# What is $\gamma\lambda_0$ as given in the chapter Relativistic Electrodynamics of Griffiths

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 $$(\gamma\lambda)$$?

• What do you mean “what is” it? It’s the product of $\gamma$ and $\lambda_0$. And that’s equal to $\lambda$. Mar 17 at 17:39
• Very helpful @Ghoster Mar 17 at 18:55
• 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. Mar 17 at 23:21
• Yes that's true. But I got my answer here: physics.stackexchange.com/a/138411/340228 Mar 18 at 4:47