So I had a huge debate about this with my friends. Imagine that you are in a non-inertial frame of reference. For simplicity, assume that frame is accelerated along x-axis. You have held a charge in your hand and hence it is rigidly bound to your body. I am in an inertial frame and I see you accelerated and so I see the radiation coming out from the charge. Would you also see the radiation coming out from the charge even though it is at rest with respect to you?

In my opinion, if one uses the correct transformations, then the radiation in the accelerated frame should simply disappear and you would not be able to detect your acceleration just by looking at the charge. However, I am not sure about this.

There is another dimension to this whole issue in my opinion. Suppose that the correct transformations actually produce some extra term giving rise to the radiation in the accelerated frame. But even then, doesn't it seem more logical that this is solely due to the fact that you would then be accelerated with respect to the matter in the universe and there wouldn't be any such extra term in an universe with no matter at all?

  • $\begingroup$ My knee-jerk guess is that the observer at rest with respect to the charged particle in the non-inertial frame will only observe the electric field of the point charge but I am not entirely comfortable with that. This is an interesting question... $\endgroup$ – honeste_vivere May 25 '16 at 17:08
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    $\begingroup$ According to the principle of equivalence, the accelerated frame should be the same as a stationary observer in a gravitational field and the charged particle in a free-fall in the gravitational field. I think this question has presented difficulties but there are attempts at answers. $\endgroup$ – Peter R May 25 '16 at 17:48

Assuming that Classical Electrodynamics (Maxwell's Equations) holds, the answer is that the inertial observer would see the radiation while the non-inertial observer would NOT. The question you are asking is basically the following paradox: https://en.wikipedia.org/wiki/Paradox_of_a_charge_in_a_gravitational_field

This paradox has been analyzed and resolved in the following paper by Camila de Almeida: http://arxiv.org/pdf/physics/0506049%E2%80%8E

I will try to give a summary of the argument:

Radiation From An Accelerating Charge

(This very helpful picture above is taken from Almeida's paper.)

The curved hyperbolic trajectory is the trajectory of the accelerating charge. Suppose that the charge emits some radiation, then the radiation can only be detected within future light cones of any point on the trajectory. Which means that the radiation can only be detected in regions I and II (if there is any radiation at all!)

Now we can also use the curved hyperbolic trajectory to represent the accelerated observer's worldline. The range of events that can be detected by this observer must lie within the past light cone of points on the curve. This means, the accelerated observer can only detect events from region I and region IV.

Combining the previous two points, we see that the only region in which the charge can affect the accelerated observer is in region I. However, Almeida's calculations show that in region I, the field perceived by the accelerated observer is a static field, which means radiation is not perceived (although the inertial observer DOES see a zigzag radiation field.) So the paradox is resolved. The difference in experimental results arises from different perspectives.

P.S. Thanks to @Peter R for pointing out that, if we modify Maxwell's Equations, the situation is different! In fact, in Feynman's modification, no radiation is observed in any reference frame and all our previous analysis fails. But neither Feynman's nor Almeida's analysis have been tested in experiments. So the answer to your question remains unknown.

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    $\begingroup$ Feynman did not think that a uniformly accelerating charge would radiate in any reference frame. He did quite a bit of work in this area. $\endgroup$ – Peter R May 25 '16 at 20:26
  • $\begingroup$ @PeterR : Please elaborate $\endgroup$ – Peaceful May 26 '16 at 3:03
  • $\begingroup$ @PeterR : Sorry just now I saw the Edit by ZhengyanShi . $\endgroup$ – Peaceful May 26 '16 at 3:09
  • $\begingroup$ @ZhengyanShi- Shouldn't the trajectory of a charged particle in a uniformly accelerating frame in the z-direction be parabolic instead of hyperbolic ( like in the figure of your answer, which is uniformly accelerating to the to the left). Or at least near the z-axis, where the charged particle has a non-relativistic velocity? $\endgroup$ – descheleschilder Apr 11 '17 at 7:03
  • $\begingroup$ @descheleschilder The trajectory is a hyperbola. But near the z axis, you can approximate it as a parabola. How would that change the analysis though? $\endgroup$ – Zhengyan Shi Apr 20 '17 at 2:30

The radiation emitted by an accelerated charge depends on the boundary conditions on the fields at infinity. When one takes this into account properly, then accelerated observers will agree with inertial observers about the emitted radiation (after trivial transforms are applied). Any treatment which purports to show that in the accelerated observer's frame there is no radiation, cannot be correct, because you can always consider the 4-momentum of the charge and calculate the emitted radiation using conservation of energy and momentum. However, this requires considering the interaction of the charge with its own electromagnetic field. A rigorous way to treat the self-force problem was only recently obtained.

  • $\begingroup$ Okay then what is the answer according to you? $\endgroup$ – Peaceful May 26 '16 at 4:44

Very intuitive: I think it has to do with how the charged particle is accelerated. In a uniformly accelerating frame, it's actually we who are accelerating towards the particle, while the electron just seems to be accelerating in free fall towards us. The particle "feels" no forces acting on it, just as we would feel no force if we are in free falling in a uniformly accelerated frame.The same holds for a charged particle (or for us) in a real gravitational field.

If the particle is accelerated by means of an electric field though, the particle "feels" the electric force (like we would feel it if we were attached to a sufficient charged object), because of its (or our) inertia (the "resistance" to being accelerated). There is no resistance to the acceleration of a charged particle in a uniformly accelerating frame (or in a gravitational field) which is the cause why we don't feel the acceleration.

So the particle in the accelerated frame (or in a gravitational field) is actually standing stil with us accelerating towards it, whlile the charged particle in an electric field really accelerates, causing it to radiate.

So I agree with Camila de Almeida and Feynman.


My opinion is that to delete the radiation by choosing the coordinate system is like to eliminte problems burying the head in the sand. You can adjust your velocity with one point-charge, but by no means with the entire field. On the other hand there is a basic mistake in the concept of to accelerate a point charge. A point charge must have an infinite mass and therefore can not be accelerated.


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