The Spectrum Of Riemannium

Home - Page 24 << 1 2 … 22 23 24 25 >>

LOG#019. Triangle mnemonics.

Posted on 2012/06/17 by amarashikiAugust 30, 2019

Today, we are going to learn some interesting mnemomic tricks using the celebrated Pythagorean Theorem from your young years at the school. We will be using some simple triangles to remember some of the wonderful formulae of Special Relativity. It is quite nice and surprising that basic euclidean trigonometry tools are very helpful in the abstract realm of relativistic theories, a theory based on non-euclidean geometry! It is one of the great powers of Mathematics. Its amazing ability to model big things with simple pictures and equations. There is no other language like the mathematical language, and despite its terrible looking sometimes, it is awesome and beautiful most of the time. Of course, like it happens in Arts, mathematical beauty can be very subjective. So, you have to be trained in order to admire and love its great features.

The first triangle we are going to study is this one:

If we apply the Pythagorean Theorem to this triangle, we get:

$X^2+(c\tau)^2=(ct)^2$

This equation is indeed related to the square or “norm” of the spacetime vector (a.k.a., spacetime event)

$\mathbb{X}\cdot \mathbb{X}=-c^2\tau^2=\mathbf{X}^2-c^2t^2$

and thus, in complete agreement with the above triangle

$\mathbf{X}^2+(c\tau)^2=(ct)^2$

And it can be easily stated with words or concepts as:

$\boxed{(\mbox{SPACE})^2+(\mbox{PROPER TIME})^2=(\mbox{TIME})^2}$

You also get:

$\sin \varphi= \dfrac{\tau}{t}$

$\cos \varphi=\dfrac{X}{ct}$

$\tan \varphi = \dfrac{X}{c\tau}=\dfrac{v}{c}=\beta$

i.e.

$\boxed{\tan \varphi=\beta}$

Moreover,

$\gamma^2=\dfrac{1}{1-\beta^2}$

Elementary trigonometry yields:

$1-\tan ^2\varphi=2-\sec^2\varphi=\dfrac{2\cos^2\varphi-1}{\cos^2\varphi}=\dfrac{\cos 2\varphi}{\cos^2\varphi}$

$\boxed{\gamma^2=\dfrac{\cos^2\varphi}{\cos 2\varphi}}$

In the same way, we can draw other cool triangle:

Now, we can relate the Pythagorean theorem with the squared momenergy vector:

$\mathbb{P}\cdot \mathbb{P}=-(mc)^2=\mathbf{P}^2-\dfrac{E}{c^2}$

$(mc^2)^2+(\mathbf{P}c)^2=E^2$

or, in agreement with the last triangle:

$(mc^2)^2+(Pc)^2=E^2$

In words, it means simply that

$\boxed{(\mbox{MASS})^2+(\mbox{MOMENTUM})^2=(\mbox{ENERGY})^2}$

There, $K(v)$ is the relativistic kinetic energy we have studied before in this blog. From this triangle, simple trigonometry provides:

$\sin \phi = \dfrac{1}{\gamma}$

$\cos \phi=\dfrac{Pc}{E}$

$\tan \phi=\dfrac{mc}{P}$

$\dfrac{E}{mc^2}=\gamma=\dfrac{1}{\sin \phi}$

$\dfrac{K}{mc^2}=\gamma - 1=\dfrac{1-\sin \phi}{\sin \phi}$

$\dfrac{P}{mc}=\dfrac{1}{\tan \phi}$

A more refined version of the above triangle is given by the following drawing ( the relativistic kinetic energy is now written as T instead of K):

In summary, triangles are cool!

Remark(I): The relationship $(mc^2)^2+(\mathbf{P}c)^2=E^2=E^2(P,m)$ is also called dispersion relation in SR. Some theories beyond the Standard Model of Particle Physics and/or “extended” relativies can modify this equation. However, any known experiment seems to be consistent with SR and this dispersion relationship as far as I know.

Remark(II): Massless particles with $m=0$ , i.e., particles with rest mass equal to zero, satisfy

$\boxed{E=Pc}$

They are called ultra-relativistic particles. This is the case of massless gauge bosons like photons, gluons and likely gravitons. It was thought that the neutrinos were massless too. In recents years, though, we have managed conclusive evidence that neutrinos are not massless and they have a very tiny mass. They are yet ultra-relativistic particles, since we can indeed use yet the equation $E=Pc$ up to a great degree of precision, but they are ultimately MASSIVE particles, and, for them, the exact mass-energy equation reads

$\boxed{E=\sqrt{(mc^2)^2+(\mathbf{P}c)^2}}$

This equation is of course the energy-mass general rule for any massive particle like neutrinos, massive fermions, massive gauge bosons, the Higgs particle, and so on.

LOG#018. Momenergy (III).

Posted on 2012/06/15 by amarashikiAugust 30, 2019

In this last article dedicated to momenergy we are going to learn:

1st) How momenergy transform under Lorent transformations, i.e., the Lorentz transformations of momenergy.

2nd) Some extra stuff on relativistic energy and other alternative deductions of some formulae we just derived in the last posts.

PART(I): LORENTZ TRANSFORMATIONS OF MOMENERGY

Momenergy is a spacetime vector. Indeed, we noted already that somehow energy is the four component of momenergy just in the same way time is the four component of spacetime events. Thus, suppose a mass point with rest mass $m$ and it is moving at $\mathbf{u}=(u,0,0)$ in the S-frame, and $\mathbf{u}'=(u',0,0)$ in the S’-frame. The relative velocity between the two inertial frames will be given by $\mathbf{v}=(v,0,0)$ . In the S-frame, the total relativistic energy reads:

$E=Mc^2=m\gamma c^2=\dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}$

and the relativistic momentum is

$P_x=Mu=\gamma m u=\dfrac{E v}{c^2}$

From the addition theorem of velocities, we also have:

$u'=\dfrac{u-v}{1-\dfrac{uv}{c^2}}$

Squaring this equation, we get the following result:

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

and if we make some algebra calculations

$1-\left(\dfrac{u'^2}{c^2}\right)=\dfrac{1-\dfrac{2uv}{c^2}+\dfrac{u^2v^2}{c^4}-\dfrac{u^2}{c^2}+\dfrac{2uv}{c^2}-\dfrac{v^2}{c^2}}{\left(1-\dfrac{uv}{c^2}\right)^2}=\dfrac{\left[1-\dfrac{u^2}{c^2}\right]\left[1-\dfrac{v^2}{c^2}\right]}{\left(1-\dfrac{uv}{c^2}\right)}$

Inverting and taking the square root, we can deduce:

$\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}=\dfrac{1-\dfrac{uv}{c^2}}{\sqrt{1-\dfrac{u^2}{c^2}}\sqrt{1-\dfrac{v^2}{c^2}}}$

Thus, we have proved the useful relationship:

$\boxed{\gamma (u')=\gamma (u) \gamma (v) \left(1-\dfrac{uv}{c^2}\right)}$

In the case of arbitrary velocities, we could indeed generalize this last equation into:

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

We calculate the momenergy in the S’-frame mutatis mutandis. Firstly, the energy will be:

$E'=\gamma (u')mc^2=\gamma (u)\gamma (v) \left(1-\dfrac{uv}{c^2}\right)mc^2=\gamma (u)\gamma (v) (mc^2-muv)=\gamma (v) \left(E-vP_x\right)$

Secondly, the momentum is easily computed as well:

$P'_x=\gamma (u') mu'=\gamma (u) \gamma (v) \left(1-\dfrac{uv}{c^2}\right) \left(1-\dfrac{uv}{c^2}\right) m\dfrac{(u-v)}{ \left(1-\dfrac{uv}{c^2}\right)}$

i.e.

$P'_x=\gamma (u) \gamma (v)(mu-mv)=\gamma (v)\left(P_x-\dfrac{Ev}{c^2}\right)$

In summary, we can completely write the Lorentz transformation (boost) of momenergy and its components from the next table:

$\boxed{\mbox{Momenergy}\; S\rightarrow S' \begin{cases}P'_0=\dfrac{E'}{c}=\gamma (v) \left (\dfrac{E}{c}-\beta (v) P_x \right)\\ P'_x=\gamma (v) \left(P_x -\beta (v) \dfrac{E}{c}\right)\\ P'_y=P_y \\ P'_z=P_z\end{cases}}$

$\boxed{\mbox{Momenergy}\; S'\rightarrow S \begin{cases}P_0=\dfrac{E}{c}=\gamma (v) \left (\dfrac{E'}{c}+\beta (v) P'_x \right)\\ P_x=\gamma (v) \left(P'_x +\beta (v) \dfrac{E'}{c}\right)\\ P_y=P'_y \\ P_z=P'_z\end{cases}}$

The general Lorentz transformation of momenergy is given (for the case of non-parallel motion to the x-axis) by:

$\boxed{\mbox{Momenergy}\; S\rightarrow S' \begin{cases}P'_0=\dfrac{E'}{c}=\gamma (v) \left( \dfrac{E}{c}-\vec{\beta}(v)\cdot \mathbf{p}\right)\\ \mathbf{p}'=\mathbf{p}+(\gamma (v)-1)\dfrac{(\vec{\beta}(v)\cdot \mathbf{p})\vec{\beta}(v)}{\beta^2(v)}-\gamma (v)\vec{\beta}(v)\dfrac{E}{c}\\ \gamma (v)=\dfrac{1}{\sqrt{1-\beta^2 (v)}}=\dfrac{1}{\sqrt{1-(\beta^2_x+\beta_y^2+\beta_z^2)}}\\ \vec{\beta}(v)=(\beta_x,\beta_y,\beta_z)=(v_x/c,v_y/c,v_z/c)\end{cases}}$

$\boxed{\mbox{Momenergy}\; S'\rightarrow S \begin{cases}P_0=\dfrac{E}{c}=\gamma (v) \left( \dfrac{E'}{c}+\vec{\beta}(v)\cdot \mathbf{p}\right)\\ \mathbf{p}=\mathbf{p}'-(\gamma (v)-1)\dfrac{(\vec{\beta}(v)\cdot \mathbf{p})\vec{\beta}(v)}{\beta^2(v)}+\gamma (v)\vec{\beta}(v)\dfrac{E}{c}\\ \gamma (v)=\dfrac{1}{\sqrt{1-\beta^2 (v)}}=\dfrac{1}{\sqrt{1-(\beta^2_x+\beta_y^2+\beta_z^2)}}\\ \vec{\beta}(v)=(\beta_x,\beta_y,\beta_z)=(v_x/c,v_y/c,v_z/c)\end{cases}}$

Indeed, we could have obtained these equations, of course, simply plugging the momenergy vector to the general Lorentz transformation in matrix form. That is, the above Lorent transformations are:

$\mathbb{P}'=\mathbb{L}(v)\mathbb{P}$

and the inverse transformations are thus

$\mathbb{P}=\mathbb{L}^{-1}(v)\mathbb{P}'$

PART(II): ADDITIONAL CALCULATIONS ABOUT RELATIVISTIC ENERGY.

The work done when a force $F_x$ is applied on a point mass along a path length $dx$ is given by $dW=F_xdx=dK$ . That is, as a force is applied, the supplied energy is stored as kinetic energy $dK$ . When the point particle moves with $\mathbf{u}=(u,0,0)$ and it has rest mass $m$ , we get

$dK=F_xdx=\gamma^3 (u)m\dfrac{du}{dt}dx=\gamma^3 (u) m\dfrac{dx}{dt}du=\dfrac{mudt}{\left(1-\beta^2(u)\right)^{3/2}}=mc^2\dfrac{\beta (u)d\beta (u)}{\left(1-\beta^2(u)\right)^{3/2}}$

If the particle begins the motion with $u=0$ ( null velocity), integrating this last equation provides

$K=mc^2\int_0^{\beta} \dfrac{\beta (u)d\beta (u)}{\left(1-\beta^2(u)\right)^{3/2}}=mc^2\dfrac{1}{\sqrt{1-\beta^2 (u)}} \bigg| _0^{\beta(u)}$

and then

$K=\dfrac{mc^2}{\sqrt{1-\beta^2(u)}}-mc^2=\left(\gamma (u)-1\right)mc^2=Mc^2-mc^2=(M-m)c^2=\delta M c^2$

with

$\delta M \equiv M-m$

being the difference between the total relativistic mass and the rest mass. This remarkable equation was derived before, and it says that, in special relativity and relativistic theories based on the Lorentz group of transformations, the kinetic energy is the difference of two energies: the total energy and the rest energy. The total energy is, of course:

$E=Mc^2=m\gamma c^2=\dfrac{mc^2}{\sqrt{1-\beta^2 (u)}}$

and the rest energy is

$e=mc^2$

If we differentiate the function $K(u)$ , the relativistic kinetic energy, with respect to time, we obtain:

$\dfrac{dK}{dt}=\dfrac{dM}{dt}c^2$

Writing down the relativistic definition of force

$\mathbf{F}=\dfrac{d\mathbf{P}}{dt}=\dfrac{d}{dt}(M\mathbf{u})=\dfrac{dM}{dt}\mathbf{u}+M\dfrac{d\mathbf{u}}{dt}$

Plugging the kinetic energy:

$\mathbf{F}=\dfrac{1}{c^2}\dfrac{dK}{dt}\mathbf{u}+M\dfrac{d\mathbf{u}}{dt}$

but we know that $M=m\gamma$ and $\mathbf{a}=\dfrac{d\mathbf{u}}{dt}$ , so we can recast the equation in an equivalent form as follows

$\gamma m\mathbf{a}=\mathbf{F}-\dfrac{1}{c^2}\dfrac{dK}{dt}\mathbf{u}$

and we get the previously obtained equation

$\boxed{P=\dfrac{dK}{dt}=\mathbf{F}\cdot\mathbf{u}}$

In conclusion, the rate of change of the kinetic energy of a body with respect to time, i.e., the power P it absorbs, is equal to the scalar product of the force and the velocity. This is a well known result from classical mechanics and special relativity does not change it at all, since the dot product is a good invariant.

LOG#017. Momenergy (II).

Posted on 2012/06/13 by amarashikiAugust 30, 2019

Imagine you own a set of n different particle species, and you make them to collide. In general, even in classical mechanics, you can get some particles loose their identities and become “new particles”. Suppose you make m different “new” species of particles. The collision can be represented as the following chemical reaction:

$\displaystyle{\boxed{\sum_{i=1}^{n}A_i \rightarrow \sum_{j=1}^{m}B_j}}$

$\boxed{A_1+\cdots +A_n \rightarrow B_1+\cdots +B_m}$

Now, we will study what happens in this type of collsions in the classical mechanics realm and in the SR framework.

Let us begin with Classical Mechanics. In absence of external forces, Newton’s second law implies that the momentum must be conserved, i.e., momentum does not change in time if there is no applied force. Therefore, the momentum is the same before and after the collision:

$\boxed{\mathbf{p}_{\mbox{before}}=\mathbf{p}_{\mbox{after}}}$

or equivalently

$\boxed{\mathbf{p}_{A_1}+\cdots +\mathbf{p}_{A_n}=\mathbf{p}_{B_1}+\cdots +\mathbf{p}_{B_m}}$

Classical momentum is $\mathbf{p}=m\mathbf{v}$ , so we get

$\boxed{m_{A_1}\mathbf{v}_{A_1}+... +_{A_n}\mathbf{v}_{A_n}=m_{B_1}\mathbf{v}_{B_1}+...+m_{B_m}\mathbf{v}_{B_m}}$

This equation holds in an inertial frame S. If we consider other inertial frame S’ moving at constant speed $\mathbf{V}$ with respect to S, conservation of momentum provides:

$\boxed{m_{A_1}\mathbf{v}^{'}_{A_1}+\cdots +m_{A_n}\mathbf{v}^{'}_{A_n}=m_{B_1}\mathbf{v}^{'}_{B_1}+\cdots +m_{B_m}\mathbf{v}^{'}_{B_m}}$

where

$\mathbf{v'}=\mathbf{v}-\mathbf{V}$

Substracting the last two boxed equations yields:

$m_{A_1}(\mathbf{v}_{A_1}-\mathbf{v}^{'}_{A_1})+...+m_{A_n}(\mathbf{v}_{A_n}-\mathbf{v}^{'}_{A_n})=m_{B_1}(\mathbf{v}_{B_1}-\mathbf{v}^{'}_{B_1})+...+m_{B_m}(\mathbf{v}_{B_1}-\mathbf{v}^{'}_{B_m})$

i.e.,

$\left(m_{A_1}+\cdots +m_{A_1}\right)\mathbf{V}= \left(m_{B_1}+\cdots +m_{B_m}\right)\mathbf{V}$

Therefore, the total inertial mass must be conserved:

$\boxed{\left(m_{A_1}+\cdots +m_{A_1}\right) =\left(m_{B_1}+\cdots +m_{B_m}\right)\leftrightarrow m_{\mbox{before}}^{\mbox{total}}=m_{\mbox{after}}^{\mbox{total}}}$

Conclusion: TOTAL (INERTIAL) MASS IS CONSERVED!

Do you recognize this fact/law? Yes: Newton’s second law plus the no force condition provide the celebrated Lavoisier’s law of Chemistry (the conservation of mass) if we consider the transformation between different inertial frames in Classical Mechanics. Lavoisier’s law remained exact until the discovery of radioactive substances. Special Relativity, thanks to the equivalence between mass and energy, could explain the mysterious origin of the radioactivity that Classical Mechanics could not understand.

Let us go back to the relativistic domain. The relativistic momentum is:

$\mathbf{P}=M\mathbf{v}=m\gamma \mathbf{v}$

Again the condition of no external forces provide the conservation of the relativistic momentum $\mathbb{P}$ :

$\boxed{\mathbb{P}_{\mbox{before}}=\mathbb{P}_{\mbox{\mbox{after}}}}$

Firstly, we will focus on the spatial part of the last four dimensional equation. It yields:

$\boxed{\mathbf{P}_{\mbox{before}}=\mathbf{P}_{\mbox{after}}}$

$\boxed{\left( \sum_i \gamma (v_{A_i})m_{A_i} \mathbf{v}_{A_i}\right)_{\mbox{before}}=\left( \sum_j \gamma (v_{B_j})m_{B_j} \mathbf{v}_{B_j}\right)_{\mbox{after}}}$

It can be written shortly as a feynmanity:

$\boxed{\Delta \left(\sum \gamma (v) m \mathbf{v} \right)=0= \left(\sum_j \gamma (v_{B_j})m_{B_j} \mathbf{v}_{B_j} - \sum_i \gamma (v_{A_i})m_{A_i} \mathbf{v}_{A_i} \right)}$

This last equation is valid in certain inertial frame S. In other inertial frame S’, the same conservation law should hold:

$\boxed{\Delta \left(\sum \gamma (v') m \mathbf{v}' \right)=0}$

From previous logs, we do know that the relativistic transformation between velocities of different frames is:

$\mathbf{v}'=-\mathbf{V}\biguplus \mathbf{v}=\dfrac{\mathbf{v}_\parallel+\gamma^{-1}(V)\mathbf{v}_\perp -\mathbf{V}}{1-\dfrac{\mathbf{V}\cdot \mathbf{v}}{c^2}}$

Using the identity:

$\gamma (v')=\gamma (V)\gamma (v)\left(1-\dfrac{\mathbf{V}\cdot \mathbf{v}}{c^2}\right)$

we deduce

$\Delta \left(\Sigma \gamma (v')m\mathbf{v}' \right)=\Delta \left(\Sigma \gamma (v)m\mathbf{v}\right)-\gamma (V)\Delta \left(\Sigma \gamma (v)m\right)\mathbf{V}+\left(\gamma (V)-1\right)\Delta \left(\Sigma \gamma (v)m\mathbf{v}_\parallel\right)$

Due to the conservation of the relativistic momentum, and specially its parallel projection into $\mathbf{V}$ as well, the lefthanded side and the first and third terms in the righthanded side vanish, so we are left with:

$\gamma (V)\Delta \left(\Sigma \gamma (v)m\right)\mathbf{V}=0$

If the relative velocity is not equal to zero, the momentum in the S’-frame will be conserved if and only if:

$\boxed{\Delta \left(\Sigma \gamma (v)m\right)=0\leftrightarrow \left( \Sigma \gamma (v)m\right)_{\mbox{before}}=\left( \Sigma \gamma (v)m\right)_{\mbox{after}}}$

But this last equation means that the relativistic mass must be conserved in order to conserve the full momenergy. Multiplying by $c^2$ we get indeed the conservation of the total relativistic mass during the collision! Mathematically speaking:

$\boxed{\Delta Mc^2=0\leftrightarrow E_{\mbox{before}}=E_{\mbox{after}}}\leftrightarrow \boxed{\left( \Sigma \gamma (v)mc^2\right)_{\mbox{before}}=\left( \Sigma \gamma (v)mc^2\right)_{\mbox{after}}}$

In the case of our reaction above, we obtain from these last relationship:

$\boxed{\gamma (v_{A_1})m_{A_1}c^2+\cdots+\gamma (v_{A_n})m_{A_n}c^2=\gamma (v_{B_1})m_{B_1}c^2+\cdots+\gamma (v_{B_m})m_{B_m}c^2}$

This equation can be also written using the relativistic mass and relativistic kinetic energy variations as follows:

$\boxed{\left(\Delta \sum m\right)c^2+\Delta \sum K =0}$

In the low velocity limit, this last equation provides:

$(m_{A_1}+...+m_{A_n})c^2+\dfrac{1}{2}(m_{A_1}v_{A_1}^2+...+m_{A_n}v_{A_n}^2)=(m_{B_1}+...+m_{B_m})c^2++\dfrac{1}{2}(m_{B_1}v_{B_1}^2+...+m_{B_m}v_{B_m}^2)$

We have to be careful to apply this last equation since in the approximation we used, the kinetic energies have small numeric values compared with the rest masses. Generally, this equation is more conveniently recasted into:

$\boxed{\Delta \sum m c^2+\Delta\sum \dfrac{1}{2}mv^2=0}$

Remark: In the special case that the initial and final objects match, we get the conservation of the rest mass (a.k.a. Lavoisier’s law) as a trivial identity AND the commonly known kinetic conservation law. They are “emergent” from SR in certain limit. When we say that the initial and the final objects match, we mean that they are completely identical, i.e., a complete identity. If, for instance, in the collision some body gets some heat (i.e., it increases its temperature) or if some body change its inner rotational state, or if some body is “excited”,…Or generally, if the inner energetic content “changes” in any of the colliding objects, the internal energy changes, therefore its inert mass too, and the collision is not elastic.

LOG#016. Momenergy (I).

Posted on 2012/06/12 by amarashikiAugust 30, 2019

We have seen that space and time are merged into the spacetime in Special Relativity (SR). Morever, in a similar way, we have also deduced that momentum and energy are merged into an analogue concept: the momenergy.

Mathematically, it is easily understood looking at the spacetime and momenergy vectors:

$\mathbb{X}=X^\mu e_\mu=(x^0,x^1,x^2,x^3)=(ct,x,y,z)=(ct,\mathbf{r})=(ct,x,y,z)$

and

$\mathbb{P}=P^\mu e_\mu=(P^0,P^1,P^2,P^3)=\left(\dfrac{E}{c},\mathbf{P}\right)=\left(\dfrac{E}{c},P_x,P_y,P_z\right)=(Mc,P_x,P_y,P_z)$

$\mathbb{P}=P^\mu e_\mu=\left(m\gamma c, m\gamma v_x, m\gamma v_y, m\gamma v_z\right)$

Now, we are going to understand better these concepts using basic undergraduate Physics, showing that everything we get from SR is consistent with newtonian physics and the whole theory as a whole.

