, and we will study the interstellar trip (in the frame of special relativity) with the following conditions:

1st. Spacetime is locally minkovskian (i.e., spacetime is flat). The solution we will expose would not be valid in the case of an interstellar trip to a very far away quasar, or a very very long distance (about thousand millions of lightyears) where we should take into account the effect of the expansion of the Universe, i.e., that on large scales, spacetime is curved (in particular, accordingly to the current data, it is pseudoriemannian). Thus, we can use special relativity in order to calculate distances, velocities and accelerations from the purely kinematical sense. After all, it is logical, since General Relativity says that locally, in small enough regions of spacetime, spacetime is described by a minkovskian metric.

2nd. We select a 4 stage accelerated motion with our rocket. We will assume that our rocket is 100% efficient in the sense it uses photons as propellant particles. The four stages are: acceleration from rest to g (acceleration step 1), decceleration to rest with -g until we approach the destination (that would be the one-way trip), acceleration with -g (seen from the S-frame) and decceleration with +g ( with respect to the S-frame) to reach Earth again in rest at the end of the round-way trip. Schematically, we can draw a spacetime sketch of this 4 stage journey:

Please, note the symmetry of the procedure and the different travel steps.

3rd. We set g=9.8m/s², or, as we saw in one of our previous posts, so we use g=1.03lyr/yr² (in units where c=1).

4th. We can not refuel during the travel.

5th. In the case we return to Earth, we proceed to come back after we stop at the destination immediately. In this case, there is no refuel of the starship or rocket in any point of the trip.

6th. We neglect any external disturbance which can stop us or even destroy us, e.g., micrometeorites, cosmic radiations, comets, and any other body that could alter our route. Indeed, this kind of stuff has to be seriously considered in any realistic travel, but we want to solve an ideal problem consisting in an ideal interstellar trip according to the current knowledge.

The main quantities we have to compute are:

1. In the S’-frame of the rocket $\tau, 2\tau,4\tau$, in years. They are, respectively, the time we are accelerating with +g, the time we are decelerating with -g, and the total time accelerating supposing we return to earth immediately from our destination target.

2. Distance in which we are accelerating (x, in light years), seen from the S-frame (we will not discuss the problem in the S’-frame, since it involves lenght contraction and it is more subtle in the calculations). According to our previous studies, we have

$\boxed{x=\dfrac{c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}$

3. Maximum depth in space D=2x. This allows us to select our target or destination to set the remaining parameters of the trip. Due to the symmetry of our problem, we have

$\boxed{D=2x=\dfrac{2c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}$

4. Total length travelled by the spaceship/rocket (in the S’frame) in the roundtrip ( to the destination and back).

$\boxed{L=4x=\dfrac{4c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}$

5. Maximum speed to an specific destination, after accelerating in a given proper time. It reads:

$\boxed{V=c\tanh \left(\dfrac{g\tau}{c}\right)}$

Indeed, the relativistic gamma factor is $\gamma =\cosh \dfrac{g\tau}{c}$

6. S-frame duration t(years) of the stage 1 (acceleration phase 1). It is

$\boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

7. One way duration of the trip (according to the S-frame):

$\boxed{t'=2t=\dfrac{2c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

8. Total duration of the trip in the S-frame ( round way trip):

$\boxed{t''=4t=\dfrac{4c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

9. Mass ratio of the final mass with the initial mass of the spaceship after 1 stage, 2 stages ( one way trip) and the 4 stages total round way trip:

$\boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{g\tau}{c}\right)}$

$\boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{2g\tau}{c}\right)}$

$\boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{4g\tau}{c}\right)}$

10. Fuel mass vs. payload mass ratio (FM/PM) after 1 stage, 2 stages (until the destination) and the total trip:

$\boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{g\tau}{c}\right)-1}$

$\boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{2g\tau}{c}\right)-1}$

$\boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{4g\tau}{c}\right)-1}$

We can obtain the following data using these expressions varying the proper time:

Here in the last two entries, data provided to be out of the limits of our calculator ( I used Libre Office to compute them). Now, we can make some final observations:

1st. We can travel virtually everywhere in the observable universe in our timelife using a photon rocket, a priori. However, it shows that the mass-ratio turns it to be theoretically impossible.

2nd. Remember that in the case we select very long distances, these calculations are not valid since we should use General Relativity to take into account the expansion of spacetime.

3rd. Compare $\tau$ with t, $2\tau$ with $2t$, $4\tau$ with $4t$. For instance, for $4\tau=40$ then we get 57700 years, and v=0.99999999773763c, FM/PM=30000 (multiplying por m-kilograms gives the fuel mass).

