1
$\begingroup$

With respect to special relativity, I was wondering how the spatial dimensions would differ between the rest and LAB frame of an electron beam.

System: Electron bunch/beam traveling in linear motion. How large/long would the bunch appear in the point of view of an electron (rest frame). How large/long would the bunch look in the lab frame?

How would the answers change if the electron bunch/beam was accelerating?

$\endgroup$
2
$\begingroup$

Due to length contraction, the bunch will appear shorter in the lab frame than it would in its own reference frame. Specifically, if the bunch had a length of $L$ in its own frame of reference, in the lab (primed) frame it would have a length of

$$ L^\prime = L/\gamma = L \sqrt{1-v^2/c^2}$$

Note that the length is only contracted in the direction of motion, so that if the bunch was spherical in the rest frame, it would appear more like a pancake moving broad-side forward in the lab frame.

Additionally, in the rest frame the bunch of electrons would very quickly spread out, since like charges repel. However in the lab frame, due to time dilation, the bunch is observed to stay fairly tight as it moves past because the spreading takes much longer in the lab frame. If it took a time $t$ to spread out by a certain amount in the rest frame, then in the lab frame the same amount of spreading would take a time

$$ t^\prime = \gamma t = \frac{t}{\sqrt{1-v^2/c^2}}$$

These are just the length contraction and time dilation formulas, respectively.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Assuming the electrons are moving at the same velocity the beam would look the same because the relative velocity is zero between the electrons. However, according to the scientist(lab frame), the beam gets shrunk (and according to the electrons the scientist gets shrunk).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.