In non relativistic Mechanics, the notion of kinetic energy is introduced as the work necessary to change the speed of some object, i.e., the work necessary in order to change the velocity. From the general definition of “work” (energy) through some path $\Gamma$ :

$\boxed{\mbox{WORK}=W=\int_{\Gamma} F=\int_1^2 \mathbf{F}\cdot d\mathbf{r}}$

we can insert the Newton’s second law for the force, $\mathbf{F}=m\mathbf{a}$ , and we obtain:

$W=\int_1^2 m\mathbf{a}\cdot d\mathbf{r}=\int_1^2 m \mathbf{a}\cdot \dfrac{\mathbf{dr}}{dt} dt=\int_1^2 m\mathbf{v}\cdot d\mathbf{v}$

We can now use the following formal differential to perform the previous integral:

$d(\mathbf{v}\cdot \mathbf{v})=d \vert \mathbf{v}\vert ^2= d v^2= 2 \mathbf{v}\cdot d\mathbf{v}\rightarrow \mathbf{v}\cdot d\mathbf{v}=\dfrac{1}{2}d\vert \mathbf{v}\vert ^2$

$W=\int_1^2 m\mathbf{v}\cdot d\mathbf{v}=m\int_1^2 \dfrac{1}{2}d\vert \mathbf{v}\vert ^2=\dfrac{1}{2}m\int_1^2 d\vert \mathbf{v}\vert ^2$

and thus, we get the classical result that the work is the variation of the kinetic energy function

$\boxed{W=\dfrac{1}{2}m v^2 \bigg| _1^2=\dfrac{1}{2}m v_2^2-\dfrac{1}{2}m v_1^2=\Delta E_k}$

with

$\boxed{E_k=\dfrac{1}{2}m v^2}$

What happens in SR? Indeed, something similar! Recall that the relativistic generalization of force is:

$\mathbf {F}_ {rel}=\dfrac{d(m\gamma \mathbf{v})}{dt}$

Therefore, the work is now:

$W=\int_{\Gamma} F_{rel}=\int_1^2 \mathbf{F}_{rel}\cdot d\mathbf{r}=\int_1^2 \dfrac{d(m\gamma \mathbf{v})}{dt}\cdot d\mathbf{r}=\int_1^2 d(m\gamma \mathbf{v})\cdot \mathbf{v}=\int_1^2 \mathbf{v}\cdot d(m\gamma \mathbf{v})$

To perform this integral we need to integrate by parts using the rule:

$\boxed{d(\mathbf{u}\cdot \mathbf{v})=d\mathbf{u}\cdot\mathbf{ v} + \mathbf{u}\cdot d\mathbf{v}}$

and the formal differential of the inverse relativistic dilatation factor $\gamma^{-1}$ . Since,

$\gamma = \dfrac{1}{\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}}\rightarrow \gamma^{-1}=\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}$

it implies that

$d(\gamma^{-1})=-\dfrac{\mathbf{v}\cdot d\mathbf{v}}{c^2\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}}=-\dfrac{\gamma\mathbf{v}\cdot d\mathbf{v}}{c^2}$

We will also need this auxiliary result:

$\gamma^2=1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}$

Then, we can now integrate by parts the relativistic version or work, obtaining firstly:

$W_{rel}= m\gamma \mathbf{v}^2 \bigg| _1^2-\int_1^2 (m\gamma \mathbf{v}) \cdot d\mathbf{v}=m\gamma \mathbf{v}^2 \bigg| _1^2+\int_1^2 (-m\gamma \mathbf{v}) \cdot d\mathbf{v}$

or equivalently

$W_{rel}=m\gamma \mathbf{v}^2 \bigg| _1^2+mc^2\int_1^2 \dfrac{1}{c^2}(-\gamma \mathbf{v}) \cdot d\mathbf{v}$

that is

$W_{rel}=m\gamma \mathbf{v}^2 \bigg| _1^2+mc^2\int_1^2 d(\gamma^{-1})=\left[m\gamma \mathbf{v}^2+\dfrac{mc^2}{\gamma}\right] \bigg| _1^2$

But after some algebra we get:

$W_{rel}=\left[m\gamma \left( \mathbf{v}^2+\dfrac{c^2}{\gamma^2}\right)\right] \bigg| _1^2=\left[m\gamma \left( \dfrac{\gamma^2\dfrac{\mathbf{v}^2}{c^2}+1}{\dfrac{\gamma^2}{c^2}}\right)\right] _1^2=\left[mc^2\gamma \left(\dfrac{1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}}{\dfrac{\gamma^2}{1}}\right)\right] _1^2$

Using the last auxiliary result from the previous discussion, i.e.,

$\gamma^2=1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}\rightarrow \dfrac{1+\dfrac{\gamma^2\mathbf{v}^2}{c^2}}{\gamma^2}=1$

we have

$W_{rel}=\left[ m\gamma c^2 \right] _1^2= \Delta E_{rel}^{total}$

where

$\boxed{\Delta E_{rel}^{total}=W_{rel}=\gamma c^2 \Delta m= \gamma c^2 (m_2-m_1)= \Delta M c^2 = (M_2-M_1)c^2}$

and of course, the total relativistic energy is the celebrated equation

$\boxed{E=Mc^2=m\gamma c^2 \equiv E_{rel}^{total}=\mbox{TOTAL RELATIVISTIC ENERGY}}$

A interesting question is now, what is the kinetic energy in the relativist domain? Well, we can define the relativistic kinetic energy $K=K(v)$ as some complex function but how can we determine it? Indeed, it is easy. We can define the relativistic kinetic energy as the total relatistic energy MINUS the rest energy. The rest energy is:

$\boxed{e=mc^2 \equiv \mbox{RELATIVISTIC REST ENERGY}}$

Sometimes it is written, with $m_0$ as the invariant mass we called “m” in our notation, as follows:

$E_0=m_0c^2$

The relativistic kinetic energy and the total energy are then:

$\boxed{K=K(v)=E-mc^2=\gamma mc^2-mc^2=(\gamma - 1)mc^2}$

$E=mc^2+K(v)$

In the small velocity limit, we have the following approximations:

$\boxed{K(v)=\dfrac{1}{2}mv^2+\dfrac{3}{8}\dfrac{mv^4}{c^2}+\mathcal{O}\left( \dfrac{v^6}{c^4}\right)\approx \dfrac{1}{2}mv^2}$

$\boxed{E=mc^2+\dfrac{1}{2}mv^2+\dfrac{3}{8}\dfrac{mv^4}{c^2}+\mathcal{O}\left( \dfrac{v^6}{c^4}\right)\approx mc^2+\dfrac{1}{2}mv^2}$

Therefore, the relativistic kinetic energy approaches the classical result in the limit of small velocities, as we expected, and the total relativistic energy agrees with the classical result, excepting a constant we called the rest energy and further corrections we neglect if $v<<c$ .

LOG#015. Time of flight.

Posted on 2012/06/08 by amarashikiAugust 29, 2019

Suppose we get a beam made of massive particles. The rest mass (the names invariant mass or proper mass are also popular) is $m$ . The particle travels a distance L in its inertial frame. The particle has an energy E in that frame. Therefore, the so-called “time of flight” from two points at x=0 and x=L will be:

$T=\dfrac{L}{v}=\dfrac{m \gamma L}{m \gamma v}=\dfrac{ m \gamma L}{p}=\dfrac{m \gamma c^2 L}{pc^2}$

where we use the relativistic definition of momentum $p=m \gamma v$ . Now, knowing that the relativistic energy is $E=m \gamma c^2$ , using the time that a massless light beam uses to travel the proper distance L, easily calculated to be $\tau = L/c$ , we will get:

$T=\dfrac{E L}{pc^2}=\dfrac{E \tau}{pc}=\dfrac{\tau}{pc/E}$

But the dispersion relation in Special Relativity is a function $E=E(p,m)$ , and the squared of the relativistic momentum or momenergy yields the well known relationship

$p^2c^2 = E^2-m^2c^4$

i.e.,

$\dfrac{p^2c^2}{E^2}=1-\dfrac{m^2c^4}{E^2}$

and then,

$\dfrac{pc}{E}=\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}$

Thus, the time of flight is finally written as follows:

$\boxed{T=\dfrac{\tau}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}}$

This equation is very important in practical applications. Specially in Astrophysics and baseline beam experiments, like those involving the neutrinos! Indeed, we usually calculate the difference between the photon (or any other massless) time of arrival and that of massive particles, e.g. the neutrinos. Several neutrino experiments can measure this difference using a well designed experimental set-up. The difference between those times of flight (the neutrino time of flight minus the photon time of flight) is:

$T-\tau=\dfrac{\tau}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}-\tau$

or equivalently

$\Delta t=T-\tau=\tau \left( \dfrac{1}{\sqrt{1-\left( \dfrac{mc^2}{E}\right)^2}}-1\right)$

or as well

$\displaystyle{ \boxed{Q=\dfrac{\Delta t}{\tau}= \dfrac{1}{\sqrt{1-\left( \frac{mc^2}{E}\right)^2}}-1}}$

This last expression can also be expressed in terms of the speed of light and neutrinos, since $c=L/\tau$ and $v_\nu=L/T$ , so

$\dfrac{ T/L-\tau/L}{\tau/L}=\dfrac{1/v_\nu-1/c}{1/c}=\dfrac{c-v_\nu}{v_\nu}$

Therefore,

$\boxed{Q=\dfrac{\Delta t}{\tau}=\dfrac{c-v_\nu}{v_\nu}}$

In the case of know light left-handed neutrinos, the rest mass is tiny (likely sub-eV and next to the meV scale), and then we can make a Taylor expansion for Q ( with $x=(mc/E)<<1$ ):

$Q\approx \dfrac{x^2}{2}+\mathcal{O}(x^4)$

Then, we would expect, accordingly to special relativity, of course, that

$\dfrac{c-v_\nu}{v_\nu}=\dfrac{T-\tau}{\tau}= \dfrac{1}{2}\left(\dfrac{m_\nu c^2}{E}\right)^2$

We can guess how large it is plugging “typical” values for the neutrino mass and energy. For instance, taking $l m_\nu \sim meV$ and $E\sim GeV$ , the Q value is about

$Q\sim \left( meV/GeV\right)^2= 10^{-24}$

Nowadays, we have no clock with this precision, so the neutrino mass measurement using this approach is impossible with current technology. However, it is clear that if we could make clocks with that precision, we would measure the neutrino mass with this “time of flight” procedure. It is a challenge. We can not do that in these times (circa 2012), and thus we don’t measure any time delay in baseline experiments. Then, neutrinos move with $v_\nu \approx c$ and since there is no observed delay (beyond the OPERA result, already corrected), neutrinos are, thus, ultra-relativistic particles, and for them $E=pc$ with great accuracy.

Moreover, from supernova SN1987A we do observe a delay due to the fact that stellar models predict that neutrinos are emitted before the star explodes! During the collapse, and without going deeper in technical details on the nuclear physics of dense matter, neutrinos scape almost instantaneously (in about some seconds) from the collapsing star but the photons find theirselves blocked during some hours due to the pressure of the dense matter. After some time passes, the block-out ends and photons are released. Historically, this picture was not the physical portrait of collapsing stars around the eighties of the past century, and the observations of supernovae entered, crucially, into scene in order to solve the puzzle of why neutrinos arrived before the photons (indeed, superluminal neutrinos were already tried and postulated to solve that paradox). However, SN1987A was helpful and useful due to the insight it offered about the nontrivial subject of collapsing stars and dense nuclear matter. Before SN1987A there were some assumptions about the nuclear processes inside collapsing stars that proved to be wrong after SN1987A. The puzzle was solved soon with the aid of SN1987A data taken by experimental set-ups like (Super)Kamiokande, and the interaction and collaboration between nuclear physicists, astrophysicists and high energy physicists. Generally speaking, that was the path towards the currently accepted models of nuclear physics in the inner shells of exploding stars.

In summary, photons find a several hours long block-out due to the dense collapsing material from some inner shells of the star, while neutrinos come out, and when the block-out ends, photons are shot into the sky (Bang! A supernova is born). The delayed time is consistent with the above expressions. The neutrinos arrive before the photons not because they are faster but due to the fact they are emitted before and they don’t find any resistence unlike photons; light is blocked by some time due the nuclear conditions of the matter inside the imploding star, but it ends after some hours. Furthermore, since neutrinos travel almost at speed of light, light generally is not able to reach them, even at astrophysical distances!

Remark (I): Generally, experimental physicists use a different formula in order to measure the time difference between the arrivals of neutrinos and photons. We will derive it easily:

$\boxed{\dfrac{c - v_\nu }{v_\nu}=\dfrac{\Delta t}{\tau}}$

This equation implies that

$\boxed{\dfrac{c - v_\nu }{c}=\dfrac{v_\nu \Delta t}{c\tau}=\dfrac{v_\nu T-v_\nu \tau}{c\tau}=\dfrac{L-L'}{L}}$

where L is the distance from CERN to Gran Sasso and L’ is the path length traveled by the neutrinos in the proper time ( $\tau$ ). Therefore, we write

$\dfrac{c- v_\nu }{c}=\dfrac{(T-\tau)}{\tau(c/v_\nu)}$

And now, since

$T=\dfrac{L}{v_\nu},\tau =\dfrac{L}{c}\longrightarrow \dfrac{T}{\tau}=\dfrac{c}{v_\nu}$

Thus,

$\dfrac{c - v_\nu}{c}=\dfrac{(T-\tau)}{\tau (T/\tau)}$

$\boxed{Q'=\dfrac{c - v_\nu }{c}=\dfrac{T-\tau}{T}=\dfrac{\Delta t}{T}}$

Obviously, it is an analogue of the previous formula Q. Q’ provides a measure of how much the propagation velocity of neutrinos differs from the speed of light. Experiments are consistent with a null result (the difference in velocity is measured to be zero). In conclusion, the neutrinos move almost at speed of light (with current experiments not being able to distinguish the tiny difference between both velocities, that of neutrino and the speed of light).

