## LOG#039. Relativity: Examples(III).

Example 1. Absorption of a photon by an atom.

In this process, we have from momenergy conservation:

$P^\mu_a P_{a\mu}+2P^\mu_a P_{b\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}$

If the atom is the rest frame, before absorption we get

$P^\mu_a =(m_a c,0,0,0)$

$P^\mu_b=(\dfrac{E_b}{c},p_{bx},0,0)=(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0)=(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0)$

Description: an atom “a” at rest with mass $m_a$ absorbs a photon “b” propagating in the x-direction turning itself into an excited atom “c”, moving in the x-axis (suffering a “recoil” after the photon hits it).

The atom after aborption has

$P^\mu_c=(m_c c,0,0,0)$

Therefore, since the photon verifies:

$P^\mu_b P_{b\mu}=\left(\dfrac{h\nu}{c}\right)^2-\left(\dfrac{h\nu}{c}\right)^2=0$

and it is true in every inertial frame. Thus,

$(m_a c)^2+2m_a c\dfrac{h\nu}{c}+0=(m_a c)^2$

Then,

$\boxed{m_c=\sqrt{m_a^2+2m_a\dfrac{h\nu}{c}}=m_a\sqrt{1+2\dfrac{h\nu}{m_a c^2}}}$

Expanding the square root

$m_c\approx m_a\left[ 1+\dfrac{h\nu}{m_a c^2}-\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c}\right)^2+\mathcal{O}(h^3)\right]$

In this case,

$m_c\approx m_a+\dfrac{h\nu}{ c^2}-\dfrac{1}{2}\left(\dfrac{h\nu}{m_a c}\right)^2$

Atom’s rest mass increases by an amount $\dfrac{h\nu}{c^2}$ up to first order in the Planck’s constant, and it decreases up to second order in h a quanity $-\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2$ due to motion ( “recoil”). Therefore,

$\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2=\dfrac{1}{2}m_a\left( \dfrac{u_a}{c}\right)^2$

In this way, identifying terms: $u_c=\dfrac{h\nu}{m_a c}$

In the laboratory frame, the excited atom velocity is calculated by momenergy conservation. It is simple:

$\dfrac{h\nu}{c}=\gamma_c m_c u_c$

$m_a c^2+h\nu=\gamma_c m_c c^2$

$\dfrac{h\nu}{u_c c}=m_a+\dfrac{h\nu}{c^2}$

Then, we obtain that:

$u_c=\dfrac{h\nu c}{m_a c^2+h\nu}=\dfrac{c}{\left(\dfrac{m_a c^2}{h\nu}\right)+1}\approx \dfrac{h\nu}{m_a}$

where we have used in the last step $m_ac^2>>h\nu$.

Example 2. Emission of a photon by an atom.

An atom c at rest, with $m_c$ the rest mass, emits a photon b with frequency $\dfrac{E_b}{h}=\nu$ in the x-direction, turning itself into a non-excited atom “a”, with $m_a$. What is the energy shift $\Delta E=E_c-E_a$?

$P^\mu_a P_{a\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}$

where

$P^\mu_a$ is $P^\mu_a =(m_a,0,0,0)$ in the rest frame of “a”.

$P^\mu_a$ in the rest from “c” reads $P^\mu_a =\left(\gamma_a m_a c, -\dfrac{E_b}{c},0,0\right)$

$P^\mu_b$ in the rest frame of “c” is $P^\mu_b =\left(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0\right)$

$P^\mu_a$ in the rest frame of “c” is $P^\mu_c=(m_c c,0,0,0)$

In this way, we have

$(m_a c)^2+\gamma _a m_a E_b +\left( \dfrac{E_b}{c}\right)^2=\gamma _am_am_c c^2$

$\gamma_a m_a c^2+E_b=m_c c^2$

$m_a^2c^2+\dfrac{E_b}{c^2}\left(m_c c^2-E_b\right)^2=m_c^2c^2-m_c E_b$

$m_a^2c^2+m_cE_b=m^2_c c^2-m_c E_b$

$m_a^2c^2+m_cE_b=m_c^2c^2-m_cE_b$

From this equations we deduce that

$E_b=\dfrac{m_c c^2-m_a^2c^2}{2m_c}=\dfrac{E_c^2-E_a^2}{2E_c}=(E_c-E_a)\left(1-\dfrac{E_c-E_a}{2E_c}\right)$

And from the definition $E_c-E_a$ we get $E_b=\Delta E\left(1-\dfrac{\Delta E}{2E_c}\right)$

Note: the photon’s energy IS NOT equal to the difference of the atomic rest energies but it is less than that due to the emission process. This fact implies that the atom experieces “recoil”, and it gains kinetic energy at the expense of the photon. There is a good chance for the photon not to be absorbed by an atom of some kind. However, “resonance absorption” becomes problematic. The condition for recoilless resonant absorption to occur nonetheless, e.g., the reabsorption of gamma ray photons by nuclei of the some kind were investigated by Mössbauer. The so-called Mössbauer effect has been important not only to atomic physics but also to verify the theory of general relativity. Furthermore, it is used in materials reseach in present time as well. In 1958, Rudolf L. Mössbauer reported the 1st reoilless gamma emission. It provided him the Nobel Prize in 1961.

