## Black hole radiation: energy of emitted particles

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 16; Box 16.1.

In Schwarzschild (S) space-time, the energy of a particle is given by

$\displaystyle e=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau} \ \ \ \ \ (1)$

Outside the event horizon ${r>2GM}$ and since the S coordinate ${t}$ represents time, it must constantly increase as the proper time ${\tau}$ increases, so ${dt/d\tau>0}$, with the result that ${e>0}$. Inside the event horizon ${r<2GM}$, but time and space swap round so ${dt/d\tau}$ can be either positive or negative, since ${t}$ is now a space coordinate. The the energy can be positive or negative inside the event horizon. For a single particle moving along a geodesic, ${e}$ is a constant of the motion, so since ${1-\frac{2GM}{r}}$ becomes zero at the event horizon, ${dt/d\tau}$ becomes infinite which results from the fact that the S time becomes infinite at the horizon. After crossing the horizon, ${1-\frac{2GM}{r}<0}$ so ${dt/d\tau}$ must also be negative to keep ${e}$ constant. However, it is possible for the particle to interact with another particle inside the horizon, which could cause ${dt/d\tau}$ to change sign, resulting in a negative energy.

The idea behind black hole radiation is actually quite simple. Quantum field theory predicts that vacuum fluctuations can occur, in which a particle-antiparticle pair spontaneously appears, essentially out of nothing, provided that the energies of the two particles are equal and opposite; that is, one particle has positive energy ${E}$ while the other has negative energy ${-E}$. This phenomenon can occur only for a very short time interval of the order of ${\Delta t\sim\hbar/E}$ (at this stage, you can just accept all this as god-given, since we haven’t studied quantum field theory yet), after which the particle recombines with its antiparticle.

Now suppose this pair creation event occurs very close to the event horizon, and the negative energy particle crosses the horizon before it has a chance to recombine, and that the positive energy particle therefore escapes to infinity. The energy of the black hole is thus reduced by ${E}$ and the energy ${E}$ is radiated away to infinity. To see how this works in a simplified model, suppose the pair creation event occurs at some small distance ${\epsilon}$ above the event horizon. We need ${\epsilon}$ to be small enough that the negative energy particle can cross the horizon before it recombines, so we need to estimate how long it takes the particle to travel to the horizon. Since we’re outside the horizon, the S coordinate ${r}$ is still spatial, so we can use the following expression to estimate the proper time required:

$\displaystyle \frac{dr}{d\tau}=-\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)\left(1+\frac{\ell^{2}}{r^{2}}\right)} \ \ \ \ \ (2)$

We’ve used the minus sign to indicate that ${r}$ is decreasing. If the particle falls radially, the angular momentum is ${\ell=0}$, so the proper time required is

$\displaystyle \Delta\tau=-\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)}} \ \ \ \ \ (3)$

To go further, we need to know ${e}$. It’s unlikely that the pair is created at rest, but suppose we’re in a locally flat reference frame that is released from rest with its origin at ${r=2GM+\epsilon}$ and that the pair creation event occurs at this location. An object in the observer’s frame has an energy per unit mass of

$\displaystyle e=\sqrt{1-\frac{2GM}{2GM+\epsilon}} \ \ \ \ \ (4)$

so for such an object the integral becomes

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle -\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{1-\frac{2GM}{2GM+\epsilon}-\left(1-\frac{2GM}{r}\right)}}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int_{2GM+\epsilon}^{2GM}\frac{dr}{\sqrt{\frac{2GM}{r}-\frac{2GM}{2GM+\epsilon}}} \ \ \ \ \ (6)$

This integral can be done using Maple although the result is a bit too complex to write out here, and then a series expansion of the result around ${\epsilon=0}$ gives the leading terms

$\displaystyle \Delta\tau=2\sqrt{2GM\epsilon}+\frac{5}{6}\sqrt{\frac{2}{GM}}\epsilon^{3/2}+\mathcal{O}\left(\epsilon^{5/2}\right) \ \ \ \ \ (7)$

If we want just the first term, we can transform the integral a bit using ${\rho\equiv r-2GM}$:

$\displaystyle \Delta\tau=\int_{0}^{\epsilon}\frac{d\rho}{\sqrt{\frac{2GM}{\rho+2GM}-\frac{2GM}{2GM+\epsilon}}} \ \ \ \ \ (8)$

If we expand each term in the square root in a series and keep terms up to first order in ${\rho}$ and ${\epsilon}$ we get

$\displaystyle \frac{2GM}{\rho+2GM}-\frac{2GM}{2GM+\epsilon}=\frac{1}{2GM}\left(\epsilon-\rho\right)+\mathcal{O}\left(\epsilon^{2}\right)+\mathcal{O}\left(\rho^{2}\right) \ \ \ \ \ (9)$

so to this order, we have

$\displaystyle \Delta\tau=\int_{0}^{\epsilon}\frac{d\rho}{\sqrt{\left(\epsilon-\rho\right)/2GM}} \ \ \ \ \ (10)$

We can use another substitution ${u\equiv\epsilon-\rho}$ to get

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \int_{0}^{\epsilon}\frac{du}{\sqrt{u/2GM}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\sqrt{2GM\epsilon} \ \ \ \ \ (12)$

This is the time required for an object released from rest at ${r=2GM+\epsilon}$ to reach the event horizon. Presumably if the particle is not at rest when it is created, but is heading towards the event horizon, it will require less time than this to reach it.

Plugging this into the field theory estimate above, we can get an estimate of the energy of the radiated particle:

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \frac{\hbar}{E}\ \ \ \ \ (13)$ $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2\sqrt{2GM\epsilon}} \ \ \ \ \ (14)$

I have to admit I’m not particularly comfortable with this calculation, since ${\Delta\tau}$ is calculated for an object released at rest from ${r=2GM+\epsilon}$ whereas the pair of particles would probably not be produced at rest. However, as an order of magnitude estimate, I suppose it’s reasonable.

## Painlevé-Gullstrand coordinates: derivation using a local flat frame

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problem 15.1.

