Relation of Magnetic field and electric field [closed]

Question

There is a wire in which the protons are non-moving with a charge density of $\lambda_+$ and the electrons move at a velocity u to the right with a charge density $\lambda_-$ in a rest frame. A particle of q+ is r away from this wire and moves at a velocity of v to the right.

Q1. What is the total force on the particle in the rest frame?

Q2. What is the total force on the particle in the moving particles non moving frame?

Answer for Q1.

Since in the rest frame the total charge is $0$ it is solely a magnetic force giving me

\begin{equation} F = \frac{q v \mu_{0} \lambda^{-} u}{2 \pi r} \end{equation}

Attempt for Q2

I began with finding the relative velocities of the positive and negative charges.

For positive since it is originally non-moving i got

\begin{equation} v'=-v \end{equation}

Hovever for negative charges i used einstiens additive velocity law and got

\begin{equation} u'=\frac{u-v}{1-\frac{uv}{c^2}} \end{equation}

Then using length contraction got the equation

\begin{equation} \lambda'=\lambda \gamma \end{equation}

Plugging in the velocities into lorentz factor led me getting the equations

\begin{equation} \lambda_+'=\frac{\lambda_+}{1-\frac{v^2}{c^2}} \end{equation}

and

\begin{equation} \lambda_-'=\lambda_-\left(\frac{1-\frac{uv}{c^2}}{\sqrt{(1-\frac{v^2}{c^2})(1-\frac{u^2}{c^2})}}\right) \end{equation}

Solving for total charge denisty gave me

\begin{equation} \lambda'=\lambda_- \gamma_+ \left(\sqrt{\frac{(1 - \frac{uv}{c^2})}{(1-\frac{v^2}{c^2})(1-\frac{u^2}{c^2})}-1}\right) \end{equation}

From my understanding the answer for q1 and q2 should be the same except one will have a factor of gamma_+ because of the transformation of forces between reference frames and the concept that magnetic fields are a product of special relativity and the electric field. However i am unable to get this to work out

Answer 1

Short answer

As you have already stated in your answer, the force in the moving frame should be the same as the force in the lab frame times a factor of gamma, so it is perfectly acceptable to write out $F' = \gamma F$ as a one-line answer for Q2.

Long answer

However, it seems like you want to give a more detailed answer by showing what physically happens to the charges in question when the Lorentz boost is applied, so let's look at what you've done so far.

For Q1, the expression you derived for the force on the moving particle only works if the charge densities $\lambda_+$ and $\lambda_-$ are equal and opposite (i.e. $\lambda_+ = -\lambda_-$); otherwise the moving particle will experience an electric force as well, and a different expression will describe the net force. Make sure you double check whether or not the charge densities actually cancel each other out before moving on to Q2.

Concerning Q2, all of the statements you have made for your answer up until

$$\lambda_+'=\frac{\lambda_+}{1-\frac{v^2}{c^2}}$$

and

$$\lambda_-'=\lambda_-\left(\frac{1-\frac{uv}{c^2}}{\sqrt{(1-\frac{v^2}{c^2})(1-\frac{u^2}{c^2}})}\right)$$

are correct. For $\lambda_+'$, the term $1-v^2/c^2$ should be square-rooted. I'm not quite sure how you ended up with the extra factor of $(1-u^2/c^2)$ in the expression for $\lambda_-'$, but it should not be there. The correct transformation for the 4-current describing the charges in the wire should give charge densities of,

$$\lambda_+'=\frac{\lambda_+}{\sqrt{1-\frac{v^2}{c^2}}} = \gamma\lambda_+$$

$$\lambda_-'=\lambda_-\left(\frac{1-\frac{uv}{c^2}}{\sqrt{(1-\frac{v^2}{c^2})}}\right) = \gamma\lambda_-\left(1-\frac{uv}{c^2}\right)$$

and the net charge density of the wire in the moving frame should be,

$$\lambda' = \lambda_+' + \lambda_-' = \gamma \left(\lambda_+ + \lambda_-\left(1-\frac{uv}{c^2}\right)\right)$$

From here, you should be able to use Gauss' Law to construct the electric field $E'$ acting on the particle. Since the particle is at rest in the comoving frame, you may ignore the presence of the magnetic field $B'$.