Example 3. Decay of two particles at rest.

The process we are going to study is the reaction $C\rightarrow AB$

The particle C decays into A and B. It is the inverse process of the completely inelastic collision we studied in a previous example.

From the conservation of the tetramomentum

$(m_A c)^2+2\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)+(m_Bc)^2=(m_C c)^2$

Choose the frame in which the following equation holds

$P^\mu_AP_{A\mu}+P^\mu_BP_{A\mu}=P^\mu_{C}P_{A\mu}$

Let $u_A,u_B$ be the laboratory frame velocities in the rest frame of “C”. Then, we deduce

$P^\mu_A=(\gamma_Am_Ac,\gamma_Am_Au_A,0,0)$

$P^\mu_B=(\gamma_Bm_Bc,\gamma_Bm_Bu_B,0,0)$

$P^\mu_C=(m_C c,0,0,0)$

From these equations $(m_Ac)^2+\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)=\gamma_Am_Am_C c^2$

$-(m_Ac)^2+(m_Bc)^2=(m_Cc)^2-2\gamma_Am_Am_Cc^2$

$2E_Am_C=(m_A^2+m_C^2-m_B^2)c^2$

$\boxed{E_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}}$

$\boxed{E_A^{kin}=T_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}-m_Ac^2=\dfrac{\left((m_C-m_A)^2-m_B^2\right)c^2}{2m_C}}$

$\boxed{E_B^{kin}=T_B=\dfrac{\left((m_C-m_B)^2-m_A^2)\right)c^2}{2m_C}}$

$\boxed{E_{kin}=T=T_A+T_B=(m_C-m_A-m_B)c^2}$

Therefore, the total kinetic energy of the two particles A,B is equal to the mass defect in the decay of the particle.

Example 4. Pair production by a photon.

Suppose the reaction $\gamma \rightarrow e^+e^-$, in which a single photon ($\gamma$)decays into a positron-electron  pair.

That is, $h\nu\rightarrow e^+e^-$.

$P^\mu_a=P^\mu_c+P^\mu_d$

Squaring the momenergy in both sides:

$P^\mu_a P_{\mu a}=P^\mu_c P_{\mu c}+2P^\mu_c P_{\mu d}+P^\mu_d P_{\mu d}$

In the case of the photon: $P^\mu_a P_{a\mu}=0$

In the case of the electron and the positron: $P^\mu_c P_{\mu c}=P^\mu_d P_{\mu d}=-(m_e c)^2=-(mc)^2$

We calculate the components of momenergy in the center of mass frame:

$P^\mu_c=\left( \dfrac{E_c}{c},p_{cx},p_{cy},p_{cz}\right)$

$P^\mu_d=\left( \dfrac{E_d}{c},-p_{dx},-p_{dy},-p_{dz}\right)$

with $E_c=E_d=mc^2$. Therefore,

$2 P^\mu_c P_{\mu d}=-2\left( \dfrac{E_c^2}{c^2}+p_x^2+p_y^2+p_z^2\right)$

so

$-2(mc)^2-2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0$

$2(mc)^2+2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0$

This equation has no solutions for any positive solution of the photon energy! It’s logical. In vacuum, it requires the pressence of other particle. For instance $\gamma \gamma \rightarrow e^+e^-$ is the typical process in a “photon collider”. Other alternative is that the photon were “virtual” (e.g., like QED reactions $e^+e^-\rightarrow \gamma^\star\rightarrow e^+e^-$). Suppose, alternatively, the reaction $AB\rightarrow CDE$. Solving this process is hard and tedious, but we can restrict our attention to the special case of three particles $C,D,E$ staying together in a cluster C. In this way, the real process would be instead $AB\rightarrow C$. In the laboratory frame:

$P^\mu_A+P^\mu_B=P^\mu_C$

we get

$P^\mu_A=\left( \dfrac{E_A}{c},\dfrac{E_A}{c},0,0\right)$

$P^\mu_B=\left( Mc,0,0,0\right)$

and in the cluster reference frame

$P^\mu_C=\left((M+2m)c,0,0,0\right)$

Squaring the momenergy:

