# Waveform produced by a collapsing magnetic field

Gold Member
A thought experiment:

A electron is moving in a straight line at velocity v. It instantly stops dead. It doesn't move another femtometer.

Obviously its magnetic field collapses and produces light. What is the waveform of the light produced?

Is it something like this: https://en.wikipedia.org/wiki/Dirac_delta_function

But then again it will take time for the magnetic field to collapse. Hmm.

Mentor
2021 Award
Just plug it into the Lienard Wiechert field equations.

Homework Helper
Gold Member
Your setup implies infinite acceleration (actually deceleration) of the electron. All accelerated charged particles emit EM waves and such is the case with this electron. However because the acceleration in this setup is infinite, the frequency and amplitude of the produced EM wave will be infinite as well.

em3ry
Gold Member
2021 Award
The acceleration is some ##\delta## distribution in time. I'm not sure, whether this extreme simplification of "stopping" will produce a finite result for the bremsstrahlung em. field. As @Dale said, just try it out plugging it into the Lienart-Wiechert (retarded) potentials.

Do we need to do anything so complex as the LW potentials for this? Surely the infinite acceleration makes this easy: write down the EM field of a charge moving at constant velocity and write down the EM field of a stationary charge. The former applies everywhere outside the future lightcone of the speed change event. The latter applies inside. The field on the future lightcone isn't well defined because of the discontinuous velocity. Or am I oversimplifying?

Dale and Nugatory
Gold Member
2021 Award
I think so, because there must be a radiation field on top of the (boosted) Coulomb field because of the acceleration, but I'm not sure about the em. response to a ##\delta##-distribution singularity in the acceleration of the charge.

I think so, because there must be a radiation field on top of the (boosted) Coulomb field because of the acceleration, but I'm not sure about the em. response to a ##\delta##-distribution singularity in the acceleration of the charge.
Wikipedia says that the covariant form of the Lienard-Wiechert potentials is$$A^\mu(X^\mu)= -\frac{\mu_0qc}{4\pi}\left(\frac{U^\mu}{R_\mu U^\mu}\right)_{t_r}$$where ##A^\mu## is the potential at an event ##X##, ##U^\mu## is the four velocity of the charge, and ##R^\mu## is a vector from the source to ##X##, and the source position and velocity is evaluated at the retarded time ##t_r##.

I assume that the two ##\mu## in the denominator should be ##\nu## (or something) and the left hand side shouldn't really have an index on the ##X##.

It's easiest to work in Einstein coordinates with their origin at the deceleration event. Then ##R^\nu=X^\nu-(R_s)^\nu##, where ##R_s## is the position of the source at the retarded time. Since the charge is stationary at the spatial origin for all ##t>0## we find that for all events with positive ##t_r## (i.e. all events inside the future lightcone), ##R_\nu U^\nu=R_t U^t=(t-t_r)=r##, where ##r## is the distance of event ##X## from the spatial origin. So everywhere inside the lightcone you have ##\phi\propto 1/r## and ##\vec A=0##.

You can make the same argument outside the lightcone by working in the initial rest frame of the source, or simply observe that causality requires that the field outside the lightcone be the Lorentz boosted Coulomb field.

So I don't think I am oversimplifying - unless there's more wrong with the Wiki article than dodgy indexes. I think the radiation field, as far as one can define it in this case, lies on the future lightcone where the behaviour isn't well defined. I would imagine that to consider this formally, one should consider a finite period of proper acceleration ##\alpha## linking the two periods of inertial motion and take the limit as ##\alpha\rightarrow\infty## while keeping the velocity change constant.

Dale, em3ry, etotheipi and 1 other person
Gold Member
2021 Award
The formula in the Wiki is correct, and I also thin your arguments.

But where is the radiation field, which must be there, right? There must be some singular contribution along the light cone. Usually such apparently "simple idealizations" turn out to be much more complicated. A famous example is the radiation from a charge in (eternal) hyperbolic motion ;-)).

em3ry and Ibix
Gold Member
Do we need to do anything so complex as the LW potentials for this? Surely the infinite acceleration makes this easy: write down the EM field of a charge moving at constant velocity and write down the EM field of a stationary charge. The former applies everywhere outside the future lightcone of the speed change event. The latter applies inside. The field on the future lightcone isn't well defined because of the discontinuous velocity. Or am I oversimplifying?
It looks to me that during the period of deceleration we are doing work on the electron, as we are decelerating it against its own field. This energy appears as radiation. The electric field before and after deceleration is radial; only during deceleration does it have a tangential component, so constituting radiation. I think the radiated energy will have a spectrum corresponding to the Fourier analysis of the deceleration curve. Clearly we cannot supply the necessary energy in zero time as it would correspond to infinite power.

em3ry
Staff Emeritus
2021 Award
I'm kind of puzzled by this. If I have infinities in the problem description - and the electron has an infinite force applied to it for zero time - why is it surprising that there infinities in the solution?

vanhees71 and em3ry
Mentor
2021 Award
The field on the future lightcone isn't well defined because of the discontinuous velocity.
The LW fields will give that also, plus they are good to know since they answer all similar questions too.

