# Electromagnetic force in different inertial frames

1. Feb 19, 2015

### MathewsMD

I've attached an image to a conceptual problem I'm having a little trouble understanding. It shows part a, b, c, and d (but d is cut off). a and b are one inertial frame while c and d are another. c and d makes sense, but I'm having a little trouble with a and b. Based on the image for c, there is length contraction for the + charges in the wire causing there to be more localized charge and thus an electric field. This makes sense since the + charges are moving in this inertial frame. Now I don't quite see why this doesn't occur in part a/b (i.e. why is there no length contraction of the - charges? If the + charges are at rest while the - charges are moving at speed v in this case, shouldn't there be a localization of - charges also causing for an electric force AND magnetic force in a/b? Any clarification on why the length contraction and thus electric force is not present in a and b would be very helpful!

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2. Feb 19, 2015

### Orodruin

Staff Emeritus
This does happen, it is just that in order for the conductor to be electrically neutral in the original system, then the minus charges must be the same distance from each other as the positive ones (it is clearly possible to arrange for this to be the case). This of course means that there is more distance between the minus charges in their rest frame than there is between the positive charges in their rest frame.

3. Feb 19, 2015

### MathewsMD

Okay, so length contraction does occur in case a/b? I may be misinterpreting something here since I don't see how the particular arrangement given in the attachment achieves this.

I agree the conductor should be electrically neutral in this case for there to be no net electric force, but I am just a little confused on how this is achieved in case a/b if there is length contraction...wouldn't this cause unequal separation?

For the conductor to be electrically neutral, shouldn't the charges in the conductor be moving +v/2 and -v/2 respectively (i.e. of equal speed) in the inertial frame?

4. Feb 19, 2015

### Orodruin

Staff Emeritus
Only if the distance between the positive charges in their rest frame is the same as the distance between the negative charges in theirs. Your problem seems to be that you are implicitly assuming that this is the case.

5. Feb 19, 2015

### MathewsMD

Yes, I was assuming that. If you don't mind explaining, why isn't this the case? Since there are an equal amount of + and - charges in the wire, can't we assume equal distribution? Thus, if the both charges travel at speed v in the other charge's inertial frame, wouldn't this assumption be true?

I think I'm starting to get what you're saying (i.e. this possible arrangement can be achieved) but I'm just failing to see why it is achieved in the particular situation given.

6. Feb 19, 2015

### Orodruin

Staff Emeritus
This statement is frame dependent. You have to define in which frame you want to have as many + and - charges. Why would it be the case? The entire situation is dynamic and I am suspecting the idea from the beginning was to have an electrically neutral wire transmitting a current. There are several ways of doing that. The main point is to realise that making it neutral in one frame will result in it being charged in another.

7. Feb 19, 2015

### MathewsMD

Okay. So we're assuming that in the inertial frame given for case a/b that this wire HAS to be neutral and is thus equally distributed in this particular frame. That makes sense. Thank you for all the help! Please feel free to add on anything you think I may still be misinterpreting. :)

8. Feb 19, 2015

### Orodruin

Staff Emeritus
Correct. You can also achieve this by, as you said, having the + charges going with velocity +v/2 and the - charges with velocity -v/2. The overall charge and current would be the same. But if you want the wire to be neutral in the rest frame of the + charges, then there is only one way of arranging that.