Does Coulomb’s law change at relativistic speeds?

[Bastion]

TL;DR Summary

Does Coulomb’s law change at relativistic speeds

Would the force (as predicted by Coulomb’s law) exerted on two charged particles q1 and q2, separated by distance r, and at rest relative to frame S change, if q1 and q2, began moving relative to S but we’re still at rest relative to each other?

 

[Ibix]

Yes. In general you will also need to factor in magnetic forces from the moving charges, and force (the three-force, anyway) is not frame invariant in relativity anyway.

 

[PeroK]

TL;DR Summary: Does Coulomb’s law change at relativistic speeds

Would the force (as predicted by Coulomb’s law) exerted on two charged particles q1 and q2, separated by distance r, and at rest relative to frame S change, if q1 and q2, began moving relative to S but we’re still at rest relative to each other?

Yes, but also because of quantum mechanics. Using Coulomb's law results in Rutherford scattering, which is a low-energy approximation of the full QM scattering formula, where the mass of the particle is also a factor. See, for example, Bhabha scattering of the electron-positron interaction:

http://www.phys.ufl.edu/~korytov/phz6355/note_A08_TheoristTalking_3.pdf

 

[Bastion]

What I really want to know is would the force on q1 and q2 increase as the distance r decreases due to length contraction, as predicted by special relativity.

 

[Ibix]

I would characterise the answer to this as "it's not that simple". Are you holding the charges at a fixed separation or are they free to move? A fixed separation in what frame? Are they undergoing proper acceleration or are we? What direction is the motion compared to the axis between the charges? Do you care about an answer during the acceleration (if the particles are the ones accelerating) or are you happy with an answer after the acceleration has stopped? All these things affect a detailed answer.

 

[Bastion]

I would characterise the answer to this as "it's not that simple". Are you holding the charges at a fixed separation or are they free to move? A fixed separation in what frame? Are they undergoing proper acceleration or are we? What direction is the motion compared to the axis between the charges? Do you care about an answer during the acceleration (if the particles are the ones accelerating) or are you happy with an answer after the acceleration has stopped? All these things affect a detailed answer.

Suppose q1 and q2 are opposite each on a one dimensional line x, seperated by distance r. Suppose that q1 and q2 are initially at rest relative to x, then Coulomb’s law predicts that q1 and q2 should experience a force related to the magnitude of distance r, and that force would be the same as measured using x as the reference frame or using q1 or q2 as the reference frame. Suppose that q1 and q2 begin moving relative to x, but that q1 and q2 remain stationary relative to each other I.e distance r remains fixed relative to q1 and q2. Special relativity predicts that distance r should decrease relative to x due to length contraction, but r would not decrease relative to q1 and q2. Does that mean that when q1 and q2 start moving relative to x, that relative to x, the force on q1 and q2 increases while relative to q1 and q2 the force does not increase? By all means discuss the accelerated phase, but wondering if the force acting on q1 and q2 is frame dependant is the essence of my interest. Any observations would be appreciated.

 

[jbriggs444]

What I really want to know is would the force on q1 and q2 increase as the distance r decreases due to length contraction, as predicted by special relativity.

So you are considering $q_1$ and $q_2$ as two charges momentarily at rest with respect to one another.

In their mutual rest frame, they are subject to a purely vanilla Coulomb force and accelerate under this force in the expected way.

Now we shift to a reference frame in which both charges are moving (for the moment) in lock step. The distance between the charges is reduced. The coulomb force will be increased. But the effect of that force on the coordinate acceleration of the two charges will be reduced.

Force is not a relativistic invariant.

This will not be the only relativistic effect. For instance, the relativity of simultaneity kicks in as soon as we try to identify "the moment" when the two charges are moving at the same speed. As @Ibix points out, it is complicated. But at the end of the day, the two reference frames will predict the same results. They may explain those results in different ways, but the results will be identical.

 

[hilbert2]

If you want to see how complicated electromagnetism can get within classical mechanics, google "relativistic magnetohydrodynamics".

 

[Ibix]

Does that mean that when q1 and q2 start moving relative to x, that relative to x, the force on q1 and q2 increases while relative to q1 and q2 the force does not increase?

As you aren't interested in the acceleration phase we can ignore it and just consider the "before" and "after" inertial phases. That way we don't have to mess with Lienard-Wiechert potentials and can use Coulomb's law and the Lorentz transforms.

Working in the rest frame of the charges we place $q_1$ at the origin and $q_2$ at $x=R$ (not $r$ because I'm going to use that symbol for the generic distance). The electric field due to $q_1$ is then $(x,y,z)^T(E/r^3)$ where $r^2=x^2+y^2+z^2$ and $E=k_eq_1$ is equal to the electric field strength at $r=1$. Work in units where $c=1$ and plug this into the Faraday tensor:$$\begin{eqnarray*}
F&=&
\left(\begin{array}{cccc}
0&-E_x&-E_y&-E_z\\
E_x&0&-B_z&B_y\\
E_y&B_z&0&-B_x\\
E_z&-B_y&B_x&0
\end{array}\right)\\
&=&
\left(\begin{array}{cccc}
0&-Ex/r^3&-Ey/r^3&-Ez/r^3\\
Ex/r^3&0&0&0\\
Ey/r^3&0&0&0\\
Ez/r^3&0&0&0
\end{array}\right)
\end{eqnarray*}$$Now we can write the Lorentz transforms as matrices:$$\Lambda=
\left(\begin{array}{cccc}
\gamma&-\gamma v&0&0\\
-\gamma v&\gamma&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}\right)$$where $\gamma=1/\sqrt{1-v^2}$. Now we can easily write the electromagnetic fields in a frame where the charges are moving by transforming the Faraday tensor: $$\begin{eqnarray*}
F'&=&\Lambda.F.\Lambda^T\\
&=&\left(\begin{array}{cccc}
0&-Ex/r^3&-\gamma Ey/r^3&-\gamma Ez/r^3\\
Ex/r^3&0&\gamma v Ey/r^3&\gamma vEz/r^3\\
\gamma Ey/r^3&\gamma v Ey/r^3&0&0\\
\gamma Ez/r^3&\gamma v Ez/r^3&0&0
\end{array}\right)
\end{eqnarray*}$$Note that I've only transformed the field components and not the coordinates. Formally I would need to also transform coordinates and there are risks in not doing so, but in this case it doesn't matter because I'm not going to do anything except transform $x=R$ and $y=z=0$ and plug that in, and there are no issues of simultaneity because the field components don't change at $q_2$'s location. So I can just plug in $x=R$ and $y=z=0$ and read off that the transformed electric field is the same as the untransformed one.

So in this case there is no change. There will be changes if the line joining the charges is not parallel to the direction of motion.

 

[vanhees71]

Of course you must not forget to also transform the spacetime arguments of the fields. The correct transformation is
$$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\hat{\Lambda}^{-1} x').$$
It's of course much simpler to formulate the Coulomb field covariantly. In the rest frame $\Sigma'$ of the charge the four-potential is
$$(A^{\prime \mu}(x'))=\begin{pmatrix} \Phi(r') \\ 0 \\ 0 \\0 \end{pmatrix}, \quad \Phi(r')=\frac{q}{4 \pi |\vec{x}^{\prime 2}|}.$$
Introducing the four-velocity $(u^{\prime \mu})=(1,0,0,0)$ of the charge this can obviously be written as
$$A^{\prime \mu}=\Phi(r') u^{\prime \mu}.$$
Further you have
$$r'=\sqrt{(u' \cdot x')^2 - x' \cdot x'}.$$
So for the frame, where the charge moves with four-velocity $u^{\mu}=\gamma (1,\vec{v})$ you have
$$(A^{\mu}) = \Phi[\sqrt{(u \cdot x)^2 - x \cdot x}]u^{\mu},$$
and you get the electromagnetic field by evaluating
$$\vec{E}=-\partial_t \vec{A} - \vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$

 

[Ibix]

Of course you must not forget to also transform the spacetime arguments of the fields.

But, given that I only want to know the field at $x=R$, what's the point? Knowing $F'(x,y,z,t)$ is actually more convenient than knowing $F'(x',y',z',t')$ in this particular case.

you get the electromagnetic field by evaluating
$$\vec{E}=-\partial_t \vec{A} - \vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$

I'll check my answer with this method later (unless you want to do it).

 

[vanhees71]

You want the field as it appears for the observer in one given reference frame, and this observer describes the field as a function of his time and space coordinates.

 

[PeroK]

Suppose q1 and q2 are opposite each on a one dimensional line x, seperated by distance r. Suppose that q1 and q2 are initially at rest relative to x, then Coulomb’s law predicts that q1 and q2 should experience a force related to the magnitude of distance r, and that force would be the same as measured using x as the reference frame or using q1 or q2 as the reference frame. Suppose that q1 and q2 begin moving relative to x, but that q1 and q2 remain stationary relative to each other I.e distance r remains fixed relative to q1 and q2. Special relativity predicts that distance r should decrease relative to x due to length contraction, but r would not decrease relative to q1 and q2. Does that mean that when q1 and q2 start moving relative to x, that relative to x, the force on q1 and q2 increases while relative to q1 and q2 the force does not increase? By all means discuss the accelerated phase, but wondering if the force acting on q1 and q2 is frame dependant is the essence of my interest. Any observations would be appreciated.

To keep things simple, let's consider motion in one-dimension.

Let's define the force (technically the three force) in a given reference frame by$$F = \frac{dp}{dt}$$Where $p$ is the relativistic momentum of the particle, as measured in that frame. Note that:
$$p = \gamma mv = \frac {mv}{\sqrt{ 1 - \frac{v^2}{c^2}}}$$Where $v$ is the instantaneous velocity of the particle, as measured in that frame.

We find that $F$ is the same in all inertial reference frames.

If we have the two charges at rest a distance $r$ apart in one frame, then we have the same force between the particles in a frame where the particles are moving.

How does this relate to Coulomb's law? In the rest frame of the particles we have:
$$F = \frac{kq_1q_2}{r^2}$$In the frame where the particles are moving, we have the same force, but $r = \gamma r'$, so: $$F' = F = \frac{kq_1q_2}{\gamma^2r'^2}$$In this frame, we have a relativistic version of Coulomb's Law, with an extra $\gamma^2$ term.

Note that this is only for the special case, where the particles are moving inertially in the direction of their separation. The general case where the particles are moving in a general direction is more complicated

Coulomb's law as it is usually written, therefore, applies only where the particles are travelling at non-relativistic speeds (and, even then, it is an approximation of the full relativistic case).

 

[Ibix]

So I've now done this three different ways: (a) by transforming the Faraday tensor but not the coordinates and subbing in $x=R$ (my original approach); (b) by transforming the Faraday tensor and the coordinates and subbing in $x'=R'+vt'$; (c) by transforming the 4-potential, deriving $\vec{E}$, and subbing in $x'=R'+vt'$. All three give the same answer - if the charges are separated by $R$ in their rest frame, and this separation is parallel to the direction of motion of the other frame, then the electric field strength at each particle (and hence the force on each) is the same in both frames. In the rest frame the field at $q_2$ is of course $q_1/(4\pi R^2)$. In the moving frame it's $q_1/(4\pi\gamma^2R'^2)$, but since $R'=R/\gamma$ this gives $q_1/(4\pi R^2)$ again.

That is a special case for the separation of the particles being parallel to the direction of motion. In other orientations, $\vec E$ changes and $\vec B\neq 0$.

I can share the maths if there's any interest (I'm on my phone, so not going to typeset it now).

 

[Meir Achuz]

The fields of a moving charge are
$${\bf E}= \frac{q{\bf r}}{\gamma^2[r^2-({\bf v\times r})^2]^{\frac{3}{2}}}$$
$${\bf B}={\bf v\times E}.$$
Then use the Lorentz force on the second moving charge.

 

[Meir Achuz]

A problem with the previous derivation is that, once the charges accelerate because of the force between them, the Lienard-Wiechert electric and magnetic field equations must be used.

 

[vanhees71]

Let's do the calculation directly for the fields. Let $\Sigma'$ be the rest frame of the point charge sitting in the origin then (in Heaviside-Lorentz units)
$$\vec{E}'(x')=\frac{q}{4 \pi |r'|^3} \vec{r}', \quad \vec{B}'(x')=0.$$
Let $\Sigma'$ move with velocity $\vec{v}=\beta \vec{c} \vec{e}_1$. Then the corresponding Boost reads
$$\vec{E}=\begin{pmatrix} E_1' \\ \gamma E_2' \\ \gamma E_3' \end{pmatrix}, \quad \vec{B}=\vec{\beta} \times \vec{E}.$$
We also have to express the $x'$ coordinates in terms of the $x$ coordinates, which is the usual Lorentz transformation
$$c t'=\gamma (c t-\beta x^1), \quad \vec{x}'=\begin{pmatrix}\gamma (x^1-v t) \\ x^2 \\ x^3 \end{pmatrix},$$
from which one gets
$$
\begin{split}
\vec{E}&=\frac{q \gamma}{4 \pi \sqrt{\gamma^2 (x^1-vt)^2+(x^2)^2 + (x^3)^2}^3} \begin{pmatrix}x^1-v t \\ x^2 \\ x^3 \end{pmatrix}, \\
\vec{B}&=\frac{q \gamma \beta}{4 \pi \sqrt{\gamma^2 (x^1-vt)^2+(x^2)^2 + (x^3)^2}^3} \begin{pmatrix}0 \\ -x^3 \\ x^2 \end{pmatrix}.
\end{split}
$$
Tha's indeed shorter than my other suggestion above in this thread ;-).

 

[Ibix]

A problem with the previous derivation is that, once the charges accelerate because of the force between them, the Lienard-Wiechert electric and magnetic field equations must be used.

That's why I asked about if the OP was interested in the acceleration period. However, I should have stated that it was only valid within the future lightcone of the end of the acceleration phase of the source charge.