## Motion under a constant Minkowski force

Reference: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.60.

The Minkowski force ${\mathbf{K}}$ is the rate of change of four-momentum with respect to proper time, and allows Newton’s law to be written in its natural form

$\displaystyle \mathbf{K}=m\boldsymbol{\alpha} \ \ \ \ \ (1)$

where ${\boldsymbol{\alpha}}$ is the proper acceleration, or second derivative of position with respect to proper time. Here we’ll investigate the behaviour of a particle subject to a constant Minkowski force in one dimension.

In terms of ordinary force, we have

$\displaystyle K=\frac{dp}{d\tau}=\frac{dp}{dt}\frac{dt}{d\tau}=\frac{1}{\sqrt{1-u^{2}/c^{2}}}F \ \ \ \ \ (2)$

The ordinary momentum ${p}$ is

$\displaystyle p=\frac{mu}{\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (3)$

so its derivative is

$\displaystyle \frac{dp}{dt}=\frac{m}{\sqrt{1-u^{2}/c^{2}}}\frac{du}{dt}+\frac{mu^{2}}{\left(1-u^{2}/c^{2}\right)^{3/2}}\frac{du}{dt} \ \ \ \ \ (4)$

Inserting this into 2 we get

$\displaystyle \frac{K}{m}dt=\frac{du}{1-u^{2}/c^{2}}+\frac{u^{2}\; du}{\left(1-u^{2}/c^{2}\right)^{2}} \ \ \ \ \ (5)$

We can integrate both sides (using software, or integral tables) to get

$\displaystyle \frac{K}{m}t+C=\frac{c}{4}\ln\left[\frac{c+u}{c-u}\right]+\frac{c^{2}}{4}\left[\frac{1}{c-u}-\frac{1}{c+u}\right] \ \ \ \ \ (6)$

where ${C}$ is a constant of integration. If the initial conditions are ${u=0}$ at ${t=0}$, then ${C=0}$ and we have

$\displaystyle \frac{K}{m}t=\frac{c}{4}\ln\left[\frac{c+u}{c-u}\right]+\frac{c^{2}}{4}\left[\frac{1}{c-u}-\frac{1}{c+u}\right] \ \ \ \ \ (7)$

This is an implicit equation for the speed of the particle as a function of time. If we want the position as a function of time, we need a relation between ${u}$ and ${x}$. Returning to 2 and 3 we have

$\displaystyle \sqrt{1-u^{2}/c^{2}}\frac{K}{m}=\frac{d}{dt}\left(\frac{u}{\sqrt{1-u^{2}/c^{2}}}\right) \ \ \ \ \ (8)$

We can use the chain rule to convert the derivative on the RHS to a derivative with respect to ${x}$ by multiplying both sides by ${dt/dx}$

$\displaystyle \frac{dt}{dx}\sqrt{1-u^{2}/c^{2}}\frac{K}{m}=\frac{dt}{dx}\frac{d}{dt}\left(\frac{u}{\sqrt{1-u^{2}/c^{2}}}\right)=\frac{d}{dx}\left(\frac{u}{\sqrt{1-u^{2}/c^{2}}}\right) \ \ \ \ \ (9)$

Now ${dx/dt=u}$ so ${dt/dx=1/u}$ and

$\displaystyle \frac{\sqrt{1-u^{2}/c^{2}}}{u}\frac{K}{m}=\frac{d}{dx}\left(\frac{u}{\sqrt{1-u^{2}/c^{2}}}\right) \ \ \ \ \ (10)$

If we call the expression in the parentheses on the RHS ${A}$, then we can integrate with respect to ${x}$ (since ${K/m}$ is a constant):

 $\displaystyle A$ $\displaystyle \equiv$ $\displaystyle \frac{u}{\sqrt{1-u^{2}/c^{2}}}\ \ \ \ \ (11)$ $\displaystyle \frac{1}{A}\frac{K}{m}$ $\displaystyle =$ $\displaystyle \frac{dA}{dx}\ \ \ \ \ (12)$ $\displaystyle \frac{K}{m}x+C$ $\displaystyle =$ $\displaystyle \frac{1}{2}A^{2} \ \ \ \ \ (13)$

Again, starting from rest at the origin we have ${u=0}$ when ${x=0}$ so ${A=0}$ also, and therefore ${C=0}$, so we have

$\displaystyle A=\frac{u}{\sqrt{1-u^{2}/c^{2}}}=\sqrt{\frac{2Kx}{m}} \ \ \ \ \ (14)$

At this point we could get a relation between ${x}$ and ${t}$ by solving 14 for ${u}$ in terms of ${x}$ and then substituting this into 7. For reference, we get

$\displaystyle u=\sqrt{\frac{2Kx}{m}}\frac{1}{\sqrt{1+2Kx/mc^{2}}} \ \ \ \ \ (15)$

so substituting will give something of a mess. To get the answer given in Griffiths requires a bit of algebra, but here is how I did it. Griffiths defines the quantity ${z}$ as

 $\displaystyle z$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{2Kx}{mc^{2}}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{A}{c}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{u}{c\sqrt{1-u^{2}/c^{2}}} \ \ \ \ \ (18)$

The quantities appearing in Griffiths’s answer are

 $\displaystyle \sqrt{1+z^{2}}$ $\displaystyle =$ $\displaystyle \frac{c}{\sqrt{c^{2}-u^{2}}}\ \ \ \ \ (19)$ $\displaystyle z\sqrt{1+z^{2}}$ $\displaystyle =$ $\displaystyle \frac{u}{c\left(1-u^{2}/c^{2}\right)} \ \ \ \ \ (20)$

We can rewrite 7 to get

$\displaystyle \frac{2Kt}{mc}=\frac{1}{2}\ln\left[\frac{c+u}{c-u}\right]+\frac{c}{2}\left[\frac{1}{c-u}-\frac{1}{c+u}\right] \ \ \ \ \ (21)$

We’ll deal with the logarithm first. Its argument is

 $\displaystyle \frac{c+u}{c-u}$ $\displaystyle =$ $\displaystyle \frac{\left(c+u\right)^{2}}{c^{2}\left(1-u^{2}/c^{2}\right)}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2u}{c\left(1-u^{2}/c^{2}\right)}+\frac{u^{2}+c^{2}}{c^{2}-u^{2}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2u}{c\left(1-u^{2}/c^{2}\right)}+\frac{c^{2}-u^{2}+2u^{2}}{c^{2}-u^{2}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2u}{c\left(1-u^{2}/c^{2}\right)}+1+\frac{2u^{2}}{c^{2}\left(1-u^{2}/c^{2}\right)} \ \ \ \ \ (25)$

Now we also have

 $\displaystyle \left(z+\sqrt{1+z^{2}}\right)^{2}$ $\displaystyle =$ $\displaystyle 2z^{2}+2z\sqrt{1+z^{2}}+1\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2u^{2}}{c^{2}\left(1-u^{2}/c^{2}\right)}+\frac{2u}{c\left(1-u^{2}/c^{2}\right)}+1\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{c+u}{c-u} \ \ \ \ \ (28)$

Therefore

 $\displaystyle \frac{1}{2}\ln\left[\frac{c+u}{c-u}\right]$ $\displaystyle =$ $\displaystyle \ln\sqrt{\frac{c+u}{c-u}}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \ln\left(z+\sqrt{1+z^{2}}\right) \ \ \ \ \ (30)$

For the second term in 21, we have

 $\displaystyle \frac{c}{2}\left[\frac{1}{c-u}-\frac{1}{c+u}\right]$ $\displaystyle =$ $\displaystyle \frac{c}{2}\frac{2u}{c^{2}\left(1-u^{2}/c^{2}\right)}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{u}{c\left(1-u^{2}/c^{2}\right)}\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle z\sqrt{1+z^{2}} \ \ \ \ \ (33)$

Putting it all together, we have

$\displaystyle \frac{2Kt}{mc}=\ln\left(z+\sqrt{1+z^{2}}\right)+z\sqrt{1+z^{2}} \ \ \ \ \ (34)$

## Elastic collision of two identical particles

Reference: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.59.