But I agree that your simple method works everywhere else

vanhees71 and em3ry
Gold Member
But where is the radiation field, which must be there, right? There must be some singular contribution along the light cone. Usually such apparently "simple idealizations" turn out to be much more complicated. A famous example is the radiation from a charge in (eternal) hyperbolic motion ;-)).

Below shows an image of the electric field with a kink in it as would be expected after a sudden deceleration. Clearly the curl at A and B is nonzero but the curl at C is zero. The curl of A is of the opposite sign of the curl at B.

According to the Feynman link I just posted above, this kink which is due to the acceleration of the charge, falls off as 1/x (think conservation of energy) while the electric field itself falls off as 1/x^2. When the electric field has fallen off so much as to be negligible this kink will still be there but will consist of electric (and magnetic) fields going around in circles (curl > 0). In other words it will be light.

Edit: This is actually the electric field of a stationary particle that suddenly moves and then stops again immediately but its enough to illustrate my point about curl.

Last edited:
Gold Member
I notice this diagram portrays radiation field components pointing only to the right. Of course, we cannot have EM radiation with a DC component. So other accelerations must have occurred, or will occur, so that the radiated waveform is balanced. The timing of these events (the initial switch-on etc), seems to dictate the lowest frequency component.

Gold Member
You don't seem to understand. The curl is the radiation. The curl is there but doesn't look like curl (doesnt look like field lines going around in circles).

The circular field lines only become visible after the electric field due to the charge becomes negligible

Last edited:
Below shows an image of the electric field with a kink in it as would be expected after a sudden deceleration.
That isn't a sudden deceleration in the sense you were using it earlier. Rather, it's the result of a short period of deceleration. Furthermore, it's a snapshot of a small part of the field very far from the source (which is why the field lines are nearly parallel).

I think that answers @tech99's objection - the overall picture is one of radial field lines with a kink at radius ##c\Delta t##, where ##\Delta t## is the time since the deceleration and ##c\Delta t## is much greater than the width of the diagram. The above picture is an exceedingly zoomed-in diagram of a small part of the field.

em3ry
Gold Member
OK so I think I've got my answer now. Thanks everyone.

To sum up: A charge experiencing a very sudden acceleration would emit very high frequency radiation.

Last edited:
Homework Helper
elow shows an image of the electric field with a kink in it as would be expected after a sudden deceleration.
Here's an interesting image for the E field which seems correct to me (acceleration instead of decceleration) Then there will also be the B field. Food for thought.

em3ry, vanhees71 and Ibix
Gold Member
Here's an interesting image for the E field which seems correct to me (acceleration instead of decceleration) Then there will also be the B field. Food for thought.
I think the B radiation field is something a stationary observer sees as the electric wave passes. I am not sure about generation of the B radiation field by the kinking of field lines.

Homework Helper
Not sure I understand your point. The "electric wave" occurs as the kinks in the E field pass and marks the edge of the B field. This is then an outgoing radiation pulse representing the frequency content of the acceleration pulse.

vanhees71
Gold Member
2021 Award
I started to calculate the field.

Of course the above solution given by @Ibix above, i.e., in the open region excluding the light cone at the origin of the Minkowski space (with the particle moving such that it arrives at ##\vec{x}=0## at time ##t=0## and is then suddenly stopped). So it's a boosted Coulomb field for a charge at rest in the entire region ##t<0## and outside the forward lightcone, i.e., for ##|x|< c t## a "simple Coulomb field of a charge at rest in the origin" in the interior of the forward lightcone, i.e., for ##|x|>c t##.