Remark (II): In the slides of OPERA-like experiments, you find the equation

$\boxed{\dfrac{v_\nu -c}{c}=\dfrac{\delta t}{TOF_c -\delta t}}$

and there, it is usually defined

$\delta t= TOF_c -TOF_\nu \longleftrightarrow TOF_\nu = TOF_c -\delta t$

There is a complete mathematical equivalence between my notation and those of experimental HEP neutrino talks handling the time of flight, if you note that:

$\delta t=-\Delta t \;\;\;\;\;\;\;\;\;\;\;\;\;TOF_c = \tau \;\;\;\;\;\;\;\;\;\;\;\;\; TOF_\nu=T$

Then, all the mathematical equations match.

Remark (III): An alternative way to calculate the relationship between neutrino speed and photon speed (in vacuum) is given by the next arguments. The total relativistic energy for the neutrino reads

$E_\nu=\dfrac{m_\nu c^2}{\sqrt{1-\dfrac{v_\nu^2}{c^2}}}$

Then, algebra allows us to write

$E^2=\dfrac{(m_\nu c^2)^2}{1-\dfrac{v_\nu^2}{c^2}}$

$E^2-(m_\nu c^2)^2=\dfrac{v_\nu^2 E^2}{c^2}$

$c^2E^2-c^2(m_\nu c^2)^2=v_\nu^2 E^2$

$v^2_\nu=c^2\left(\dfrac{c^2E^2-(m_\nu c^2)^2}{E^2}\right)=c^2\left( 1-\left(\dfrac{m_\nu c^2}{E}\right)^2\right)$

$v_\nu=\pm c\sqrt{1-\left(\dfrac{m_\nu c^2}{E}\right)^2}$

That is,

$\boxed{\dfrac{\vert v_\nu -c \vert}{c}=\bigg| \sqrt{1-\left(\dfrac{m_\nu c^2}{E}\right)^2}-1\bigg|}$

In the case of neutrinos, ultra-relativistic particles, and with high energy E compared with their rest mass, we get

$\dfrac{\vert v_\nu -c \vert}{c}\approx \dfrac{1}{2}\left(\dfrac{mc^2}{E}\right)^2$

as we obtained before.

__________________________________________________________________________________

PS: This article was semifinalist in the 3quarksdaily 2012 online contest. Here the logo picture I got from the official semifinalist list http://www.3quarksdaily.com/3quarksdaily/2012/06/3qd-science-prize-semifinalists-2012.html

LOG#014. Vectors in spacetime.

Posted on 2012/06/08 by amarashikiAugust 29, 2019

We are going to develop the mathematical framework of vectors in (Minkowski) spacetime. Vectors are familiar oriented lines in 3d calculus courses. However, mathematically are a more general abstract entity: you can add, substract and multiply vectors by some number. We will focus here on the 4D world of usual SR, but the discussion can be generalized to any other D-dimensional spacetime. I have included a picture of a vector OQ above these lines. It is a nice object, isn’t it?

First of all, remark that the conventional kinematical variables to describe the motion in classical mechanics are the displacement vector (or the position vector), and its first and second “derivatives” with respect to time. These magnitudes are called velocity (its magnitude is the speed or modulus of the velocity) and acceleration (in 3d):

$\mathbf{r}(t)=(x,y,z)=\begin{pmatrix}x \\ y \\ z\end{pmatrix}\equiv\mbox{POSITION VECTOR}$

$\mathbf{v}=\dfrac{d\mathbf{r}}{dt}=\left(\dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right)=\begin{pmatrix}\dfrac{dx}{dt}\\ \; \\ \dfrac{dy}{dt}\\ \; \\ \dfrac{dz}{dt}\end{pmatrix}\equiv\mbox{VELOCITY VECTOR}$

$\mathbf{a}=\left(\dfrac{dv_x}{dt},\dfrac{dv_y}{dt},\dfrac{dv_z}{dt}\right)=\begin{pmatrix}\dfrac{dv_x}{dt}\\ \; \\ \dfrac{dv_y}{dt}\\ \; \\\dfrac{dv_z}{dt}\end{pmatrix}\equiv\mbox{ACCELERATION VECTOR}$

In Classical (nonrelativistic) Mechanics, you learn that velocity and acceleration are enough (in general) to fully describe the dynamics and motion of objects. Velocity can be thought as the rate of change of the displacement vector in a tiny amount of time. Acceleration is similarly the rate of change of the velocity vector in some small time interval. You can studied motion con those three vectors, and the notion of “force” is introduced since Newton times as the object that can change the state of “rest” or “uniform” motion (with constant velocity) of any material body. Indeed, Newton’s fundamental law of Dynamics ( it is also called Newton second law) says that the total force (also a vector additive quantity) IS the rate of change of the linear “momentum” vector with respect to time,

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

where the momentum is simply defined as the vector

$\mathbf{p}=m\mathbf{v}\equiv \mbox{MOMENTUM}$

and $m$ is the so called “inertial mass” a measure of the inertia of bodies saying how hard to change its velocity is. Larger the mass is, larger the force necessary to change its velocity is. The total foce is denoted as a sum

$\sum \mathbf{F}=F_1+F_2+\ldots\equiv \mathbf{F}_t \longleftrightarrow \mbox{TOTAL FORCE}$

May the Force be with you! In the case that the mass is constant ( the most general case in Mechanics), Newton second law can also be written as:

$\sum \mathbf{F}=\dfrac{d(m\mathbf{v})}{dt}=m\dfrac{d\mathbf{v}}{dt}$

i.e.,

$\boxed{\mathbf{F}_t=m\mathbf{a}}$

The squared length of vectors is defined as the scalar (dot) product:

$\mathbf{V}\cdot{\mathbf{V}}=V^2=\vert \mathbf{V}\vert^2=V^2_x+V^2_y+V^2_z\equiv \mbox{squared length of any vector V}$

You can also define some interesting differential operators (some king of “gadgets” making “other” stuff form some stuff when they act on vectors somehow):

$\nabla=e^i \partial_i =(\partial_x,\partial_y,\partial_z)=\left(\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z}\right)\equiv \mbox{NABLA OPERATOR}$

$\nabla^2=\nabla \cdot \nabla=\partial^i \partial_i=\partial^2_x +\partial^2_y +\partial^2_z=\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}\equiv \mbox{LAPLACIAN OPERATOR}$

OK, that is a good crash course on elementary Dynamics and vectors. Back to the future! Back to relativity! You can extend this formalism into the spacetime with care, since we do know time is not an invariant any more in relativistic Physics. Fortunately, we saw that there is an invariant quantity in special relativity: the proper time $\tau$ ! We remember the important result:

$\dfrac{dt}{d\tau}=\gamma$

So, we proceed to the spacetime generalization of the previous stuff. Firstly, events in spacetime are given by:

$\boxed{\mathbb{X}=X^\mu e_\mu=(ct,x,y,z)=\begin{pmatrix}ct \\ \; \\ x \\ \; \\ y \\ \; \\ z\end{pmatrix}\equiv \mbox{SPACETIME EVENT}}$

The “spacetime length” is got using the following dot product

$\boxed{S^2=-(ct)^2+x^2+y^2+z^2=\mathbb{X}\cdot \mathbb{X}=X^\mu X_\mu \equiv \mbox{SPACETIME INVARIANT}}$

so we have the so called contravariant components

$X^\mu=(ct,\mathbf{r})$

and the covariant components

$X_\mu=(-ct,\mathbf{r})$

The spacetime interval bewtween two arbitrary events, e.g., A and B, will be:

$\Delta \mathbb{X}_{AB}= \mathbb{X}_B - \mathbb{X}_A= c(t_B -t_A)+ (x_B-x_A)+ (y_B-y_A)+(z_B-z_A)$

The squared spacetime “length” separation (an invariant) between those two events is:

$\Delta \mathbb{X}^2_{AB}=\Delta \mathbb{X}\cdot \Delta \mathbb{X}=\Delta X^\mu \Delta X_\mu$

or equivalenty

$\Delta \mathbb{X}^2=-c^2 (t_B -t_A)^2+ (x_B-x_A)^2+ (y_B-y_A)^2+ (z_B-z_A)^2$

You can also write the last two equations in a differential ( or infinitesimal) way:

$d \mathbb{X}^2=d \mathbb{X}\cdot d \mathbb{X}=dX^\mu d X_\mu$

$d \mathbb{X}^2=-c^2 d t^2+d x^2+d y^2+d z^2$

The spacetime velocity will be then:

$\boxed{\mathbb{U}=U^\mu e_\mu=\dfrac{d\mathbb{X}}{d\tau} \equiv \mbox{SPACETIME VELOCITY}}$

We now use the trick based on the chain rule of differential calculus:

$\dfrac{d\mathbb{X}}{d\tau}=\dfrac{d\mathbb{X}}{dt}\dfrac{dt}{d\tau}$

Therefore,

$\mathbb{U}=\left( c\dfrac{dt}{d\tau},\dfrac{d\mathbf{r}}{dt}\dfrac{dt}{d\tau}\right)= (\gamma c,\gamma\mathbf{v})$

So, we get

$\mathbb{U}=U^\mu e_\mu \leftrightarrow U^0=\gamma c \;\;\;\; U^i=\gamma v^i$

i.e.

$\boxed{\mathbb{U}=\dfrac{d\mathbb{X}}{d\tau}=(c\gamma,\gamma \mathbf{v})}$

The spacetime velocity has a nice property:

$\mathbb{U}^2=\mathbb{U}\cdot\mathbb{U}=-c^2\gamma^2+\gamma^2 v^2=-c^2\gamma^2(1-\frac{v^2}{c^2})=-c^2$

i.e.

$\boxed{\mathbb{U}^2=U^\mu U_\mu=-c^2}$

The spacetime momentum that we will call momenergy ( the reasons will be seen later) is defined as:

$\boxed{\mathbb{P}=P^\mu e_\mu=m\mathbb{U}=(mc\gamma,\gamma m\mathbf{v})\equiv\mbox{MOMENERGY}}$

The momenergy components are given by

$\boxed{P^0=mc\gamma \;\;\;\; P^i=m\gamma v^i}$

In the same way that time is the “fourth” coordinate in the spacetime, we can identify the zeroth component of the spacetime momentum, $P^0$ , as the total relativistic energy in the following way:

$P^0\equiv \dfrac{E}{c}$

$P^0=mc\gamma=\dfrac{E}{c}\longleftrightarrow \boxed{E=P^0 c=m\gamma c^2}$

$\boxed{E=Mc^2}$

where the relativistic mass is defined as

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

Define the relativistic (spacelike) momentum as

$\mathbf{P}=\gamma \mathbf{p}$

Now we can understand why the spacetime momentum was called momenergy, since it can be rewritten as:

$\boxed{\mathbb{P}=\left(Mc, \mathbf{P}\right)=\left(\dfrac{E}{c},\mathbf{P}\right)=\left(\gamma mc, \gamma m\mathbf{v}\right)}$

We can make an invariant from the squared momenergy:

$\mathbb{P}^2=\mathbb{P}\cdot \mathbb{P}=P^\mu P_\mu=-M^2+\mathbf{P}^2=-m^2c^2\gamma^2\left( 1-\dfrac{v^2}{c^2}\right)=-m^2c^2$

i.e.

$\boxed{\mathbb{P}^2=\mathbb{P}\cdot \mathbb{P}=P^\mu P_\mu=-m^2c^2 \equiv (\mbox{MOMENERGY})^2 \; \mbox{INVARIANT}}$

We can use this result to write the most general relation between energy, momentum and mass in SR:

$\mathbb{P}^2=\mathbb{P}\cdot \mathbb{P}=P^\mu P_\mu=\mathbf{P}^2-\dfrac{E^2}{c^2}=-m^2c^2$

Then, we have

$\dfrac{E^2}{c^2}=\mathbf{P}^2+m^2c^2$

or equivalenty

$\boxed{E^2=\mathbf{P}^2c^2+m^2c^4 \longleftrightarrow E^2=(\mathbf{P}c)^2+(mc^2)^2}$

Setting units with c=1, we get

$\boxed{(\mbox{ENERGY})^2=(\mbox{MOMENTUM)}^2+(\mbox{MASS})^2}$

Wonderful! Momenergy allows us to unify the concepts of mass, momentum and energy in a 4D spacetime world!

What else? We can also calculate the spacetime acceleration, but it is a bit more complicated. Indeed, we need an auxiliary result from calculus:

$\dfrac{d\gamma}{dt}=\dfrac{d}{dt}\left(1-\dfrac{v^2}{c^2}\right)^{-1/2}=-\dfrac{1}{2}\left(1-\dfrac{v^2}{c^2}\right)^{-3/2}(-2)\dfrac{\mathbf{v}\cdot\mathbf{a}}{c^2}=\dfrac{\gamma^3}{c^2}\mathbf{v}\cdot \mathbf{a}$

that is

$\boxed{\dfrac{d\gamma}{dt}=\dfrac{\gamma^3}{c^2}\mathbf{v}\cdot \mathbf{a}}$

Now, we proceed to the calculations:

$\boxed{\mathbb{A}=A^\mu e_\mu=\dfrac{d\mathbb{U}}{d\tau}}$

and with the same trick we saw before

$\dfrac{d\mathbb{U}}{d\tau}=\dfrac{d\mathbb{U}}{dt}\dfrac{dt}{d\tau}=\gamma \dfrac{d\mathbb{U}}{dt}$

and

$\mathbb{A}=\gamma \left(\dfrac{d\gamma}{dt} (c, \mathbf{v}) +\gamma \dfrac{d}{dt}(c,\mathbf{v})\right)$

$\mathbb{A}=\gamma \left( \dfrac{\gamma^3 \mathbf{v}\cdot \mathbf{a}}{c^2}(c, \mathbf{v}) +\gamma (0,\mathbf{a})\right)$

i.e.

$\mathbb{A}=\left( \dfrac{\gamma^4 \mathbf{v}\cdot \mathbf{a}}{c^2}(c, \mathbf{v}) +\gamma^2 (0,\mathbf{a})\right)$

Therefore:

$\boxed{\mathbb{A}=(A^0,\mathbf{A})\equiv \begin{pmatrix}PROPER\\ ACCELERATION\end{pmatrix}\longleftrightarrow A^0=\dfrac{\gamma^4 \mathbf{v}\cdot \mathbf{a}}{c}\;\; A^i=\dfrac{\gamma^4 \mathbf{v}\cdot \mathbf{a}}{c^2}v^i+\gamma^2 a^i }$

We can also calculate the proper acceleration invariant, i.e., the squared of the spacetime acceleration:

$\mathbb{A}^2=\mathbb{A}\cdot \mathbb{A}=\dfrac{\gamma^8}{c^4}\left[ v^2(\mathbf{v}\cdot \mathbf{a})^2-c^2(\mathbf{v}\cdot \mathbf{a})^2\right]+2\dfrac{\gamma^6}{c^2}(\mathbf{v}\cdot \mathbf{a})^2+\gamma^4(\mathbf{a}\cdot \mathbf{a})$

i.e.,

$\boxed{ \mathbb{A}^2=\mathbb{A}\cdot \mathbb{A}=\dfrac{\gamma^6}{c^2}(\mathbf{v}\cdot \mathbf{a})^2+\gamma^4 a^2=\gamma^4\left(\dfrac{\gamma^2}{c^2}(\mathbf{v}\cdot \mathbf{a})^2+a^2\right)=\alpha^2\equiv \begin{pmatrix}\mbox{SQUARED PROPER} \\ \mbox{ACCELERATION}\end{pmatrix}}$

We can calculate some important particular cases of the last equation. If the acceleration is linear, $\mathbf{v}\parallel \mathbf{a}$ , the last equation provides:

$\alpha^2=\gamma^6\beta^2a^2+\gamma^4 a^2=\gamma^4 a^2(\gamma^2\beta^2+1)=\gamma^6 a^2$

i.e. linear accelerations have proper acceleration $\alpha =\gamma^3 a$ . In the case of circular motions, where $\mathbf{v}\perp\mathbf{a}$ , we get proper acceleration $\alpha=\gamma^2 a$ . Thus, the centripetal proper acceleration, e.g. in a storage ring, is $\gamma^2\dfrac{v^2}{r}\approx \gamma^2 \dfrac{c^2}{r}$ .

Finally, we calculate the so-called Power-Force vector, sometimes it is also referred as Minkowski force:

$\mathbb{F}=\mathcal{F}^\mu e_\mu=(\mathcal{F}^0,\mathcal{F}^i)=\dfrac{d\mathbb{P}}{d\tau}=\gamma \left( \dfrac{1}{c}\dfrac{dE}{dt},\dfrac{d\mathbf{P}}{dt}\right)=\gamma \left( \dfrac{1}{c}\dfrac{dE}{dt},\mathbf{F}\right)\equiv \left(\mbox{POWER-FORCE}\right)$

where

$\mathbf{F}=\dfrac{d\mathbf{P}}{dt}=\dfrac{d(M\mathbf{v})}{dt}=\dfrac{d(\gamma m\mathbf{v})}{dt}=\gamma m\mathbf{a}+\gamma^3\dfrac{m}{c^2}(\mathbf{v}\cdot \mathbf{a})\mathbf{v}=\gamma m\left(\mathbf{a}+\dfrac{\gamma^2}{c^2}(\mathbf{v}\cdot \mathbf{a})\mathbf{v}\right)$

It is called Power-Force due to the dimensions of physical magnitudes there

$\boxed{\mathcal{F}^0= \dfrac{ \gamma}{c}\dfrac{dE}{dt}}$

and

$\boxed{\mathcal{F}^i=\gamma F^i}$

Recall the previously studied spacetime and momenergy vectors. Here, it is often defined the transversal mass as $M_T=\gamma m=M$ (mass-like coefficient for force perpendicular to the motion) and also the so-called longitudinal mass $M_L=\gamma^3 m=\gamma^2 M$ (mass-like term for the force parallel to the direction of motion).

Moreover, we could obtain the acceleration from the force in the following way:

$\mathbf{F}\cdot \mathbf{v}=\gamma m \left[\mathbf{v}\cdot \mathbf{a}\right]\left(1+\dfrac{\gamma^2 v^2}{c^2}\right)=\gamma m \left[\mathbf{v}\cdot \mathbf{a}\right]\left(1+\dfrac{\beta^2}{1-\beta^2}\right)=\gamma m \left[\mathbf{v}\cdot \mathbf{a}\right]\left(\dfrac{1}{1-\beta^2}\right)$

i.e.

$\mathbf{F}\cdot \mathbf{v}=\gamma^3 m\left[\mathbf{v}\cdot \mathbf{a}\right]$

and thus

$\mathbf{F}=\gamma m \mathbf{a}+\dfrac{1}{c^2}\left(\mathbf{F}\cdot\mathbf{v}\right)\mathbf{v}$

and therefore

$\boxed{\mathbf{a}=\dfrac{1}{\gamma m}\left(\mathbf{F}-\dfrac{\mathbf{F}\cdot\mathbf{v}}{c^2}\mathbf{v}\right)=\dfrac{1}{M}\left(\mathbf{F}-\dfrac{\mathbf{F}\cdot\mathbf{v}}{c^2}\mathbf{v}\right)}$

We observe that Newton’s secon law is not generally valid in SR, since

$\mathbf{F}\neq m\mathbf{a}$ . However, we can “generalize” the Newton’s second law to be covariantly valid in SR in this neat form:

$\boxed{\mathbb{F}=m\mathbb{A}}$

The non-trivial expression for the spacetime acceleration and the mathematical consistency of the theory do the rest of the work for us.

Finally, we can make some nice invariant operator as well. We define the spacetime generalization of nabla as follows:

$\boxed{\square =e^\mu \partial_\mu =\left(\dfrac{1}{c}\partial_t, \partial_x,\partial_y,\partial_z\right)=\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z}\right)=\left(\dfrac{1}{c}\dfrac{\partial}{\partial t},\nabla \right)}$

The spacetime analogue of the laplacian operator is sometimes called D’Alembertian (or “wave” operator):

$\boxed{\square^2=\square \cdot \square = \partial^\mu \partial_\mu= -\dfrac{\partial^2_t}{c^2}+\partial^2_x +\partial^2_y +\partial^2_z=-\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}+\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}}$

or equivalently, in a short hand notation

$\boxed{\square^2=-\dfrac{1}{c^2}\dfrac{\partial^2}{\partial t^2}+\nabla^2}$

Remark: Some books and authors define the D’Alembertian as $\square =\square \cdot \square$ . While that notation is indeed valid/possible, and despite it can safe time sometimes, it is obviously a tricky and non-intuitive, somewhat unlucky, notation. We warn you about it, and we highly recommend transparent and powerful notations like the one introduced here, keeping in mind the physical concepts behind all this framework.

Remark(II): The mathematical framework using vectors in spacetime can be easily generalized to spacetime with an arbitrary number of dimensions, e.g., $D=d+1$ , where $d$ is the number of space-like dimensions or, even we could consider $D=d+q=s+t$ , where we have an arbitrary number $d$ (or $s$ ) of space-like coordinates and an arbitrary number $q$ (or $t$ ) of time-like coordinates. However, it is not so easy to handle with these generalizations by different reasons (both mathematical and physical). A useful quantity in that case is the spacetime “signature”, defined as the the difference between the number of space-like and time-like dimensions, that is,

$\mbox{Signature of spacetime}=(s-t)=(d-q)$

The easiest ( the most studied so far) case with a higher dimensional spacetime is that of signature $2$ , the SR case, or its most natural generalization with signature $(d-1)=(D-2)$ .

LOG#013. Spacetime.

Posted on 2012/06/07 by amarashikiAugust 29, 2019

“(…)The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality(…)”

– Hermann Minkowski, 1908

We have elucidated some amazing results from the Special Relavity (SR) postulates and more stuff is to come. Indeed, the own structure of Lorentz transformations is hinting some hidden symmetry between the concepts of space and time that are seen as independent from each other in Classical Mechanics. Einstein’s work on the principle of relativity and the electrodynamics of moving objects was showinga new symmetry of Nature, and it was pointed out by the old known Maxwell’s equations. Nobody realized it until Einstein’s papers were published.

Herman Minkowski was a professor of A. Einstein in Zurich. He was surprised by the fact Einstein were so deep in the problem of Electrodynamics. Then, Minkovski himself did a new advance on Einstein’s relativity and exposed his own works in 1908. Minkowski suggested that relativity meant that space and time are no longer independent entities, as in fact the Lorentz transformations show. Lorentz transformations mix space and time coordinates. Relativity was saying, according to Einstein, that space and time as absolute entities did not exist. Minkowski was a step further and suggested that relativity is just geometry of space and time merged together, a.k.a., that Lorentz transformations and relativity were… Physics in spacetime! Physics processes are then labelled by coordinates in “space” and “time”, or in “spacetime” for short. SR is just a set of rules or geometry handling with transformation between differente events in spacetime. Events in spacetime are labelled by some set of coordinates of space and time. If we restrict to the commonly known 3 dimensions of space and the single dimension of time we seem to observe, Minkowski exposed some mathematical framework to work in a 4-dimensional (D=3+1) spacetime. His tools can indeed be generalized to any arbitrary spacetime with D-dimension as well, but we will no go further in that direction today. We will only discuss 4D spacetime in this post.

Question: if time and space are relative, as Einstein suggested,…Does it mean that we have nothing “invariant” to study? Geometry in Minkowskian spacetime is the answer it. Fortunately, mathematicians in the 19th century had studied non-euclidean geometries and it was just rediscovered by Minkowski that non-euclidean geometries could fit the new theory of relativity.

Minkowski argued that events en spacetime E are given by four coordinate set of numbers:

$\mathbb{X}=(ict,x,y,z)$

where $i=\sqrt{-1}$ . However, in modern language, physicists avoid the imaginary time (or equivalently, the so-called quaternionic formulation of SR) using a gadget called spacetime “metric”. Using the metric, you don’t need an imaginary time and you write:

$\mathbb{X}=(ct,x,y,z)=(x^0,x^1,x^2,x^3)$

The Minkovski spacetime was very helpful in the building of the relativistic theory of gravitation, a.k.a., general relativity in spite Einstein himself put critics on the spacetime formalism. Soon, he changed his view and he learned the Minkowki spacetime stuff.

Well, now, how can be the spacetime structure help in relativity? It is quite easy. It is true that time or space are not invariant by theirselves, as Lorentz transformations show but, it can be shown that, the “space-time” interval is invariant. What is a space-time interval? Easy! Take two events $E_1, E_2$ in spacetime. The squared spacetime interval between those events is defined as:

$(\Delta S)^2=(\Delta x)^2+(\Delta y)^2+(\Delta z)^2-(\Delta ct)^2$

or equivalenty, writing explicetly the coordinates of the two events in the S-frame $E_1(ct_1,x_1,y_1,z_1)$ and $E_2(ct_2,x_2,y_2,z_2)$ , we get

$\boxed{(\Delta S)^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2-(ct_2-ct_1)^2=invariant}$

In the S’-frame, by the other hand, we will have another spacetime interval:

$\boxed{(\Delta S')^2=(x'_2-x'_1)^2+(y'_2-y'_1)^2+(z'_2-z'_1)^2-(ct'_2-ct'_1)^2=invariant}$

If the S-frame and the S’-frame are related by Lorentz transformations, the invariant is the same. That is the key of spacetime! The invariant is called proper time, i.e., the time measured on the clock travelling attached to the reference frame, or, equivalently, the time measured by an observer in motion with the frame. Any other frame will not be “invariant” and it has to be Lorentz transformed in order to agree on clock measurements with any other observer in other different frame.

Thus, relativity unifies the classical notions of space and time into a wider and more general notion: spacetime. The speed of light is indeed the conversion factor between units of space and time. Sometimes, physicists use units in which the speed of light is set to the unity $c=1$ . Common units can be recovered carefully reintroducing “c”. If we go from the interval to an infinitesimal variation of spacetime coordinates in our events, the proper time is defined as

$dS^2\equiv -c^2d\tau^2=dx^2+dy^2+dz^2-c^2dt^2$

or, from the S’-frame

$dS'^2\equiv -c^2d\tau^2=dx'^2+dy'^2+dz'^2-c^2dt'^2$

Please, note that, the speed of light is invariant as it should, accordingly to the SR postulates.

The proper time can be related to the usual time with some algebraic manipulations:

$-c^2\left(\dfrac{d\tau}{dt}\right)^2=\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2+\left(\dfrac{dz}{dt}\right)^2-c^2\left(\dfrac{dt}{dt}\right)^2$

so, knowing that the velocity in space (or 3d velocity) is indeed $v=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\dfrac{dz}{dt}\right)$ . we get

$-c^2\left(\dfrac{d\tau}{dt}\right)^2=v^2-c^2=-c^2\left(1-\dfrac{v^2}{c^2}\right)=-\dfrac{c^2}{\gamma ^2}$

where

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

is the well know relativistic dilation factor we found in our previous studies. We have then obtained the relation between proper time and usual time, an important mathematical relationship that is very useful in order to simplify calculation in Minkowski spacetime (even if it is D-dimensional):

$\boxed{\dfrac{d\tau}{dt}=\dfrac{1}{\gamma }\longleftrightarrow \dfrac{dt}{d\tau}=\gamma}$

Minkovski ideas were proved useful and important for general relativity and the own mathematical framework of Special Relativity. The legacy of H. Minkowski is with us today yet:

1) The use of spacetime vectors. In the case of a D=4=3+1 spacetime, spacetime vectors are called 4-vectors and they represent events in the continuum spacetime geometry. This hyperbolic non-euclidean geometry was indeed studied by Gauss, Lobachevski and other bright mathematicians during the 19th century. The use of these vectors simplifies long calculations.

2) The use of Minkowski diagrams. Minkowski suggested that the causal structure of the geometrical spacetime was given by the structure of the “light-cones”. In $\mathbb{R}^4$ , two hyperboloids (“hyperplanes”) can intersect in one point and it settles the physical events. The use of Minkowski diagrams is intuitive and it helps to visualize and solve physical problems too.

You can see a light-cone in Minkowski space above, fixing the causal structure of spacetime.

LOG#012. Michelson-Morley.

Posted on 2012/06/07 by amarashikiAugust 29, 2019

During the 19th century, the electromagnetic theory of Maxwell assumed that electromagnetic waves travelled in a medium called ether. The Michelson-Morley experiment was an experiment devoted to detect the ether. We can think about the electromagnetic waves like an analogue of waves in a medium (for instance, water). Look at this picture:

Imagine a boat sailing in this river. The river flows from the left side to the right side at constant velocity $v$ relative to the banks. Now, the trick of this analogy. Imagine that the boat is light, i.e., the boat are our loved electromagnetic waves ( whatever they are, and it implies they propagate in some kind of elastic medium like the water in the river; this medium was called ether by the physicists during the 19th century). When the boat ( remember it is an electromagnetic field hidden in the analogue model) is sailing downstream (or upstream), its velocity with respect to the banks will be $c+v$ downstream and $c-v$ upstream. If you have to cover a distace $l_1$ from the points A to B and back to A, it will spend a time that can be easily computed:

$t_1=T_1+T_2=\dfrac{l_1}{c+v}+\dfrac{l_1}{c-v}=\dfrac{2l_1 c}{c^2-v^2}=\dfrac{2l_1}{c}\dfrac{1}{1-\dfrac{v^2}{c^2}}$

By the other hand, if the boat travels at right angles to the river ( remember it is playing the role of the ether), we can also calculate the time required by the boat to travel the distance $l_2$ from C to D and come back to C. The velocity in this case is calculated using the Pythagorean theorem (or doing the calculus by components) and it is equal to $V=\sqrt{c^2-v^2}$ . In this way, the time will be then:

$t_2=\dfrac{2l_2}{V}=\dfrac{2l_2}{\sqrt{c^2-v^2}}$

The time difference between these two times is calculated as well in a straightforward way:

$\Delta t = t_1-t_2=\dfrac{2l_1}{c}\left(\dfrac{1}{1-\dfrac{v^2}{c^2}}\right)-\dfrac{2l_2}{\sqrt{c^2-v^2}}$

or equivalently, the fundamental formula for the Michelson-Morley light-dragging through the ether reads:

$\boxed{\Delta t =\dfrac{2l_1}{c}\left(\dfrac{1}{1-\dfrac{v^2}{c^2}}\right) - \dfrac{2l_2}{c} \left( \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\right)}$

Michelson and Morley guessed an experimental set-up to measure the drag velocity $v$ of light in the ether. From the hypherphysics project page, hyperphysics.phy-astr.gsu.edu/hbase/hframe.html, we get this nice illustration of their device:

How is this related to the previous calculation? Well, it is pretty nice and simple. Michelson and Morley built what is called an stellar interferometer for light. Suppose that we write $l_1=l_2=L$ in out main boxed equation above. L is the arm lenght of our stellar interferometer. Similar changes can be done in the formulae for $t_1$ ( $t'$ in the picture above) and $t_2$ ( $t''$ in the picture). In that case the time difference will be:

$\Delta t =\dfrac{2L}{c}\left(\dfrac{1}{1-\dfrac{v^2}{c^2}}-\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\right)$

This mathematical expression is complicated. But if we suppose that the drag velocity is small compared with the speed of light, we can make an approximation ( the technical “magic words” would be Taylor series):

$\Delta t\approx \dfrac{2L}{c}\left(1+\dfrac{v^2}{c^2}\right)- \dfrac{2L}{c}\left(1+\dfrac{v^2}{2c^2}\right)=\dfrac{2L}{c}\left(\dfrac{v^2}{2c^2}\right)=\dfrac{L}{c}\left(\dfrac{v^2}{c^2}\right)$

As it is show in the figure as well, for a typical low drag velocity, the interferometer ( thanks to the wave character of light) could indeed measure tiny time separations searching for “moved fringes”. The fringe “shift” due to the rotation of the arms of the interferometer can be easily calculated. After a 90º rotation, the time diference $\Delta t$ flips its sign, so the light should get a phase difference $2\Delta t$ . We know that the period of light is given by the relationship $T=\lambda/c$ . Then, the fringe shift pattern that Michelson and Morley expected to obtain was:

$\boxed{\Delta N=\dfrac{2\Delta t}{T}\approx \dfrac{2L}{c}\left(\dfrac{v}{c}\right)^2}$

When Michelson did his first experiment in 1881, he had got $L=1.2m$ and $\lambda \sim 5 \cdot 10^{-7}m$ , and thus the expected fringe shift provided the minimum $\Delta N \sim 1/20$ . The shift was not observed, and performing some improvements, Michelson and his collaborator Morley, in 1887, got $L=11m$ and achieved a better ressolution in the fringe shift pattern. But, surprinsingly, there was no fringe shift in both cases!

In 1892, George Fitzgerald and H. Lorentz proposed (before A. Einstein true explanation with his relativity) that objects in the aether were contracted in the same fashion (with the same formula) than we saw in relativity. In their conception, they indeed derived the Lorentz (or Lorentz-Fitzgerald) transformations but they intepreted (wrongly) in the context ot the electromagnetic ether theory. In this way, bodies were contrated in their motion through the aether and it could explain the null result of the Michelson-Morley experiment and other similar experiments. It changed radically when Einstein published his articles on relativity with the correct physical insight and consequences of their transformations, that Einstein himself derived from two simple principles, and more remarkably, neglecting the own existence of the ether!

The Michelson-Morley is likely one of the null experiments most famous in the history of Physics. Indeed, it advanced the rising of the special theory of relativity and, in perspective, it was saying that the ether hypothesis was not necessary for the electromagnetic field and its waves to exist and propagate “in vacuum”.

Of course, from the modern viewpoint, relativity moved the question of the ether into another…Nobody questions today how can light travel at speed of light in vaccum always at the same speed, independly from the source. Nobody questions that light can self-sustain their own oscillations in vacuum in space and time everywhere in the known Universe. Simply, we do know that it can and it does…Light and vacuum are entangled somehow. Indeed, the questions about the nature and properties of the ether has shifted now into the question what the vacuum is...That is, now we ask about what are the properties and nature of the vacuum and its properties. But that is another story that advanced the rising of the theory and methods of Quantum Physics and Quantum Field Theory. Too far for this modest post today!

Finally, I would like to remark that similar experiments to the Michelson-Morley experiment have been performed. Modern experiments are based on the concept of optical resonators and they have a set-up like this:

Every result of Michelson-Morley type experiments has been null. The ether as the 19th century physicists imagined does not exist.

LOG#011. Relativistic accelerations.

Posted on 2012/06/06 by amarashikiAugust 29, 2019

Imagine the S’-frame moves at constant velocity (see the frames above this line):

$\mathbf{v}=(v,0,0)$

relative to the S-frame. In the S’-frame an object moves with acceleration

$\mathbf{a}'=(a'_x,a'_y,a'_z)=\left(\dfrac{du'_x}{dt'},\dfrac{du'_y}{dt'},\dfrac{du'_z}{dt'}\right)$

QUESTION: What is the acceleration in the S-frame?