Here’s another example the kinematics of particle collisions in relativity. This time, we’ll look at the common first year physics problem of elastic scattering of two identical masses, with one mass ${m}$ coming in along the ${x}$ axis with speed ${v}$ and hitting another mass ${m}$ at rest. In classical physics, the two masses always fly off at right angles after the collision, but things are more complicated in relativity. As usual, it’s easiest to convert to the centre of momentum frame to do the calculations and then convert back at the end.

The speed of the centre of momentum frame is (I’ll set ${c=1}$ and use the standard symbol ${\beta=v/c}$ to make notation easier; we can reinsert the ${c}$ at the end):

 $\displaystyle \bar{\beta}$ $\displaystyle =$ $\displaystyle \frac{\sum p_{i}}{\sum E_{i}}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\gamma\beta m}{m+\gamma m}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\gamma\beta}{1+\gamma} \ \ \ \ \ (3)$

where the quantities on the RHS refer to the lab frame. Centre of momentum (COM) quantities have a bar over them. Since mass 2 is at rest in the lab, its velocity in the COM frame is

$\displaystyle \bar{\beta}_{2}=-\frac{\gamma\beta}{1+\gamma} \ \ \ \ \ (4)$

and by conservation of momentum, the COM velocity of mass 1 must be

$\displaystyle \bar{\beta}_{1}=\frac{\gamma\beta}{1+\gamma} \ \ \ \ \ (5)$

At this point, we can check the results using the velocity addition formula. For particle 2, its lab velocity is given by

$\displaystyle \beta_{2}=\frac{\bar{\beta}_{2}+\bar{\beta}}{1+\bar{\beta}_{2}\bar{\beta}}=0 \ \ \ \ \ (6)$

For particle 1, we have

 $\displaystyle \beta_{1}$ $\displaystyle =$ $\displaystyle \frac{\bar{\beta}_{1}+\bar{\beta}}{1+\bar{\beta}_{1}\bar{\beta}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\frac{\gamma\beta}{1+\gamma}\frac{1}{1+\left(\gamma\beta\right)^{2}/\left(1+\gamma\right)^{2}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\frac{\gamma\beta}{1+\gamma}\frac{\left(1+\gamma\right)^{2}}{\left(1+\gamma\right)^{2}+\left(\gamma\beta\right)^{2}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\gamma\beta\left(1+\gamma\right)}{1+2\gamma+\gamma^{2}+\left(\gamma\beta\right)^{2}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\gamma\beta\left(1+\gamma\right)}{\gamma^{2}\left(1-\beta^{2}\right)+2\gamma+\gamma^{2}+\left(\gamma\beta\right)^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\gamma\beta\left(1+\gamma\right)}{2\gamma+2\gamma^{2}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \beta \ \ \ \ \ (13)$

where in the fifth line we used ${\gamma^{2}\left(1-\beta^{2}\right)=1}$. Thus both velocities convert correctly back to the lab frame. Returning to our main calculation, what we’d like to do is find the scattering angle in the lab frame, given the angle ${\bar{\phi}}$ between particle 1’s initial and final directions in the COM frame. As the total momentum in COM is zero, we know that the two particles must fly off in opposite directions, so we can write out the ${x}$ and ${y}$ components for both particles as

 $\displaystyle \bar{\beta}_{1_{x}}$ $\displaystyle =$ $\displaystyle \bar{\beta}\cos\bar{\phi}\ \ \ \ \ (14)$ $\displaystyle \bar{\beta}_{1_{y}}$ $\displaystyle =$ $\displaystyle \bar{\beta}\sin\bar{\phi}\ \ \ \ \ (15)$ $\displaystyle \bar{\beta}_{2_{x}}$ $\displaystyle =$ $\displaystyle -\bar{\beta}\cos\bar{\phi}\ \ \ \ \ (16)$ $\displaystyle \bar{\beta}_{2_{y}}$ $\displaystyle =$ $\displaystyle -\bar{\beta}\sin\bar{\phi} \ \ \ \ \ (17)$

Using the velocity addition formulas, we can convert these back to the lab frame, using ${\bar{\gamma}\equiv1/\sqrt{1-\bar{\beta}^{2}}}$.

 $\displaystyle \beta_{1_{x}}$ $\displaystyle =$ $\displaystyle \frac{\bar{\beta}_{1_{x}}+\bar{\beta}}{1+\bar{\beta}_{1_{x}}\bar{\beta}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\bar{\beta}\cos\bar{\phi}+\bar{\beta}}{1+\bar{\beta}^{2}\cos\bar{\phi}}\ \ \ \ \ (19)$ $\displaystyle \beta_{1_{y}}$ $\displaystyle =$ $\displaystyle \frac{\bar{\beta}_{1_{y}}}{\bar{\gamma}\left(1+\bar{\beta}_{1_{x}}\bar{\beta}\right)}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\bar{\beta}\sin\bar{\phi}}{\bar{\gamma}\left(1+\bar{\beta}^{2}\cos\bar{\phi}\right)}\ \ \ \ \ (21)$ $\displaystyle \beta_{2_{x}}$ $\displaystyle =$ $\displaystyle \frac{-\bar{\beta}\cos\bar{\phi}+\bar{\beta}}{1-\bar{\beta}^{2}\cos\bar{\phi}}\ \ \ \ \ (22)$ $\displaystyle \beta{}_{2_{y}}$ $\displaystyle =$ $\displaystyle \frac{-\bar{\beta}\sin\bar{\phi}}{\bar{\gamma}\left(1-\bar{\beta}^{2}\cos\bar{\phi}\right)} \ \ \ \ \ (23)$

In the lab frame, the angle ${\phi_{1}}$ that the incident particle makes with its incoming direction is given by

 $\displaystyle \tan\phi_{1}$ $\displaystyle =$ $\displaystyle \frac{\beta_{1_{y}}}{\beta_{1_{x}}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sin\bar{\phi}}{\bar{\gamma}\left(1+\cos\bar{\phi}\right)} \ \ \ \ \ (25)$

Similarly for particle 2

 $\displaystyle \tan\phi_{2}$ $\displaystyle =$ $\displaystyle \frac{\beta_{2_{y}}}{\beta_{2_{x}}}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{-\sin\bar{\phi}}{\bar{\gamma}\left(1-\cos\bar{\phi}\right)} \ \ \ \ \ (27)$

Since ${\phi_{1}>0}$ and ${\phi_{2}<0}$ (that is, in the lab frame, particle 1 scatters upwards from the ${x}$ axis while particle 2 scatters downwards; obviously we could interchange the roles of the two particles but to conserve ${y}$ momentum, the two particles must scatter on opposite sides of the ${x}$ axis), the total angle between the two particles is ${\phi=\phi_{1}-\phi_{2}}$. Using the formula for the tangent of the difference of two angles, we get

 $\displaystyle \tan\phi$ $\displaystyle =$ $\displaystyle \frac{\tan\phi_{1}-\tan\phi_{2}}{1+\tan\phi_{1}\tan\phi_{2}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sin\bar{\phi}}{\bar{\gamma}}\left[\frac{1}{1+\cos\bar{\phi}}+\frac{1}{1-\cos\bar{\phi}}\right]\left[1-\frac{\sin^{2}\bar{\phi}}{\bar{\gamma}^{2}\left(1+\cos\bar{\phi}\right)\left(1-\cos\bar{\phi}\right)}\right]^{-1}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sin\bar{\phi}}{\bar{\gamma}}\left(\frac{2}{1-\cos^{2}\bar{\phi}}\right)\left[1-\frac{\sin^{2}\bar{\phi}}{\bar{\gamma}^{2}\left(1-\cos^{2}\bar{\phi}\right)}\right]^{-1}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\bar{\gamma}\sin\bar{\phi}\left(1-1/\bar{\gamma}^{2}\right)}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\bar{\gamma}\bar{\beta}^{2}\sin\bar{\phi}}\ \ \ \ \ (32)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\left(\frac{2}{\bar{\gamma}\bar{\beta}^{2}\sin\bar{\phi}}\right) \ \ \ \ \ (33)$

[This answer is the same form as that given in Griffiths’s question, except that he seems to use lab coordinates for ${\gamma}$ and ${\beta}$. However, after checking my solution I can’t see anything wrong, so hopefully I’ve got it right.]

In the classical limit, ${\beta}$ becomes very small and ${\gamma\rightarrow1}$ so from 3, ${\bar{\beta}}$ also becomes very small and ${\bar{\gamma}\rightarrow1}$, so the argument of the arctan tends to infinity and ${\phi\rightarrow\pi/2}$ as required.

## Collision of a pion and a proton

Reference: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.58.

As another example of using conservation of energy and momentum to work out the kinematics of particle collisions, suppose we fire a pion at a proton at rest. One possible outcome of such a collision is the conversion of the pion and proton into kappa and sigma particles, but this can only occur if the momentum of the pion is high enough, since the rest energies of the kappa plus sigma are greater than those of the pion plus proton. We can find the minimum pion momentum (called the threshold momentum), as measured in the lab, at which this reaction can occur. It’s easiest to convert to the centre of momentum frame to do the calculations and then convert back at the end.

In the centre of momentum frame, the pion and proton head towards each other with equal and opposite momenta, so using the usual relativistic notation, and expressing rest energy in MeV (so we can ignore the ${c^{2}}$ factor):

$\displaystyle \gamma_{\pi}\beta_{\pi}m_{\pi}=\gamma_{p}\beta_{p}m_{p} \ \ \ \ \ (1)$

At the threshold momentum, the pion and proton collide and produce a ${K}$ and ${\Sigma}$ at rest, so from conservation of energy

$\displaystyle \gamma_{\pi}m_{\pi}+\gamma_{p}m_{p}=m_{K}+m_{\Sigma} \ \ \ \ \ (2)$

Using Griffiths’s approximate values for the rest energies, we have (in MeV)

 $\displaystyle m_{\pi}$ $\displaystyle =$ $\displaystyle 150\ \ \ \ \ (3)$ $\displaystyle m_{p}$ $\displaystyle =$ $\displaystyle 900\ \ \ \ \ (4)$ $\displaystyle m_{K}$ $\displaystyle =$ $\displaystyle 500\ \ \ \ \ (5)$ $\displaystyle m_{\Sigma}$ $\displaystyle =$ $\displaystyle 1200 \ \ \ \ \ (6)$

so from 1 and 2

 $\displaystyle 150\gamma_{\pi}\beta_{\pi}$ $\displaystyle =$ $\displaystyle 900\gamma_{p}\beta_{p}\ \ \ \ \ (7)$ $\displaystyle \gamma_{\pi}\beta_{\pi}$ $\displaystyle =$ $\displaystyle 6\gamma_{p}\beta_{p}\ \ \ \ \ (8)$ $\displaystyle 150\gamma_{\pi}+900\gamma_{p}$ $\displaystyle =$ $\displaystyle 1700\ \ \ \ \ (9)$ $\displaystyle \gamma_{\pi}+6\gamma_{p}$ $\displaystyle =$ $\displaystyle \frac{34}{3} \ \ \ \ \ (10)$

From these equations, we get

 $\displaystyle \gamma_{\pi}^{2}$ $\displaystyle =$ $\displaystyle \left(\frac{34}{3}-6\gamma_{p}\right)^{2}=\frac{1}{1-\beta_{\pi}^{2}}\ \ \ \ \ (11)$ $\displaystyle \beta_{\pi}^{2}$ $\displaystyle =$ $\displaystyle 1-\left(\frac{34}{3}-6\gamma_{p}\right)^{-2}\ \ \ \ \ (12)$ $\displaystyle \left(\gamma_{\pi}\beta_{\pi}\right)^{2}$ $\displaystyle =$ $\displaystyle \left(\frac{34}{3}-6\gamma_{p}\right)^{2}-1\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 36\gamma_{p}^{2}\beta_{p}^{2} \ \ \ \ \ (14)$

where we used 8 to get the last line. We can now solve the last two equations to find ${\gamma_{p}}$:

 $\displaystyle 36\gamma_{p}^{2}\beta_{p}^{2}$ $\displaystyle =$ $\displaystyle \left(\frac{34}{3}-6\gamma_{p}\right)^{2}-1\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{34}{3}\right)^{2}-1-136\gamma_{p}+36\gamma_{p}^{2}\ \ \ \ \ (16)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \left(\frac{34}{3}\right)^{2}-1-136\gamma_{p}+36\gamma_{p}^{2}\left(1-\beta_{p}^{2}\right)\ \ \ \ \ (17)$ $\displaystyle 0$ $\displaystyle =$ $\displaystyle \left(\frac{34}{3}\right)^{2}-1+36-136\gamma_{p}\ \ \ \ \ (18)$ $\displaystyle \gamma_{p}$ $\displaystyle =$ $\displaystyle 1.202\ \ \ \ \ (19)$ $\displaystyle \beta_{p}$ $\displaystyle =$ $\displaystyle 0.555 \ \ \ \ \ (20)$

We can now get the values for the pion in the centre of momentum frame from 11 and 12:

 $\displaystyle \gamma_{\pi}$ $\displaystyle =$ $\displaystyle \frac{34}{3}-6\gamma_{p}=4.123\ \ \ \ \ (21)$ $\displaystyle \beta_{\pi}$ $\displaystyle =$ $\displaystyle \sqrt{1-\left(\frac{34}{3}-6\gamma_{p}\right)^{-2}}=0.970 \ \ \ \ \ (22)$

The speed of the proton in the centre of momentum frame is also the speed of the centre of momentum frame relative to the lab frame, so we can use a Lorentz transformation on the pion’s four-momentum to get back to the lab frame:

 $\displaystyle p_{\pi}^{1}$ $\displaystyle =$ $\displaystyle \gamma_{p}\left(\bar{p}_{\pi}^{1}+\beta_{p}\bar{p}_{\pi}^{0}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.202\left(150\gamma_{\pi}\beta_{\pi}+0.555\times150\gamma_{\pi}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1133\mbox{ MeV}/c \ \ \ \ \ (25)$

Notice that we must use ${\gamma_{p}}$ and ${\beta_{p}}$ (that is, the values for the proton, not the pion) in doing the Lorentz transformation, since it’s the speed of the proton that determines the relative speed of the two frames.

## Lorentz transformation in two dimensions

Reference: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.57.

Although we’ve looked at the Lorentz transformations for a general 3-d motion of one frame relative to another, we’ll have a look here at the slightly more specialized case of general 2-d motion.

