# Maximum magnetic field from moving charge

1. Sep 27, 2014

### KeeperOfKeys

Looking at the fields that result from a uniformly moving electric charge, derived from the Lienard-Wiechert potentials, you find that an observer at a right angle to the direction of travel will see a magnetic field proportional to $\beta (1 - \beta^2)$ suggesting a maximum field at $\beta = 1/\sqrt 3$.
Can someone help me understand on an intuitive level why it doesn't continue to increase with velocity?

2. Sep 28, 2014

### pervect

Staff Emeritus
I haven't seen that result before, and I'm a bit suspicious of it, so the first think I'd do is verify it.

The E and B fields when combined properly form a tensor. That means you don't have to recompute the E and B fields when you change your frame of reference from a stationary to a moving one, you simply have to transform them. If you know the E and B fields at a point in one reference frame, you can find the E and B fields at the same point in another reference frame.

Wiki gives the transformation laws in http://en.wikipedia.org/w/index.php...gnetism_and_special_relativity&action=history, other sources should give equivalent laws. There's a tensor form of the law, but it may be easier to work with the non-tensor version.

Finding the perpendicular component of E and B in a frame stationary with the charge is easy, E points away from the charge, and B is zero. Applying the transform, I don't see how one would get the result you describe above.

Not being familiar with your result, my first impulse would be to look at it more closely. It's possible I've made some silly mistake,, though I think there are other reasons to suspect your result, I don't see how it could be consistent with the Biot-Savart law.

3. Sep 28, 2014

### KeeperOfKeys

Hmmm, I’m having trouble spotting my error, but it seems my calculation from the LW potential doesn’t agree with the field transformation approach. I’ll be the first to admit my knowledge of relativistic EM is shaky.
I’ll walk through my derivation and maybe you can help me catch my mistake.

My starting point is here:
http://en.wikipedia.org/wiki/Liénard–Wiechert_potential#Equations
Specifically the equation for B
Right off the bat I drop the second term since I assume velocity is constant, leaving
$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \left(\frac{q c(\boldsymbol{\beta} \times \mathbf{n})}{\gamma^2 (1-\mathbf{n} \cdot \boldsymbol{\beta})^3 |\mathbf{r} - \mathbf{r}_s|^2} \right)_{t_r}$

I am going to assume $\bf{n}$ and $\bf{\beta}$ are perpendicular giving:
$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \left(\frac{q c \beta}{\gamma^2 |\mathbf{r} - \mathbf{r}_s|^2} \right)_{t_r}$

And, just to make life easier, I can choose my radius from the direction of travel to be small so $t_r = t$

Express the $\gamma$ with $\beta$ and you get the offending result.

4. Sep 28, 2014

### Staff: Mentor

This was interesting, I had not been aware of this before, but I looked into the Lienard-Wiechert potentials in detail and found exactly what you described.

The effect is due to relativistic aberration. The maximum field increases without bound as $\beta \to 1$, but the direction where that maximum field points becomes more and more tightly collimated towards the direction of travel. So if you fix the direction then you will get a speed beyond which the maximum field has "passed" and gone to a tighter angle.

Last edited: Sep 28, 2014
5. Sep 28, 2014

### Staff: Mentor

For easy reference, the angle of the maximum field is:
$$\theta = \arccos \left( \frac{-1+\sqrt{1+24 \beta}}{4 \beta} \right)$$
which goes to 0 as $\beta \to 1$

The magnitude of the magnetic field at that angle is:
$$\sqrt{\frac{512 \left(\beta ^2-1\right)^2 \left(-4 \beta ^2+\sqrt{24 \beta ^2+1}-1\right)}{\left(\sqrt{24 \beta ^2+1}-5\right)^6}}$$

Which increases without bound as $\beta \to 1$.

Last edited: Sep 28, 2014
6. Sep 28, 2014

### pervect

Staff Emeritus
I'm not postiive yet I'm not the one making a mistake ...
Are $\bf{n}$ and $\bf{\beta}$ perpendicular at the retarded time $t_r$?

I.e suppose we use cartesian coordinates $\bf{r} =$ (x,y,z), with $\hat{x},\hat{y},\hat{z}$ being unit vectors pointing in the various spatial directions.. Let (beta) be a scalar and $\bf{\beta}$ be a vector. If we assume the motion is in the z direction, then we can write $z =$ (beta) t, and the vector value of $\bf{\beta}$ will be (beta) $\hat{z}$,

Now assuming we are trying to calculate the value of the B-field B at x=$x_0$, y=0,z=0, t=0 (did I understand the problem properly? Is this what we are trying to calculate?). Assuming this is what we're after, then $t_r$ will not be zero, it will be around but not quite $-x_0/c$. Thus $\bf{r}-\bf{rs}$ will be in the $\hat{x}$ direction at t=0 and thus perpendicular to $\beta$ which is in the $\hat{z}$ direction, but at $t=t_r$ $\bf{\beta}$ and $\bf{r}-\bf{rs}$ won't be perpendicular. Note that $\bf{n}$ points in the direction of $\bf{r}-\bf{rs}$, it's just normalized to unit length.

Last edited: Sep 28, 2014
7. Sep 28, 2014

### davek

Steve Carlip showed in his paper "Aberration and the Speed of Gravity" (http://arxiv.org/abs/gr-qc/9909087) that the aberration effect cancels out for static fields like this one. Isn't that a contradiction?

8. Sep 28, 2014

### Staff: Mentor

No. Carlip showed that at each point the electric field points towards the non-retarded position of the source. He did not even calculate which direction had the maximum magnetic field magnitude. Conversely, I showed which direction had the maximum magnetic field magnitude. I did not even calculate the direction the electric field points.

9. Sep 29, 2014

### harrylin

My knowledge of this is a bit "rusty", but it looks rather suspect to me. And you even don't need to compare it with special relativity, just Heaviside will do (and SR agrees with his results). For 90° angle I find that B is proportional to $v/\sqrt (1-\beta^2)$.

This was based on my textbook, directly from the equation for B. Compare also with for example:
https://en.wikipedia.org/wiki/Relativistic_electromagnetism#The_field_of_a_moving_point_charge
That description has an explanation for why there is no net aberration effect.
And from the equation there for E, using that B = v x E/c2 , I find again the above result.
But of course, I may also be mistaken! :p So I'll be gladly corrected too. :)

Last edited: Sep 29, 2014
10. Sep 30, 2014

### Meir Achuz

Working with the L-W fields is tricky, because of the retarded time. What is perpendicular at the retarded time, is not perpendicular now. For uniform velocity, a simpler direct equation for B is given in textbooks that results in $B=qv/(r^2\sqrt{1-v^2})$ for r perpendicular to v.
Note: I just noticed Harrylin's post. Everything he says is correct