The trouble is to get what happens precisely at the light cone, where the field is singular, and this singularity represents somehow the radiation field, i.e., seen in position space it's a singular spherical wave pulse exactly on the sphere ##|\vec{x}|=c t## (##t>0##). Of course, when calculating the fields the derivatives of the involved Heaviside unitstep functions from the LW four-potential give the appropriate Dirac ##\delta## distribution "peaked" at the light cone. I don't think that it's worth the effort to really calculate it, because as I guessed before, this complication is the result of overidealizing a real "braking process" of a charged particle, which always takes a finite time over a finite distance and is smooth. So if it's done rapidly enough, i.e., when the overidealization is not too bad, you get some spherical wave packet with a smooth distribution peaked around the sphere ##|\vec{x}|=c t## (of course superimposed with the Coulomb field and boosted Coulomb field as described above).

hutchphd
Even with a finite braking time the field between the lightcones appears to be... complicated to describe.

vanhees71
Gold Member
2021 Award
Well, while the "sudden stop" idealization is at least calculable in closed form, I guess it's hard to find a more realistic motion of a point charge of this kind such that you can solve for the LW potential analytically. The LW potential only looks simple. The problem is to calculate the retarded time (or the "retarded proper" time when using the relativistic covariant formulation of particle motion).

What's easier is the Fourier decomposition and/or multipole expansion (if the particle motion is restricted to a finite region in space). See Landau Lifshitz vol. 2.

em3ry and hutchphd
It's not enormously difficult, but you end up with a nasty polynomial mess. I can post my methodology later.

etotheipi, hutchphd and vanhees71
Gold Member
2021 Award
Which model have you used?

Staff Emeritus
2021 Award
Please do! I love looking at nasty polynomial messes!

When they aren't mine, of course.

Nugatory and Ibix
Which model have you used?
Constant proper acceleration.
Please do! I love looking at nasty polynomial messes!
Enjoy!

Start in the frame ##S## where the charge starts in motion in the -x direction, decelerates at proper acceleration ##a##, and comes to rest at time ##t=0## at ##x=1/a##. Note that this means that the worldline during the acceleration is part of a hyperbola that's invariant under Lorentz boost. Pick an event ##E^\mu=(T,X,Y,Z)## that's between the future light cones of the start and end of the acceleration phase. We will calculate the EM field at this event.

Transform to the frame ##S'## where the charge is instantaneously at rest at the retarded time associated with ##E##. In this frame, ##E^\mu=(T',X',Y',Z')## and the charge is at ##(0,1/a,0,0)##. Thus we can immediately write that, in this frame, ##U^\mu=(1,0,0,0)## and ##R^\mu=(T',X'-1/a,Y',Z')##, and hence that ##A^\mu(E)=(Q/T',0,0,0)##, where ##Q=-qc\mu_0/4\pi## is the prefactor from the Lienard-Wiechert potential. Then we can transform back to ##S##, and find that ##A^\mu(E)=\left(\frac{Q}{T-vX},\frac{Qv}{T-vX},0,0\right)##. Lowering the index on ##A##, that leads to the Faraday tensor$$\begin{eqnarray*} F_{\mu\nu}&=&\partial_\nu A_\mu-\partial_\mu A_\nu\\ &=&\left(\begin{array}{cccc} 0&Q\frac{X(\partial v/\partial X)}{(vX-T)^2}&Q\frac{X(\partial v/\partial Y)}{(vX-T)^2}&Q\frac{X(\partial v/\partial Z)}{(vX-T)^2}\\ -Q\frac{X(\partial v/\partial X)}{(vX-T)^2}&0&Q\frac{T(\partial v/\partial Y)}{(vX-T)^2}&Q\frac{T(\partial v/\partial Z)}{(vX-T)^2}\\ -Q\frac{X(\partial v/\partial Y)}{(vX-T)^2}&-Q\frac{T(\partial v/\partial Y)}{(vX-T)^2}&0&0\\ -Q\frac{X(\partial v/\partial Z)}{(vX-T)^2}&-Q\frac{T(\partial v/\partial Z)}{(vX-T)^2}&0&0 \end{array}\right)\end{eqnarray*}$$You can read off the E and B fields from there. Where the nasty polynomials comes in is the calculation of ##v##. It's easy enough to deduce - you know that ##E## is null separated from ##(0,1/a,0,0)## in ##S'##, so you just require that ##\eta_{\mu\nu}R^\mu R^\nu=0## and solve for ##v##. That turns out to be$$v=\frac{4a^2TX\left(a^2\rho^2+1-a^2T^2\right)\sqrt{-4a^2X^2+a^4T^4+\left(2a^2-2a^4\rho^2\right)T^2+(a^2\rho^2+1)^2}}{a^4T^4+\left(2a^2-2a^4\rho^2\right)T ^2+(a^2\rho^2+1)^2}$$where ##\rho^2=X^2+Y^2+Z^2##. Have fun substituting that and its derivatives into the Faraday tensor!

Last edited:
vanhees71, em3ry, Vanadium 50 and 1 other person
Hm. I've just realized that $$\begin{eqnarray*} \rho^2-T^2&=&\rho'^2-T'^2\\ &=&X'^2+Y'^2+Z'^2-T'^2 \end{eqnarray*}$$from the invariance of the interval. But we know that ##E^\mu## is null separated from ##(0,1/a,0,0)## in ##S'##, so:$$\begin{eqnarray*} 0&=&(X'-1/a)^2+Y'^2+Z'^2-T'^2\\ \frac{2X}{a}-\frac{1}{a^2}&=&\rho'^2-T'^2\\ &=&\rho^2-T^2 \end{eqnarray*}$$
Substituting this into ##v##:$$v=\frac{2TX}{X^2+T^2}$$...which has no dependence on ##Y## or ##Z##. So I've done something wrong somewhere.

vanhees71 and em3ry
...no, I think I've just hidden my ##Y## and ##Z## dependence inside ##T##. So if I use ##\frac{2X}{a}-\frac{1}{a^2}=\rho^2-T^2## to eliminate ##T## from the expression for ##v## instead of using it to eliminate ##\rho##, I get:$$v=\frac{2aX\sqrt{a^2\rho^2+1-2aX}}{a^2X^2+a^2\rho^2+1-2aX}$$which isn't anywhere near so horrible. It's still going to be messy when you insert it into the components of the Faraday tensor, though.

I think that's enough thinking aloud for one day.

vanhees71 and em3ry
Last post, I promise! In case anyone wants to play around and/or criticise, here's the Maxima (free symbolic maths package) batch file I used for all of the above, with comments.
Code:
/* Deduce the field from a charge decelerating with proper acceleration a, */
/* passing through x=1/a at t=0. Uses the Lienard-Wiechert potentials.     */

/* Lorentz transform, speed v, and Minkowski metric */
g:1/sqrt(1-v^2);
L:matrix([g,-v*g,0,0],[-g*v,g,0,0],[0,0,1,0],[0,0,0,1]);
eta:matrix([1,0,0,0],[0,-1,0,0],[0,0,-1,0],[0,0,0,-1]);

/* Event E=[T,X,Y,Z] transformed to a frame where we assert that the charge */
/* is at [0,1/a,0,0] at the retarded time associated with E.                */
Ep:ratsimp(L.[T,X,Y,Z]);

/* Calculate the four vectors R and U used in Lienard-Wiechert, and hence */
/* the potential, all working in the transformed frame.                   */
Rp:Ep-[0,1/a,0,0];
Up:[1,0,0,0];
Ap:Q*Up/ratsimp(transpose(Up).eta.Rp); /*Q=-c*mu_0*q/(4*pi), from LW potential */

/* Inverse transform the potential to the original frame */
A:ratsimp(invert(L).Ap);

/* Calculate the Faraday tensor from A */
F(A):=block ( /* A is assumed to be a one-form */
(f,coords,i,j),
coords:[T,X,Y,Z],
f:matrix([0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]),
for i:1 thru 4 do block (
for j:1 thru 4 do block (
f[i,j]:ratsimp(diff(A[i,1],coords[j])-diff(A[j,1],coords[i]))
)
),
f
);
depends(v,[X,Y,Z]);
F(eta.A); /* Lower the index on A */

/* Solve for v, by noting that E is null-separated from the charge location */
/* at the retarded time, and hence Rp is null.                              */
ratsimp(transpose(Rp).eta.Rp)=0;
solve(%,sqrt(1-v^2))[1];
solve(%^2,v);
vc:rhs(%[2]);

/* Noting that T^2-X^2-Y^2-Z^2=T'^2-X'^2-Y'^2-Z'^2 and that Rp is null, can */
/* deduce that 2*X/a-1/a^2=X^2+Y^2+Z^2-T^2. Use this to simplify vc.        */
assume(a>0);
ratsimp(substitute(sqrt(rho^2-X^2-Y^2),Z,vc));
vc:ratsimp(substitute(sqrt(rho^2-2*X/a+1/a^2),T,%));

Last edited:
vanhees71, etotheipi and em3ry
Wow

To try to begin to understand your method I drew a picture, I wondered if it agrees with what you had in mind (or whether it's completely off )? The red line is the worldline of the charge, and at ##t' = 0##, its 4-velocity is parallel to the ##t'## axis, so the charge is instantaneously at rest at this time in ##S'##. Also, the vector ##R## joining this event to ##E## is null, and thus parallel to the yellow lightcones.

How does that look?

vanhees71 and Ibix
That's it. Thanks - I was planning to draw a Minkowski diagram myself, but got side-tracked with the simplification stuff.

Looking at that, I wonder if I've got the wrong sign on my solution for ##v##, since it should be negative. I'll have a look tomorrow.

vanhees71 and etotheipi