Earlier, we worked out the basis vectors in a locally flat frame for a freely falling observer near a black hole. These basis vectors are worked out by considering the four-velocity in two frames: the local, flat frame, and the Schwarzschild (S) frame. In particular, in the flat frame, ${\mathbf{u}=\mathbf{o}_{t}=\left[1,0,0,0\right]}$ so in the other frame, the four-velocity is the transformed time basis vector: ${\mathbf{u}^{\prime}=\mathbf{o}_{t}^{\prime}}$. Using this argument, we worked out ${\mathbf{o}_{t}^{\prime}}$ in the S frame for a freely falling observer and got

$\displaystyle \mathbf{o}_{t}^{\prime}=\left[\left(1-\frac{2GM}{r}\right)^{-1},-\sqrt{\frac{2GM}{r}},0,0\right] \ \ \ \ \ (1)$

In the flat frame, we can write the interval between two events as ${d\mathbf{s}=\left[d\tau,dx,dy,dz\right]}$. In the Painlevé-Gullstrand system, the time coordinate ${\mathring{t}}$ is just the proper time of a freely falling observer, so ${d\mathring{t}=d\tau}$. Still in the flat frame, we have therefore

$\displaystyle d\mathring{t}=-\eta_{ij}o_{t}^{i}ds^{j}=-\mathbf{o}_{t}\cdot d\mathbf{s} \ \ \ \ \ (2)$

since only the component ${o_{t}^{t}}$ is non-zero, and ${\eta_{tt}=-1}$ in flat space. Since this is a scalar product, it has the same value in any coordinate system, such as the S system where we have

 $\displaystyle d\mathring{t}$ $\displaystyle =$ $\displaystyle \mathbf{o}_{t}^{\prime}\cdot d\mathbf{s}^{\prime}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ij}o_{t}^{\prime i}ds^{\prime j} \ \ \ \ \ (4)$

In the S system, we have

$\displaystyle d\mathbf{s}^{\prime}=\left[dt,dr,d\theta,d\phi\right] \ \ \ \ \ (5)$

so

 $\displaystyle d\mathring{t}$ $\displaystyle =$ $\displaystyle -\left[-\left(1-\frac{2GM}{r}\right)\right]\left(1-\frac{2GM}{r}\right)^{-1}dt-\left(1-\frac{2GM}{r}\right)^{-1}\left(-\sqrt{\frac{2GM}{r}}\right)dr\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle dt+\left(1-\frac{2GM}{r}\right)^{-1}\sqrt{\frac{2GM}{r}}dr \ \ \ \ \ (7)$

This agrees with the earlier result for Painlevé-Gullstrand.

## Kruskal-Szekeres diagrams: another space ship disaster

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problem 15.8.

In this example, we have a spaceship originally at some fixed distance from a black hole, but the engines fail and the ship starts to fall vertically downward (tail first) towards the black hole.

In the KS diagram, the tail of the ship follows the black world line and the front of the ship follows the parallel magenta world line. The diagonal lines with slope ${+1}$ indicate photons emitted by the tail end (with the green line being also the event horizon). The violet photon is emitted before the tail crosses the horizon and is received by the front before it crosses the horizon. A photon emitted just as the tail crosses the horizon follows the green world line, and since this line also corresponds to a constant ${r=2GM}$, the photon remains at the event horizon and is received by the front when the front crosses the horizon. The brown photon leaves the tail after it crosses the horizon and is received by the front after it too crosses the horizon. Thus there isn’t any time during the period where the ship is crossing the horizon that the front receives no photons from the tail, but any photons received by the front are emitted by the tail when both the front and the tail are on the same side of the horizon. When the front is outside the horizon and the tail is inside, the front still sees the tail, but it is seeing the tail as it was before it crossed the horizon.

Closer to the singularity at ${r=0}$, however, there is a cutoff point beyond which any photons (such as the short red line) emitted by the tail are absorbed at ${r=0}$ before reaching the front, so the front end will not be able to see the tail hit ${r=0}$.

## Kruskal-Szekeres diagrams: saving a space shuttle

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problems 15.7 15.9.

In this example (Moore, problem P15.9), we have a spaceship maintaining a constant distance ${R}$ from a black hole, such that the Kruskal-Szekeres (KS) coordinates satisfy

$\displaystyle u^{2}-v^{2}=\left(\frac{R}{2GM}-1\right)e^{R/2GM}=1$

At Schwarzschild (S) time ${t=0}$ (corresponding to ${v=0}$), a shuttle leaves the spaceship and gets pulled towards the black hole along the line ${u=1}$ in the KS diagram. If the mother ship is capable of speeds up to light speed, what is the latest time that it can leave its orbit to intercept the shuttle before the shuttle crosses the event horizon?

In the diagram, the shuttle’s world line is shown in yellow and the ship’s world line is the grey hyperbola. If the ship suddenly breaks out of its orbit and travels at light speed towards the black hole, and intercepts the shuttle just before it crosses the event horizon, the ship will have to follow the turquoise diagonal. (Remember that photon world lines have slopes of ${\pm1}$ on a KS diagram.) We therefore need to find the intersection of the turquoise line and the grey hyperbola, and then find the S time ${t_{r}}$ at which the ship starts on its rescue mission. On a KS diagram, curves of constant ${t}$ are straight lines through the origin, so the time is the thin green line in the diagram.

The turquoise line has slope ${-1}$ and passes through the point ${\left(1,1\right)}$ so its equation is

 $\displaystyle v-1$ $\displaystyle =$ $\displaystyle -\left(u-1\right)$ $\displaystyle v$ $\displaystyle =$ $\displaystyle -u+2$

The intersection with the hyperbola is

 $\displaystyle u^{2}-\left(-u+2\right)^{2}$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle 4u-4$ $\displaystyle =$ $\displaystyle 1$ $\displaystyle u$ $\displaystyle =$ $\displaystyle \frac{5}{4}$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{3}{4}$

The S time is

 $\displaystyle t_{r}$ $\displaystyle =$ $\displaystyle 2GM\ln\frac{u+v}{u-v}$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2GM\ln\frac{8}{2}$ $\displaystyle$ $\displaystyle =$ $\displaystyle 4GM\ln2$ $\displaystyle$ $\displaystyle \approx$ $\displaystyle 2.77GM$

As a slight variant on this problem (Moore, problem P15.7), suppose you are working on the outside of the spaceship hovering at radius ${R}$ and you drop a valuable piece of equipment, which then falls towards the black hole. Even though an observer at infinity would say that the object gets stuck at the event horizon (since the S ${t}$ coordinate becomes infinite there), the KS diagram shows that you have a limited time (as measured by your own proper time) to go after the object if you are to catch it before it crosses the horizon. In this case, the dropped object has a world line similar to the yellow line on the diagram (not quite the same line, since the dropped object would follow a geodesic, which isn’t a straight line here), and to catch it you’d have to leave your orbit at light speed along a turquoise diagonal that intercepts the object just before it crosses the horizon.

## Kruskal-Szekeres metric: more fun with photons

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problems 15.5, 15.6.

Here’s another example of using a Kruskal-Szekeres (KS) diagram. We have the same observer in a space ship falling into a black hole as before, but this time he is firing photon torpedoes radially outward. If these torpedoes follow photon world lines is there any danger that they will hit the ship before it reaches ${r=0}$? It might seem this is possible, since everything (even light) moves inwards inside the event horizon, so both the torpedoes and the ship are heading for ${r=0}$. However, looking at the KS diagram, we see this won’t happen:

The ship is the heavy black curve. Photons fired radially outwards will have world lines with slope +1, so the three heavy red lines represent three torpedoes fired from the ship after it enters the event horizon. We can see that these torpedoes won’t hit the ship, although they will get absorbed by the singularity at ${r=0}$.

As a variant on this problem, suppose that the black curve represents the surface of a collapsing star instead of a falling space ship. After the star collapses through its event horizon, it can still emit photons radially outwards. If another ship fell into the event horizon as the star is collapsing, it might follow the thick purple world line.

Since the ship’s world line intersects the lines of the emitted photons, this observer would still see light from the surface of the star as it collapses, so the interior of a black hole isn’t dark, at least until the star has finished collapsing.

## Kruskal-Szekeles metric: what can you see as you fall into a black hole?

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Problem 15.4.

The Kruskal-Szekeres (KS) metric is an alternative to the Schwarzschild (S) metric that makes many phenomena easier to visualize by drawing a diagram in which world lines are plotted on a graph of the coordinate ${v}$ versus ${u}$. For radial motion, the KS metric is

$\displaystyle ds^{2}=\frac{32\left(GM\right)^{3}}{r}e^{-r/2GM}\left(du^{2}-dv^{2}\right)$

and the relation between the KS coordinates ${u}$ and ${v}$ and the S coordinates ${r}$ and ${t}$ is

 $\displaystyle u^{2}-v^{2}$ $\displaystyle =$ $\displaystyle \left(\frac{r}{2GM}-1\right)e^{r/2GM}$ $\displaystyle 2GM\ln\left|\frac{u+v}{u-v}\right|$ $\displaystyle =$ $\displaystyle t$

As an example of using a KS diagram, suppose we have an observer in a space ship that is falling into a black hole. As we saw in the last post, any massive object’s world line must have a slope with a magnitude greater than 1 at all of its points in the KS diagram. One possible world line is shown as the heavy black curve in the diagram:

The motion of the observer is upwards in the diagram. In S coordinates, an observer at infinity sees the ship slow down as it approaches the event horizon, eventually freezing at the horizon, since the S ${t}$ coordinate becomes infinite there (as shown in the diagram by the heavy green line in the upper right quadrant). However, as measured by the observer in the ship, it takes a finite time to move from a point outside the horizon to a point inside it; there is no divergence at ${r=2GM}$. This is shown in the KS diagram as the black curve simply continues across the green line without anything unusual happening.

Does the fact that the ship crosses the ${t=\infty}$ line mean that the observer in the ship can see the entire future of the outside world as he crosses ${r=2GM}$? To answer this, we first note that in order for the observer to see an event from the outside world, a photon from that event must be able to reach him before he gets to ${r=0}$, where his own world line ends. Since photons travel along lines with slopes of ${\pm1}$, the only photons that can reach him after he crosses the horizon are those between the two heavy red lines shown. Thus he can see into the distant future (that is, very large ${t}$ values) only for events that happen just outside the horizon (that is, events just to the lower right of the green line, between the two red lines). Events that occur at ${t=0}$ lie on the ${u}$ axis, so he can see only those events that lie between approximately ${u=0.75}$ and ${u=1.1}$. The ${r}$ values of these events are determined by the set of hyperbolas that cross the ${u}$ axis between these points (some of the hyperbolas of constant ${r}$ are shown as dashed grey curves).

Thus even though he is inside the event horizon, the observer can still see some events that happened outside the horizon, but he cannot see all events that happen at any time in the future.

## Kruskal-Szekeres coordinates and the event horizon

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Box 15.4.

The Painlevé-Gullstrand (PG, also known as the global rain system) metric eliminates some of the problems in the Schwarzschild metric (S metric), but the PG metric is not diagonal and as a result, can be a bit difficult to work with. Another metric developed by Martin Kruskal and George Szekeres, and thus known as the Kruskal-Szekeres (KS) metric avoids this problem, although at the cost of introducing coordinates whose meanings aren’t exactly intuitive. However, the KS system has a number of other benefits which often make qualitative interpretation of happenings around the event horizon easier.

KS coordinates retain the usual angular coordinates ${\theta}$ and ${\phi}$, but the ${r}$ and ${t}$ coordinates of the S metric are replaced with new coordinates ${u}$ and ${v}$ which can be defined as:

 $\displaystyle u$ $\displaystyle =$ $\displaystyle \sqrt{\frac{r}{2GM}-1}e^{r/4GM}\cosh\frac{t}{4GM}$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \sqrt{\frac{r}{2GM}-1}e^{r/4GM}\sinh\frac{t}{4GM}$

for ${r>2GM}$ and

 $\displaystyle u$ $\displaystyle =$ $\displaystyle \sqrt{1-\frac{r}{2GM}}e^{r/4GM}\sinh\frac{t}{4GM}$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \sqrt{1-\frac{r}{2GM}}e^{r/4GM}\cosh\frac{t}{4GM}$

for ${r<2GM}$. (As I said, they’re non-intuitive!) These equations can be inverted to give

 $\displaystyle u^{2}-v^{2}$ $\displaystyle =$ $\displaystyle \left(\frac{r}{2GM}-1\right)e^{r/2GM}$ $\displaystyle 2GM\ln\left|\frac{u+v}{u-v}\right|$ $\displaystyle =$ $\displaystyle t$

Because the equation for ${r}$ is transcendental, it’s not possible to solve explicitly for ${r}$ in terms of ${u}$ and ${v}$ using simple functions, so ${r}$ is retained in the expression for the metric, but it should be regarded as a function ${r\left(u,v\right)}$ rather than as an independent coordinate.