Of course it has to be something like this

$\mathbf{a}=(a_x,a_y,a_z)=\left(\dfrac{du_x}{dt},\dfrac{du_y}{dt},\dfrac{du_z}{dt}\right)$

In order to get the relationship between both accelerations (and frames) we have to use the addition law of velocities from the previous post. Without loss of generality, we will use the rule (I) and we leave the general case as an exercise for the eager reader. We obtain:

$a_x=\dfrac{du_x}{dt}=\dfrac{\dfrac{du_x}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{d}{dt'}\left[\dfrac{u'_x+v}{1+\dfrac{u'_xv}{c^2}}\right]}{\dfrac{1}{c}\dfrac{d}{dt'}\left[\gamma(ct'+\beta x')\right]}=\dfrac{\dfrac{du'_x}{dt'}\left(1+\dfrac{u'_xv}{c^2}\right)-(u'_x+v)\dfrac{v}{c^2}\dfrac{du'_x}{dt'}}{\left(1+\dfrac{u'_x v}{c^2}\right)^2 \gamma \left(\dfrac{dt'}{dt'}+\dfrac{1}{c}\beta \dfrac{dx'}{dt'}\right)}$

Therefore,

$a_x=\dfrac{\left(1-\dfrac{v^2}{c^2}\right)a'_x}{\left(1+\dfrac{u'_x v}{c^2}\right)^2 \gamma \left(1+\dfrac{u'_x v}{c^2}\right)}=\dfrac{1}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^3}a'_x$

Thus we get the first transformed component transformation for the acceleration:

$\boxed{a_x=\dfrac{a'_x}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^3}}$

For the transverse components, say $a_y$ ( an analogue symmetrical calculation provides $a_z$ ), we calculate

$a_y=\dfrac{du_y}{dt}=\dfrac{\dfrac{du_y}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{d}{dt'}\left[\dfrac{u'_y}{\gamma \left(1+\dfrac{u'_x v}{c^2}\right)}\right]}{\dfrac{1}{c}\dfrac{d}{dt'}\left[(ct'+\beta x')\right]}=\dfrac{\dfrac{du'_y}{dt'}\gamma \left(1+\dfrac{u'_x v}{c^2}\right)-u'_y \gamma \dfrac{v}{c^2}\dfrac{du'_x}{dt'}}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^2 \gamma \left(\dfrac{dt'}{dt'}+\dfrac{1}{c}\beta\dfrac{dx'}{dt'}\right)}$

Some basic algebra manipulations allow us to get:

$a_y=\dfrac{\gamma \left(1+\dfrac{u'_x v}{c^2}\right)a'_y-\gamma\dfrac{u'_y v}{c^2}a'_x}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^2\left(1+\dfrac{u'_x v}{c^2}\right)}=\dfrac{a'_y}{\gamma^2 \left(1+\dfrac{u'_x v}{c^2}\right)^2}-\dfrac{\dfrac{u'_y v}{c^2}a'_x}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^3}$

Thus, the complete transformation of accelerations between frames from S’ to S(and from S to S’) are given by the following tables:

$\boxed{\mbox{Transf.of acc.in SR:} S'\rightarrow S \begin{cases}a_x=\dfrac{a'_x}{\gamma^3\left(1+\dfrac{u'_x v}{c^2}\right)^3}\\ \; \\ a_y=\dfrac{a'_y}{\gamma^2 \left(1+\dfrac{u'_x v}{c^2}\right)^2}-\dfrac{\dfrac{u'_y v}{c^2}a'_x}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^3}\\ \; \\ a_z=\dfrac{a'_z}{\gamma^2 \left(1+\dfrac{u'_x v}{c^2}\right)^2}-\dfrac{\dfrac{u'_z v}{c^2}a'_x}{\gamma^2\left(1+\dfrac{u'_x v}{c^2}\right)^3}\end{cases}}$

$\boxed{\mbox{Transf.of acc.in SR:} S\rightarrow S' \begin{cases}a'_x=\dfrac{a_x}{\gamma^3\left(1-\dfrac{u_x v}{c^2}\right)^3}\\ \; \\ a'_y=\dfrac{a_y}{\gamma^2 \left(1-\dfrac{u_x v}{c^2}\right)^2}+\dfrac{\dfrac{u_y v}{c^2}a_x}{\gamma^2\left(1-\dfrac{u_x v}{c^2}\right)^3}\\ \; \\ a'_z=\dfrac{a_z}{\gamma^2 \left(1-\dfrac{u_x v}{c^2}\right)^2}+\dfrac{\dfrac{u_z v}{c^2}a_x}{\gamma^2\left(1-\dfrac{u_x v}{c^2}\right)^3}\end{cases}}$

where to obtain the second table we used the usual trick to map primed variables to unprimed variables and to map $v$ into $-v$ .

One important comment is necessary in order to understand the above transformations. One could argue that the mathematical description of accelerated motions is beyond the scope of the special theory of relativity since it is a theory of inertial reference frames. THAT IDEA IS WRONG! Special Relativity (SR) IS based rather on the concept that every reference frame used for the calculations IS an inertial reference frame. It has NOTHING TO DO with accelerations between frames, and as we have showed you here, they can be calculated in the framework of SR!

In conclusion: the description of accelerated motions relative to inertial reference frames makes sense in SR and IS NOT subject to any limitation (unless, of course, you change/modify/extend the basic ideas and/or postulates of SR).

LOG#010. Relativistic velocities.

Posted on 2012/06/06 by amarashikiAugust 29, 2019

In our daily experience, we live in a “non-relativistic” world with a very high degree of accuracy. Thus, if you see a train departing from you ( you are at rest relative to it) with speed $V$ (in the positive direction of the x-axis), you move with relative velocity $V-u$ respect to the train if you run in pursuit of it with speed $u$ , or maybe you can also run with relative speed $V+u$ if you run away from it in the opposite direction of motion.

However, light behavior is diferent to material bodies. Light, a.k.a. electromagnetic waves, is weird. I hope you have realized it from previous posts. We will see what happen with an SR analogue gedanken experiment of the previously mentioned “non-relativistic” train(S’-frame)-track(S-frame) experiment and that we have seen lot of times in our ordinary experience. We will discover that velocities close to the speed of light add in a different way, but we recover the classical result ( like the above) in the limit of low velocities ( or equivalently, in the limit $c\rightarrow \infty$ ).

Problem to be solved:

In the S’-frame, an object (or particle) moves at constant velocity $\vec{v}=\mathbf{v}=(v,0,0)$ relative to the S-frame. In the S’-frame, the object/particle moves with velocity

$\vec{u}\,'=\mathbf{u}'=(u'_x,u'_y,u'_z)=\left( \dfrac{dx'}{dt'},\dfrac{dy'}{dt'},\dfrac{dz'}{dt'}\right)$

The question is: what is the velocity in the S-frame

$\vec{u}=\mathbf{u}=(u_x,u_y,u_z)=\left( \dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt}\right)$

CAUTION ( important note): v is constant (a relative velocity from one frame into another). u is not constant, since it is a vector describing the motion of a particle in some frame.

From the definiton of velocity, and the Lorentz transformation for a parallel motion, we have

$u_x=\dfrac{dx}{dt}=\dfrac{\dfrac{dx}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{d}{dt'}\left[\gamma (x'+\beta ct')\right]}{\dfrac{d}{dt'}\left[\gamma (t'+\frac{\beta }{c}x')\right]}=\dfrac{\dfrac{dx'}{dt'}+\beta c \dfrac{dt'}{dt'}}{\dfrac{dt'}{dt'}+\dfrac{\beta}{c}\dfrac{dx'}{dt'}}$

and thus we get the addition law of velocities in the direction of motion

$\boxed{u_x=\dfrac{u'_x+v}{1+\dfrac{u'_x v}{c^2}}}$

We can also calculate the transformation of the transverse components to the velocity in the sense of motion. We only calculate the component $u_y$ since the remaining one would be identical but labelled with other letter(the z-component indeed):

$u_y=\dfrac{dy}{dt}=\dfrac{\dfrac{dy}{dt'}}{\dfrac{dt}{dt'}}=\dfrac{\dfrac{dy'}{dt'}}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}$

$\boxed{u_y=\dfrac{u'_y}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}}$

Therefore, the transverse velocity also changes in that way! There is an alternative deduction of the above formula using space and time coordinates. We will proceed in two important cases only.

The first case is when the motion happens with parallel relative velocity. Suppose two inertial frames S and S’. S is moving relative to S’ with velocity $V$ along the X-axis. Moreover, suppose an object that is moving parallel to OX, with velocity $v$ . Imagine two “frozen pictures” of the object at two different times according to S, e.g., fix two times $t_1$ and $t_2$ . The two events have coordinates of space and time given by

$E_1(t_1,x_1,y_1,z_1)$

and

$E_2(t_2,x_2,y_2,z_2)$

But we do know that $t_2=t_1+\Delta t$ and so,

$(t_2,x_2,y_2,z_2)=(t_1+\Delta t,x_1+v\Delta t,y_1,z_1)$

What does the S’-frame observe? Using the Lorentz boost with speed $-V$ , we get

$(t'_1,x'_1,y'_1,z'_1) = (\gamma (V)(t_1+Vx_1/c^2),\gamma (V) (x_1+Vt_1),y_1,z_1)$

and

$(t'_2,x'_2,y'_2,z'_2)=(\gamma (V)(t_2+Vx_2/c^2),\gamma (V)(x_2+Vt_2),y_2,z_2)$

Therefore, the velocity of the object according to the S’-frame will be:

$v'=\dfrac{\Delta x'}{ \Delta t'}=\dfrac{x'_2-x'_1}{t'_2-t'_1}=\dfrac{x_2+Vt_2-(x_1+Vt_1)}{t_2+(V/c^2)x_2-(t_1+(V/c^2)x_1)}$

and in this way, we obtain, dividing by $t_2-t_1$

$v'=\dfrac{\dfrac{(x_2-x_1)+V(t_2-t_1)}{(t_2-t_1)}}{\dfrac{(t_2-t_1)+(V/c^2)(x_2-x_1)}{(t_2-t_1)}}$

or equivalently

$\boxed{v'=\dfrac{v+V}{1+\dfrac{V \cdot v}{c^2}}}$

as before.

The second case is when in the velocities between the frames are orthogonal (perpendicular) can be also calculated in this way. Suppose that some object is moving in the orthogonal ( perpendicular) direction to the OX axis. For instance, we can suppose it moves along the y-axis (OY axis) with velocity v measured in the S-frame. We proceed in the same fashion that the previous calculation. We take two “imaginary pictures” of the body at $t_1$ and $t_2=t_1+\Delta t$ . We write the coordinates of space and time of this object as

$E_1(t_1,x_1,y_1,z_1)$

and

$E_2(t_2,x_2,y_2,z_2)$

But we do know that $t_2=t_1+\Delta t$ and so,

$(t_2,x_2,y_2,z_2)=(t_1+\Delta t,x_1,y_1+v\Delta t,z_1)$

We make the corresponding Lorentz boost on those coordinates

$(t'_1,x'_1,y'_1,z'_1) = (\gamma (V)(t_1+Vx_1/c^2),\gamma (V) (x_1+Vt_1),y_1,z_1)$

$(t'_2,x'_2,y'_2,z'_2)=(\gamma (V)(t_2+Vx_2/c^2),\gamma (V)(x_2+Vt_2),y_2,z_2)$

and now, the two components of this motion in the S’-frame will be given by ( note than our two set of coordinates have $x_2=x_1$ and $y_2\neq y_1$ in this particular case):

$v'_{x'}=\dfrac{\Delta x'}{\Delta t'}=\dfrac{x_2+Vt_2-(x_2+Vt_1)}{t_2+(V/c^2)x_2-(t_1+(V/c^2)x_1)}=V$

$v'_{y'}=\dfrac{\Delta y'}{\Delta t'}=\dfrac{y'_2-y'_1}{\gamma (V)(t_1+(V/c^2)x_2-(t_1+(V/c^2)x_1))}=\dfrac{\dfrac{(y_2-y_1)}{(t_2-t_1)}}{\gamma (V)}=\dfrac{v}{\gamma (V)}$

and so, in summary, in the orthogonal relative motion we have

$\boxed{ v'_{x'}=V \;\; v'_{y'}=\dfrac{v}{\gamma (V)}}$

Indeed, these two cases are particular cases of the general transformations we got before.

The complete transformation of the velocity components and their inverses (obtained with the simple rule of mapping v into -v, and primed variables into unprimed variables) can be summarized by these formulae:

$\mbox{SR: Adding velocity(I)}\begin{cases}u_x=\dfrac{u'_x+v}{1+\dfrac{u'_x v}{c^2}} \; \; u_y=\dfrac{u'_y}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}\; \; u_z=\dfrac{u'_z}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}\\ \; \\ u'_x=\dfrac{u_x-v}{1-\dfrac{u_x v}{c^2}} \; \; u'_y=\dfrac{u_y}{\gamma \left( 1-\dfrac{u_x v}{c^2}\right)}\; \; u'_z=\dfrac{u_z}{\gamma \left( 1-\dfrac{u_x v}{c^2}\right)}\\ \; \\ \mathbf{u}=(u_x,u_y,u_z)\;\; \mathbf{u}'=(u'_x,u'_y,u'_z)\;\; \gamma =\dfrac{1}{\sqrt{1-\beta ^2}}\;\; \beta =\dfrac{v}{c}\end{cases}$

Of course, these transformations are valid in the case of a parallel relative motion between S and S’. What are the transformations in the case of non-parallel motion? Suppose that

$\vec{\beta} =\dfrac{\mathbf{v}}{c}=(\beta_x,\beta_y,\beta_z)$

and $\mathbf{u}=(u_x,u_y,u_z), \; \mathbf{u}'=(u'_x,u'_y,u'_z)$ as before. Then, using the most general Lorentz transformations

$\mathbf{u}'=\dfrac{d\mathbf{r}'}{dt'}=\dfrac{d\mathbf{r}+(\gamma - 1)\dfrac{\left(\vec{\beta}\cdot \mathbf{u}\right)\vec{\beta}}{\beta^2}-\gamma \vec{\beta}cdt}{\gamma dt - \dfrac{1}{c}\gamma \vec{\beta}\cdot d\mathbf{r}}$

then, using the same trick as above

$\mathbf{u}'=\dfrac{d\mathbf{r}'}{dt'}=\dfrac{\dfrac{d\mathbf{r}'}{dt}}{\dfrac{dt'}{dt}}$

