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According to Griffiths (Chapter: Electrodynamics and Relativity, Section: Relativistic Electrodynamics)-
We have a string with positive line charge $\lambda$. And on the same string a negative line charge $-\lambda$. The positive charges moving towards right with velocity $v$, and negative charges moving towards left with velocity $v$. So we have a net current to the right, of magnitude $I=2\lambda v$.

If we now look at the setting from a reference frame that is moving towards right, with speed $u$, the velocities of the positive and negative line charges are $$v_{\pm}=\frac{v\mp u}{1\mp vu/c^2}.$$

After this Griffiths has written -
From this new frame the wire seems to carry a net negative charge! In fact, $$\lambda_{\pm}=\pm(\gamma_{\pm})\lambda_0,$$ with $$\gamma_{\pm}=\frac{1}{\sqrt{1-v_{\pm}^2/c^2}},$$ with $\lambda_0$ being the charge density of the +ve line charge in its own rest frame.

My question is --
What is $(\gamma\lambda_0)$? I know $l=l_0 /\gamma$, $t=t_0\gamma$. But if $\lambda=charge/length$, then what is this $(\gamma\lambda)$?

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    $\begingroup$ What do you mean “what is” it? It’s the product of $\gamma$ and $\lambda_0$. And that’s equal to $\lambda$. $\endgroup$
    – Ghoster
    Mar 17 at 17:39
  • $\begingroup$ Very helpful @Ghoster $\endgroup$
    – QuestionTheAnswer
    Mar 17 at 18:55
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    $\begingroup$ Your question is unclear. The point of my comment was to get you to make it clear, so that it doesn’t get downvoted and/or closed for lack of clarity. $\endgroup$
    – Ghoster
    Mar 17 at 23:21
  • $\begingroup$ Yes that's true. But I got my answer here: physics.stackexchange.com/a/138411/340228 $\endgroup$
    – QuestionTheAnswer
    Mar 18 at 4:47

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