From here, we calculate the differentials of these two equations and then substitute for ${dr}$ and ${dt}$ in the S metric to get the KS metric. For the ${r}$ equation:

 $\displaystyle 2u\; du-2v\; dv$ $\displaystyle =$ $\displaystyle e^{r/2GM}\left(\frac{1}{2GM}+\frac{r}{\left(2GM\right)^{2}}-\frac{1}{2GM}\right)dr$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{r/2GM}\frac{r}{4\left(GM\right)^{2}}dr$ $\displaystyle dr$ $\displaystyle =$ $\displaystyle \frac{8\left(GM\right)^{2}}{r}e^{-r/2GM}\left(u\; du-v\; dv\right)$

For the ${t}$ equation, there are two possible cases, due to the absolute value: the argument of the absolute value can be positive or negative. If it’s positive, then

 $\displaystyle dt$ $\displaystyle =$ $\displaystyle 2GM\frac{u-v}{u+v}\left[\frac{\left(du+dv\right)\left(u-v\right)-\left(u+v\right)\left(du-dv\right)}{\left(u-v\right)^{2}}\right]$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2GM\left[\frac{du+dv}{u+v}-\frac{du-dv}{u-v}\right]$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2GM}{u^{2}-v^{2}}\left[\left(du+dv\right)\left(u-v\right)-\left(du-dv\right)\left(u+v\right)\right]$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2GM}{r/2GM-1}e^{-r/2GM}\left[2u\; dv-2v\; du\right]$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\left(GM\right)^{2}}{r-2GM}e^{-r/2GM}\left[u\; dv-v\; du\right]$

where in the fourth line, we used the ${r}$ equation above to get rid of ${u^{2}-v^{2}}$.

If the argument is negative, then we are finding the differential of ${\ln\left[\frac{v+u}{v-u}\right]}$, that is, we have just interchanged ${u}$ and ${v}$. By looking at the second line, we can see this gives the same result for ${dt}$. Now it’s just a matter of inserting these expressions for ${dr}$ and ${dt}$ into the S metric, which is, for radial motion:

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2} \ \ \ \ \ (1)$

We get

 $\displaystyle dr^{2}$ $\displaystyle =$ $\displaystyle \frac{64\left(GM\right)^{4}}{r^{2}}e^{-r/GM}\left(u^{2}du^{2}+v^{2}dv^{2}-2uv\; du\; dv\right)$ $\displaystyle \left(1-\frac{2GM}{r}\right)^{-1}dr^{2}$ $\displaystyle =$ $\displaystyle \frac{r}{r-2GM}\frac{64\left(GM\right)^{4}}{r^{2}}e^{-r/GM}\left(u^{2}du^{2}+v^{2}dv^{2}-2uv\; du\; dv\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{64\left(GM\right)^{4}}{r\left(r-2GM\right)}e^{-r/GM}\left(u^{2}du^{2}+v^{2}dv^{2}-2uv\; du\; dv\right)$ $\displaystyle dt^{2}$ $\displaystyle =$ $\displaystyle \frac{64\left(GM\right)^{4}}{\left(r-2GM\right)^{2}}e^{-r/GM}\left(u^{2}dv^{2}+v^{2}du^{2}-2uv\; du\; dv\right)$ $\displaystyle -\left(1-\frac{2GM}{r}\right)dt^{2}$ $\displaystyle =$ $\displaystyle -\frac{r-2GM}{r}\frac{64\left(GM\right)^{4}}{\left(r-2GM\right)^{2}}e^{-r/GM}\left(u^{2}dv^{2}+v^{2}du^{2}-2uv\; du\; dv\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{64\left(GM\right)^{4}}{r\left(r-2GM\right)}e^{-r/GM}\left(u^{2}dv^{2}+v^{2}du^{2}-2uv\; du\; dv\right)$

Putting this together, we get

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle \frac{64\left(GM\right)^{4}}{r\left(r-2GM\right)}e^{-r/GM}\left[\left(u^{2}-v^{2}\right)\left(du^{2}-dv^{2}\right)\right]$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{64\left(GM\right)^{4}}{r\left(r-2GM\right)}e^{-r/GM}\left(\frac{r}{2GM}-1\right)e^{r/2GM}\left(du^{2}-dv^{2}\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{32\left(GM\right)^{3}}{r}e^{-r/2GM}\left(du^{2}-dv^{2}\right)$

This metric allows a few things to be seen fairly easily. For example, a photon world line, with ${ds^{2}=0}$, must have ${du=\pm dv}$. That is, on a plot of ${v}$ versus ${u}$, photon world lines always have slope ${\pm1}$, just like they do in special relativity space-time diagrams. From the ${r}$ equation above, if ${r}$ is a constant ${r_{0}}$, then

$\displaystyle u^{2}-v^{2}=\left(\frac{r_{0}}{2GM}-1\right)e^{r_{0}/2GM}=const$

This is the equation of a hyperbola which crosses the ${u}$ axis and opens to the right if ${r_{0}>2GM}$ or a hyperbola crossing the ${v}$ axis and opening upwards if ${r_{0}<2GM}$. In particular, ${r_{0}=0}$ gives the hyperbola ${u^{2}-v^{2}=-1}$.

Finally, if ${t}$ is a constant ${t_{0}}$, then

$\displaystyle 2GM\ln\left|\frac{u+v}{u-v}\right|=t_{0}$

from which we get

 $\displaystyle \frac{u+v}{u-v}$ $\displaystyle =$ $\displaystyle const\equiv a$ $\displaystyle u+v$ $\displaystyle =$ $\displaystyle au-av$ $\displaystyle v$ $\displaystyle =$ $\displaystyle \frac{a-1}{a+1}u$

Thus curves of constant ${t}$ are straight lines through the origin.