$\mathbf{u}'=\dfrac{\dfrac{1}{\gamma}\mathbf{u}+\left(1-\dfrac{1}{\gamma}\right)\dfrac{\left(\vec{\beta}\cdot \mathbf{u}\right)\vec{\beta}}{\beta^2}-\vec{\beta} c}{1-\dfrac{1}{c}\vec{\beta} \cdot \mathbf{u}}$

We have got the following transformations (we apply the same recipe to obtain the inverse transformations, also included in the box below):

$\mbox{SR: Adding velocity(II)}\begin{cases} \mathbf{u}'=\dfrac{1}{1-\dfrac{\mathbf{u}\cdot\mathbf{v}}{c^2}}\left[\dfrac{\mathbf{u}}{\gamma}+\left[\left(1-\dfrac{1}{\gamma}\right)\dfrac{\mathbf{u}\cdot\mathbf{v}}{v^2}-1\right]\mathbf{v}\right]\\ \;\\ \mathbf{u}=\dfrac{1}{1+\dfrac{\mathbf{u}'\cdot\mathbf{v}}{c^2}}\left[\dfrac{\mathbf{u}'}{\gamma}-\left[-\left(1-\dfrac{1}{\gamma}\right)\dfrac{\mathbf{u}'\cdot\mathbf{v}}{v^2}-1\right]\mathbf{v}\right]\end{cases}$

We observe that these equations are a non-linear addition of velocities. Equivalently, they can be rewritten as follows after some elementary algebra using a mathematical structure called gyrovector (or gyrovector addition):

$\mbox{gyrovector law}\begin{cases}\mathbf{u}'\equiv -\mathbf{v}\biguplus_{REL}\mathbf{u}=\dfrac{1}{1-\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}\left[-\mathbf{v}+\dfrac{\mathbf{u}}{\gamma_{\mathbf{v}}}+\dfrac{1}{c^2}\left(\dfrac{\gamma_{\mathbf{v}}}{\gamma_{\mathbf{v}}+1}\right)\left(\mathbf{v}\cdot \mathbf{u}\right)\mathbf{v}\right]\\ \; \\ \mathbf{u}\equiv \mathbf{v}\biguplus_{REL}\mathbf{u}'=\dfrac{1}{1+\dfrac{\mathbf{u}'\cdot \mathbf{v}}{c^2}}\left[\mathbf{v}+\dfrac{\mathbf{u}'}{\gamma_{\mathbf{v}}}+\dfrac{1}{c^2}\left(\dfrac{\gamma_{\mathbf{v}}}{\gamma_{\mathbf{v}}+1}\right)\left(\mathbf{v}\cdot \mathbf{u}'\right)\mathbf{v}\right]\end{cases}$

These 2 cases can be seen as particular examples in the addition rule of velocities as a “gyrovector sum”, the nonlinear addition rule given by the formula:

$\mathbf{u}\biguplus_{REL}\mathbf{v}=\dfrac{1}{1+\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}\left[\mathbf{u}+\dfrac{\mathbf{v}}{\gamma_{\mathbf{u}}}+\dfrac{1}{c^2}\left(\dfrac{\gamma_{\mathbf{u}}}{\gamma_{\mathbf{u}}+1}\right)\left(\mathbf{u}\cdot \mathbf{v}\right)\mathbf{u}\right]$

This formula is usually written in a more intuitive expression with the following arguments. Suppose some object moves with velocity $\mathbf{v}$ in some inertial frame S. S is moving itself with relative velocity $\mathbf{V}$ respect to another frame S’. In the S’-frame, the velocity is given by:

$\boxed{\mathbf{v}'\equiv \mathbf{V}\biguplus_{REL}\mathbf{v}=\dfrac{\mathbf{v}_\parallel+\gamma^{-1}(V)\mathbf{v}_\perp + \mathbf{V}}{1+\dfrac{\mathbf{V}\cdot \mathbf{v}}{c^2}}}$

and where we have defined the projections of $\mathbf{v}$ in the direction parallel and orthogonal to $\mathbf{V}$ . They are given by:

$\boxed{\mathbf{v}_\parallel =\dfrac{(\mathbf{V}\cdot \mathbf{u})\mathbf{V}}{V^2}\;\;\;\;\;\; V^2=\vert \mathbf{V}\vert^2\;\;\;\;\;\mathbf{v}_\perp =\mathbf{v}-\mathbf{v}_\parallel}$

We will talk about gyrovectors more in a future post. They have a curious mathematical structure and geometry, and they are not well known by physicists since they are not in the basic curriculum and background of SR courses. Of course, the non-associative composition rule for velocities is not a standard formula you can find in books about relativity, so I will write it here:

$\boxed{\mathbf{u}\boxplus\mathbf{v}=\dfrac{\mathbf{u}+\mathbf{v}}{1+\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}+\dfrac{\gamma_{\mathbf{u}}}{c^2(\gamma_{\mathbf{u}}+1)}\dfrac{\mathbf{u}\times\left(\mathbf{u}\times\mathbf{v}\right)}{1+\dfrac{\mathbf{u}\cdot \mathbf{v}}{c^2}}}$

and where we used the previous formula for $\mathbf{u}\biguplus_{REL}\mathbf{v}$ and after some algebra we used the known relationship for the cross product of three vectors, two being the same,

$u\times(u\times v)=(u\cdot v)u-(u\cdot u)v$

Let’s go back to the beginning. Suppose now we imagine a train (our S’-frame) travelling at velocity $v$ . Suppose that Special Relativity matters now. We are in the tracks as observers “in relative rest” with respect to the train ( we are the S-frame) and suppose that we take into account the SR corrections above to the addition of velocities. Inside the train some object is being thrown with velocity $u'_x$ in the direction of motion. What is the velocity $u_x$ in the S-frame? That is, what is the velocity we observe in the tracks? In this simple example, we use the easier addition rule of velocities ( named addition rule SR(I) above). Firstly, we note some expected features from the mathematical structure of the relativistic addition rule of velocities (valid propterties as well in the general case (II) with a suitable generalization):

1st. For low velocities, i.e., if $u'_x/c<<1$ and/or $\beta=\dfrac{v}{c}<<1$ , the result approaches the nonrelativistic “ordinary life” experience: $u_x=u'_x+v$ .

2nd. For positive velocities $u'_x>0$ and a positive relative velocity between frames $v>0$ , the addition of velocities is generally $u_x<u'_x+v$ , i.e., we get a velocity smaller that in the non-relativistic (ordinary or “common” experience) limit.

Now, some easy numerical examples to see what is going on bewteen the train (S’) and the track (S) where we are:

Example 1. Moderate velocity case. We have, e.g., velocities $v=u'_x=30 km\cdot s^{-1}=10^{-4}c$ . This gives, using (I):

$u_x=\dfrac{2 \cdot 10^{-4}c}{1+10^{-8}}\approx 60 km\cdot s^{-1} - 0.6 mm\cdot s^{-1}$

Then, the deviation with respect to the non-relativistic value ( 60km/s) is negligible for all the practical purposes! This typical velocity, 30km/s, is about the typical velocities in 20th and early 21st century space flight. So, our astronauts can not note/observe relativistic effects. The addition theorem in SR is not practical in current space travel (20th/early 21st century).

Example 2. Case velocities are “close enough” to the speed of light. E.g.: One quarter and one half of the speed of light. In the first case,

$v=u'_x=0.25c$

and then

$u_x=\dfrac{0.25c+0.25c}{1+0.0625}=\dfrac{0.5c}{1.0625}\approx 0.47c$

In the seconde case, we write $v=u'_x=0.5c$ . This provides

$u_x=\dfrac{0.5c+0.5c}{1+0.25}=\dfrac{c}{1.25}=0.8c$

Thus, we observe that a higher velocities, e.g., those in particle accelerators, some processes in the Universe, etc, the relativistic effects of the non-linear addition of velocities can NOT be neglected. The effect is important and becomes increasingly important when the velocity increases itself ( you can note how large the SR effect is if you compare the 0.25c and 0.5c examples above).

Example 3. Speed of light case. Extreme case: we are trying to exceed the velocity of light. Suppose now, that the train could move with relative velocity equal to c. The object is thrown with relative speed $u'_x=c$ and $u'_y=u'_z=0$ . What we do see on the track. Naively, ordinary life would suggest the answer 2c, but we do know that velocities transform non-linearly, so, we plug the values in the formula to get the answer:

$u_x=\dfrac{c+c}{1+1}=c$ and $u_y=u_z=0$ . Therefore, if a train is travelling at the speed of light, and inside the train an object is thrown forward at c, we DO NOT observe/measure a 2c velocity, we observe/measure it has velocity $c$ !!!! if we stand at rest on the track. Amazing!

Suppose we try to do it in a “transverse way”. That is, suppose that the velocities are now $v=c$ , $u'_x=0$ and the transverse speed is not $u'_y=c$ . This case results in the numbers:

$u_x=c$

and

$u_y=\dfrac{c}{\gamma}=0$

since

$\gamma \rightarrow \infty$

and thus $u=c$ .

Therefore, if the train travels at c, an inside of the train an object is launched at right angles to the direction of motion at the velocity c, the object itself is measured/seen to have velocity c measured from a rest observer placed beside the tracks. Amazing, surprise again!

Example 4. Case: Superluminal relative motion. Suppose, somehow, the relative motion between the two frames provides $v=2c$ (even you can plug $v=nc$ with $n>1$ if you wish). Suppose the object is measured to have the extremal limit speed $u'_x=c$ (imagine we consider a light beam/flash, for instance). Again, using the addition law we would get:

$u_x=\dfrac{3c}{1+2}=c$

and

$u_x=\dfrac{n+1}{1+n}c=c$

Even if the inertial frames move at superluminal velocities relative to each other, a light beam would remain c in the S-frame if SR holds! Surprise, again!

In this way, we can conclude one of the most important conclusions of special relativity ( something that it is ignored by many Sci-fi writers, and that we would like to be able to overcome somehow if we have to master the interstellar travel/interstellar communications as Sci-fi fans, or as an interstellar civilization, you should get some trick to avoid/”live with” it.):

$\boxed{\mbox{The speed of light can never be exceeded by adding velocities in SR.}}$

If SR holds, the velocity (or speed) of light is the maximum speed attainable in the Universe. You can like it or hate it, but if SR is true, you can not avoid this conclusion.

There is another special case of motion important in practical applications: two dimensional motion. I mean, imagine that in the S’-frame, an object has the velocity $\mathbf{u}'=(u'_x,u'_y,0)$ . The velocity subtends an angle $\theta '$ with the x’-axis. See the figure below:

What is the angle $\theta$ in the frame that we observe between $\mathbf{u}$ and the x-axis? For the S-frame we find:

$\tan \theta= \dfrac{u_y}{u_x}=\dfrac{\dfrac{u'_y}{\gamma \left( 1+\dfrac{u'_x v}{c^2}\right)}}{\dfrac{u'_x+v}{1+\dfrac{u'_x v}{c^2}}}$

and after some easy algebraic manipulations we get the important result

$\boxed{\tan \theta =\dfrac{1}{\gamma}\dfrac{u'_y}{u'_x+v}}$

We observe that, according to this last equation, with teh exception of $l v=0$ , $\tan \theta$ is smaller in the S frame than $\tan \theta '=\dfrac{u'_y}{u'_x}$ in the S’-frame. In the non-relativist limit, we recover the result that our ordinary intuition and experience provides ( $\gamma \rightarrow 1$ ):

$\theta_{nonrel}=\dfrac{u'_y}{u'_x+v}$

It is logical. In the nonrelativistic limit we do know that $u'_y=u_y$ and $u_x=u'_x+v$ , so the result agrees with our experience in the low velocity realm.

The Spectrum Of Riemannium

Mobilis in mobili: Physmatics Chronicles!

LOG#019. Triangle mnemonics.

Like this:

LOG#018. Momenergy (III).

Like this:

LOG#017. Momenergy (II).

Like this:

LOG#016. Momenergy (I).

Like this:

LOG#015. Time of flight.

Like this:

LOG#014. Vectors in spacetime.

Like this:

LOG#013. Spacetime.

Like this:

LOG#012. Michelson-Morley.

Like this:

LOG#011. Relativistic accelerations.

Like this:

LOG#010. Relativistic velocities.

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this:

Share this:

Like this: