Coulomb’s Law in a Moving Reference Frame

[LarryS]

TL;DR

Inverse square laws and relativistic length contraction

In classical electromagnetism, Coulomb’s Law gives the force between two stationary charged particles, a certain distance apart. It is an inverse square law – the density of the electric field lines is spherically symmetric. What if the same two charged particles are moving in parallel, say to the right. Is the density of the field lines no longer spherically symmetric due to the relativistic length contraction?

Thanks in advance.

 

[Vanadium 50]

Your question ignores magnetism. A moving charge produces a magnetic field. You need to consider that to make any sense of what happens.

 

[PeterDonis]

If the charged particles are moving, Coulomb's Law is no longer the correct law. You have to use the full Maxwell's Equations. Even trying to think in terms of "density of electric field lines" no longer works because the field is not purely electric.

 

[LarryS]

Your question ignores magnetism. A moving charge produces a magnetic field. You need to consider that to make any sense of what happens.

I was not asking about the TOTAL force on the particles. I am only asking about the electrical component of the Lorentz Force. Does Coulomb's Law remain a simple inverse square law in light of the relativistic length contraction?

 

[PeterDonis]

the electrical component of the Lorentz Force

Which still isn't "Coulomb's Law", or even "Coulomb's Law with relativistic length contraction", because the electric field (i.e., the electric part of the electromagnetic field) isn't being determined by just one equation. It's being determined by all of Maxwell's Equations.

 

[PeterDonis]

Coulomb’s Law gives the force between two stationary charged particles, a certain distance apart.

Note also that, if these two particles are all that's present in the scenario, the scenario will be time-dependent; the particles will either accelerate away from each other (like charges) or towards each other (unlike charges). That also complicates the solution since the particles are not stationary.

 

[PeroK]

I was not asking about the TOTAL force on the particles. I am only asking about the electrical component of the Lorentz Force. Does Coulomb's Law remain a simple inverse square law in light of the relativistic length contraction?

There are many resources online that explain how the electric and magnetic fields transform between inertial reference frames. E.g.

https://en.m.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

Or, search for "transforming the electromagnetic fields in special relativity" for lots more.

 

[Ibix]

I was not asking about the TOTAL force on the particles. I am only asking about the electrical component of the Lorentz Force. Does Coulomb's Law remain a simple inverse square law in light of the relativistic length contraction?

The problem is that your question is asking two separate things (which do have answers) but conflating them. And your implied reasoning isn't right.

You can certainly ask what the electric field of a moving charge looks like. However, this is not in any sense a Coulomb field because it isn't the transform of the Coulomb field, and is neither the simple $1/r^2$ nor is it just a length contracted version of it.

You could also ask what Coulomb's Law looks like in a frame where the charge is moving. The answer is that the field has both electric and magnetic components and is axisymmetric, not spherically symmetric. Whether you call that field a Coulomb field or not is, I think, a matter if personal choice. I would prefer to call it the transform of a Coulomb field.

However, you asked about how Coulomb's Law behaves in the moving frame implying you were interested in the EM field, but also said you were only interested in the electric field, which is contradictory. I believe that you are actually interested only in the electric component of the transformed field. You are correct that this is no longer spherically symmetric, but this is not simply due to length contraction. Broadly speaking, length contraction arises from the fact that four vectors transform as $\vec v'=\Lambda \vec v$, where $\Lambda$ is the Lorentz transforms in matrix form. However, the EM field is a (1,1) tensor (usually denoted $F$) not a vector, and transforms as $F'=\Lambda F\Lambda^T$, and its dependence on coordinates must also be transformed. Then you may extract the electric field components from the tensor. That's a rather more involved process than "because of length contraction", I'm afraid.

 

[strangerep]

@LarryS: you want "Lienard-Wiechert Potential". If you google that, lots of source material is available.

 

[Sagittarius A-Star]

What if the same two charged particles are moving in parallel, say to the right. Is the density of the field lines no longer spherically symmetric due to the relativistic length contraction?

Yes, for example the electric field lines in transverse direction are more dense than in the un-primed rest frame: $E'_{y}=\gamma E_{y}$.

Say, charge $\alpha$ is at rest at $x=y=z=0$ and test-charge $\beta$ is at rest at $y=y_B, x=z=0$.
Then at the location of test-charge $\beta$, the $E$-field from $\alpha$ has only an $y$ component and there is no $B$-field.
$E_x=E_z=B_x=B_y=B_z=0\ \ \ \ \$(1)

Say, this frame moves with $-v$ in $x'$ direction in the primed frame ($ct', x', y', z'$).
Now use the relativistic transformation formulas:
$E'_{y}=\gamma \left(E_{y}-vB_{z}\right)$, $\ \ \ \ \ \ \ B'_{z}=\gamma \left(B_{z}-{\frac {v}{c^{2}}}E_{y}\right)$.

With equation (1) follows:
$E'_{y}=\gamma E_{y}$ $\ \ \ \ \ \ \ B'_{z}=\gamma \left(-{\frac {v}{c^{2}}}E_{y}\right)$.

Lorentz-Force:
$F'_y= q (E'_{y}+vB'_{z})=q (\gamma E_{y}- v\gamma{\frac {v}{c^{2}}}E_{y})=qE_{y}\gamma(1-\frac{v^2}{c^2})=F_y/ \gamma$.

Compare to the last equations in:
http://www.sciencebits.com/Transformation-Forces-Relativity

 

[PeterDonis]

at the location of test-charge $\beta$, the $E$-field from $\alpha$ has only an $y$ component and there is no $B$-field.

But you can't do a Lorentz transform just at the location of test charge $\beta$. You have to do it everywhere. You aren't doing that.

 

[hutchphd]

May I suggest the Feynman Lectures II 13, 14. Nobody does it better IMHO

 

[Sagittarius A-Star]

But you can't do a Lorentz transform just at the location of test charge $\beta$. You have to do it everywhere. You aren't doing that.

I am only interested in the fields at the location of the test charge $\beta$ to calculate the Lorentz force on this charge. The transformation formulas I used, calculate the fields in the primed frame, expressed as function of the un-primed coordinates.

 

[PeterDonis]

I am only interested in the fields at the location of the test charge to calculate the Lorentz force on this charge.

That's not what the OP is asking about. He is asking about the "density of field lines", i.e., the distribution of the field in space. He is not just asking about the field at one point.

Also, as I noted before, with just two charged particles the scenario will be time dependent.

Also, you are assuming that the direction of relative motion of the two frames lies along the line separating the charges. The OP nowhere specified that.

 

[Sagittarius A-Star]

That's not what the OP is asking about. He is asking about the "density of field lines", i.e., the distribution of the field in space. He is not just asking about the field at one point.

Yes. I gave only a specific, simple example, a subset of what the OP asked for.

Also, as I noted before, with just two charged particles the scenario will be time dependent.

I assume, that both charges have a fixed location in the un-primed frame. That implies, that they must be held at their places with forces, counteracting the Lorentz-forces.

Also, you are assuming that the direction of relative motion of the two frames lies along the line separating the charges. The OP nowhere specified that.

I assume, that the direction of relative motion of the two frames ($x$) lies perpendicular to the line separating the charges ($y$).

 

[LarryS]

Also, you are assuming that the direction of relative motion of the two frames lies along the line separating the charges. The OP nowhere specified that.

My mistake. I should have specified that. That is what I was thinking.

 

[Sagittarius A-Star]

TL;DR Summary: Inverse square laws and relativistic length contraction

Is the density of the field lines no longer spherically symmetric due to the relativistic length contraction?

Here is a good description:

We can use the same kind of argument to find the e and b fields of a point charge q moving at uniform velocity v through the lab frame S, say along the x axis.
...
Note that the instantaneous electric field is still radial [in spite of the fact that it was “caused” before the charge arrived at the origin!] and that its strength still falls off as the inverse square of the distance along any radius. But the field is now stronger at the sides than in front and back. In fact, if we use the standard field-strength representation by the number of field lines crossing unit area (as is permitted by Gauss’s outflux theorem), one can prove the interesting result that the field line pattern of the moving charge is simply the isotropic Coulomb pattern shrunk by a Lorentz factor – as if by length contraction!

Source:
http://www.scholarpedia.org/article...e.7D_.5C.29_and_.5C.28_.5Cmathbf.7Bb.7D.5C.29

 

[pervect]

TL;DR Summary: Inverse square laws and relativistic length contraction

In classical electromagnetism, Coulomb’s Law gives the force between two stationary charged particles, a certain distance apart. It is an inverse square law – the density of the electric field lines is spherically symmetric. What if the same two charged particles are moving in parallel, say to the right. Is the density of the field lines no longer spherically symmetric due to the relativistic length contraction?

Thanks in advance.

The way you are asking your question is a bit confusing, since to measure the electric field of a charge in some frame, one examines the force on a stationary charge, not the force on a moving charge.

If we interpret this question as having a stationary frame of reference, in which a single charge is moving, and we wish to measure the electric field lines of said moving charge, a graphic of the field lines would look something like this:

My favorite images for this seem to have disappeared, I'm not sure how long this particular image will be around, nor can I vouch for the quality of any text associated with this particular image.

The lengths of the field lines in the diagram are not to scale and may be confusing. Given a configuration of field lines, the rule is that the magnitude of the electric force is proportional to the density of the field lines. The diagram shows (from memory) the direction of the field lines correctly, but it doesn't show the magnitude, which is given by the above rule.

A more mathematical approach will give the magnitude numerically (rather than by the rule that it's proportional to the density of the field lines) in addition to the direciton, and one will find that the result obeys Gauss' law as expected. The transverse field of the moving chargeis greater than than the field would be if the charge were stationary, while the longitudinal field will be smaller. Gauss' law is still satisfied.

I believe several other posters have described the math, so I'm only giving a rough diagram, with some explanation, and the simple answer to the original question (as interpreted by me) which is "No, there will not be circular symmetry, the curve normal to the field lines will be an ellipse".

 

[PeterDonis]

That is what I was thinking.

Then you should be aware that many statements about this special case will not generalize.

 

[vanhees71]

You can simply express the solution for a point particle at rest in an inertial reference frame in a covariant way. In the rest frame of the particle the four-potential is given by
$$(A^{\mu})=\begin{pmatrix} \Phi(\vec{x}) \\0 \\0 \\0 \end{pmatrix},$$
where (in Heavyside-Lorentz units, which are most convenient here)
$$\Phi(\vec{x})=\frac{q}{4 \pi |\vec{x}|}.$$
To write this in a manifestly covariant way, just introduce the four-velocity of this rest frame of the charge, $u^{\mu}=(1,0,0,0)$. Then
$$A^{\mu}=\Phi u^{\mu}.$$
Now $\Phi$ is a scalar field, and all you have to do is to express it also in a manifestly covariant way. That's simple, because
$$\vec{x}^2=(x^0)^2-x \cdot x = (u \cdot x)^2-x \cdot x$$,
where
$$a \cdot b=\eta_{\mu \nu} a^{\mu} b^{\nu}=a^0 b^0-\vec{a} \cdot \vec{b}$$
is the Minkowski product between four-vectors.

So writing
$$\Phi(x)=\frac{1}{4 \pi \sqrt{(u \cdot x)^2-x \cdot x}},$$
you can now set $u=\gamma (1,\vec{\beta})$ to describe the field seen from an arbitrary inertial frame, where the charge moves with a constant velocity $\vec{v}=c \vec{\beta}$ and $\gamma=1/\sqrt{1-\vec{\beta}^2}$. Finally you get the fields in the usual way
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} A^0 = \frac{q}{4 \pi} \frac{\gamma(\vec{x}-\vec{v} t)}{\sqrt{(u \cdot x)^2-x \cdot x}}$$
and
$$\vec{B}=\vec{\nabla} \times \vec{A}=\vec{\beta} \times \vec{E}.$$
For a calculation, using the retarded potentials for a uniformly moving point charge, see Sect. 4.8 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Vanadium 50]

B=this, B=that, you guys keep saying that. Did you hear the OP? NOI MAGNETISM!!

Of course, one might argue that a moving charge creates a magnetic field, and a moving magnetic field creates an electric field. Or that different observers will disagree on what fraction of the force is electric and what is magnetic. But our instructions are "mo magnetism!"

 

[PeroK]

But our instructions are "mo magnetism!"

Isn't "mo magnetism" the opposite of "no magnetism"?

 

[Vanadium 50]

Mo better magnetism!

 

[vanhees71]

B=this, B=that, you guys keep saying that. Did you hear the OP? NOI MAGNETISM!!

Of course, one might argue that a moving charge creates a magnetic field, and a moving magnetic field creates an electric field. Or that different observers will disagree on what fraction of the force is electric and what is magnetic. But our instructions are "mo magnetism!"

At least I gave a clear answer, and of course, there's a magnetic field as soon as the charge is moving with respect to the observer. Note that, of course, the split into electric and magnetic field components is frame dependent. The covariant quantity describing the electromagnetic field is the antisymmetric Faraday tensor. The electric and magnetic components as observed by an arbitrary (inertial) observer can be covariantly defined by the four-vectors
$$E^{\mu}=F^{\mu \nu} u_{\nu}, \quad B^{\mu} =\frac{1}{2} \epsilon^{\mu \nu \rho \sigma} F_{\rho \sigma} u_{\nu},$$
where $u^{\mu}$ is the dimensionless four-velocity of the observer, $u^{\mu} = \mathrm{d} x^{\mu}/\mathrm{d} \tau /c$, such that $u_{\mu} u^{\mu}=1$.

 

[LarryS]

Thanks everyone for the very thorough answers!