Electric force due to a moving charge

[LCSphysicist]

Homework Statement

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Relevant Equations

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I am having some problems involving the force that a source moving with speed v along the x-axis would exert on a test charge at the x axis.

Moving to the frame of the source charge, we got that the electric field it exerts is $$E' = kq/x'²$$

Now, moving back to the lab frame, and considering that $x'$ above is contracted, since it was at the frame of a moving charge, we got $\vec{E'}_{ \parallel} = \vec{E}_{ \parallel}$, we got that:

$$ E = kq/x'² = kq \frac{\gamma² }{x² } \implies F = kq Q\frac{\gamma² }{x² }$$

This is wrong, but i don't know why :/ Where is my error? How to solve this question?

 

[PeroK]

If the charge is moving you have a time dependent electric field.

 

[LCSphysicist]

If the charge is moving you have a time dependent electric field.

I think i didn't get your point. The answer should be divided by square of gamma, not multiplied. Why?

 

[PeroK]

I think i didn't get your point. The answer should be divided by square of gamma, not multiplied. Why?

Perhaps I don't understand what you are doing. The electric field due to a moving charge is time-dependent. You need one variable for the position of the charge and one variable for the position at which you are measuring the electric field.

 

[ergospherical]

There's a few ways of going about such a problem, the most direct being i) Lorentz transform the 4-potential $A^{\mu}$ or ii) Lorentz transform the EM tensor $F^{\mu \nu}$.

For example, in the rest frame of the source $\phi = kq/r^2$ whilst $\mathbf{A} = 0$. Because $A^{\mu} = (\phi, \mathbf{A})$ is a 4-vector, then a boost by velocity $v$ (i.e. $x' = \gamma(x-vt)$) transforms these quantities to
\begin{align*}
\phi' = \gamma \phi &= \dfrac{\gamma k q}{r^2} \\ \\
\mathrm{and} \ \ \ \mathbf{A}' &= \phi \mathbf{v}
\end{align*}
Upon re-writing $r^2 = x^2 + y^2 + z^2$ in terms of the primed co-ordinates, you can find the electric field $\mathbf{E}'$ in the moving frame from the definition $\mathbf{E}' = -\nabla \phi' - \dfrac{\partial \mathbf{A}'}{\partial t}$.

The alternative is to notice that $E^i = F^{i0}$, so you can work through writing out all the tems in\begin{align*}
F'^{i0} = {\Lambda^i}_{\mu} {\Lambda^0}_{\nu} F^{\mu \nu}
\end{align*}

 

[TSny]

Now, moving back to the lab frame, and considering that $x'$ above is contracted

The mistake is assuming that $x'$ is the Lorentz contraction of $x$ (i.e., $x' = x/\gamma$). You can discover that this is not the case by working through the following.

The observer in the unprimed frame can be considered to be located at the unprimed origin and the charge can be considered as being located at the primed origin. The charge passes the unprimed origin at $t = t' = 0$. Let the unprimed observer measure $E$ at the particular instant $t$.

What is the distance $x$ between the charge and the unprimed observer according to the unprimed frame when the measurement is made? Express in terms of $v$ and $t$.

The measurement of $E$ by the unprimed observer can be considered as a spacetime event. At what time $t'$ does this event take place in the primed frame?

According to the primed frame, what is the distance $x'$ between the unprimed observer and the charge when the measurement is made by the unprimed observer? You should find that $x'$ is greater than $x$ rather than a contraction of $x$.

 

[ergospherical]

Thought it might be useful to illustrate @TSny's nice idea,

The event $p$ is the measurement of $E$ by the unprimed observer.

In the unprimed frame, $p$ is simultaneous with the point $a$ on the charge's worldline.
In the primed frame, $p$ is simultaneous with the point $b$ on the charge's worldline.

Then, in terms of the quantities that @TSny has defined, you have $x = x(a) - x(p) = x(a)$ whilst $x' = x'(b) - x'(p) = -x'(p)$.

 

[TSny]

Here's another approach using length contraction. Consider a rod at rest in the primed frame and oriented along the x'-axis. Let x' denote the length of the rod as measured in the primed frame. Label the ends of the rod $a$ and $b$. Let the charge $q$ be fixed to the rod at end $a$.

So, the electric field at end $b$ in the primed frame will have magnitude $E'_b = \large \frac{ kq}{x'^2}$.

As you noted in the OP, electric fields that are parallel to the relative motion of the frames have the same value in both frames. So, the magnitude of the electric field at $b$ in the unprimed frame is $E_b = E'_b$,

Consider the distance x between the charge at $a$ and the point $b$ in the unprimed frame at any instant in the unprimed frame.