Discrepancy in Co-Moving Charges Magnetic Force?

[greswd]

Consider two identical charges separated by some distance. Both are moving together with the same velocity, and the velocity vectors are perpendicular to a straight line drawn between the charges.

Each charge has charge q, speed v and they are separated by a distance of r.

In the rest frame of both charges, the E-field at either charge is $\frac{1}{4π ε_0}\frac{q}{r^2} = E_0$

Based on the Biot-Savart Law, the B-field at either charge is $\frac{μ_0}{4π}\frac{qv}{r^2} = B_1$

$\frac{μ_0}{4π}\frac{qv}{r^2} = [v ⋅μ_0⋅ε_0] ⋅ [\frac{1}{4πε_0}\frac{q}{r^2}]$

$B_1 = \frac{v}{c^2}⋅ E_0$


Let's switch to the relativistic transformation for E and B fields: $B_y'=γ(B_y+\frac{v}{c^2}⋅E_z)$

There is no magnetic field in the rest frame of both charges, hence $B_y = 0$

Then $B_y'=γ⋅\frac{v}{c^2}⋅E_z$

$E_z$ is the E-field in the rest frame of both charges, hence $E_z = E_0$

We thus obtain $B_y'=γ⋅\frac{v}{c^2}⋅E_0$



We would expect $B_y'=B_1$ , but that is clearly not the case.

What is the explanation for the discrepancy?

 

[Dale]

We would expect B′y=B1

Why would you expect that?

 

[nrqed]

Consider two identical charges separated by some distance. Both are moving together with the same velocity, and the velocity vectors are perpendicular to a straight line drawn between the charges.

Each charge has charge q, speed v and they are separated by a distance of r.

In the rest frame of both charges, the E-field at either charge is $\frac{1}{4π ε_0}\frac{q}{r^2} = E_0$

Based on the Biot-Savart Law, the B-field at either charge is $\frac{μ_0}{4π}\frac{qv}{r^2} = B_1$

$\frac{μ_0}{4π}\frac{qv}{r^2} = [v ⋅μ_0⋅ε_0] ⋅ [\frac{1}{4πε_0}\frac{q}{r^2}]$

$B_1 = \frac{v}{c^2}⋅ E_0$


Let's switch to the relativistic transformation for E and B fields: $B_y'=γ(B_y+\frac{v}{c^2}⋅E_z)$

There is no magnetic field in the rest frame of both charges, hence $B_y = 0$

Then $B_y'=γ⋅\frac{v}{c^2}⋅E_z$

$E_z$ is the E-field in the rest frame of both charges, hence $E_z = E_0$

We thus obtain $B_y'=γ⋅\frac{v}{c^2}⋅E_0$



We would expect $B_y'=B_1$ , but that is clearly not the case.

What is the explanation for the discrepancy?

Notice that for non relativistic speeds, $\gamma \approx 1$ and the two agree. The key point is that Biot-Savart law is an approximation, valid if we neglect corrections of order $v^2/c^2$.

 

[greswd]

Notice that for non relativistic speeds, $\gamma \approx 1$ and the two agree. The key point is that Biot-Savart law is an approximation, valid if we neglect corrections of order $v^2/c^2$.

Thanks, is there a relativistic form of the Biot-Savart Law?

 

[greswd]

Why would you expect that?

I thought that the laws of electromagnetism are compatible with SR.

 

[greswd]

Okay, so, before SR, we had these E and B transformations.

https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dd649e1f88760a1b79f004798ad7dae9f5a9cdee
https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5215bf1a0ae6503a923bef5e6c85288494e5290b

The SR transformations follow a similar idea, but with relativistic corrections.


For this thread, I was inspired by Purcell's explanation of such transformations by combining electromagnetics and length contraction.

I was also inspired by these UIUC lecture notes which derive the SR E&B transformations by using parallel plate capacitors and cylindrical solenoids.
http://web.hep.uiuc.edu/home/serrede/P436/Lecture_Notes/P436_Lect_18p5.pdf

 

[Dale]

I thought that the laws of electromagnetism are compatible with SR.

Yes, they are. But the classical laws of EM are Maxwell's equations.

 

[greswd]

Yes, they are. But the classical laws of EM are Maxwell's equations.

Doesn't the Biot-Savart Law form part of the basis of Maxwell's equations?

For Gauss' Law for Magnetism, and also for Ampere's Law. Maxwell used Ampere's Law to as a basis for the mathematics of the Displacement current too.

Also, does the B-S Law have a relativistic form?

On a side note, what do you think of Purcell's and Errede's (UIUC) derivations?

