How Does the Electric Field of a Moving Point Charge Differ from Its Rest Frame?

[nos]

Hi everyone,

Consider two frames: the rest frame that has a stationary point charge and the moving frame which has a moving point charge.
In the rest frame the point charge has electric field lines pointing radically outward. The charge can be imagined to be surrounded by a spherical shell. In the moving frame, by length contraction, this sphere is deformed into a spheroid. The field lines still points directly away from the charge, but the perpendicular components of the field are enhanced by the gamma factor.

1. If I understand correctly, the electric field and so electric force will tend to be more vertical and also stronger in the moving frame than in the rest frame, except for the field at the x-axis which is the same in both frames?
2. To calculate this electric field, can I simply use pythagorean theorem to calculate this electric field? For example, to calculate the components of the electric field in the rest frame at an angle of 30°, we use sine and cosine to determine the components of the field. Now that I know the vertical component of the field, I also know the vertical component in the moving frame(enhcnaed by gamma factor). The electric field parallel will be the same in both frames. So now that I know vertical component and horizontal component in the moving frame, can I use pythagorean theorem to calculate the resulting field in the moving frame?
3. I suppose this does not apply to gravitational field of moving masses?

 

[Bill_K]

1. Yes
2. Yes
3. Extra factor of gamma

 

[Q-reeus]

1. If I understand correctly, the electric field and so electric force will tend to be more vertical and also stronger in the moving frame than in the rest frame, except for the field at the x-axis which is the same in both frames?

Last bit is not correct - unless you think comparing the field strengths at different radii in a given frame is 'fair'. For a fixed r in the moving frame, on-axis field is reduced by factor (gamma)² - see eqn's (10a), (10b), and following comments here: http://www.physicsinsights.org/moving_charge_1.html

2. To calculate this electric field, can I simply use pythagorean theorem to calculate this electric field? For example, to calculate the components of the electric field in the rest frame at an angle of 30°, we use sine and cosine to determine the components of the field. Now that I know the vertical component of the field, I also know the vertical component in the moving frame(enhcnaed by gamma factor).

Only if your ruler contracts along x-axis by gamma factor! But if you want to compare apples with apples, ruler should be reading the same in all directions.

The electric field parallel will be the same in both frames.

Same comment as above.

So now that I know vertical component and horizontal component in the moving frame, can I use pythagorean theorem to calculate the resulting field in the moving frame?

No - just use the appropriate transformation eqn's (8) given in above link, or equivalent.

 

[Bill_K]

What I think is fair is comparing the values of E in different reference frames at the same point. You are comparing the functional forms, E'(r') vs E(r), and of course you get a different answer since the point where r = r₀ is not the same as the point r' = r₀.

As noh pointed out, the transformation is simply E_x' = E_x, E_y' = γ E_y, the E' vector points toward the origin, and the magnitude of E' is given by the Pythagorean relation E' = √(E_x'² + E_y'²).

 

[pervect]

Hi everyone,

Consider two frames: the rest frame that has a stationary point charge and the moving frame which has a moving point charge.
In the rest frame the point charge has electric field lines pointing radically outward. The charge can be imagined to be surrounded by a spherical shell. In the moving frame, by length contraction, this sphere is deformed into a spheroid. The field lines still points directly away from the charge, but the perpendicular components of the field are enhanced by the gamma factor.

1. If I understand correctly, the electric field and so electric force will tend to be more vertical and also stronger in the moving frame than in the rest frame, except for the field at the x-axis which is the same in both frames?
2. To calculate this electric field, can I simply use pythagorean theorem to calculate this electric field? For example, to calculate the components of the electric field in the rest frame at an angle of 30°, we use sine and cosine to determine the components of the field. Now that I know the vertical component of the field, I also know the vertical component in the moving frame(enhcnaed by gamma factor). The electric field parallel will be the same in both frames. So now that I know vertical component and horizontal component in the moving frame, can I use pythagorean theorem to calculate the resulting field in the moving frame?
3. I suppose this does not apply to gravitational field of moving masses?

1. Yes
2. Yes
3. Extra factor of gamma

A really good and accurate answer to this problem is tricky - Bill K's answer is a very good approximation. The extra factor of gamma comes from the fact that the total energy of the moving particle is increased by a factor of gamma, while the total charge of the moving particle is not increased.

However, if you use Bill K's answer to calculate the deflection of starlight, you'll find that the predicted deflection is off by a factor of 2.

Unfortunately, it's not easy to "fix" this error without an extended and deep discussion. A very brief and imprecise summary is that Bill K assumed that space (not space-time, just space!) was flat in computing this answer, and that it's not really flat.

Textbooks tend to not discuss this issue, alas, so it's hard to give a definitive and precise answer to the question. My own take on the issue is that the concept of curved space-time is more general than the concept of a force - while one can identify certain components of the tensors that describe the curvature of space-time as being equivalent to classical "forces" - as in Newton Cartan theory - there are other components to the tensors that describe the curvature of space-time that are not readily interpretable as a classical forces.

Thus forces can be considered to be a "subset" of curved-space time. But curved space-time includes things that can't be described numerically solely by forces.

One might also mention in passing that the magnetic forces in the electric-field case and their equivalent in the gravitational case have not been discussed at all, but are needed for the puprose of general covariance.

 

[nos]

Thank you all for replying:)

 

[Q-reeus]

What I think is fair is comparing the values of E in different reference frames at the same point. You are comparing the functional forms, E'(r') vs E(r), and of course you get a different answer since the point where r = r₀ is not the same as the point r' = r₀.

If you think I used the usual retarded field eqn's for general motion (e.g. eqn's (17), (19), here: http://fermi.la.asu.edu/PHY531/larmor/index.html); they give curved field lines even for constant velocity motion, so that is not correct. I was assuming the standard instantaneous present position form for E field of an unaccelerated moving charge - for which E field lines are always straight and directed from the charge's assumed present position.

As noh pointed out, the transformation is simply E_x' = E_x, E_y' = γ E_y, the E' vector points toward the origin, and the magnitude of E' is given by the Pythagorean relation E' = √(E_x'² + E_y'²).

And that's why I suspected a non-standard 'velocity axis contracted' version was being used, because specifying E_x' = E_x automatically implies a contracted separation distance along x-axis by factor γ seen in the moving frame. I suppose it's ok to do that provided the reader is made aware this amounts to evaluating the field strength on the surface of an oblate spheroid and not a sphere. The latter geometry is normally assumed when one is specifying the field as a function of angle - radius r is taken to be held constant.