Suppose that frame ${\bar{\mathcal{S}}}$ moves relative to ${\mathcal{S}}$ with velocity

$\displaystyle \mathbf{v}=\beta c\left(\cos\phi\hat{\mathbf{x}}+\sin\phi\hat{\mathbf{y}}\right) \ \ \ \ \ (1)$

that is, its direction makes an angle ${\phi}$ with the ${x}$ axis. To get the Lorentz transformation here, we use the fact that distances perpendicular to ${\mathbf{v}}$ are unaffected, and distances parallel to ${\mathbf{v}}$ transform using the regular 1-d Lorentz transformation. It’s therefore easiest to transform to a lab frame ${\mathcal{S}'}$ that is rotated relative to ${\mathcal{S}}$ by ${\phi}$, so that the basis vectors are

 $\displaystyle \hat{\mathbf{x}}'$ $\displaystyle =$ $\displaystyle \cos\phi\hat{\mathbf{x}}+\sin\phi\hat{\mathbf{y}}\ \ \ \ \ (2)$ $\displaystyle \hat{\mathbf{y}}'$ $\displaystyle =$ $\displaystyle -\sin\phi\hat{\mathbf{x}}+\cos\phi\hat{\mathbf{y}} \ \ \ \ \ (3)$

A point with coordinates ${\left[x',y'\right]}$ in ${\mathcal{S}'}$ therefore has coordinates in terms of ${\mathcal{S}}$ coordinates of

 $\displaystyle x'$ $\displaystyle =$ $\displaystyle x\cos\phi+y\sin\phi\ \ \ \ \ (4)$ $\displaystyle y'$ $\displaystyle =$ $\displaystyle -x\sin\phi+y\cos\phi \ \ \ \ \ (5)$

In this system, the Lorentz transformation is

$\displaystyle \Lambda'=\left[\begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (6)$

We can write this out explicitly for each coordinate. The time coordinate is the same both ${\mathcal{S}}$ and ${\mathcal{S}'}$ since the two frames are at rest relative to each other, so

 $\displaystyle \bar{t}'$ $\displaystyle =$ $\displaystyle \gamma\left(t-\beta x'\right)\ \ \ \ \ (7)$ $\displaystyle \bar{t}$ $\displaystyle =$ $\displaystyle \gamma t-\gamma\beta x\cos\phi-\gamma\beta y\sin\phi \ \ \ \ \ (8)$

The ${x'}$ coordinate transforms as follows:

 $\displaystyle \bar{x}'$ $\displaystyle =$ $\displaystyle \gamma\left(x'-\beta t\right)\ \ \ \ \ (9)$ $\displaystyle \bar{x}\cos\phi+\bar{y}\sin\phi$ $\displaystyle =$ $\displaystyle \gamma x\cos\phi+\gamma y\sin\phi-\gamma\beta t \ \ \ \ \ (10)$

And ${y'}$ transforms as

 $\displaystyle \bar{y}'$ $\displaystyle =$ $\displaystyle y'\ \ \ \ \ (11)$ $\displaystyle -\bar{x}\sin\phi+\bar{y}\cos\phi$ $\displaystyle =$ $\displaystyle -x\sin\phi+y\cos\phi \ \ \ \ \ (12)$

Multiplying 10 by ${\cos\phi}$ and 12 by ${-\sin\phi}$ and adding, we get

$\displaystyle \bar{x}=-\beta\gamma t\cos\phi+\left(\gamma\cos^{2}\phi+\sin^{2}\phi\right)x+\left(\gamma-1\right)y\sin\phi\cos\phi \ \ \ \ \ (13)$

Multiplying 10 by ${\sin\phi}$ and 12 by ${\cos\phi}$ and adding, we get

$\displaystyle \bar{y}=-\beta\gamma t\sin\phi+\left(\gamma-1\right)x\sin\phi\cos\phi+\left(\gamma\sin^{2}\phi+\cos^{2}\phi\right)y \ \ \ \ \ (14)$

Combining 8, 13 and 14 (along with ${\bar{z}=z}$) into a matrix, we get

$\displaystyle \Lambda=\left[\begin{array}{cccc} \gamma & -\gamma\beta\cos\phi & -\gamma\beta\sin\phi & 0\\ -\beta\gamma\cos\phi & \gamma\cos^{2}\phi+\sin^{2}\phi & \left(\gamma-1\right)\sin\phi\cos\phi & 0\\ -\beta\gamma\sin\phi & \left(\gamma-1\right)\sin\phi\cos\phi & \gamma\sin^{2}\phi+\cos^{2}\phi & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (15)$

This reduces to 6 when ${\phi=0}$, so that the axes of the two frames are parallel. This 2-d transformation matrix is also symmetric.

## Symmetry of a rank 2 tensor is preserved under Lorentz transformation

Reference: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.49.

We’ve seen that the electric and magnetic fields can be represented as components of an anti-symmetric rank 2 tensor. In fact, any symmetric or anti-symmetric tensor retains its symmetry property under Lorentz transformation. We can show this by some index juggling as usual in relativity.

First, suppose the tensor ${T}$ is symmetric so that ${T^{ij}=T^{ji}}$. Then under Lorentz transformation, we have

 $\displaystyle \bar{T}^{ij}$ $\displaystyle =$ $\displaystyle \Lambda_{\; k}^{i}\Lambda_{\; l}^{j}T^{kl}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Lambda_{\; k}^{i}\Lambda_{\; l}^{j}T^{lk}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Lambda_{\; l}^{j}\Lambda_{\; k}^{i}T^{lk}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \bar{T}^{ji} \ \ \ \ \ (4)$

If ${T}$ is anti-symmetric, then ${T^{ij}=-T^{ji}}$ and

 $\displaystyle \bar{T}^{ij}$ $\displaystyle =$ $\displaystyle \Lambda_{\; k}^{i}\Lambda_{\; l}^{j}T^{kl}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\Lambda_{\; k}^{i}\Lambda_{\; l}^{j}T^{lk}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\Lambda_{\; l}^{j}\Lambda_{\; k}^{i}T^{lk}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\bar{T}^{ji} \ \ \ \ \ (8)$

If you want to use full matrix notation, you can write

$\displaystyle \bar{T}=\Lambda T\Lambda^{T} \ \ \ \ \ (9)$

Taking the transpose of a matrix product reverses the order of terms and takes the transpose of each term, so

 $\displaystyle \bar{T}^{T}$ $\displaystyle =$ $\displaystyle \left(\Lambda T\Lambda^{T}\right)^{T}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Lambda T^{T}\Lambda^{T} \ \ \ \ \ (11)$

[Sorry about using a ${T}$ for the matrix and a superscript ${T}$ for transpose, but hopefully you can keep them separate.]

Now for a symmetric matrix ${T^{T}=T}$, so ${\bar{T}^{T}=\bar{T}}$ and for an anti-symmetric matrix ${T^{T}=-T}$, so ${\bar{T}^{T}=-\bar{T}}$, showing that the symmetry property is preserved. In fact, this latter proof shows that the symmetry or anti-symmetry of ${T}$ is preserved no matter what matrix ${\Lambda}$ is.

## Relativistic transformation of electromagnetic waves; the Doppler effect

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.47.

Let’s apply the transformation equations for electromagnetic fields to an electromagnetic wave. Suppose we have a plane EM wave with frequency ${\omega}$ in the lab frame. The wave is polarized in the ${y}$ direction and is travelling in the ${x}$ direction, so its fields are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle E_{0}e^{i\left(kx-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (1)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{E_{0}}{c}e^{i\left(kx-\omega t\right)}\hat{\mathbf{z}} \ \ \ \ \ (2)$

where

$\displaystyle k=\frac{\omega}{c} \ \ \ \ \ (3)$

To see how this wave looks in a frame moving with velocity ${v}$ in the ${x}$ direction, we can use the transformation equations

 $\displaystyle \bar{E}_{y}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{y}-vB_{z}\right)\ \ \ \ \ (4)$ $\displaystyle \bar{B}_{z}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{z}-\frac{v}{c^{2}}E_{y}\right) \ \ \ \ \ (5)$

We get

 $\displaystyle \bar{\mathbf{E}}$ $\displaystyle =$ $\displaystyle \gamma E_{0}\left(1-\frac{v}{c}\right)e^{i\left(kx-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{0}\sqrt{\frac{1-v/c}{1+v/c}}e^{i\left(kx-\omega t\right)}\hat{\mathbf{y}}\ \ \ \ \ (7)$ $\displaystyle \bar{\mathbf{B}}$ $\displaystyle =$ $\displaystyle \gamma\frac{E_{0}}{c}\left(1-\frac{v}{c}\right)e^{i\left(kx-\omega t\right)}\hat{\mathbf{z}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{E_{0}}{c}\sqrt{\frac{1-v/c}{1+v/c}}e^{i\left(kx-\omega t\right)}\hat{\mathbf{z}} \ \ \ \ \ (9)$

To get the final forms, we used

 $\displaystyle \gamma\left(1-\frac{v}{c}\right)$ $\displaystyle =$ $\displaystyle \frac{1-v/c}{\sqrt{1-v^{2}/c^{2}}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-v/c}{\sqrt{\left(1-v/c\right)\left(1+v/c\right)}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{1-v/c}{1+v/c}} \ \ \ \ \ (12)$

The amplitude of the wave gets smaller as the speed of the frame increases, becoming zero as ${v\rightarrow c}$.

To express this in the moving frame’s coordinates, we use the (inverse) Lorentz transformations:

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \gamma\left(\bar{x}+v\bar{t}\right)\ \ \ \ \ (13)$ $\displaystyle t$ $\displaystyle =$ $\displaystyle \gamma\left(\bar{t}+\frac{v\bar{x}}{c^{2}}\right) \ \ \ \ \ (14)$

giving

 $\displaystyle kx-\omega t$ $\displaystyle =$ $\displaystyle \gamma\left(k-\frac{\omega v}{c^{2}}\right)\bar{x}-\gamma\left(\omega-kv\right)\bar{t}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \gamma\frac{\omega}{c}\left(1-\frac{v}{c}\right)\bar{x}-\gamma\omega\left(1-\frac{v}{c}\right)\bar{t}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\omega}{c}\sqrt{\frac{1-v/c}{1+v/c}}\bar{x}-\omega\sqrt{\frac{1-v/c}{1+v/c}}\bar{t} \ \ \ \ \ (17)$

Thus in the moving frame, the frequency of the wave is

$\displaystyle \bar{\omega}=\omega\sqrt{\frac{1-v/c}{1+v/c}} \ \ \ \ \ (18)$

and the wavelength is

 $\displaystyle \bar{\lambda}$ $\displaystyle =$ $\displaystyle \frac{2\pi}{\bar{k}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi c}{\bar{\omega}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi c}{\omega}\sqrt{\frac{1+v/c}{1-v/c}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \lambda\sqrt{\frac{1+v/c}{1-v/c}} \ \ \ \ \ (22)$

That is, the wavelength gets longer, approaching infinity as ${v\rightarrow c}$, while the frequency gets smaller, approaching zero as ${v\rightarrow c}$. This is the Doppler effect for light. If ${v>0}$ so that we are moving in the direction of propagation of the wave, the wavelength gets longer resulting in a red-shift. If ${v<0}$ so that we are moving against the direction of propagation, the wavelength gets shorter and we have a blue-shift.

However, note that the speed of the wave is

$\displaystyle \bar{c}=\frac{\bar{\lambda}\bar{\omega}}{2\pi}=c \ \ \ \ \ (23)$

so not surprisingly, the speed of the wave remains the same in the moving frame.

The intensity of the wave in the lab frame is given by

$\displaystyle I=\frac{E_{0}^{2}}{2}c\epsilon_{0} \ \ \ \ \ (24)$

In the moving frame this becomes

$\displaystyle \bar{I}=\frac{E_{0}^{2}}{2}c\epsilon_{0}\frac{1-v/c}{1+v/c} \ \ \ \ \ (25)$

so the ratio is

$\displaystyle \frac{\bar{I}}{I}=\frac{1-v/c}{1+v/c} \ \ \ \ \ (26)$

and the intensity drops to zero as ${v\rightarrow c}$.

## Relativistic invariants involving electromagnetic fields

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.46.

Using the transformation equations for electromagnetic fields, we can show that there are a couple of quantities that are invariant under transformations between inertial frames.

The transformation equations are

 $\displaystyle \bar{E}_{x}$ $\displaystyle =$ $\displaystyle E_{x}\ \ \ \ \ (1)$ $\displaystyle \bar{E}_{y}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{y}+vB_{z}\right)\ \ \ \ \ (2)$ $\displaystyle \bar{E}_{z}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{z}-vB_{y}\right)\ \ \ \ \ (3)$ $\displaystyle \bar{B}_{x}$ $\displaystyle =$ $\displaystyle B_{x}\ \ \ \ \ (4)$ $\displaystyle \bar{B}_{y}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{y}-\frac{v}{c^{2}}E_{z}\right)\ \ \ \ \ (5)$ $\displaystyle \bar{B}_{z}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{z}+\frac{v}{c^{2}}E_{y}\right) \ \ \ \ \ (6)$

First, we calculate the dot product ${\mathbf{E}\cdot\mathbf{B}}$ and transform it:

 $\displaystyle \bar{\mathbf{E}}\cdot\bar{\mathbf{B}}$ $\displaystyle =$ $\displaystyle E_{x}B_{x}+\gamma^{2}\left[\left(E_{y}+vB_{z}\right)\left(B_{y}-\frac{v}{c^{2}}E_{z}\right)+\left(E_{z}-vB_{y}\right)\left(B_{z}+\frac{v}{c^{2}}E_{y}\right)\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{x}B_{x}+\gamma^{2}\left[\left(E_{y}B_{y}+E_{z}B_{z}\right)\left(1-\frac{v^{2}}{c^{2}}\right)\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{x}B_{x}+E_{y}B_{y}+E_{z}B_{z}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{E}\cdot\mathbf{B} \ \ \ \ \ (10)$

Second, we can find another invariant:

 $\displaystyle \bar{E}^{2}-c^{2}\bar{B}^{2}$ $\displaystyle =$ $\displaystyle E_{x}^{2}-c^{2}B_{x}^{2}+\gamma^{2}\left[\left(E_{y}+vB_{z}\right)^{2}+\left(E_{z}-vB_{y}\right)^{2}-c^{2}\left[\left(B_{y}-\frac{v}{c^{2}}E_{z}\right)^{2}+\left(B_{z}+\frac{v}{c^{2}}E_{y}\right)^{2}\right]\right]\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E_{x}^{2}-c^{2}B_{x}^{2}+\gamma^{2}\left[\left(E_{y}^{2}+E_{z}^{2}-c^{2}B_{y}^{2}-c^{2}B_{z}^{2}\right)\left(1-\frac{v^{2}}{c^{2}}\right)\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle E^{2}-c^{2}B^{2} \ \ \ \ \ (13)$

These invariants put constraints on the forms given electric and magnetic fields can have in different frames. For example, if ${\mathbf{B}=0}$ and ${\mathbf{E}\ne0}$ in one frame, then ${E^{2}-c^{2}B^{2}>0}$ in all frames, so it’s impossible to find a frame in which ${\mathbf{E}=0}$. The first invariant also tells us that if ${\mathbf{E}}$ and ${\mathbf{B}}$ are perpendicular in one frame (as they are in an electromagnetic wave, for example) then they are perpendicular in all frames.

## Relativistic transformation of electric and magnetic fields: an example

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.45.

As an example of the transformation equations for electromagnetic fields, consider the following situation. In the lab frame ${A}$, we have a charge ${-q}$ moving at speed ${v}$ in the ${+x}$ direction, and a charge ${+q}$ moving at the same speed ${v}$ but in the ${-x}$ direction, with ${-q}$ following the path ${y=0}$ and ${+q}$ following ${y=+d}$. Their positions are such that their closest approach occurs when they cross the ${y}$ axis.

First, we can work out the fields and the force on ${+q}$ at this point in the lab frame. The fields produced by a moving point charge are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}\gamma^{2}}\frac{1}{\left(1-\beta^{2}\sin^{2}\theta\right)^{3/2}}\frac{\hat{\mathbf{R}}}{R^{2}}\ \ \ \ \ (1)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}c^{2}}\frac{qv\left(1-v^{2}/c^{2}\right)\sin\theta}{\left[1-\left(v^{2}/c^{2}\right)\sin^{2}\theta\right]^{3/2}}\frac{\hat{\boldsymbol{\phi}}}{R^{2}} \ \ \ \ \ (2)$

where ${\mathbf{R}}$ is the vector from the moving charge to the observer and the direction of ${\hat{\boldsymbol{\phi}}}$ is found from using the right-hand rule on the particle’s velocity ${\mathbf{v}}$, as usual. The angle ${\theta}$ is the angle between ${\mathbf{R}}$ and ${\mathbf{v}}$. In our case, at the point where the charges are at their closest approach ${\theta=\pi/2}$ and ${R=d}$ so we get

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{q\gamma}{4\pi\epsilon_{0}d^{2}}\hat{\mathbf{y}}\ \ \ \ \ (3)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{q\gamma v}{4\pi\epsilon_{0}c^{2}d^{2}}\hat{\mathbf{z}} \ \ \ \ \ (4)$

The force on ${+q}$ with velocity ${\mathbf{v}=-v\hat{\mathbf{x}}}$ can be found from the Lorentz force law:

 $\displaystyle \mathbf{F}$ $\displaystyle =$ $\displaystyle q\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{q^{2}\gamma}{4\pi\epsilon_{0}d^{2}}\left(\hat{\mathbf{y}}+\frac{v^{2}}{c^{2}}\left(-\hat{\mathbf{x}}\times\hat{\mathbf{z}}\right)\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{q^{2}\gamma}{4\pi\epsilon_{0}d^{2}}\left(1+\frac{v^{2}}{c^{2}}\right)\hat{\mathbf{y}} \ \ \ \ \ (7)$

Now suppose we switch to frame ${B}$ in which ${+q}$ is at rest. This frame is moving with velocity ${-v\hat{\mathbf{x}}}$ with respect to ${A}$, so the transformation equations are

 $\displaystyle \bar{E}_{x}$ $\displaystyle =$ $\displaystyle E_{x}\ \ \ \ \ (8)$ $\displaystyle \bar{E}_{y}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{y}+vB_{z}\right)\ \ \ \ \ (9)$ $\displaystyle \bar{E}_{z}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{z}-vB_{y}\right)\ \ \ \ \ (10)$ $\displaystyle \bar{B}_{x}$ $\displaystyle =$ $\displaystyle B_{x}\ \ \ \ \ (11)$ $\displaystyle \bar{B}_{y}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{y}-\frac{v}{c^{2}}E_{z}\right)\ \ \ \ \ (12)$ $\displaystyle \bar{B}_{z}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{z}+\frac{v}{c^{2}}E_{y}\right) \ \ \ \ \ (13)$

[Note that this is a special case where the speed ${v}$ of the frame ${B}$ happens to be the same as the speed of the charge in the original lab frame ${A}$, so we can use the same symbol for both. In the more general case, the ${v}$ in the above 6 equations would be different from the ${v}$ in equations 1 and 2.]

Only ${E_{y}}$ and ${B_{z}}$ are non-zero, so we get

 $\displaystyle \bar{\mathbf{E}}$ $\displaystyle =$ $\displaystyle -\frac{q\gamma^{2}}{4\pi\epsilon_{0}d^{2}}\left(1+\frac{v^{2}}{c^{2}}\right)\hat{\mathbf{y}}\ \ \ \ \ (14)$ $\displaystyle \bar{\mathbf{B}}$ $\displaystyle =$ $\displaystyle -\frac{2q\gamma^{2}v}{4\pi\epsilon_{0}d^{2}c^{2}}\hat{\mathbf{z}} \ \ \ \ \ (15)$

[The ${y}$ and ${\bar{y}}$, and ${z}$ and ${\bar{z}}$ axes are parallel so we can use the unit vectors from frame ${A}$ or ${B}$.] The force seen in frame ${B}$ (where the velocity of ${+q}$ is ${\mathbf{v}=0}$ so there is no magnetic force) is thus

 $\displaystyle \bar{\mathbf{F}}$ $\displaystyle =$ $\displaystyle q\bar{\mathbf{E}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{q^{2}\gamma^{2}}{4\pi\epsilon_{0}d^{2}}\left(1+\frac{v^{2}}{c^{2}}\right)\hat{\mathbf{y}} \ \ \ \ \ (17)$

Note that the force in frame ${B}$ is larger by a factor of ${\gamma}$ than the force in frame ${A}$. As we saw earlier, an object experiences its maximum force in the frame in which it’s at rest.

Finally, let’s look at things in frame ${C}$ where ${-q}$ is at rest. This can be found from the results from frame ${A}$ by transforming to a frame moving at ${+v}$ relative to ${A}$, so the transformation equations are equations 9 and 13 with ${v}$ replaced by ${-v}$, giving

 $\displaystyle \bar{\mathbf{E}}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{y}-vB_{z}\right)\hat{\mathbf{y}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{q\gamma^{2}}{4\pi\epsilon_{0}d^{2}}\left(1-\frac{v^{2}}{c^{2}}\right)\hat{\mathbf{y}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{q}{4\pi\epsilon_{0}d^{2}}\hat{\mathbf{y}}\ \ \ \ \ (20)$ $\displaystyle \bar{\mathbf{B}}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{z}-\frac{v}{c^{2}}E_{y}\right)\hat{\mathbf{z}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (22)$

Since ${-q}$ is at rest in frame ${C}$, its electric field is just the Coulomb field from electrostatics, and there is no magnetic field. The force on ${+q}$ is therefore just the Coulomb force

$\displaystyle \bar{\mathbf{F}}=-\frac{q^{2}}{4\pi\epsilon_{0}d^{2}}\hat{\mathbf{y}} \ \ \ \ \ (23)$

## Relativistic transformation of force, electric and magnetic fields

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problems 12.41, 12.44.

Griffiths gives fairly complete derivations of the relativistic transformation laws for force and electromagnetic fields as we move from one inertial frame to another in his sections 12.2.4 and 12.3.2, so I won’t grind through the whole thing again here. I’ll just recap the ideas behind these derivations to give a flavour of what’s being done.

‘Ordinary’ force is defined as the derivative of the four-momentum (or at least its spatial part) with respect to ‘ordinary’ (not proper) time. Since four-momentum is a four-vector, it transforms using Lorentz transformations, as does ‘ordinary’ time. The problem with force is that its transformation is then the ratio of two Lorentz-transformed objects, so the transformation equations get a bit messy. The derivation of the transformations is done in much the same way as the derivation of the velocity addition formulas. For a frame ${\bar{\mathcal{S}}}$ moving at velocity ${v}$ in the ${x}$ direction relative to a stationary frame ${\mathcal{S}}$ we get

 $\displaystyle \bar{F}_{y}$ $\displaystyle =$ $\displaystyle \frac{d\bar{p}_{y}}{d\bar{t}}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{dp_{y}}{\gamma\left(dt-\frac{\beta}{c}dx\right)}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{dp_{y}/dt}{\gamma\left(1-\beta u_{x}/c\right)}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{F_{y}}{\gamma\left(1-\beta u_{x}/c\right)} \ \ \ \ \ (4)$

where ${\mathbf{u}}$ is the object’s velocity in ${\mathcal{S}}$. Remember that the velocity ${v}$ used in the calculation of ${\gamma}$ and ${\beta}$ is the relative velocity of ${\mathcal{S}}$ and ${\bar{\mathcal{S}}}$ and ${\mathbf{u}}$ is the velocity of the object relative to ${\mathcal{S}}$.

The calculation of the other two components is done in a similar way, with the result

 $\displaystyle \bar{F}_{x}$ $\displaystyle =$ $\displaystyle \frac{F_{x}-\beta\left(\mathbf{u}\cdot\mathbf{F}\right)/c}{1-\beta u_{x}/c}\ \ \ \ \ (5)$ $\displaystyle \bar{F}_{y}$ $\displaystyle =$ $\displaystyle \frac{F_{y}}{\gamma\left(1-\beta u_{x}/c\right)}\ \ \ \ \ (6)$ $\displaystyle \bar{F}_{z}$ $\displaystyle =$ $\displaystyle \frac{F_{z}}{\gamma\left(1-\beta u_{x}/c\right)} \ \ \ \ \ (7)$

Example 1 We have a charge ${q_{A}}$ at rest at the origin of ${\mathcal{S}}$ and another charge ${q_{B}}$ moving at speed ${v}$ in the ${+x}$ direction along the line ${y=d}$. When ${q_{B}}$ crosses the ${y}$ axis, it feels only an electric field (since ${q_{A}}$ is at rest it generates no magnetic field in ${\mathcal{S}}$), so the force on it is

$\displaystyle \mathbf{F}=\frac{q_{A}q_{B}}{4\pi\epsilon_{0}d^{2}}\hat{\mathbf{y}} \ \ \ \ \ (8)$

If we now switch to ${q_{B}}$‘s frame ${\bar{\mathcal{S}}}$ which is moving relative to ${\mathcal{S}}$ at speed ${v}$ along the ${x}$ axis, we can find the force experienced by ${q_{B}}$ in this frame using 6. In this case, ${\mathbf{u}=\mathbf{v}}$ (the object’s velocity in ${\mathcal{S}}$ is the same as the relative velocity of ${\mathcal{S}}$ and ${\bar{\mathcal{S}}}$) so

 $\displaystyle \bar{F}_{y}$ $\displaystyle =$ $\displaystyle \frac{F_{y}}{\gamma\left(1-\beta v/c\right)}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{F_{y}}{\gamma\left(1-v^{2}/c^{2}\right)}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \gamma F_{y}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\gamma q_{A}q_{B}}{4\pi\epsilon_{0}d^{2}}\hat{\mathbf{y}} \ \ \ \ \ (12)$

That is, the force experienced by ${q_{B}}$ is greater in the frame at which it is at rest.

In fact we can see from the transformation equations above that all components of force perpendicular to the motion are at a maximum in an object’s rest frame. In that frame ${\mathbf{u}=0}$ so the equations become

 $\displaystyle \bar{F}_{x}$ $\displaystyle =$ $\displaystyle F_{x}\ \ \ \ \ (13)$ $\displaystyle \bar{F}_{y}$ $\displaystyle =$ $\displaystyle \frac{F_{y}}{\gamma}\ \ \ \ \ (14)$ $\displaystyle \bar{F}_{z}$ $\displaystyle =$ $\displaystyle \frac{F_{z}}{\gamma} \ \ \ \ \ (15)$

To transform the electric and magnetic fields we can use a similar approach to that for the electric field, in which we considered a parallel plate capacitor with charge densities of ${\pm\sigma_{0}}$ (in the capacitor’s rest frame ${\mathcal{S}_{0}}$) on the two plates. In ${\mathcal{S}_{0}}$ however, since the charge is at rest, there is no magnetic field so we can’t use that system as it stands to derive the general transformation rules for electromagnetic fields. What we can do is introduce two other frames ${\mathcal{S}}$ (moving at speed ${v_{0}}$ relative to ${\mathcal{S}_{0}}$) and ${\bar{\mathcal{S}}}$ (moving at speed ${v}$ relative to ${\mathcal{S}}$). From the velocity addition formula, the velocity of ${\bar{\mathcal{S}}}$ relative to ${\mathcal{S}_{0}}$ is then

$\displaystyle \bar{v}=\frac{v+v_{0}}{1+vv_{0}/c^{2}} \ \ \ \ \ (16)$

Since we now have the velocities of both ${\mathcal{S}}$ and ${\bar{\mathcal{S}}}$ relative to ${\mathcal{S}_{0}}$ (where the charge is at rest, remember), we can eliminate ${\mathcal{S}_{0}}$ and express the fields in ${\bar{\mathcal{S}}}$ in terms of those in ${\mathcal{S}}$. Griffiths goes through the details, with the results

 $\displaystyle \bar{E}_{x}$ $\displaystyle =$ $\displaystyle E_{x}\ \ \ \ \ (17)$ $\displaystyle \bar{E}_{y}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{y}-vB_{z}\right)\ \ \ \ \ (18)$ $\displaystyle \bar{E}_{z}$ $\displaystyle =$ $\displaystyle \gamma\left(E_{z}+vB_{y}\right)\ \ \ \ \ (19)$ $\displaystyle \bar{B}_{x}$ $\displaystyle =$ $\displaystyle B_{x}\ \ \ \ \ (20)$ $\displaystyle \bar{B}_{y}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{y}+\frac{v}{c^{2}}E_{z}\right)\ \ \ \ \ (21)$ $\displaystyle \bar{B}_{z}$ $\displaystyle =$ $\displaystyle \gamma\left(B_{z}-\frac{v}{c^{2}}E_{y}\right) \ \ \ \ \ (22)$

Notice how the electric and magnetic field components perpendicular to the motion get tangled up with each other when we transform frames. This shows how it’s possible for a test charge to experience only an electric field in one frame, but a combination of electric and magnetic fields in another frame (and vice versa).

Example 2 Now that we have the general transformation rules, we can return to the special case of frames ${\mathcal{S}}$ in which the capacitor plates are at rest and ${\bar{\mathcal{S}}}$ where they are moving in the ${x}$ direction at speed ${v}$. When the plates are at rest, there is only an electric field (in the ${y}$ direction, assuming that the plates are parallel to the ${xz}$ plane). From these equations, we see that in any other frame moving in the ${x}$ direction, the electric field will have only a ${y}$ component and the magnetic field will have only a ${z}$ component. [I’m not certain this is the approach Griffiths wants in his problem 12.41 in which he asks why ${\bar{E}_{z}=0}$ in the moving frame, but it does appear to answer the question.]

Example 3 We can also revisit Example 1 and calculate the force felt by ${q_{B}}$ in its own rest frame by applying the field transformations. When ${q_{B}}$ crosses the ${\bar{y}}$ axis (which we’ll assume coincides with the ${y}$ axis at the time when ${q_{B}}$ crosses it) we had, from above,

$\displaystyle E_{y}=\frac{q_{A}}{4\pi\epsilon_{0}d^{2}} \ \ \ \ \ (23)$

with ${E_{x}=E_{z}=0}$, and ${\mathbf{B}=0}$. In ${q_{B}}$‘s frame, we have

 $\displaystyle \bar{E}_{y}$ $\displaystyle =$ $\displaystyle \gamma E_{y}=\frac{\gamma q_{A}}{4\pi\epsilon_{0}d^{2}}\ \ \ \ \ (24)$ $\displaystyle \bar{B}_{z}$ $\displaystyle =$ $\displaystyle -\gamma\frac{v}{c^{2}}E_{y}=-\frac{\gamma q_{A}v}{4\pi\epsilon_{0}d^{2}c^{2}} \ \ \ \ \ (25)$

with all other components equal to zero. The force felt by ${q_{B}}$ is again entirely electric (since it’s not moving in its own rest frame, it experiences no magnetic force), and we get

$\displaystyle \bar{F}_{y}=q_{B}\bar{E}_{y}=\frac{\gamma q_{A}q_{B}}{4\pi\epsilon_{0}d^{2}} \ \ \ \ \ (26)$

which is the same as 12.

## Gauss’s law for a relativistic point charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Chapter 12, Problem 12.43.

In Griffiths’s example 12.13, he rederives the formula for the electric field due to a moving charge, this time using relativity instead of retarded potentials. The result, which we’ve examined before, is

 $\displaystyle \mathbf{E}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{1-v^{2}/c^{2}}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}\frac{\hat{\mathbf{R}}}{R^{2}}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}\gamma^{2}}\frac{1}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}\frac{\hat{\mathbf{R}}}{R^{2}}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}\gamma^{2}}\frac{1}{\left(1-\beta^{2}\sin^{2}\theta\right)^{3/2}}\frac{\hat{\mathbf{R}}}{R^{2}} \ \ \ \ \ (3)$

where

$\displaystyle \mathbf{R}\equiv\mathbf{r}-\mathbf{v}t \ \ \ \ \ (4)$

is the vector from the particle’s present (not retarded) position to the observer (assuming the particle passes through the origin at ${t=0}$) and ${\theta}$ is the angle between ${\mathbf{R}}$ and ${\mathbf{v}}$. We can verify that Gauss’s law holds for this moving charge by integrating ${\mathbf{E}\cdot d\mathbf{a}}$ over a sphere of radius ${R}$.

$\displaystyle \int\mathbf{E}\cdot d\mathbf{a}=\frac{q}{2\epsilon_{0}\gamma^{2}}\int_{0}^{\pi}\frac{\sin\theta d\theta}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}} \ \ \ \ \ (5)$

The integral is nasty because we’re missing a ${\cos\theta}$ in the numerator that would make the integral easy. Maple handles it easily enough, but for those interested in how to derive it, I worked backward from Maple’s answer to figure out how to do it. First, we can split the integrand into the sum of two terms (I’ll use ${\beta\equiv v/c}$ to simplify the notation):

 $\displaystyle \frac{\sin\theta}{\left(1-\beta^{2}\sin^{2}\theta\right)^{3/2}}$ $\displaystyle =$ $\displaystyle \frac{\sin\theta}{\left(1-\beta^{2}+\beta^{2}\cos^{2}\theta\right)^{3/2}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(1-\beta^{2}\right)}\frac{\sin\theta\left(1-\beta^{2}+\beta^{2}\cos^{2}\theta-\beta^{2}\cos^{2}\theta\right)}{\left(1-\beta^{2}+\beta^{2}\cos^{2}\theta\right)^{3/2}}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left(1-\beta^{2}\right)}\frac{\sin\theta}{\sqrt{1-\beta^{2}+\beta^{2}\cos^{2}\theta}}-\frac{\beta^{2}}{\left(1-\beta^{2}\right)}\frac{\cos^{2}\theta\sin\theta}{\left(1-\beta^{2}+\beta^{2}\cos^{2}\theta\right)^{3/2}} \ \ \ \ \ (8)$

We can now integrate the second term by parts:

 $\displaystyle -\frac{\beta^{2}}{\left(1-\beta^{2}\right)}\int\frac{\cos^{2}\theta\sin\theta}{\left(1-\beta^{2}+\beta^{2}\cos^{2}\theta\right)^{3/2}}d\theta$ $\displaystyle =$ $\displaystyle -\frac{\beta^{2}}{\left(1-\beta^{2}\right)}\int\left(\cos\theta\right)\left[\frac{\cos\theta\sin\theta}{\left(1-\beta^{2}+\beta^{2}\cos^{2}\theta\right)^{3/2}}\right]d\theta\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\cos\theta}{\left(1-\beta^{2}\right)\sqrt{1-\beta^{2}+\beta^{2}\cos^{2}\theta}}-\frac{1}{\left(1-\beta^{2}\right)}\int\frac{\sin\theta}{\sqrt{1-\beta^{2}+\beta^{2}\cos^{2}\theta}}d\theta\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{\left(1-\beta^{2}\right)}\frac{\cos\theta}{\sqrt{1-\beta^{2}\sin^{2}\theta}}-\frac{1}{\left(1-\beta^{2}\right)}\int\frac{\sin\theta}{\sqrt{1-\beta^{2}+\beta^{2}\cos^{2}\theta}}d\theta \ \ \ \ \ (11)$

The second term in the last line now cancels the integral of the first term in 8, so we’re left with

 $\displaystyle \int_{0}^{\pi}\frac{\sin\theta d\theta}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}$ $\displaystyle =$ $\displaystyle -\frac{1}{\left(1-\beta^{2}\right)}\left.\frac{\cos\theta}{\sqrt{1-\beta^{2}\sin^{2}\theta}}\right|_{0}^{\pi}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{1-\beta^{2}}=2\gamma^{2} \ \ \ \ \ (13)$

[There might be an easier way to do this, but I couldn’t see any obvious substitutions that worked.]

Plugging this back into 5 we get

$\displaystyle \int\mathbf{E}\cdot d\mathbf{a}=\frac{q}{\epsilon_{0}} \ \ \ \ \ (14)$

so Gauss’s law is satisfied.

We can also calculate the Poynting vector by using the magnetic field of a point charge in uniform motion, which Griffiths works out in his example 12.14:

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{1}{c^{2}}\mathbf{v}\times\mathbf{E}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\frac{qv\left(1-v^{2}/c^{2}\right)\sin\theta}{\left[1-\left(v^{2}/c^{2}\right)\sin^{2}\theta\right]^{3/2}}\frac{\hat{\boldsymbol{\phi}}}{R^{2}}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}qc\beta\sin\theta}{4\pi\gamma^{2}\left(1-\beta^{2}\sin^{2}\theta\right)^{3/2}}\frac{\hat{\boldsymbol{\phi}}}{R^{2}} \ \ \ \ \ (17)$

where the ${\phi}$ direction is that found by using the right-hand rule with the thumb pointing in the direction of ${\mathbf{v}}$, the particle’s motion. The Poynting vector is

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B} \ \ \ \ \ (18)$

If we restrict ourselves to a charge moving in the ${+z}$ direction, so that ${\mathbf{v}=v\hat{\mathbf{z}}}$ and at the time when the charge passes through the origin, then ${\mathbf{R}}$ in 3 becomes the radial coordinate in spherical coordinates and ${\phi}$ is the azimuthal coordinate. In that case, ${\hat{\mathbf{R}}\times\hat{\boldsymbol{\phi}}=-\hat{\boldsymbol{\theta}}}$ so

$\displaystyle \mathbf{S}=-\frac{q^{2}}{16\pi^{2}\epsilon_{0}\gamma^{4}}\frac{c\beta\sin\theta}{\left(1-\beta^{2}\sin^{2}\theta\right)^{3}}\frac{\hat{\boldsymbol{\theta}}}{R^{4}} \ \ \ \ \ (19)$