These features can be illustrated on a KS diagram, as shown:

The ${r=0}$ hyperbola is shown in blue at the top of the diagram. A few hyperbolas for ${r_{0}>2GM}$ are shown as grey dashed curves on the right, and a few lines of constant ${t_{0}}$ are the green dashed lines meeting at the origin. The red diagonal line with slope ${-1}$ represents ${t=-\infty}$ and the heavy green line with slope ${+1}$ is ${t=+\infty}$. This green line and the upper half of the red line also represent ${r=2GM}$ (the event horizon), since the hyperbola reduces to ${u^{2}-v^{2}=0}$ in this case. The region between the ${r=2GM}$ lines and the ${r=0}$ hyperbola represents space inside the event horizon, but notice that neither the ${u}$ nor ${v}$ coordinate does anything pathological at the event horizon boundary, so it is clear that it’s easy to cross from outside to inside the event horizon. All world lines move upwards in the diagram, and must end when they reach the ${r=0}$ hyperbola.

All photon world lines must be parallel to either the red or green diagonal line. This means that any massive particle’s world line must always have a slope whose magnitude is greater than 1 in the diagram, just as in special relativity. In particular, once a particle crosses into the area inside the event horizon, it cannot get out again, since that would require some part of its world line to have a slope less than 1.

It might seem that the condition that any massive particle’s world line has a slope greater than 1 would mean that any such particle must eventually cross the event horizon, but that’s not quite true. For example, if ${r}$ is a constant, the world line is one of the grey hyperbolas in the diagram. Such a hyperbola always has a slope greater than 1 (in the upper right quadrant) and yet it never crosses the event horizon. If an object wishes to move away from the black hole, it can manage this by maintaining a world line whose slope is slightly less than the slope of the constant-${r}$ hyperbola at each point, yet is still a slope greater than 1. This will continually shift the object to larger values of ${r}$. Since all the constant-${r}$ hyperbolas have the same asymptote, any desired distance from the black hole can be reached if we carry on this process long enough, since any two constant-${r}$ hyperbolas will get infinitesimally close eventually. Because the KS diagram squashes together the curves of ${r}$ and ${t}$ far from the black hole, it’s not the best diagram for analyzing motion far away; its main strength is in letting us see what happens near and inside the event horizon.

## Painlevé-Gullstrand metric: photon paths inside the event horizon

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Box 15.3, Problem P15.3.

The Painlevé-Gullstrand (PG, also known as the global rain system) metric is a different way of describing empty space surrounding a spherically symmetric mass that avoids some of the problems encountered with the Schwarzschild metric (S metric). The PG metric is

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)d\mathring{t}^{2}+2\sqrt{\frac{2GM}{r}}d\mathring{t}\; dr+dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}$

where ${\mathring{t}}$ is the PG time coordinate, and is the proper time as measured by a clock falling along a radial line from rest at infinity. The ${\mathring{t}}$ coordinate, since it is measured as the proper time on a clock, is always a time coordinate, so the PG metric avoids the switch between distance and time coordinates that occurs in the S metric. It might seem that this causes problems for ${r<2GM}$, since there the metric component ${g_{\mathring{t}\mathring{t}}=-\left(1-\frac{2GM}{r}\right)}$ becomes positive, so it might seem that all the terms are positive, resulting in ${ds^{2}>0}$ for a time-like world line (contrary to what it should be). However, the off-diagonal metric component ${g_{\mathring{t}r}=2\sqrt{\frac{2GM}{r}}}$ comes to the rescue here, since for ${r<2GM}$, if we have ${dr<0}$ we can still get ${ds^{2}<0}$. That is, for a time-like world line, we must have ${dr<0}$ once we cross the event horizon, which is the same condition that we got from the S metric, except there it was done by swapping the meanings of ${r}$ and ${t}$.

For a radial world line ${d\theta=d\phi=0}$ and we have

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)d\mathring{t}^{2}+2\sqrt{\frac{2GM}{r}}d\mathring{t}\; dr+dr^{2}$

which we can write as

$\displaystyle \left(\frac{ds}{d\mathring{t}}\right)^{2}=-\left(1-\frac{2GM}{r}\right)+2\sqrt{\frac{2GM}{r}}\frac{dr}{d\mathring{t}}+\left(\frac{dr}{d\mathring{t}}\right)^{2}$

Photons travelling radially will have ${\left(\frac{ds}{d\mathring{t}}\right)^{2}=0}$, so we can find ${\frac{dr}{d\mathring{t}}}$ by solving the quadratic equation on the RHS.

 $\displaystyle \frac{dr}{d\mathring{t}}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[-2\sqrt{\frac{2GM}{r}}\pm\sqrt{\frac{8GM}{r}+4\left(1-\frac{2GM}{r}\right)}\right]$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{2GM}{r}}\pm1$

The minus sign corresponds to an inward-moving photon, and the plus sign to an outward-moving one. Leaving aside the obvious problem that, for ${r<2GM}$, both these speeds can exceed 1 in magnitude, we can solve this equation to find ${\mathring{t}}$ as a function of ${r}$. We get, for the inward-moving photon

$\displaystyle -\int\frac{dr}{\sqrt{\frac{2GM}{r}}+1}=\int d\mathring{t}$

Using Maple for the integral, we get

$\displaystyle \mathring{t}=-r+2\sqrt{2GMr}-4GM\ln\left[\sqrt{2GM}+\sqrt{r}\right]+C \ \ \ \ \ (1)$

where ${C}$ is the constant of integration. If we start the photon at ${r=2GM}$ at ${\mathring{t}=0}$, then

 $\displaystyle C$ $\displaystyle =$ $\displaystyle -2GM+4GM\ln\left(2\sqrt{2GM}\right)$ $\displaystyle \frac{\mathring{t}}{GM}$ $\displaystyle =$ $\displaystyle 2\sqrt{\frac{2r}{GM}}-\frac{r}{GM}-2-4\ln\left[\sqrt{2GM}+\sqrt{r}\right]+4\ln\left(2\sqrt{2GM}\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\sqrt{\frac{2r}{GM}}-\frac{r}{GM}-2-4\ln\left[\frac{1}{2}\left(1+\sqrt{\frac{r}{2GM}}\right)\right]$

When ${r=0}$, we get

$\displaystyle \frac{\mathring{t}}{GM}=-2+4\ln2=0.7726$

A plot of ${\frac{\mathring{t}}{GM}}$ versus ${\frac{r}{GM}}$ is as shown:

The particle starts off at ${r/GM=2}$ and its world line travels up and to the left until it hits ${r=0}$.

For an outward moving photon, we get

$\displaystyle -\int\frac{dr}{\sqrt{\frac{2GM}{r}}-1}=\int d\mathring{t}$

This time, we get for the integral

$\displaystyle \mathring{t}=r+2\sqrt{2GMr}+4GM\ln\left[\sqrt{2GM}-\sqrt{r}\right]+C$

In this case, we can’t start an outward-moving photon at ${r=2GM}$, since this gives an infinite logarithm. In fact, if we emit an outward-moving photon at some ${r<2GM}$, it will actually move inward (since we require ${dr<0}$), although if we emit the photon near the event horizon (from the inside), it will take a very long time to move inwards from its starting point. To make things definite, suppose ${\mathring{t}=0}$ at ${r=GM}$. Then

 $\displaystyle C$ $\displaystyle =$ $\displaystyle -\left(1+2\sqrt{2}\right)GM-4GM\ln\left(\left(\sqrt{2}-1\right)\sqrt{GM}\right)$ $\displaystyle \frac{\mathring{t}}{GM}$ $\displaystyle =$ $\displaystyle \frac{r}{GM}+2\sqrt{\frac{2r}{GM}}-\left(1+2\sqrt{2}\right)+4\ln\left[\frac{\sqrt{2}-\sqrt{r/GM}}{\sqrt{2}-1}\right]$

A plot looks like this:

Again, the motion along the red curve is up and to the left. A photon emitted outward at ${r=GM}$ takes ${\mathring{t}=1.0833GM}$ to reach ${r=0}$, while an inward moving photon takes ${\mathring{t}=0.3108GM}$ (which we get by evaluating ${\mathring{t}}$ for ${r=GM}$ in 1 and taking the difference between that and 0.7726) .

## Painlevé-Gullstrand (global rain) coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 15; Boxes 15.1, 15.2.

See also Karl Martel and Eric Poisson, Am. J. Phys. 69, 476 (2001) for a different approach to this problem.

(Statistics check: this is the 700th post…phew.)

It’s time to take another look at the problems with the Schwarzschild metric at the event horizon.

The Schwarzschild metric (which we’ll call the S metric) is:

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (1)$

and it’s clear that problems arise when ${r=2GM}$, which defines the event horizon. Let’s take a step back and look at the two dimensional surface of a sphere, which can be described by the metric

$\displaystyle ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (2)$

We can define the two angular coordinates ${\theta}$ and ${\phi}$ uniquely anywhere on the sphere except at the two poles. At that point ${\theta=0}$ so ${\sin\theta=0}$ and the ${d\phi^{2}}$ term disappears, no matter what ${\phi}$ is, so we can’t assign a meaningful value to ${\phi}$ when ${\theta=0}$, yet there is nothing special about the sphere at that point; the problem lies entirely with the coordinates we’re using.

It looks as though the time coordinate ${t}$ suffers the same fate in the S metric. When ${r=2GM}$, the ${dt^{2}}$ term vanishes, so at this point we can’t assign a meaningful value to ${t}$. As we showed in the earlier post, however, there doesn’t appear to be anything special about the value ${r=2GM}$, since the distance ${\Delta s}$ and proper time interval ${\Delta\tau}$ are finite as we cross the event horizon. It therefore appears that the problem lies with the coordinates.

Another way of looking at it is to recall the relation between the S time coordinate ${t}$ and the proper time ${\tau}$:

$\displaystyle d\tau=\sqrt{1-\frac{2GM}{r}}dt$

The two times are equal at infinity, and for a given proper time interval, the corresponding interval ${dt}$ gets larger and larger the closer we get to the event horizon, eventually becoming infinite at the horizon. Thus again we see that we can’t assign a unique value to ${dt}$ at the event horizon.

Moore (in his Chapter 14) also states that a particle that tries to be at rest at ${r=2GM}$ must have a light-like worldline (that is, ${ds^{2}=0}$), since if the particle is at rest, ${dr=d\theta=d\phi=0}$. However, if ${r=2GM}$, the metric component ${g_{rr}=1/0}$, so we’re dividing zero by zero, which is indeterminate, so I wouldn’t put too much weight on that argument.

In any event, as we saw earlier, when we cross the event horizon, the ${t}$ and ${r}$ coordinates swap their meanings, with the result that a particle must continuously move inward (in the direction of decreasing ${r}$) as time advances. In other words, it’s impossible for a particle to remain at rest when it crosses the event horizon.

It’s worth pointing out that we can still define a surface of constant ${r}$ and ${t}$ within the event horizon by setting ${dr=dt=0}$. This gives us the metric for the surface of a sphere in 2, so ${r}$ still does define a radial distance in the sense that we can take the circumference of a shell within the event horizon and divide by ${2\pi}$ to get ${r}$. The key point is that a particle cannot be at rest on such a surface since the ${r}$ coordinate is a timelike coordinate.

We therefore would like to find a coordinate system in which we can define a meaningful time at every point. However, merely changing coordinates cannot, of course, change any of the physical predictions that we got from the S metric. In particular, it must still be true that once we cross the event horizon, we are forced to keep travelling inwards. We would just like this to happen without having a singularity at ${r=2GM}$, and without the ${r}$ and ${t}$ coordinates swapping their meaning.

There are several alternative coordinate systems that achieve this, but we’ll have a look at only one of them in this post: the Painlevé-Gullstrand system (sometimes called the global rain coordinate system; we’ll refer to it as the PG system). The PG system introduces a new time coordinate ${\mathring{t}}$ (that’s a ${t}$ with a little circle on top to represent a raindrop), while keeping the ${r}$, ${\theta}$ and ${\phi}$ coordinates from the S metric. To avoid the problem of a time singularity at the event horizon, we use the following scheme. Along each radial line, we drop (from rest) a sequence of clocks in from the infinity towards the black hole. Since the clocks start from infinity, their proper time is equal to the S time when they are dropped. To find the PG time ${\mathring{t}}$ of some event at a given spatial location (that is, with given values of ${r}$, ${\theta}$ and ${\phi}$) we find the clock that is falling down that particular radial line and passes the location ${r}$ at the same time as the event occurs.

To work out the PG metric, we need to find ${\mathring{t}}$ in terms of the S coordinates so we can do the transformation. From symmetry, all directions in space are equivalent, so ${\mathring{t}}$ cannot depend on ${\theta}$ or ${\phi}$, so it must be a function of ${t}$ and ${r}$ only: ${\mathring{t}=\mathring{t}\left(t,r\right)}$. We therefore seek the derivatives in the equation

$\displaystyle d\mathring{t}=\frac{\partial\mathring{t}}{\partial t}dt+\frac{\partial\mathring{t}}{\partial r}dr$

The idea is that if we can determine the two partial derivatives, we can solve this equation for ${dt}$ in terms of ${d\mathring{t}}$ and ${dr}$ and substitute the result into the S metric to get the PG metric.

Let’s consider the derivative ${\partial\mathring{t}/\partial t}$ first. This term occurs when ${dr=0}$, that is, when we look at two events that occur at the same spatial location but are separated by the S time interval ${dt}$. In terms of the PG system, ${\mathring{t}}$ is measured as the proper time on a clock falling in on a given radial line. The only way we can get two such clocks passing the same point but separated by a time ${dt}$ is if they were dropped at slightly different times from the same point. Since the paths followed by the two clocks are identical in space (and we’re assuming that the system is static, so the black hole isn’t moving), they must take the same time to traverse these paths. In the S system, the time at which clock ${i}$ (${i=1,2)}$ arrives at the given point is

$\displaystyle t_{i}=t_{i;0}+\Delta t$

where ${t_{i;0}}$ is the time at which clock ${i}$ was dropped, and ${\Delta t}$ is the time taken (as measured in the S system) to travel to the given point from the point at which it was dropped, and this time is the same for both clocks. Therefore

$\displaystyle dt=t_{2}-t_{1}=t_{2;0}-t_{1;0} \ \ \ \ \ (3)$

That is, the time interval between the events of the clocks being dropped and between the events of the clocks’ arrival at the point where the two events occur (as measured in the S system ) are the same.

Now consider the PG system. Here, the time interval between the dropping of the clocks is the same as in the S system, since the clocks are synchronized with the S system when they are dropped. In the PG system, however, the time ${\mathring{t}}$ is the proper time registered by the clocks as they fall. However, since the two paths are identical, both the proper times taken by the two clocks must be same as well. Thus we have

$\displaystyle \mathring{t}_{i}=t_{i;0}+\Delta\tau=t_{i;0}+\Delta\mathring{t}$

so the increment in PG time between the two events is

$\displaystyle d\mathring{t}=\mathring{t}_{2}-\mathring{t}_{1}=t_{2;0}-t_{1;0}=dt$

That is:

$\displaystyle \frac{\partial\mathring{t}}{\partial t}=1$

At first glance, we might think that this implies ${\mathring{t}=t}$, and that the PG system is identical to the S system. However, that would be true only if ${\partial\mathring{t}/\partial r=0}$ which, as we’ll now see, isn’t true. To show this, we return to the equations of motion for a particle in S coordinates. For a clock dropped from rest at infinity, the energy per unit mass is ${e=1}$ and the angular momentum is ${\ell=0}$, so we get

$\displaystyle \frac{dr}{d\tau}=-\sqrt{\frac{2GM}{r}} \ \ \ \ \ (4)$

where the minus sign indicates that we’re moving inwards (decreasing ${r}$). For the S time, we get

$\displaystyle \frac{dt}{d\tau}=\left(1-\frac{2GM}{r}\right)^{-1}$

so

$\displaystyle \frac{dr}{dt}=-\sqrt{\frac{2GM}{r}}\left(1-\frac{2GM}{r}\right) \ \ \ \ \ (5)$

To work out ${\partial\mathring{t}/\partial r}$ we need to consider two events that occur at the same ${S}$ time (${dt=0}$) but at slightly different radii. The clock that arrives at the smaller radius must have been dropped slightly earlier so that it has the extra time to fall the extra distance. However, the PG time (= proper time) to fall the extra distance is not equal to the S time to fall that same distance, so there will be two contributions to ${\partial\mathring{t}/\partial r}$: one because one clock was dropped earlier than the other, and a second due to the fact that one clock falls further than the other.

First, how much extra S time is required to fall the extra distance ${dr}$? From 5, this depends on the radius, so

$\displaystyle dt=-\left[\sqrt{\frac{2GM}{r}}\left(1-\frac{2GM}{r}\right)\right]^{-1}dr \ \ \ \ \ (6)$

That is, the clock that ends up at the lower radius must be dropped a time ${dt}$ (in S time) before the other clock. Since both time systems are synchronized when the clocks are dropped, this also represents the difference in PG time due to the different dropping times of the two clocks. In order to get the sign right, we’re defining ${dt}$ to be as in 3 above, that is, the time at which the second clock is dropped minus that at which the first clock is dropped.

Now suppose we dropped the two clocks at the same time, but let one fall a distance ${dr}$ further than the other. We need to calculate the difference in proper (that is, PG) time for these two trajectories. According to 4, this is

$\displaystyle d\tau=\sqrt{\frac{r}{2GM}}dr$

Note that we’ve dropped the minus sign, since we’re finding the difference in proper time required for the second clock minus that for the first clock. The second clock falls a shorter distance, so it takes less time, so ${d\tau<0}$ here (remember that ${dr<0}$).

The total difference in PG time ${d\mathring{t}}$ is the sum of these two effects. That is:

 $\displaystyle d\mathring{t}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{r}{2GM}}\left[1-\left(1-\frac{2GM}{r}\right)^{-1}\right]dr$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{r}{2GM}}\frac{2GM}{r-2GM}dr$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}dr$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}\left|dr\right|$

(As an aside at this point, note that if we apply the same logic to the S system, then the time difference at the dropping of the two clocks is given by 6, while the difference in the times required to fall the two different distances is just the negative of 6, giving a net ${dt=0}$, which is what we’re assuming here.)

Thus, since ${\partial\mathring{t}/\partial r}$ is the rate of change of ${\mathring{t}}$ with respect to increasing ${r}$, we have

$\displaystyle \frac{\partial\mathring{t}}{\partial r}=\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}$

and the differential is

 $\displaystyle d\mathring{t}$ $\displaystyle =$ $\displaystyle dt+\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}dr$ $\displaystyle dt$ $\displaystyle =$ $\displaystyle d\mathring{t}-\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}dr$

We can substitute this back into the S metric 1 and get

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}\right)\left(d\mathring{t}-\sqrt{\frac{2GM}{r}}\frac{1}{1-2GM/r}dr\right)^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}\right)d\mathring{t}^{2}+2\sqrt{\frac{2GM}{r}}d\mathring{t}\; dr+\left(1-\frac{2GM}{r}\right)\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}\right)d\mathring{t}^{2}+2\sqrt{\frac{2GM}{r}}d\mathring{t}\; dr+dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2}$

The last equation gives the PG metric, which we can see is not diagonal, due to the ${d\mathring{t}\; dr}$ term.

## Duality transformation for magnetic and electric charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 7.60.

Continuing our excursion in the fantasy world where magnetic monopoles exist, we can explore the duality transformations, which describe a rotation in an E-B space, as follows:

 $\displaystyle \mathbf{E}^{\prime}$ $\displaystyle =$ $\displaystyle \mathbf{E}\cos\alpha+c\mathbf{B}\sin\alpha$ $\displaystyle c\mathbf{B}^{\prime}$ $\displaystyle =$ $\displaystyle c\mathbf{B}\cos\alpha-\mathbf{E}\sin\alpha$ $\displaystyle cq_{e}^{\prime}$ $\displaystyle =$ $\displaystyle cq_{e}\cos\alpha+q_{m}\sin\alpha$ $\displaystyle q_{m}^{\prime}$ $\displaystyle =$ $\displaystyle q_{m}\cos\alpha-cq_{e}\sin\alpha$

where ${\alpha}$ is an arbitrary angle and ${c=1/\sqrt{\mu_{0}\epsilon_{0}}}$. Currents and charge densities transform the same way as the corresponding point charges.

It’s a straightforward, though tedious, exercise to verify that Maxwell’s equations (including magnetic charge) are invariant under these transformations. The two equations that are different are:

 $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu_{0}\rho_{m}$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J}_{m}-\frac{\partial\mathbf{B}}{\partial t}$

A few sample calculations should show how it goes. We’ll do one involving a divergence:

 $\displaystyle c\nabla\cdot\mathbf{B}^{\prime}$ $\displaystyle =$ $\displaystyle c\nabla\cdot\mathbf{B}\cos\alpha-\nabla\cdot\mathbf{E}\sin\alpha$ $\displaystyle$ $\displaystyle =$ $\displaystyle c\mu_{0}\rho_{m}\cos\alpha-\frac{\rho_{e}}{\epsilon_{0}}\sin\alpha$ $\displaystyle$ $\displaystyle =$ $\displaystyle c\mu_{0}\left(\rho_{m}\cos\alpha-c\rho_{e}\sin\alpha\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c\mu_{0}\rho_{m}^{\prime}$

And another involving a curl:

 $\displaystyle \nabla\times\mathbf{E}^{\prime}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{E}\cos\alpha+c\nabla\times\mathbf{B}\sin\alpha$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\mu_{0}\mathbf{J}_{m}-\frac{\partial\mathbf{B}}{\partial t}\right)\cos\alpha+c\left(\mu_{0}\mathbf{J}_{e}+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}\right)\sin\alpha$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu_{0}\left(\mathbf{J}_{m}\cos\alpha-c\mathbf{J}_{e}\sin\alpha\right)-\frac{\partial\mathbf{B}}{\partial t}\cos\alpha+\frac{1}{c}\frac{\partial\mathbf{E}}{\partial t}\sin\alpha$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J}_{m}^{\prime}-\frac{\partial\mathbf{B}^{\prime}}{\partial t}$

The other two equations transform similarly.

If we take the force law for electric and magnetic charges to be

$\displaystyle \mathbf{F}=q_{e}\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)+q_{m}\left(\mathbf{B}-\frac{1}{c^{2}}\mathbf{v}\times\mathbf{E}\right)$

then the transformed law is

 $\displaystyle \mathbf{F}^{\prime}$ $\displaystyle =$ $\displaystyle q_{e}^{\prime}\left(\mathbf{E}^{\prime}+\mathbf{v}\times\mathbf{B}^{\prime}\right)+q_{m}^{\prime}\left(\mathbf{B}^{\prime}-\frac{1}{c^{2}}\mathbf{v}\times\mathbf{E}^{\prime}\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(q_{e}\cos\alpha+\frac{q_{m}}{c}\sin\alpha\right)\times$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\mathbf{E}\cos\alpha+c\mathbf{B}\sin\alpha+\mathbf{v}\times\left(\mathbf{B}\cos\alpha-\frac{1}{c}\mathbf{E}\sin\alpha\right)\right]+$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(q_{m}\cos\alpha-cq_{e}\sin\alpha\right)\times$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\mathbf{B}\cos\alpha-\frac{1}{c}\mathbf{E}\sin\alpha-\frac{1}{c^{2}}\mathbf{v}\times\left(\mathbf{E}\cos\alpha+c\mathbf{B}\sin\alpha\right)\right]$ $\displaystyle$ $\displaystyle =$ $\displaystyle q_{e}\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)+q_{m}\left(\mathbf{B}-\frac{1}{c^{2}}\mathbf{v}\times\mathbf{E}\right)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{F}$

To get the penultimate line it is just a matter of multiplying out the first expression and using ${\cos^{2}\alpha+\sin^{2}\alpha=1}$, and cancelling off terms.