Total Force on a Point Charge in Motion

[ARoyC]

Homework Statement

An infinite rod of charge density λ lies parallel to the x-axis. A point charge q is stationary at a distance r from the rod. An observer is moving with a velocity -vx̂. What is the force on the point charge from that observer's frame?

Relevant Equations

F = F_E + F_B
B of an Infinitely Long Current carrying Wire = µI/2πr
F_B = q(vxB)
F_E = qE
E of an Infinitely Long Uniformly Charged Rod = λ/2πεr r̂

As the observer is moving, there will be a magnetic force.

Electric Field of the Rod = λ/2πεr r̂
Electric Force on the Point Charge = qλ/2πεr r̂
Magnetic Force on the Point Charge = q(vxB) = qvB n̂ = qv(µI/2πr) n̂ = qv(µλv/2πr) n̂
= µqλv²/2πr n̂

Total Force = Electric Force + Magnetic Force

 

[kuruman]

I think that for this problem you need to use the relativistic transformation equations. Find what the fields that might exert a force on the particle look like in the rest frame of the particle and transform to the rest frame of the observer.

 

[ARoyC]

I think that for this problem you need to use the relativistic transformation equations. Find what the fields that might exert a force on the particle look like in the rest frame of the particle and transform to the rest frame of the observer.

Isn't Magnetic Force just a Relativistic Effect?

 

[kuruman]

Isn't Magnetic Force just a Relativistic Effect?

It is, but you are confusing the two frames. When you say

Electric Field of the Rod = λ/2πεr r̂

that's the electric field in the frame of charged wire. Then you say

Magnetic Force on the Point Charge = q(vxB) = qvB n̂ = qv(µI/2πr) n̂ = qv(µλv/2πr) n̂
= µqλv²/2πr n̂

There is no magnetic force in the frame of the wire because the particle is at rest in that frame.

The problem clearly asks you to find the force on the point charge in the observer's frame. That is why you need to transform the field(s) to that frame.

 

[TSny]

As the observer is moving, there will be a magnetic force.

Electric Field of the Rod = λ/2πεr r̂
Electric Force on the Point Charge = qλ/2πεr r̂
Magnetic Force on the Point Charge = q(vxB) = qvB n̂ = qv(µI/2πr) n̂ = qv(µλv/2πr) n̂
= µqλv²/2πr n̂

Total Force = Electric Force + Magnetic Force

Your work looks good except you haven't taken into account the effect of length contraction on $\lambda$.

Also, you need to give the direction of the unit vector $\hat n$.

With your approach, you will not need to know the relativistic transformation equations for the fields.

 

[ARoyC]

Your work looks good except you haven't taken into account the effect of length contraction on $\lambda$

If I multiply the Electric Field expression by Gamma, will the expression of the Magnetic Force remain the same?

Also, you need to give the direction of the unit vector $\hat n$.

How would I find that as no position of q has been given?

 

[TSny]

If I multiply the Electric Field expression by Gamma, will the expression of the Magnetic Force remain the same?

What is the justification for multiplying the electric field by gamma?

I'm not sure what you mean when you ask if the expression for B will remain the same.

How would I find that as no position of q has been given?

You should be able to specify the direction of $\hat n$ in relation to the direction of $\hat r$.

 

[ARoyC]

What is the justification for multiplying the electric field by gamma?

Transformation of the Electric Field to the Moving Frame?

I'm not sure what you mean when you ask if the expression for B will remain the same

Isn't F = qE + qvB, where is the Electric Field from the rest frame?

You should be able to specify the direction of $\hat n$ in relation to the direction of $\hat r$.

How to do that?

 

[vanhees71]

Why don't you just solve for the electrostatic problem in the frame, where the rod is at rest and then simply transform the electromagnetic field (or simpler the four-potential!) with a Lorentz transformation? Then you get the magnetostatic solution in the reference frame, where the rod moves, and from this the force on a test charge.

 

[ARoyC]

Why don't you just solve for the electrostatic problem in the frame, where the rod is at rest and then simply transform the electromagnetic field (or simpler the four-potential!) with a Lorentz transformation? Then you get the magnetostatic solution in the reference frame, where the rod moves, and from this the force on a test charge.

Yes, I can do it. But I want to understand the mistake in my solution.

 

[TSny]

Why don't you just solve for the electrostatic problem in the frame, where the rod is at rest and then simply transform the electromagnetic field (or simpler the four-potential!) with a Lorentz transformation? Then you get the magnetostatic solution in the reference frame, where the rod moves, and from this the force on a test charge.

Yes, you can certainly solve it this way. But, the OP's approach in the first post will give the correct answer without having to know the general transformation equations for the fields. The only correction needed is to use length contraction to get an expression for $\lambda$ in the observer's frame. I think this approach has pedagogical advantages.

 

[vanhees71]

Well, then you have to know the transformation laws for line-charge densities...

 

[TSny]

Well, then you have to know the transformation laws for line-charge densities...

Yes, but that is just elementary relativity (length contraction) along with invariance of charge.

 

[TSny]

Transformation of the Electric Field to the Moving Frame?

If you want to solve the problem by transforming the fields to the moving frame as suggested by @kuruman and @vanhees71, then that is a good approach. But, the method in your first post will also work, and it does not require knowing the general transformation equations of the fields.

In the observer's frame (primed frame), there is an infinitely long line charge moving parallel to itself. The charge density in this frame is $\lambda'$, which differs from $\lambda$. Gauss' law is still valid in this frame. So, the expression for the E-field in this frame is the same as the expression for the E-field of an infinite line charge at rest except that you need to use $\lambda'$ instead of $\lambda$.

In the observer's frame, $B'$ is just the field of a an infinitely long straight current where the current, $I'$, can be expressed in terms of $\lambda'$ and $v$.

 

[TSny]

I think the quickest way to the answer would be to use the relativistic transformation equations for force.

 

[kuruman]

But, the OP's approach in the first post will give the correct answer without having to know the general transformation equations for the fields.

I would like to see how that is done specifically the equation for the magnetic field in the observer's frame. It does not match what one gets with the field transformation equation.

I think the quickest way to the answer would be to use the relativistic transformation equations for force.

Tsk, tsk, tsk. What happened to solutions with pedagogical advantages?

 

[TSny]

Tsk, tsk, tsk. What happened to solutions with pedagogical advantages?

Exactly!

 

[ARoyC]

If you want to solve the problem by transforming the fields to the moving frame as suggested by @kuruman and @vanhees71, then that is a good approach. But, the method in your first post will also work, and it does not require knowing the general transformation equations of the fields.

In the observer's frame (primed frame), there is an infinitely long line charge moving parallel to itself. The charge density in this frame is $\lambda'$, which differs from $\lambda$. Gauss' law is still valid in this frame. So, the expression for the E-field in this frame is the same as the expression for the E-field of an infinite line charge at rest except that you need to use $\lambda'$ instead of $\lambda$.

In the observer's frame, $B'$ is just the field of a an infinitely long straight current where the current, $I'$, can be expressed in terms of $\lambda'$ and $v$.

Got it. Thank you.

 

[vanhees71]

It's even simpler. You don't need any Lorentz transformations.

In the restframe of the wire you have an electrostatic situation, i.e.,
$$A^{*\mu}(x^*)=\begin{pmatrix} \Phi(x^*) \\0 \\0 \\0 \end{pmatrix}.$$
This you can write in a manifestly covariant way as
$$A^{*\mu}(x^*)=u^{*\mu} \Phi(x^*).$$
Now you only need to write this in a general frame, where $u$ is an arbitrary four-velocity of the wire:
$$A^{\mu}(x)=u^{\mu} \Phi(x^*),$$
and then you simply need to express the $x^*$ by $x$, using the Lorentz transformation $x=\Lambda^{-1}(\vec{v})$ with $\vec{v}=\vec{u}/u^0$.

The other way, is also not as complicated as I first thought. By the same reasoning you have
$$\lambda^{*\mu}=\begin{pmatrix}\lambda^* \\ 0 \\0 \\ 0 \end{pmatrix},$$
where $\lambda^*$ is the charge-line-density in the restframe of the wire. In the general frame again you have
$$\lambda^{\mu}=u^{\mu} \lambda^*.$$
Then you can solve the corresponding magnetostatic problem with this given four-current-line density.