 

[jartsa]

I studied the wikipedia, and learned that Biot and Savart do not have much to do with that incorrect law for point charges:

The "Biot–Savart law for a point charge" is called that because it looks like the standard Biot-Savart law. The "Biot–Savart law for a point charge" was first derived by Oliver Heaviside in 1888. The Biot-Savart law was developed about 1820 by Biot and Savart, and that law is correct.

https://en.wikipedia.org/wiki/Biot–Savart_law

 

[greswd]

I studied the wikipedia, and learned that Biot and Savart do not have much to do with that incorrect law for point charges:

The "Biot–Savart law for a point charge" is called that because it looks like the standard Biot-Savart law. The "Biot–Savart law for a point charge" was first derived by Oliver Heaviside in 1888. The Biot-Savart law was developed about 1820 by Biot and Savart, and that law is correct.

https://en.wikipedia.org/wiki/Biot–Savart_law

No, I think both forms of B-S Laws are correct.

I'm still thinking about the relativistic forms though.

I'm wondering why the non-relativistic forms are not co-variant although they appear to have a role in the development of the relativistic E and B transformations.

 

[jartsa]

No, I think both forms of B-S Laws are correct.

If there's a discrepancy between formulas, then some formula must be incorrect... Oh, now I see that there is an exact version and an approximate version of the Biot-Savart law for a point charge... and you were using the inexact version as far as I can see.

EDIT: No. Sorry, I saw incorrectly.
EDIT2: But did you use the correct formula for E? It seems to be a more complicated formula for E in the wikipedia.

I was looking at this:
https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity

 

[Dale]

Doesn't the Biot-Savart Law form part of the basis of Maxwell's equations?

No.

Do you know how to derive the Biot Savart law from Maxwell's equations? Maxwell's equations are the classical laws of EM. For anything else it is important to know the derivation. Any derivation includes certain boundary conditions or assumptions, so they don't apply when those assumptions are violated.

Both Coulomb's law and the Biot Savart law can be derived as simplifications to the Lienard Wiechert potentials for point charges. The simplifications assume that the acceleration is 0 and that the speed is much lower than c.

Also, does the B-S Law have a relativistic form?

I wouldn't call it a relativitistic form of Biot Savart, but what you are interested in is the Lienard Wiechert potentials. Those are the solution of Maxwell's equations for a point charge. Simplifying them gives Coulomb's law and Biot Savart law.

 

[nrqed]

I thought that the laws of electromagnetism are compatible with SR.

Yes, they are. But the point here is that the expression you used for the B field produced by a current assumes an infinite line of current while you are treating a single charge in motion. As Dale pointed out, one must be more careful and use the Liénard-Wiechert potentials.

 

[greswd]

Do you know how to derive the Biot Savart law from Maxwell's equations? Maxwell's equations are the classical laws of EM. For anything else it is important to know the derivation. Any derivation includes certain boundary conditions or assumptions, so they don't apply when those assumptions are violated.

Both Coulomb's law and the Biot Savart law can be derived as simplifications to the Lienard Wiechert potentials for point charges. The simplifications assume that the acceleration is 0 and that the speed is much lower than c.

Today a question regarding this topic popped into my head.

Are these formulas considered relativistically accurate?

https://wikimedia.org/api/rest_v1/media/math/render/svg/0f863b311a45bcf9cef19b6455265b2d3033dec0
https://wikimedia.org/api/rest_v1/media/math/render/svg/e591f53b6c36fbc3a2a82938dfa2e1731b6c5209

 

[PeterDonis]

Are these formulas considered relativistically accurate?

Please use the PF LaTeX feature.

 

[greswd]

Please use the PF LaTeX feature.

here is the formula I wanted to post:

$\mathbf{B} =\frac{\mu_0 q}{4\pi}\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}} \mathbf{v} \times \frac{\mathbf{\hat r'}}{|\mathbf r'|^2}$
from https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity

Is it considered relativistically accurate?

Also, what is the deadline for editing a post?

 

[PeterDonis]

Is it considered relativistically accurate?

Not by itself, since you have only given the formula for B. You need to include both the B and E formulas to have a set of equations that is "relativistically accurate". Even then you have to be careful how you interpret that term: it doesn't mean that B and E don't change under Lorentz transformations; it just means that the B and E given by those formulas will satisfy Maxwell's Equations after they are subjected to a Lorentz transformation (their individual values will change but the relationships between them given by Maxwell's Equations will remain valid).

what is the deadline for editing a post?

4 hours after it's originally posted. Unless you have magic Mentor powers; we can edit posts forever. Is there something you need to change?

 

[greswd]

4 hours after it's originally posted. Unless you have magic Mentor powers; we can edit posts forever. Is there something you need to change?

I wanted to edit #14 instead of creating a new post. But nevermind.

$\mathbf{B} =\frac{\mu_0 q}{4\pi}\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}} \mathbf{v} \times \frac{\mathbf{\hat r'}}{|\mathbf r'|^2}$

Both the bolded and unbolded v(s) have the same value right?

 

[PeterDonis]

Both the bolded and unbolded v(s) have the same value right?

No. The bolded v is a 3-vector. The unbolded $v$ is its magnitude. These are not the same thing.

 

[greswd]

No. The bolded v is a 3-vector. The unbolded $v$ is its magnitude. These are not the same thing.

yeah, I was thinking that they should have the same magnitude.

 

[greswd]

Do you know how to derive the Biot Savart law from Maxwell's equations? Maxwell's equations are the classical laws of EM. For anything else it is important to know the derivation.

Both Coulomb's law and the Biot Savart law can be derived as simplifications to the Lienard Wiechert potentials for point charges. The simplifications assume that the acceleration is 0 and that the speed is much lower than c.

What I understand from your statement is that after modifying the classical E and B fields to their relativistic, retarded potential forms, we can still satisfy Maxwell's displacement current equation:

$\nabla \times \mathbf{B} = \mu_0\left(\mathbf{J} + \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} \right)$

But what about J though? Should J be modified into a relativistic form?

 

[PeterDonis]

after modifying the classical E and B fields to their relativistic, retarded potential forms, we can still satisfy Maxwell's displacement current equation

And all of the other Maxwell equations as well.

what about J though? Should J be modified into a relativistic form?

The charge $\rho$ and current J have to be transformed properly under Lorentz transformations.

 

[greswd]

The charge $\rho$ and current J have to be transformed properly under Lorentz transformations.

I'm only considering a single frame, no transformations between frames.

The classical Biot-Savart Law satisfies the curl equation:

$\mathbf B = \frac{\mu_0}{4\pi}\iiint_V \ \mathbf{J} \times \frac{\mathbf{\hat r'}}{|\mathbf r'|^2} dV$

$\nabla \times \mathbf{B} = \mu_0\mathbf{J}$
$\mathbf{J}=ρ\mathbf{v}$


But then we consider the relativistic Biot-Savart law:

$\mathbf B = \frac{\mu_0}{4\pi}\iiint_V \ \frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}\mathbf{J} \times \frac{\mathbf{\hat r'}}{|\mathbf r'|^2} dV$

I wonder how it satisfies the curl equation. Maybe the curl of the field remains unaffected?

 

[PeterDonis]

The classical Biot-Savart Law satisfies the curl equation

The classical Biot-Savart law for a continuous current distribution in some volume.

the relativistic Biot-Savart law

The relativistic Biot-Savart law for a point particle. Which is not the same as a continuous current distribution in some volume. Notice that there is no integral sign in the "relativistic" expression on the Wikipedia page.

 

[greswd]

The relativistic Biot-Savart law for a point particle. Which is not the same as a continuous current distribution in some volume. Notice that there is no integral sign in the "relativistic" expression on the Wikipedia page.

yes true. so what is the Biot-Savart law for a continuous charge distribution that is moving at relativistic speeds?
Does it have any bearing on the curl of the B field when compared to the classical Biot-Savart law?

 

[Dale]

what is the Biot-Savart law for a continuous charge distribution that is moving at relativistic speeds?

The closest thing I can think of would be Jefimenko's equations.

 

[Dale]

But what about J though? Should J be modified into a relativistic form?

J is fine, you only worry about relativity for J if you are transforming to another frame.

 

[greswd]

J is fine, you only worry about relativity for J if you are transforming to another frame.

Imagine two fields of charge. One has a high charge density but slow speeds. One has very low charge density but high speeds, relativistic speeds.

Therefore both have the same J fields, numerically speaking.

Although the B fields of both are different, do they both have the same curl?

 

[Dale]

Although the B fields of both are different,

The B field of both are the same

 

[greswd]

The B field of both are the same

Why does this factor not lead to a different B field? If the proof is complex maybe you can just post a link, that's ok.

$\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$

 

[Dale]

Why does this factor not lead to a different B field? If the proof is complex maybe you can just post a link, that's ok.

$\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$

The easiest way to see that is to just look at the retarded potentials. In the retarded potentials A depends only on J and B depends only on A. So if J is identical then B is also identical.

 

[greswd]

The easiest way to see that is to just look at the retarded potentials. In the retarded potentials A depends only on J and B depends only on A. So if J is identical then B is also identical.

Wow that's so counter-intuitive. I think almost anyone looking at the problem at first glance would assume that the B-field would be affected.

Does this mean that the relativistic changes within the charge distribution somehow offset each other?

 

[Dale]

Does this mean that the relativistic changes within the charge distribution somehow offset each other?

I don't know what you mean by that. Aren't you dealing with a single frame? So what relativistic changes are you talking about in a single frame?

 

[greswd]

what relativistic changes are you talking about in a single frame?

The $\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$ factor getting larger and larger as the speeds of the charge distribution increase.

The increase in speed is matched by a decrease in charge density, as mentioned in #28, so that J remains unchanged.

 

[Dale]

The $\frac{1-v^2/c^2}{(1-v^2\sin^2\theta/c^2)^{3/2}}$ factor getting larger and larger as the speeds of the charge distribution increase.

I am not exactly sure what that factor is or where it comes from (in terms of assumptions in the derivation). If it is for a point charge then you cannot increase v without increasing J also.