- #1
PineApple2
- 49
- 0
Hello. Here is the problem:
A charge q is moving at speed v0 to the right in the lab frame. At t=0 the particle passes at the origin (0,0,0). calculate the field at (-d,0,0) at the instant it passes the origin, both in the lab frame and in the charge frame. are they equal?
My solution:
In the lab frame, the field is
[tex]
\frac{q}{\gamma^2 d^2}
[/tex]
in the negative x direction. By the transformation of parallel fields, [tex]E_{||}'=E_{||} [/tex] so the result in the charge frame should be the same.
However working in the charge frame,
[tex]
E=-\frac{q}{d'^2}, \qquad d'=d/\gamma
[/tex]
because of length contraction. therefore the field in the charge frame is
[tex]
E=-\frac{q}{(d/\gamma)^2}=\frac{\gamma^2 q}{d^2}
[/tex]
these are not equal, the gamma here is in the "wrong place".
what is going on here?
thanks.
A charge q is moving at speed v0 to the right in the lab frame. At t=0 the particle passes at the origin (0,0,0). calculate the field at (-d,0,0) at the instant it passes the origin, both in the lab frame and in the charge frame. are they equal?
My solution:
In the lab frame, the field is
[tex]
\frac{q}{\gamma^2 d^2}
[/tex]
in the negative x direction. By the transformation of parallel fields, [tex]E_{||}'=E_{||} [/tex] so the result in the charge frame should be the same.
However working in the charge frame,
[tex]
E=-\frac{q}{d'^2}, \qquad d'=d/\gamma
[/tex]
because of length contraction. therefore the field in the charge frame is
[tex]
E=-\frac{q}{(d/\gamma)^2}=\frac{\gamma^2 q}{d^2}
[/tex]
these are not equal, the gamma here is in the "wrong place".
what is going on here?
thanks.
Last edited: