Electromagnetic field in different reference frames

  • #1
Jokar
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Suppose there are two reference frames. One is rotating with respect to another with rotational velocity ##\omega##.

Now if in one of the reference frames the vector potential is $$(1, 0, 0, 0)$$ then in the other reference frame it will be $$(\sqrt{1-(\omega r)^2}, 0, \omega r, 0)$$.

Now in the first reference frame the electric field is zero. But in the second reference frame
$$E = \frac{d\phi}{dr} = \frac{\omega^2 r}{\sqrt{1-(\omega r)^2}} $$ .

Now we know if $$F_{\mu\nu} = 0$$ then under coordinate transform also it will remain zero. So why are we getting the electric field?
 
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  • #2
Jokar said:
So why are we getting the electric field?
We are not. You are not treating the rotating frame correctly. Please show your work.
 
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  • #3
Jokar said:
Now in the first reference frame the electric field is zero.
Not just the electric field, but the magnetic field as well, since that's the curl of the spatial part, which vanishes.

Jokar said:
Now we know if $$F_{\mu\nu} = 0$$
Which it is.

Jokar said:
then under coordinate transform also it will remain zero.
Yes. But transformation ##F_{\mu\nu}## is different from transforming the vector potential, because the former is gauge invariant while the latter is not. So when you transform the vector potential, you can't just accept the result as the physically significant vector potential in the new frame. You have to look at possible gauge transformations as well.

In the example you give, you should be able to find a gauge transformation in both frames (though it won't be the same one in each frame) that makes the vector potential vanish.
 
  • #4
PeterDonis said:
Yes. But transformation Fμν is different from transforming the vector potential, because the former is gauge invariant while the latter is not. So when you transform the vector potential, you can't just accept the result as the physically significant vector potential in the new frame. You have to look at possible gauge transformations as well.
That is irrelevant. The transformation of the 4-potential as a 4-vector may not satisfy the original gauge conditions (unless the gauge conditions themselves are explicitly invariant), but it will result in the same fields.

There is no ”correct” 4-potential. Only equivalence classes of 4-potentials that result in the same fields and that are related by gauge transformation.
 
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  • #5
PeterDonis said:
In the example you give, you should be able to find a gauge transformation in both frames (though it won't be the same one in each frame) that makes the vector potential vanish.
It should be obvious in the first case that (1,0,0,0) is the gradient of ##\psi = t##. Consequently, ##A_\mu = \partial_\mu t##. This holds in both frames (just that ##t## must be expressed in whatever coordinates are imposed on the rotating frame when considering that frame).

The vector potential being a gradient automatically results in the field tensor being zero.
 
  • #6
Orodruin said:
It should be obvious in the first case that (1,0,0,0) is the gradient of ##\psi = t##. Consequently, ##A_\mu = \partial_\mu t##. This holds in both frames (just that ##t## must be expressed in whatever coordinates are imposed on the rotating frame when considering that frame).

The vector potential being a gradient automatically results in the field tensor being zero.
Hi,

I am trying to understand the section 5 equation 9 of this article. ( I was just trying to copy the cutting of the section. However, some how I can't attach image.)

[Moderator's note: attachment deleted. See post #9 for link.]
 
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  • #7
Orodruin said:
The transformation of the 4-potential as a 4-vector may not satisfy the original gauge conditions (unless the gauge conditions themselves are explicitly invariant), but it will result in the same fields.

There is no ”correct” 4-potential. Only equivalence classes of 4-potentials that result in the same fields and that are related by gauge transformation.
I agree with all this. I'm just noting that it should always be possible, for those who are uncomfortable with this kind of general argument, to explicitly find a gauge transformation that makes it more "obvious" why the fields vanish.

Orodruin said:
The vector potential being a gradient automatically results in the field tensor being zero.
This seems to me to be a nicer general argument, because the vector potential being a gradient of a scalar field is frame independent. That scalar field won't be the same as the time coordinate in every frame, but it is a valid scalar field in every frame. And then the vanishing of the EM field tensor follows immediately from ##dd = 0##.
 
  • #8
Jokar said:
this article.
Where does this article come from? Can you give a link? That's the proper way to refer to a published article here.
 
  • #9
PeterDonis said:
Where does this article come from? Can you give a link? That's the proper way to refer to a published article here.


I am trying to understand the section 5 equation 9 of this article.

This is the link.
https://adsabs.harvard.edu/full/1953MNRAS.113...34S
 
  • #10
Jokar said:
I am trying to understand the section 5 equation 9 of this article.
The article provides the expression for the vector potential in the rotating frame, and it's clearly different from your expression. I'd suggest to pause here and ask yourself why the difference. If you can't figure it out yourself, then I'd suggest to post more details on how you came with the result.
 
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  • #11
Yuras said:
The article provides the expression for the vector potential in the rotating frame, and it's clearly different from your expression. I'd suggest to pause here and ask yourself why the difference. If you can't figure it out yourself, then I'd suggest to post more details on how you came with the result.

Sorry, still can't find the difference

According to the notation of the paper, in frame 1 the vector potential is $$(-1, 0, 0, 0)$$

In the rotating reference frame it is $$(-\sqrt{1+(\omega r)^2}, \omega y, -\omega x, 0)$$.

Now in the first reference frame the electromagnetic tensor is zero. So after coordinate transform also the tensor should remain zero.

So why are we getting the value of $$E$$.

I am sure I have some small issue is understanding. However, if you understood it can you please explain it so that I can get it.
 
  • #12
Again, if the field tensor is zero in one frame it will be zero in all frames. If you have a source that says this is not the case then you should generally not trust that source.

You still have not provided your computations as asked for, which makes it very difficult to help you. Please provide the definition of your rotating coordinates and your process for transforming the original 4-potential into the rotating coordinate frame.
 
  • #13
Orodruin said:
Again, if the field tensor is zero in one frame it will be zero in all frames. If you have a source that says this is not the case then you should generally not trust that source.

You still have not provided your computations as asked for, which makes it very difficult to help you. Please provide the definition of your rotating coordinates and your process for transforming the original 4-potential into the rotating coordinate frame.


Hi, I gave the article and the equation number.

I am trying to understand the section 5 equation 9 of this article.

This is the link.
https://adsabs.harvard.edu/full/1953MNRAS.113...34S

According to the notation of the paper, in frame 1 the vector potential is ##(-1, 0, 0, 0)##

In the rotating reference frame it is ##(-\sqrt{1+(\omega r)^2}, \omega y, -\omega x, 0)##.

Now lets say we are only interested in ## E_x = F_{tx}##. Of course in the first reference frame its 0. But in the second reference frame its non-zero. Why is that?

If its 0 in one frame then it should be zero in other frames.
 
  • #14
Jokar said:
In the rotating reference frame it is $$(-\sqrt{1+(\omega r)^2}, \omega y, -\omega x, 0)$$.
No, that's not what that vector potential represents. It is not the non-rotating vector potential ##(-1, 0, 0, 0)## transformed to a rotating frame. It is a rotating vector potential, due to a "rotating universe", described in the same non-rotating frame as the non-rotating potential.

In other words, the potential in your quote above is a physically different one from ##(-1, 0, 0 ,0)##. That's why it leads to a nonzero electric field.

Note also that this paper by Sciama is not about electromagnetism. He is exploring the implications of a theory of gravity based on a vector potential. He uses this simplified theory of gravity, instead of the more complete tensor theory of General Relativity, because it's simpler to analyze.
 
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  • #15
Orodruin said:
You still have not provided your computations as asked for
The vector potential the OP gave is explicitly given in the paper he referenced, as is the electric field it gives rise to. The problem is that the OP is misinterpreting the meaning of that potential and of the paper as a whole. See my post #14 just now.
 
  • #16
Yep, it's an active transformation vs a passive one. @Jokar if you actually calculate the potential in a rotating frame instead of copying it from the article, you'd find that it's still ##(-1,0,0,0)##. That's because it's orthogonal to the hyperplane ##t=0## (in the original coordinates), i.e. ##A_\mu v^\mu=0## for any vector of the form ##v=(0,v_x,v_y,v_z)##, and change of variables to rotating frame doesn't touch time, so ##A## has to be orthogonal to the hyperplane ##t=0## in the new coordinates as well. It's the same hyperplane after all!
 
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  • #17
Yuras said:
it's an active transformation vs a passive one.
I'm not sure what you mean by this; active vs. passive transformations both use the same transformation equations. The "rotating" potential in the Sciama paper is not obtained by any kind of transformation of the non-rotating one.
 
  • #18
Jokar said:
So why are we getting the electric field?
Because you are using an invalid formula to compute the electric field. Or more precisely, you use a formula which is valid in Minkowski coordinates and apply it in coordinates which are not Minkowski.

If you want to compute electric field in arbitrary coordinates, you must define electric field as a 4-vector. For that purpose see e.g. https://arxiv.org/abs/1302.5338.
 
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  • #19
Demystifier said:
you are using an invalid formula to compute the electric field.
Please note that, as I have already pointed out in post #15, the "rotating" vector potential and electric field that the OP gives are explicitly given in the Sciama paper the OP has referenced. The OP did not compute them himself; he took them directly from that reference.
 
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  • #20
PeterDonis said:
I'm not sure what you mean by this; active vs. passive transformations both use the same transformation equations. The "rotating" potential in the Sciama paper is not obtained by any kind of transformation of the non-rotating one.
You are right, thank you for the correction. I meant that it can be obtained by applying the inverse of the coordinate transformation. In fact that's how a contravariant vector ##(-1,0,0,0)## will transform under rotation. (I'm not claiming that it actually was obtain in this particular way - I didn't read the article, so I don't know.)
 
  • #21
Yuras said:
it can be obtained by applying the inverse of the coordinate transformation.
What "it" are you talking about? If you mean the "rotating" vector potential in the OP, no, that can't be obtained from ##(-1, 0, 0, 0)## the way you describe. Nor does the Sciama paper the OP referenced claim that it can.
 
  • #22
PeterDonis said:
What "it" are you talking about? If you mean the "rotating" vector potential in the OP
Yep, I meant this one.
PeterDonis said:
, no, that can't be obtained from ##(-1, 0, 0, 0)## the way you describe.
Hmm, I tried to calculate it again, and indeed the result is different. I guess I made an arithmetic errors somewhere the last time. My bad.
 
  • #24
After some cleanup, the thread is reopened.
 
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