$0+2E_AM+(Mc)^2=(M+2m)^2c^2$

$2ME_A=4Mmc^2+4m^2c^2$

$E_A=2mc^2+\dfrac{2m^2c^2}{M}$

$\boxed{E_A=2mc^2\left(1+\dfrac{m}{M}\right)}$

In the absence of an extra mass M, i.e., when $M\rightarrow 0$, the energy $E_A$ would be undefined, and it would become unphysical. The larger M is, the smaller is the additional energy requiere for pair production. If M is the electron mass, and $M=m$, the photon’s energy must be twice the size of the rest energy of the pair, four times the rest energy of the photon. It means that we would obtain

$E=4mc^2=2m_{pair}c^2$

and thus $\gamma =4\rightarrow$

$\beta^2=1-\dfrac{1}{16}=\dfrac{15}{16}$

$\beta=\sqrt{\dfrac{15}{16}}=\dfrac{\sqrt{15}}{4}\approx 0.97$

$v=\dfrac{\sqrt{15}}{4}c\approx 0.97c$

In general, if $m\neq M$ we would deduce:

$\boxed{\beta=\sqrt{1-\dfrac{1}{4\left(1+\frac{m}{M}\right)^2}}}$

Example 5. Pair annihilation of an  electron-positron couple.

Now, the reaction is the annihilation of a positron-electron pair into two photons, turning mass completely into (field) energy of light quanta. $e^+e^-\rightarrow \gamma \gamma$ implies the momenergy conservation

$P^\mu_a+P^\mu_b=P^\mu_c+P^\mu_c$

where “a” is the moving electron and “b” is a postron at rest. Squaring the identity, it yields

$P^\mu_aP_{\mu a}+2P^\mu_aP_{\mu b}+P^\mu_bP_{\mu b}=P^\mu_cP_{\mu c}+2P^\mu_cP_{\mu c}+P^\mu_dP_{\mu d}$

Then, we deduce

$P^\mu_a=\left(\dfrac{E_a}{c},p_{ax},0,0\right)$

$P^\mu_b=\left( m_e c,0,0,0\right)$

while we do know that

$P^\mu_aP_{\mu a}=P^\mu_bP_{\mu b}=-(m_e c)^2$ for the electron/positron and

$P^\mu_cP_{\mu c}=P^\mu dP_{\mu d}=0$ since they are photons. The left hand side is equal to $-2(m_e c)^2+2E_am_e$, and for the momenergy in the right hand side

$P^\mu_c=\left(\dfrac{E_c}{c},p_{cx},0,0\right)$

$P^\mu_d=\left( \dfrac{E_d}{c},p_{dx},0,0\right)$

Combining both sides, we deduce

$(m_a c)^2+E_a m_e=\dfrac{E_cE_d}{c^2}-p_{cx}p_{dx}$

The only solution to the right hand side to be not zero is when we select $p_{cx}=\pm \dfrac{E_c}{c}$ and $p_{dx}=\pm\dfrac{E_d}{c}$ and we plug values with DIFFERENT signs. In that case,

$\boxed{(mc)^2+E_a m=\dfrac{2E_cE_d}{c^2}}$

From previous examples:

$P^\mu_aP_{\mu_b}+P^\mu_bP_{\mu b}=P^\mu_c P_{b\mu}+P^\mu_aP_{\mu b}$

and we evaluate it in the laboratory frame to give

$\boxed{E_a m+(mc)^2=E_c m+E_d m}$

The last two boxed equations allow us to solve for $E_d$

$\boxed{E_d=E_a-E_c+mc^2}$

If we insert this equation into the first boxed equation of this example, we deduce that

$(mc)^2+mE_a=\dfrac{2E_c}{c^2}\left( E_a-E_c+mc^2\right)$

or

$\dfrac{1}{2}mc^2\left(mc^2+E_a\right)=-E_c^2+E_c(E_a+mc^2)$

Solving for $E_c$ this last equation

$\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{1}{2}\left(E_a+mc^2\right)\pm\sqrt{\dfrac{1}{4}\left(E_a+mc^2\right)^2-\dfrac{1}{2}mc^2\left(mc^2+E_a\right)}}$

$\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{\left(E_a+mc^2\right)\pm \sqrt{\left(E_a+mc^2\right)\left(E_a-mc^2\right)}}{2}}$

## LOG#038. Relativity: Examples(II).

Example 1. Completely inelastic collision of 2 particles $AB\rightarrow C$.

We will calculate the mass and velocity after the collision, when a cluster is formed.

$E_a+E_b=E_c$

$p_{ax}+p_{bx}=p_{cx}$

$p_{ay}+p_{by}=p_{cy}$

$p_{az}+p_{az}=p_{cz}$

From the consevation of momentum in the collision, we get $P^\mu_a+P^\mu_b=P^\mu_c$

Then,

$\left(\dfrac{E_a}{c},p_{ax},p_{ay},p_{az}\right)+\left(\dfrac{E_b}{c},p_{bx},p_{by},p_{bz}\right)=\left(\dfrac{E_c}{c},p_{ac},p_{ac},p_{ac}\right)$

Squaring both sides:

$\left(P_{\mu a}+P_{\mu b}\right)\left(P^{\mu a}+P^{\mu b}\right)=P^\mu_c P_{\mu c}$

$P^\mu_a P_{\mu a}+P^\mu_a P_{\mu b}+P^\mu_b P_{\mu a}+P^\mu_b P_{\mu b}=P^\mu_c P_{\mu c}$

$P^\mu_a P_{\mu a}+2P^\mu_a P_{\mu b}+P^\mu_b P_{\mu b}=P^\mu_c P_{\mu c}$

In the rest frame of the particle labelled with “a”:

$E_a=mc^2$ $\mathbf{p}_a=\mathbf{0}$

$P^\mu_a=(m_a c,0,0,0)$

$P^\mu_a P_{\mu a}=(-m_a^2 c^2)$

$P^\mu_b P_{\mu b}=(-m_b^2 c^2)$

$P^\mu_c P_{\mu c}=(-m_c^2 c^2)$

Now, with $E_a=\gamma _a m_a c^2$ and $E_b =\gamma _b m_b c^2$

and

$P^{\mu}_a=(\gamma _a m_a c, \gamma _a m_a u_a,0,0)$

$P^{\mu}_b=(\gamma _b m_b c, \gamma _b m_b u, 0, 0)$

$2P^\mu_a P_{\mu b}=2\gamma _a \gamma_ bm_a m_b (u_a u_b-c^2)$

And thus,

$-(m_a c)^2+2\gamma _a\gamma _b m_a m_b (u_a u_b-c^2)-(m_b c)^2=-(m_c c)^2$

$(m_a c)^2+2\gamma _a\gamma _b m_a m_b (c^2-u_a u_b)+(m_b c)^2=(m_c c)^2$

From this equation, we get

$\boxed{m_c=\sqrt{m_a^2+m_b^2+2m_am_b\gamma _a\gamma_b\left(1-\dfrac{u_au_b}{c^2}\right)}}$

and

$\gamma _a \gamma b \left(1-\dfrac{u_a u_b}{c^2}\right)=\dfrac{1-\dfrac{u_a u_b}{c^2}}{\sqrt{1-\dfrac{u_a^2}{c^2}}\sqrt{1-\dfrac{u_b^2}{c^2}}}$

If $u_a, u_b<, then $m_c=m_a+m_b$, as we should expect from classical physics ( Lavoisier’s law: the mass is conserved).

We observe that, in general, there is NO rest mass conservation law in relativistic physics ( as we have mentioned before). Momentum and energy can be transformed into rest mass and viceversa. From the next equations:

$\gamma _am_a c^2+\gamma _b m_bc^2=\gamma _cm_c c^2$

$\gamma_a m_a u_a+\gamma_b m_bu_b=\gamma_c m_cu_c$

We get

$\boxed{u_c=\dfrac{\gamma _a m_a u_a +\gamma _bm_bu_b}{\gamma _am_a+\gamma_bm_b}}$

is the center or mass velocity in the laboratory frame. If $u_b=0$ and $\gamma_b=1$, then

$u_c=\dfrac{\gamma_a m_au_a}{\gamma_a m_a+m_b}=\dfrac{u_a}{1+\dfrac{m_b}{\gamma_a m_a}}=\dfrac{u_a}{1+\dfrac{m_b}{m_a}\sqrt{1-\dfrac{u_a^2}{c^2}}}$

And if we write $u_a\rightarrow 0$, the formula provides

$u_c=\dfrac{m_a}{m_a+m_b}u_a$

And it is the classical center of mass equation for two particles, as we expected to be.

Example 2. Production of a proton-antiproton pair $p-\overline{p}$ in the chemical highly energetic process given by $pp\rightarrow p\overline{p} pp$.

If we get an extremely energetic beam of protons. Then, we collide one proton A of the beam with another proton B, at rest. If the kinetic energy is enough, we can create a proton-antiproton pair! The question is, what is the minimum (total) energy, sometimes called threshold energy?What is the minimum kinetic energy, or threshold kinetic energy? And what is the velocity of the incident proton A in order to get the pair production? We are going to solve this really nice problem.

The proton B at rest has $u_B=0$ and then $\gamma_B=1$.

Note that we have $m_A=m_B=m_{proton}=m_p=m$ and $m_C=4m_p=4m$, for the system after the collision when the pair is created.

Using the formulae we have studied

$(mc)^2+2\gamma_A(mc)^2+(mc)^2=(4mc)^2$

$2\gamma_A m^2c^2=14m^2c^2$

and thus

$\boxed{E_A^{min}=7mc^2}$

It implies $\boxed{T_A^{min}=E_A^{min}-mc^2=6mc^2}$ for the kinetic threshold energy. The necessary velocity can be obtained in a simple way:

$E_A=\dfrac{mc^2}{\sqrt{1-\dfrac{u_A^2}{c^2}}}=7mc^2\rightarrow u_A=c\sqrt{1-\dfrac{1}{7}^2}\approx 0.990c$

And thus,

$\left(\dfrac{c-u_a}{c}\approx 1\%\right)$

Energetically, we find that the collision is more “favourable”, in general, when we get a storage ring with 2 protons and we crash them with $V_A=-V_B$. In that case, the energy threshold and the kinetic energy threshold, provide:

$u_A=u$ $u_B=-u$ $\gamma_a=\gamma_b=\gamma_u$ $m_a=m_b=m$

$m_C=4m_p=4m$

$(mc)^2+2\gamma^2(c^2+u^2)+(mc)^2=(4mc)^2$

$\gamma^2(c^2+u^2)=7c^2$

$c^2+u^2=7(c^2-u^2)$

and then, the minimum velocity of each proton has to be: $u=\sqrt{\dfrac{3}{4}}c=\dfrac{\sqrt{3}}{2}c\approx 0.866c<0.990c$

Thus, the minimun threshold total energy of each proton is:

$E=\dfrac{mc^2}{\sqrt{1-\frac{3}{4}}}=\dfrac{mc^2}{\sqrt{\frac{1}{4}}}=2mc^2$

Therefore, the minimum kinetic energy, the kinetic threshold energy, is given by $T=mc^2$. For each proton PAIR, the total energy threshold is $E_{pair}^{min}=4mc^2$. This value differs from the relativistic value previously obtained by an amount $2mc^2$. Now the kinetic threshold energy to obtain the pair equals $T_{min}=4mc^2$i.e., 1/3 ($E=2mc^2$)of the total energy in the previous example ($6mc^2$) is saved smashing 2 protons in opposite directions in order to create the pair proton-antiproton.

We can generalize the problem to the creation of n-pairs proton-antiproton. The equations yield:

$E_A^{min}=\dfrac{(2n+2)^2(mc)^2-2(mc)^2}{2m}$

and then

$\boxed{E_A^{min,n-pairs}=\left(2(n+1)^2-1\right)mc^2=(2n^2+4n+1)mc^2}$

The threshold kinetic energy will be in this case:

$\boxed{T_A^{min,n-pairs}=\left(2n(n+2)\right)mc^2=(2n^2+4n)mc^2}$

and the incident minimum velocity to hit the proton at rest is now:

$u_A=c\sqrt{1-\left(\dfrac{mc^2}{E}\right)^2}=c\sqrt{1-\left(\dfrac{1}{2n^2+4n+1}\right)^2}$

In the case of a very large number of pairs, $n>>1$ we get that

$u_A\approx c\sqrt{1-\left(\dfrac{1}{2n^2}\right)^2}=c\sqrt{1-\left(\dfrac{1}{4n^4}\right)}\approx c$

In the case of a storage ring,

$\gamma_u (c^2+u^2)=(2n^2+4n+1)c^2$

and from this equation, algebra provides

$\boxed{u=\sqrt{\dfrac{2n^2+4n}{2n^2+4n+2}}c}$

and

$\boxed{E=\dfrac{mc^2}{\sqrt{1-\left(\dfrac{2n^2+4n}{2n^2+4n+2}\right)^2}}}$

## LOG#037. Relativity: Examples(I)

Problem 1. In the S-frame, 2 events are happening simultaneously at 3 lyrs of distance. In the S’-frame those events happen at 3.5 lyrs. Answer to the following questions: i) What is the relative speed between frames? ii) What is the temporal distance of events in the S’-frame?

Solution. i) $x'=\gamma (x-\beta c t)$

$x'_2-x'_1=\gamma ((x_2-x_1)-\beta c (t_2-t_1))$

And by simultaneity, $t_2=t_1$

Then $\gamma=\dfrac{x'_2-x'_1}{x_2-x_1}=\dfrac{7}{6}$

$\beta=\sqrt{1-\gamma^{-2}}\approx 0.5$

ii) $ct'=\gamma (ct-\beta x)$

$c(t'_2-t'_1)=-\gamma \beta (x_2-x_1)$

since we have simultaneity implies $t_2-t_1=0$. Then,

$c\Delta t'\approx -1.8 lyrs$

Problem 2. In S-frame 2 events occur at the same point separated by a temporal distance of 3yrs. In the S’-frame, $D'=3.5yrs$ is their spatial separation. Answer the next questions: i) What is the relative velocity between the two frames? ii) What is the spatial separation of events in the S’-frame?

Solution. i) $ct'=\gamma (ct-\beta x)$ with $x_1=x_2$

As the events occur in the same point $x_2=x_1$

$c(t'_2-t'_1)=\gamma c (t_2-t_1)$

$\gamma=\dfrac{t'_2-t'_1}{t_2-t_1}=\dfrac{7}{6}$

$\beta=\sqrt{1-\gamma^{-1}}\approx 0.5$

ii) $x'=\gamma (x-\beta c t)$

$x_1=x_2$ implies $x'_2-x'_1=-\gamma \beta c (t_2-t_1)\approx -1.8 lyrs$

Therefore, the second event happens 1.8 lyrs to the “left” of the first event. It’s logical: the S’-frame is moving with relative speed $v\approx c/2$ for $3.5 yrs$.

Problem 3. Two events in the S-frame have the following coordinates in spacetime: $P_1(x_0=ct_1,x_1=x_0)$, i.e., $E_1(ct_1=x_0,x_1=x_0)$ and $P_2(ct_2=0.5x_0, x_2=2x_0)$, i.e., $E_2(ct_2=x_0/2,x_2=2x_0)$. The S’-frame moves with velocity v respect to the S-frame. a) What is the magnitude of v if we want that the events $E_1,E_2$ were simultaneous? b) At what tmes t’ do these events occur in the S’-frame?

Solution. a) $ct'=\gamma (ct-\beta x)$

$t'_2-t'_1=0$ and then $0=\gamma (c(t_2-t_1)-\beta (x_2-x_1))$

$\beta =\dfrac{c(t_2-t_1)}{(x_2-x_1)}=-\dfrac{0.5x_0}{x_0}=-0.5$

b) $t'=\gamma ( 1-\beta x/c)=\gamma ( 1-\beta x/c)$

$t'_1=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5x_0 /c)\right)\approx 1.7x_0/c$

$t'_2=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5\cdot 2\cdot x_0/c)\right)\approx 1.7x_0/c$

Problem 4. A spaceship is leaving Earth with $\beta =0.8$. When it is $x_0=6.66\cdot 10^{11}m$ away from our planet, Earth transmits a radio signal towards the spaceship. a) How long does the electromagnetic wave travel in the Earth-frame? b) How long does the electromagnetic wave travel in the space-ship frame?

For the spaceship, $ct=x/\beta$ and for the signal $ct=x+ct_0$. From these equations, we get

$late \beta ct=x$ and $ct=x+ct_0$, and it yields $\beta ct =ct-ct_0$ and thus $t=\dfrac{t_0}{1-\beta}$ for the intersection point. But, $\beta=0.8=8/10=4/5$ and $1-\beta=1/5$. Putting this value in the intersection point, we deduce that the intersection point happens at $t_1=5t_0$. Moreover,

$t_1-t_0=4t_0=4\dfrac{x_0}{0.8c}\approx 11100 s=3.08h=3h 5min$

b) We have to perform a Lorentz transformation from $(ct_0,0)$ to $(ct_1,x_1)$, with $t'=t=0$.

$t_0=x_0/v=2775s$ and $t_1=5t_0=13875s$. Then $x_1=vt_1=5vt_0=5x_0=5\cdot 6.66\cdot 10^{11}m=3.33\cdot 10^{12}m$. And thus, we obtain that $\gamma=5/3$. The Lorentz transformation for the two events read

$(t'_2-t'_1)=\gamma (t_1-t_0)=\gamma (t_1-t_0)-\beta/c(x_1-x_0)=3700 s\approx 1.03h=1h1m40s$

Remarks: a) Note that $t_1-t_0$ and $t'_2-t'_1$ differ by 3 instead of $5/3$. This is due to the fact we haven’t got a time interval elapsing at a certain location but we face with a time interval between two different and spatially separated events.

b)The use of the complete Lorentz transformation (boost) mixing space and time is inevitable.

Problem 5. Two charged particles A and B, with the same charge q, move parallel with $\mathbf{v}=(v,0,0)$. They are separated by a distance d. What is the electric force between them?

$E'=\left(0,\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{d^2},0\right)$

$B'=\left(0,0,\gamma \dfrac{\gamma v q}{4\pi\epsilon_0d^2c^2}\right)$

In the S-frame, we obtain the Lorentz force:

$\mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)=\left(0,\gamma k_C\dfrac{q^2}{d^2}-\gamma \beta^2\dfrac{q^2}{d^2},0\right)=\left(0,\dfrac{k_Cq^2}{\gamma d^2},0\right)$

The same result can be obtained using the power-force (or forpower) tetravector performing an inverse Lorentz transformation.

Problem 6. Calculate the electric and magnetic field for a point particle passing some concrete point.

The electric field for the static charge is: $E=k_C\dfrac{q}{x'^2+y'^2+z'^2}=k_C\dfrac{q}{r'^2}$

with $\mathbf{v}=(v,0,0)$ when the temporal origin coincides, i.e., at the time $t'=t=0$.  Suppose now two points that for the rest observer provide:

$P=(0,a,0)$ and $P'=(-vt',a,0)$. For the electric field we get:

$E'=k_C\dfrac{q}{r'^3}(x',y',z')$ and $E'(t')=k_C\dfrac{q}{(\sqrt{(x'^2+y'^2+z'^2})^3}(x',y',z')$

$E'_(t')=k_C\dfrac{q}{(v^2t'^2+a^2)^{3/2}}(-vt',a,0)$

Then, $E'_p\rightarrow E_p$ implies that $t'=\gamma t=\gamma (t-\dfrac{vx}{c^2})\vert_{x=0}$

$E'_p(t)=k_C\dfrac{q}{(\gamma^2v^2t^2+a^2)^{3/2}}(-\gamma v t,a,0)$

$B'=B'_p(t')=(0,0,0)=B'_p(t)$

$E_p(t)=(E'_{p_x}(t),\gamma E'_{p_y}(t),0)=k_C\dfrac{q}{\gamma^2 v^2t^2+a^2}(-\gamma v t,\gamma a,0)$

$B_p(t)=(B_{p_x},B_{p_y},B_{p_z})=(0,0,\gamma \dfrac{\gamma v E'_p(t)}{c^2})$

$B_p(t)=\dfrac{q}{\gamma^2v^2t^2+a^2}(0,0,\gamma \dfrac{v}{c^2}a)=(0,0,\dfrac{v}{c^2}E_{p_y}(t))$

There are two special cases from the physical viewpoint in the observed electric fields:

a) When P is directly above the charge q. Then $E_p(t=0)=(0,k_C\gamma \dfrac{q}{a^2},0)$

b) When P is directly in front of ( or behind) q. Then, for a=0, $E_p(t)=(-k_C\dfrac{vt}{\gamma^2(v^2t^2)^{3/2}},0,0)$

Note that we have $\dfrac{vt}{(v^2t^2)^{3/2}}\neq \dfrac{1}{v^2t^2}$ if $t<0$.

## LOG#036. Action and relativity.

The hamiltonian formalism and the hamiltonian H in special relativity has some issues with the definition. In the case of the free particle one possible definition, not completely covariant, is the relativistic energy

$\boxed{E=\sqrt{m^2c^4+c^2p^2}=H}$

There are two others interesting scalars in classical relativistic theories. They are the lagrangian L and the action functional S. The lagrangian is obtained through a Legendre transformation from the hamiltonian:

$\boxed{L=pv-H}$

From the hamiltonian, we get the velocity using the so-called hamiltonian equation:

$\dot{\mathbf{q}}=\mathbf{v}=\dfrac{\partial H}{\partial \mathbf{p}}=c^2\dfrac{\mathbf{p}}{E}$

Then,

$L=\dfrac{E}{c^2}\mathbf{v}^2-E=E\left(\dfrac{v^2}{c^2}-1\right)=-\dfrac{E}{\gamma^2}=-\dfrac{m\gamma c^2}{\gamma^2}=-\dfrac{mc^2}{\gamma}$

and finally

$\boxed{L=-mc^2\sqrt{1-\dfrac{v^2}{c^2}}=-\dfrac{mc^2}{\gamma}=-mc\sqrt{-\dot{X}^2}}$

The action functional is the time integral of the lagrangian:

$\boxed{S=\int Ldt}$

However, let me point out that the above hamiltonian in SR has some difficulties in gauge field theories. Indeed, it is quite easy to derive that a more careful and reasonable election for the hamiltonian in SR should be zero!