4th. In SR, the technological problems are associated to the way photon rockets have, the unfavorable mass ratios (of final mass with initial mass) and fuel mass/payload rations neccesary in the voyage (supposing of course, current physics and that we can not refuel during the trip).

5th. You can choose some possible distance destinations (D=2x) and work out the different parameters of the travel with the above equations.

In this way, for instance, for some celebrated known space marks, arriving at complete stop ( you can make other assumptions and see how the answer changes varying g or varying the arrival velocity too), we can easily get:

Example 1: 4.3 ly Nearest star (Proxima Centauri) in $t=2\tau=3.6$ yrs (S’-frame).

Example 2: 27 ly Vega (Contact movie and book) star in $t=2\tau=6.6$ yrs (S’-frame).

Example 3: 30000ly, our galactic center, in $t=2\tau=20$ yrs (S’-frame).

Example 4: 2000000ly , the Andromeda galaxy, in $t=2\tau=28$ yrs(S’-frame).

Example 5: Generally, you can travel n ly anywhere (neglecting curved spacetime) in $t=2\tau=1.94 \cosh^{-1}(n/1.94+1)$ years (S’-frame).

## LOG#028. Rockets and relativity.

The second post in this special thread of 3 devoted to Neil Armstrong memory has to do with rocketry.

Firstly, for completion, we are going to study the motion of a rocket in “vacuum” according to classical physics. Then, we will deduce the relatistic rocket equation and its main properties.

CLASSICAL NON-RELATIVISTIC ROCKETS

The fundamental law of Dynamics, following Sir Isaac Newton, reads:

$\mathbf{F}=\dfrac{d\mathbf{p}}{dt}$

Suppose a rocket with initial mass $M_i$ and initial velocity $u_i=0$. It ejects mass of propellant “gas” with “gas speed” (particles of gas have a relative velocity or speed with respect to the rest observer when the rocket move at speed $\mathbf{v}$) equals to $u_0$ (note that the relative speed will be $u_{rel} and the propellant mass is$latex m_0\$. Generally, this speed is also called “exhaust velocity” by engineers. The motion of a variable mass or rocket is given by the so-called Metcherski’s equation:

$\boxed{M\dfrac{d\mathbf{v}}{dt}=-\mathbf{u_0}\dfrac{dM}{dt}+\mathbf{F}}$

where $-\mathbf{u_0}=\mathbf{v_{gas}}-\mathbf{v}$. The Metcherski’s equation can be derived as follows: the rocket changes its mass and velocity so $M'=M+dM$ and $V'=V+dV$, so the change in momentum is equal to $M'V'=(M+dM)(V+dV)$, plus an additional term $v_{gas}dv_{gas}$ and $-mV$. Therefore, the total change in momentum:

$dP=Fdt=(M+dM)(V+dV)+v_{gas}dv_{gas}-mV$

Neglecting second order differentials, and setting the conservation of mass (we are in the non-relativistic case)

$dM+dm_{gas}=0$

we recover

$MdV=v_{rel}dM+Fdt$

that represents (with the care of sign in relative speed) the Metcherski equation we have written above.

Generally speaking, the “force” due to the change in “mass” is called thrust.  With no external force, from the remaining equation of the thrust and velocity, and it can be easily integrated

$u_f=-u_0\int_{M_i}^{M_f}dM$

and thus we get the Tsiolkowski’s rocket equation:

$\boxed{\mathbf{u_f}=\mathbf{u_0}\ln \dfrac{M_i}{M_f}}$

Engineers use to speak about the so-called mass ratio $R=\dfrac{M_f}{M_i}$, although sometimes the reciprocal definition is also used for such a ratio so be aware, and in terms of this the Tsiokolski’s equation reads:

$\boxed{\mathbf{u_f}=\mathbf{u_0}\ln \dfrac{1}{R}}$

We can invert this equation as well, in order to get

$\boxed{R=\dfrac{M_f}{M_i}=\exp\left(-\dfrac{u_f}{u_0}\right)}$

Example: Calculate the fraction of mass of a one-stage rocket to reach the Earth’s orbit. Typical values for $u_f=8km/s$ and $u_0=4km/s$ show that the mass ratio is equal to $R=0.14$. Then, only the $14\%$ of the initial mass reaches the orbit, and the remaining mass is fuel.

Multistate rockets offer a good example of how engineer minds work. They have discovered that a multistage rocket is more effective than the one-stage rocket in terms of maximum attainable speed and mass ratios. The final n-stage lauch system for rocketry states that the final velocity is the sum of the different gains in the velocity after the n-th stage, so we can obtain

$\displaystyle{u_f=\sum_{i=1}^{n}u_i^f=u_1^f+\cdots+u_n^f}$

After the n-th step, the change in velocity reads

$u_i^f=c_i\ln \dfrac{1}{R_i}$

where the i-th mass ratios are definen recursively as the final mass in the n-th step and the initial mass in that step, so we have

$\displaystyle{u_f=\sum_i c_i\ln \dfrac{1}{R_i}}$

and we define the total mass ratio:

$\displaystyle{R_T=\prod_i R_i}$

If the average effective rocket exhaust velocity is the same in every step/stage, e.g. $c_i=c$, we get

$\displaystyle{u_f=c\ln \left( \prod_{i=1}^{n} R_i^{-1}\right)}$

or

$\displaystyle{u_f=c \ln \left[ \left(\dfrac{M_0}{M_f}\right)_1\left(\dfrac{M_0}{M_f}\right)_2\cdots \left(\dfrac{M_0}{M_f}\right)_n\right]=c\ln \left[\left(\dfrac{M_0}{M_f}\right)_T\right]}$

The influence of the number of steps, for a given exhaust velocity, in the final attainable velocity can be observed in the next plots:

RELATIVISTIC ROCKETS

We proceed now to the relativistic generalization of the previous rocketry. An observer in the laboratory frame observes that total momentum is conserved, of course, and so:

$M'du'=-u'_0dM'$

where $du'$ is the velocity increase in the rocket with a rest mass M’ in the instantaneous reference frame of the moving rocket S’. It is NOT equal to its velocity increase measured in the unprimed reference frame, du. Due to the addition theorem of velocities in SR, we have

$u+du=\dfrac{u+du'}{1+\dfrac{udu'}{c^2}}$

where u is the instantenous velocity of the rocket with respect to the laboratory frame S. We can perform a Taylor expansion of the denominator in the last equation, in order to obtain:

$u+du=(u+du')\left(1-\dfrac{udu'}{c^2}\right)$

and then

$u+du=u+du'\left(1-\dfrac{u^2}{c^2}\right)$

and finally, we get

$du'=\dfrac{du}{1-\dfrac{u^2}{c^2}}=\gamma^2_u du$

Plugging this equation into the above equation for mass (momentum), and integrating

$\displaystyle{\int_{0}^{u_f}\dfrac{du}{1-\dfrac{u^2}{c^2}}=-u'_0\int_{M'_0}^{M'_f}dM}$

we deduce that the relativistic version of the Tsiolkovski’s rocket equation, the so-called relativistic rocket equation, can be written as:

$\dfrac{c}{2}\ln \dfrac{1+\dfrac{u_f}{c}}{1-\dfrac{u_f}{c}}=u'_0\ln\dfrac{M'_i}{M'_f}$

We can suppress the primes if we remember that every data is in the S’-frame (instantaneously), and rewrite the whole equation in the more familiar way:

$\boxed{u_f=c\dfrac{1-\left(\dfrac{M_f}{M_0}\right)^{\frac{2u_0}{c}}}{1+\left(\dfrac{M_f}{M_0}\right)^{\frac{2u_0}{c}}}=c\dfrac{1-R^{\frac{2u_0}{c}}}{1+R^{\frac{2u_0}{c}}}}$

where the mass ratio is defined as before $R=\dfrac{M_f}{M_i}$. Now, comparing the above equation with the rapidity/maximum velocity in the uniformly accelerated motion:

$u_f=c\tanh \left(\dfrac{g\tau}{c}\right)$

we get that relativistic rocket equation can be also written in the next manner:

$u_f=c\tanh \left[ -\dfrac{u_0}{c}\ln \left(\dfrac{1}{R}\right)\right]$

or equivalently

$u_f=c\tanh \left[ \dfrac{u_0}{c}\ln R\right]$

since we have in this case

$\dfrac{g\tau}{c}=-\dfrac{u_0}{c}\ln \left(\dfrac{1}{R}\right)=\dfrac{u_0}{c}\ln R$

and thus

$R^{\frac{u_0}{c}}=\left(\dfrac{M_f}{M_i}\right)^{\frac{u_0}{c}}=\exp \left(-\dfrac{g\tau}{c}\right)$

If the propellant particles move at speed of light, e.g., they are “photons” or ultra-relativistic particles that move close to the speed of light we have the celebrated “photon rocket”. In that case, setting $u_0=c$, we would obtain that:

$\boxed{u_f=c\dfrac{1-\left(\dfrac{M_f}{M_0}\right)^{2}}{1+\left(\dfrac{M_f}{M_0}\right)^{2}}=c\dfrac{1-R^{2}}{1+R^{2}}=c\tanh \ln R}$

and where for the photon rocket (or the ultra-relativistic rocket) we have as well

$\dfrac{g\tau}{c}=-\ln \left(\dfrac{1}{R}\right)=\ln R$

Final remark: Instead of the mass ratio, sometimes is more useful to study the ratio fuel mass/payload. In that case, we set $M_f=m$ and $M_0=m+M$, where M is the fuel mass and m is the payload. So, we would write

$R=\dfrac{m}{m+M}$

so then the ratio fuel mass/payload will be

$\dfrac{M}{m}=R^{-1}-1=\exp \left(\dfrac{g\tau}{c}\right)-1$

We are ready to study the interstellar trip with our current knowledge of Special Relativity and Rocketry. We will study the problem in the next and final post of this fascinating thread. Stay tuned!

## LOG#027. Accelerated motion in SR.

Hi, everyone! This is the first article in a thread of 3 discussing accelerations in the background of special relativity (SR). They are dedicated to Neil Armstrong, first man on the Moon! Indeed,  accelerated motion in relativity has some interesting and sometimes counterintuitive results, in particular those concerning the interstellar journeys whenever their velocities are close to the speed of light(i.e. they “are approaching” c).

Special relativity is a theory considering the equivalence of every  inertial frame ( reference frames moving with constant relative velocity are said to be inertial frames) , as it should be clear from now, after my relativistic posts! So, in principle, there is nothing said about relativity of accelerations, since accelerations are not relative in special relativity ( they are not relative even in newtonian physics/galilean relativity). However, this fact does not mean that we can not study accelerated motion in SR. The own kinematical framework of SR allows us to solve that problem. Therefore, we are going to study uniform (a.k.a. constant) accelerating particles in SR in this post!

First question: What does “constant acceleration” mean in SR?   A constant acceleration in the S-frame would give to any particle/object a superluminal speed after a finite time in non-relativistic physics! So, of course, it can not be the case in SR. And it is not, since we studied how accelerations transform according to SR! They transform in a non trivial way! Moreover, a force growing beyond the limits would be required for a “massive” particle ( rest mass $m\neq 0$). Suppose this massive particle (e.g. a rocket, an astronaut, a vehicle,…) is at rest in the initial time $t=t'=0$, and it accelerates in the x-direction (to be simple with the analysis and the equations!). In addition, suppose there is an observer left behind on Earth(S-frame), so Earth is at rest with respect to the moving particle (S’-frame). The main answer of SR to our first question is that we can only have a constant acceleration in the so-called instantaneous rest frame of the particle.  We will call that acceleration “proper acceleration”, and we will denote it by the letter $\alpha$. In fact, in many practical problems, specially those studying rocket-ships, the acceleration is generally given the same magnitude as the gravitational acceleration on Earth ($\alpha=g\approx 9.8ms^{-2}\approx 10 ms^{-2}$).

Second question: What are the observed acceleration in the different frames? If the instantaneous rest frame S’ is an inertial reference frame in some tiny time $dt'$, at the initial moment, it has the same velocity as the particle (rocket,…) in the S-frame, but it is not accelerated, so the velocity in the S’-frame vanishes at that time:

$\mathbf{u}'=(0,0,0)$

Since the acceleration of the particle is, in the S’-frame, the proper acceleration, we get:

$\mathbf{a}'=(a'_x,0,0)=(\alpha,0,0)=(g,0,0)=\mbox{constant}$

Using the transformation rules for accelerations in SR we have studied, we get that the instantaneous acceleration in the S-frame is given by

$\mathbf{a}=(a_x,0,0)=\left(\dfrac{g}{\gamma^3},0,0\right)$

Since the relative velocity between S and S’ is always the same to the moving particle velocity in the S-frame, the following equation holds

$v=u_x$

We do know that

$a_x=\dfrac{du_x}{dt}=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g$

Due to time dilation

$dt'=dt/\gamma$

so in the S-frame, the particle moves with the velocity

$du_x=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g dt$

We can now integrate this equation

$\int_0^{u_x}\dfrac{1}{(c^2-u_x^2)^{3/2}}du_x=\dfrac{g}{c^3}\int_0^t dt$

The final result is:

$\boxed{u_x=\dfrac{g t}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}}$

We can check some limit cases from this relativistic result for uniformly accelerated motion in SR.

1st. Short time limit: $gt<< c\longrightarrow u_x\approx gt=\alpha t$. This is the celebrated nonrelativistic result, with initial speed equal to zero (we required that hypothesis in our discussion above).

2nd. Long time limit: $t\rightarrow \infty$. In this case, the number one inside the root is very tiny compared with the term depending on acceleration, so it can be neglected to get $u_x\approx \dfrac{gt}{gt/c}=c$. So, we see that you can not get a velocity higher than the speed of light with the SR framework at constant acceleration!

Furthermore, we can use the definition of relativistic velocity in order to integrate the associated differential equation, and to obtain the travelled distance as a function of $t$, i.e. $x(t)$, as follows

$u_x=\dfrac{dx}{dt}=\dfrac{gt}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}$

$\int_0^x dx=\int_0^t\dfrac{gt dt}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}=\int_0^t\dfrac{ctdt}{\sqrt{\dfrac{c^2}{g^2}+t^2}}$

We can perform the integral with the aid of the following known result ( see,e.g., a mathematical table or use a symbolic calculator or calculate the integral by yourself):

$\int \dfrac{ctdt}{\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}}=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}+\mbox{constant}=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}+C$

From this result, and the previous equation, we get the so-called relativistic path-time law for uniformly accelerated motion in SR:

$x=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}-\dfrac{c^2}{g}$

or equivalently

$\boxed{x=x(t)=\dfrac{c^2}{g}\left(\sqrt{1+\left(\dfrac{gt}{c}\right)^2}-1\right)}$

For consistency, we observe that in the limit of short times, the terms in the big brackets approach $1+\frac{1}{2}\left(\frac{gt}{c}\right)^2$, in order to get $x\approx \frac{1}{2}gt^2$, so we obtain the nonrelativistic path-time relationship $x=\frac{1}{2}gt^2$ with $g=a_x$. In the limit of long times, the terms inside the brackets can be approximated to $gt/c$, and then, the final result becomes $x\approx ct$. Note that the velocity is not equal to the speed of light, this result is a good approximation whenever the time is “big enough”, i.e., it only works for “long times” asymptotically!

And finally, we can write out the transformations of accelaration between the two frames in a explicit way:

$a_x=\left[1-\dfrac{\left(\dfrac{gt}{c}\right)^2}{1+\left(\dfrac{gt}{c}\right)^2}\right]^{3/2}g$

that is

$\boxed{a_x=\dfrac{1}{\left[1+\left(\dfrac{gt}{c}\right)^2\right]^{3/2}}g}$

Check 1: For short times, $a_x\approx g=\mbox{constant}$, i.e., the non-relativistic result, as we expected!

Check 2: For long times, $a_x\approx \dfrac{c^3} {g^2t^3}\rightarrow 0$. As we could expect, the velocity increases in such a way that “saturates” its own increasing rate and the speed of light is not surpassed. The fact that the speed of light can not be surpassed or exceeded is the unifying “theme” through special relativity, and it rest in the “noncompact” nature of the Lorentz group due to the $\gamma$ factor, since it would become infinity at v=c for massive particles.

It is inevitable: as time passes, a relativistic treatment is indispensable, as the next figures show

The next table is also remarkable (it can be easily built with the formulae we have seen till now with any available software):

Let us review the 3 main formulae until this moment

$\boxed{a_x=\dfrac{1}{\left[1+\left(\dfrac{gt}{c}\right)^2\right]^{3/2}}g}$ $\boxed{u_x=\dfrac{\alpha t}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}}$ $\boxed{x=x(t)=\dfrac{c^2}{g}\left(\sqrt{1+\left(\dfrac{gt}{c}\right)^2}-1\right)}$

We have calculated these results in the S-frame, it is also important and interesting to calculate the same stuff in the S’-frame of the moving particle. The proper time $\tau=t'$ is defined as:

$\boxed{d\tau=dt\sqrt{1-\left(\dfrac{u_x}{c}\right)^2}}$

Therefore,

$d\tau=dt\left[1-\dfrac{\left(\dfrac{gt}{c}\right)^2}{1+\left(\dfrac{gt}{c}\right)^2}\right]^{1/2}$

We can perform the integral as before

$\displaystyle{\int_0^\tau d\tau=\int_0^t\dfrac{dt}{\sqrt{1+\left(\dfrac{gt}{c}\right)^2}}}$

and thus

$\tau=\dfrac{c}{g}\int_0^\tau\dfrac{dt}{\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}}=\dfrac{c}{g}\ln \left(\dfrac{gt}{c}+\sqrt{\left(\dfrac{gt}{c}\right)^2+1}\right)\bigg|_0^t$

Finally, the proper time(time measured in the S’-frame) as a function of the elapsed time on Earth (S-frame) and the acceleration is given by the very important formula:

$\boxed{\tau=\dfrac{c}{g}\ln \left(\dfrac{gt}{c}+\sqrt{1+\left(\dfrac{gt}{c}\right)^2}\right)}$

And now, let us set $z=gt/c$, therefore we can write the above equation in the following way:

$\dfrac{g\tau}{c}=\ln \left( z+\sqrt{1+z^2}\right)$

Remember now, from our previous math survey, that $\sinh^{-1}z=\ln \left( z+\sqrt{1+z^2}\right)$, so we can invert the equation in order to obtain t as function of the proper time since:

$\boxed{\tau=\dfrac{c}{g}\sinh^{-1}\left(\dfrac{gt}{c}\right)}$

$\boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$

Inserting this last equation in the relativistic equation path-time for the uniformly accelerated body in SR, we obtain:

$x=x(\tau)=\dfrac{c^2}{g}\left(\sqrt{1+\sinh^2\left(\dfrac{g\tau}{c}\right)}-1\right)$

i.e.,

$\boxed{x=x(\tau)=\dfrac{c^2}{g}\left[\cosh \left(\dfrac{g\tau}{c}\right)-1\right]}$

Similarly, we can calculate the velocity-proper time law. Previous equations yield

$u_x=\dfrac{c\sinh\left(\dfrac{g\tau}{c}\right)}{\sqrt{1+\sinh^2\left(\dfrac{g\tau}{c}\right)}}=\dfrac{c\sinh \left(\dfrac{g\tau}{c}\right)}{\cosh \left(\dfrac{g\tau}{c}\right)}$

and thus the velocity-proper time law becomes

$\boxed{u_x=c\tanh \left(\dfrac{g\tau}{c}\right)}$

Remark: this last result is compatible with a rapidity factor $\varphi= \left(\dfrac{g\tau}{c}\right)$.

Remark(II): $a_x=\dfrac{du_x}{dt}=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g=\left(1-\tanh^2\left(\dfrac{g\tau}{c}\right)\right)^{3/2}g=\dfrac{1}{\cosh^3\left(\dfrac{g\tau}{c}\right)}g$. From this, we can read the reason why we said before that constant acceleration is “meaningless” unless we mean or fix certain proper time in the S’-frame since whenever we select a proper time, and this last relationship gives us the “constant” acceleration observed from the S-frame after the transformation. Of course, from the S-frame, as this function shows, acceleration is not “constant”, it is only “instantaneously” constant. We have to take care in relativity with the meaning of the words. Mathematics is easy and clear and generally speaking it is more precise than “words”, common language is generally fuzzy unless we can explain what we are meaning!

As the final part of this log entry, let us summarize the time-proper time, velocity-proper time, acceleration-proper time-proper acceleration and distance- proper time laws for the S’-frame:

$\boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}$ $\boxed{u_x=c\tanh \left(\dfrac{g\tau}{c}\right)}$ $\boxed{a_x=\dfrac{1}{\cosh^3\left(\dfrac{g\tau}{c}\right)}g}$ $\boxed{x=x(\tau)=\dfrac{c^2}{g}\left[\cosh \left(\dfrac{g\tau}{c}\right)-1\right]}$

My last paragraph in this post is related to express the acceleration $g\approx 10ms^{-2}$ in a system of units where space is measured in lightyears (we take c=300000km/s) and time in years (we take 1yr=365 days). It will be useful in the next 2 posts:

$g=10\dfrac{m}{s^2}\dfrac{1ly}{9.46\cdot 10^{15}m}\dfrac{9.95\cdot 10^{14}s^2}{1yr^2}=1.05\dfrac{lyr}{yr^2}\approx 1\dfrac{lyr}{yr^2}$

Another election you can choose is

$g=9.8\dfrac{m}{s^2}=1.03\dfrac{lyr}{yr^2}\approx 1\dfrac{lyr}{yr^2}$

so there is no a big difference between these two cases with terrestrial-like gravity/acceleration.

## LOG#026. Boosts, rapidity, HEP.

In euclidean two dimensional space, rotations are easy to understand in terms of matrices and trigonometric functions. A plane rotation is given by:

$\boxed{\begin{pmatrix}x'\\ y'\end{pmatrix}=\begin{pmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}}\leftrightarrow \boxed{\mathbb{X}'=\mathbb{R}(\theta)\mathbb{X}}$

where the rotation angle is $\theta$, and it is parametrized by $0\leq \theta \leq 2\pi$.

Interestingly, in minkovskian two dimensional spacetime, the analogue does exist and it is written in terms of matrices and hyperbolic trigonometric functions. A “plane” rotation in spacetime is given by:

$\boxed{\begin{pmatrix}ct'\\ x'\end{pmatrix}=\begin{pmatrix}\cosh \varphi & -\sinh \varphi \\ -\sinh \varphi & \cosh \varphi \end{pmatrix}\begin{pmatrix}ct\\ x\end{pmatrix}}\leftrightarrow \boxed{\mathbb{X}'=\mathbb{L}(\varphi)\mathbb{X}}$

Here, $\varphi = i\psi$ is the so-called hiperbolic rotation angle, pseudorotation, or more commonly, the rapidity of the Lorentz boost in 2d spacetime. It shows that rapidity are a very useful parameter for calculations in Special Relativity. Indeed, it is easy to check that

$\mathbb{L}(\varphi_1+\varphi_2)=\mathbb{L}(\varphi_1)\mathbb{L}\mathbb(\varphi_2)$

So, at least in the 2d spacetime case, rapidities are “additive” in the written sense.

Firstly, we are going to guess the relationship between rapidity and velocity in a single lorentzian spacetime boost. From the above equation we get:

$ct'=ct\cosh \varphi -x\sinh \varphi$

$x'=-ct\sinh \varphi +x\cosh \varphi$

Multiplying the first equation by $\cosh \varphi$ and the second one by $\sinh \varphi$, we add the resulting equation to obtain:

$ct'\cosh\varphi+x'\sinh \varphi =ct\cosh^2 \varphi -ct\sinh^2 \varphi =ct$

that is

$ct'\cosh\varphi+x'\sinh \varphi =ct$

From this equation (or the boxed equations), we see that $\varphi=0$ corresponds to $x'=x$ and $t'=t$. Setting $x'=0$, we deduce that

$x'=0=-ct\sinh \varphi +x\cosh \varphi$

and thus

$ct\tanh \varphi =x$ or $x=ct\tanh\varphi$.

Since $t\neq 0$, and the pseudorotation seems to have a “pseudovelocity” equals to $V=x/t$, the rapidity it is then defined through the equation:

$\boxed{\tanh \varphi=\dfrac{V}{c}=\beta}\leftrightarrow\mbox{RAPIDITY}\leftrightarrow\boxed{\varphi=\tanh^{-1}\beta}$

If we remember what we have learned in our previous mathematical survey, that is,

$\tanh^{-1}z=\dfrac{1}{2}\ln \dfrac{1+z}{1-z}=\sqrt{\dfrac{1+z}{1-z}}$

We set $z=\beta$ in order to get the next alternative expression for the rapidity:

$\varphi=\ln \sqrt{\dfrac{1+\beta}{1-\beta}}=\dfrac{1}{2}\ln \dfrac{1+\beta}{1-\beta}\leftrightarrow \exp \varphi=\sqrt{\dfrac{1+\beta}{1-\beta}}$

In experimental particle physics, in general 3+1 spacetime, the rapidity definition is extended as follows. Writing, from the previous equations above,

$\sinh \varphi=\dfrac{\beta}{\sqrt{1-\beta^2}}$

$\cosh \varphi=\dfrac{1}{\sqrt{1-\beta^2}}$

and using these two last equations, we can also write momenergy components using rapidity in the same fashion. Suppose that for some particle(objetc), its  mass is $m$, its energy is $E$, and its (relativistic) momentum is $\mathbf{P}$. Then:

$E=mc^2\cosh \varphi$

$\lvert \mathbf{P} \lvert =mc\sinh \varphi$

From these equations, it is trivial to guess:

$\varphi=\tanh^{-1}\dfrac{\lvert \mathbf{P} \lvert c}{E}=\dfrac{1}{2}\ln \dfrac{E+\lvert \mathbf{P} \lvert c}{E-\lvert \mathbf{P} \lvert c}$

This is the completely general definition of rapidity used in High Energy Physics (HEP), with a further detail. In HEP, physicists used to select the direction of momentum in the same direction that the collision beam particles! Suppose we select some orientation, e.g.the z-axis. Then, $\lvert \mathbf{P} \lvert =p_z$ and rapidity is defined in that beam direction as:

$\boxed{\varphi_{hep}=\tanh^{-1}\dfrac{\lvert \mathbf{P}_{beam} \lvert c}{E}=\dfrac{1}{2}\ln \dfrac{E+p_z c}{E-p_z c}}$

In 2d spacetime, rapidities add nonlinearly according to the celebrated relativistic addition rule:

$\beta_{1+2}=\dfrac{\beta_1+\beta_2}{1+\frac{\beta_1\beta_2}{c^2}}$

Indeed, Lorentz transformations do commute in 2d spacetime since we boost in a same direction x, we get:

$L_1^xL_2^x-L_2^xL_1^x=0$

with

$L_1^x=\begin{pmatrix}\gamma_1 & -\gamma_1\beta_1\\ -\gamma_1\beta_1 &\gamma_1 \end{pmatrix}$

$L_2^x=\begin{pmatrix}\gamma_2 & -\gamma_2\beta_2\\ -\gamma_2\beta_2 &\gamma_2 \end{pmatrix}$

This commutativity is lost when we go to higher dimensions. Indeed, in spacetime with more than one spatial direction that result is not true in general. If we build a Lorentz transformation with two boosts in different directions $V_1=(v_1,0,0)$ and $V_2=(0,v_2,0)$, the Lorentz matrices are ( remark for experts: we leave one direction in space untouched, so we get 3×3 matrices):

$L_1^x=\begin{pmatrix}\gamma_1 & -\gamma_1\beta_1 &0\\ -\gamma_1\beta_1 &\gamma_1 &0\\ 0& 0& 1\end{pmatrix}$

$L_2^y=\begin{pmatrix}\gamma_2 & 0&-\gamma_2\beta_2\\ 0& 1& 0\\ -\gamma_2\beta_2 & 0&\gamma_2 \end{pmatrix}$

and it is easily checked that

$L_1^xL_2^y-L_2^yL_1^x\neq 0$

Finally, there is other related quantity to rapidity that even experimentalists do prefer to rapidity. It is called: PSEUDORAPIDITY!

Pseudorapidity, often denoted by $\eta$ describes the angle of a particle relative to the beam axis. Mathematically speaking is:

$\boxed{\eta=-\ln \tan \dfrac{\theta}{2}}\leftrightarrow \mbox{PSEUDORAPIDITY}\leftrightarrow \boxed{\exp (\eta)=\dfrac{1}{\tan\dfrac{\theta}{2}}}$

where $\theta$ is the angle between the particle momentum $\mathbf{P}$  and the beam axis. The above relation can be inverted to provide:

$\boxed{\theta=2\tan^{-1}(e^{-\eta})}$

The pseudorapidity in terms of the momentum is given by:

$\boxed{\eta=\dfrac{1}{2}\ln \dfrac{\vert \mathbf{P}\vert +P_L}{\vert \mathbf{P}\vert -P_L}}$

Note that, unlike rapidity, pseudorapidity depends only on the polar angle of its trajectory, and not on the energy of the particle.

In hadron collider physics,  and other colliders as well, the rapidity (or pseudorapidity) is preferred over the polar angle because, loosely speaking, particle production is constant as a function of rapidity. One speaks of the “forward” direction in a collider experiment, which refers to regions of the detector that are close to the beam axis, at high pseudorapidity $\eta$.

The rapidity as a function of pseudorapidity is provided by the following formula:

$\boxed{\varphi=\ln\dfrac{\sqrt{m^2+p_T^2\cosh^2\eta}+p_T\sinh \eta}{\sqrt{m^2+p_T^2}}}$

where $p_T$ is the momentum transverse to the direction of motion and m is the invariant mass of the particle.

Remark: The difference in the rapidity of two particles is independent of the Lorentz boosts along the beam axis.

Colliders measure physical momenta in terms of transverse momentum $p_T$ instead of the momentum in the direction of the beam axis (longitudinal momentum) $P_L=p_z$, the polar angle in the transverse plane (genarally denoted by $\phi$) and pseudorapidity $\eta$. To obtain cartesian momenta $(p_x,p_y,p_z)$  (with the z-axis defined as the beam axis), the following transformations are used:

$p_x=P_T\cos\phi$

$p_y=P_T\sin\phi$

$p_z=P_T\sinh\eta$

Thus, we get the also useful relationship

$\vert P \vert=P_T\cosh\eta$

This quantity is an observable in the collision of particles, and it can be measured as the main image of this post shows.