In the case of the free relativistic particle, we obtain

$S=-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt$

Using the relation between time and proper time (the time dilation formula):

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

direct substitution provides

$-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt=-mc^2\int d\tau$

And defining the infinitesimal proper length in spacetime as $ds=cd\tau$, we get the simple and wonderful result:

$\boxed{S=-mc\int ds}$

Sometimes, the covariant lagrangian for the free particle is also obtained from the following argument. The proper length is defined as

$ds^2=d\mathbf{x}^2-c^2dt^2$

The invariant in spacetime is related with the proper time in this way:

$ds^2=-c^2d\tau^2=d\mathbf{x}^2-c^2dt^2$

Thus, dividing by $dt^2$

$-c^2\dfrac{d\tau^2}{dt^2}=\mathbf{v}^2-c^2$

and

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

so

$d\tau=\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}dt$

$cd\tau=ds=\sqrt{c^2-\mathbf{v}^2}dt=\sqrt{-\dot{X}^2}dt$

that is

$\boxed{ds=cd\tau=\sqrt{-\dot{X}^2}dt=\sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=\sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}$

and the free coordinate action for the free particle would be:

$\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}dt=-mc^2\int \sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}$

Note, that since the election of time “t” is “free”, we can choose $t=\tau$ to obtain the generally covariant free action:

$\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}d\tau=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}d\tau}$

Remark: the (rest) mass is the “coupling” constant for the free particle proper lenght to guess the free lagrangian

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

Now, we can see from this covariant action that the relativistic hamiltonian should be a feynmanity! From the equations of motion,

$P_\mu=\dfrac{\partial L}{\partial \dot{X}^\mu}=mc\dfrac{\dot{X}_\mu}{\sqrt{-\dot{X}^2}}$

The covariant hamiltonian $\mathcal{H}$, different from H, can be build in the following way:

$\mathcal{H}=P_\mu \dot{X}^\mu-L=mc\dfrac{\dot{X}_\mu \dot{X}^\mu}{\sqrt{-\dot{X}^2}}-mc\sqrt{-\dot{X}^2}=0$

The meaning of this result is hidden in the the next identity ( Noether identity or “hamiltonian constraint” in some contexts):

$\boxed{\mathcal{H}=P_\mu P^\mu+m^2c^2=0}$

since

$P_\mu P^\mu=m U_\mu mU^\mu=-m^2c^2$

This strange fact that $\mathcal{H}=0$ in SR, a feynmanity as the hamiltonian, is related to the Noether identity $E^\mu \dot{X}_\mu$ for the free relativistic lagrangian, indeed, a consequence of the hamiltonian constraint and the so-called reparametrization invariance $\tau'=f (\tau)$. Note, in addition, that the free relativistic particle would also be invariant under diffeomorphisms $x^{\mu'}= f^\mu (x)=f^\mu (x^\nu)$ if we were to make the metric space-time dependent, i.e., if we make the substitution $\eta_{\mu\nu}\rightarrow g_{\mu\nu} (x)$. This last result is useful and important in general relativity, but we will not discuss it further in this moment. In summary, from the two possible hamiltonian in special relativity

$H=E=\sqrt{\mathbf{p}^2c^2+(mc^2)^2}$

$\mathcal{H}=P_\mu P^\mu+m^2c^2=0$

the natural and more elegant (due to covariance/invariance) is the second one. Moreover, the free particle lagrangian and action are:

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

$\boxed{S=-mc^2\int d\tau=-mc\int ds=\int L dt}$

Remark: The true covariant lagrangian dynamics in SR is a “constrained” dynamics, i.e., dynamics where we are undetermined. There are more variables that equations as a result of a large set of symmetries ( reparametrization invariance and, in the case of local metrics, we also find diffeomorphism invarince).

The dynamical equations of motion, for a first order lagrangian (e.g., the free particle we have studied here), read for the lagrangian formalism:

$\boxed{\delta S=\delta \int L (q,\dot{q};t)dt =0\leftrightarrow\begin{cases}E(L)=0,\mbox{with E(L)=Euler operator}\\ E(L)=\dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{q}}\right)=0\end{cases}}$

By the other hand, for the hamiltonian formalism, dynamical equations are:

$\boxed{\delta S=\delta \int H (q, p; t)dt =0\leftrightarrow\begin{cases}p=\dfrac{\partial L}{\partial \dot{q}},\;\mbox{with}\; \det\left(\dfrac{\partial ^2L}{\partial \dot{q}^i\partial \dot{q}^j}\right)\neq 0\\ \;\\ \dfrac{dq}{dt}=\dot{q}=\dfrac{\partial H}{\partial p}\\ \;\\\dfrac{dp}{dt}=\dot{p}=-\dfrac{\partial H}{\partial q} \\ \;\\ \dfrac{dH}{dt}=\dot{H}=\dfrac{\partial H}{\partial t}=-\dfrac{\partial L}{\partial t}\end{cases}}$

## 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:

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

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

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:

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

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

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.