(Previous posts in this series: Part 1, Part 2, Part 3, Part 4, Part 5, Part 6, Part 7, Part 8)
We saw in the previous post how Einstein’s Principle of Equivalence explained why two masses dropped from the same height in a gravitational field will fall at the same rate and hit the ground at the same time. It also seemed to resolve the paradox about whether an electric charge will fall more slowly than a neutral particle in a gravitational field. The answer arrived at was ‘no’ because since according to the PoE, the situation with the falling accelerating charge in the frame of the Earth E was equivalent to the charge being stationary in the inertial frame of space S, the charge would not radiate energy, contrary to naive expectations that any accelerating charge would radiate. On the surface, that seems to resolve the paradox that started this series of posts. The price we have paid is that we must abandon Postulate #2 and conclude that accelerating charges do not radiate when falling freely in a gravitational field.
But that resolution, apart from being awkward, has other major problems. As I said, there is a lot of controversy over this question. I have been reading a whole slew of papers on this topic and I will provide a bibliography in a future post for people who are interested. For the present, I will follow the lead of Fritz Rohrlich who is considered one of the experts in this area and is credited with doing some of the most careful analyses of the problem, even if not everyone agrees with his conclusions. He finds many subtleties that need to be taken into account. I will use his work as the basis of the resolution, not necessarily because I agree with (or have independently confirmed) everything he says but because he is very thorough and his analyses are very clear. His conclusions will then be used to contrast with what others have said on this topic.
I will be using the following references as my starting point:
(1) Classical Radiation from a Uniformly Accelerated Charge by Thomas Fulton and Fritz Rohrlich (Annals of Physics: 9, 499-517, 1960)
(2) The Principle of Equivalence by Fritz Rohrlich (Annals of Physics: 22, 169-191, 1963)
(3) The book Classical Charged Particles (1965) by Fritz Rohrlich.
A key point Rohrlich makes is that “If one argues … that this situation involves an accelerated charge which should always radiate, the argument is erroneous, because the fact that a charge is accelerated does not necessarily imply that it radiates, unless the acceleration takes place relative to an inertial observer. A nonnertial observer uses different clocks and yard- sticks… Since radiation is not a generally covariant concept the question whether the charge really radiates is meaningless unless it is referred to a particular coordinate system.” [2]
This is important. Recall in an earlier post that we said that an ‘event’ is something that all observers can agree on, even if they do not agree on specific values associated with that event. One might think that whether or not an electric charge radiates is an event that all observers, whatever their state of motion, will agree either happened or did not happen. Not so, says Rohrlich. He says that we cannot tell purely by looking at the state of motion of the charge whether it radiates or not. What we can unambiguously tell is whether a radiation detector (either human or mechanical) registers the arrival of radiation, by a click, a track, a flash of light, or some other form of tangible signal. It is that signal that constitutes an ‘event’ that everyone can agree on. If a detector registers such a signal, we can infer that the charge emitted radiation. However, if the detector does not detect anything, we cannot immediately infer that the charge did not radiate but need to look at the situation more closely to see if theory predicts that there should be radiation or not, since there may be other reasons for the lack of detection. The state of motion of the detector needs to be taken into account.
To find out theoretically whether there is radiation or not, Rohrlich calculates what is called the Poynting vector, which is a function of both electric and magnetic fields produced by a moving charge. It is this quantity that gives us the energy and momentum in the radiation carried by electromagnetic fields. Both electric and magnetic fields must be present for radiation to exist and if they do exist in a frame at the location of the detector, then it will detect radiation. If either is zero at the location of the detector, then there will be no radiation detected. In his work, he looks at charges and detectors in various states of motion and tries to reconcile the results of all the scenarios.
The experiments to test these are very difficult because the size of the effects are so small. At this stage, the best we can hope for are theoretical calculations that give explanations that are plausible and internally consistent with known laws of physics. But experimental physicists are ingenious and in the past they have found ways to test predictions of extremely small effects that were thought to be almost impossible, such as Louis De Broglie’s idea that particles have wave-like properties and John Bell’s inequality to test his theorem concerning quantum non-locality. One hopes that they will devise clever ways to test these outcomes too so that we are not dependent only on theoretical plausibility.
The next post in this series will look at the calculations.
(To be continued.)
I’m no physicist, but I’m a little confused with the following sentence in the 6th paragraph:
Isn’t it “Both electric and magnetic fields…”
It’s been years since my electrodynamics courses, but as I understood it, a moving charge generates electric and magnetic field fluctuations in space-time. That, in fact, electric and magnetic fields are the same phenomena only mathematically orthogonal within the structure of space. But it’s been many years, and my understanding could easily be flawed.
flex, yeah, it’s ‘and’. A typo.
It’s an oversimplification to say the E (the electric field) and B (the magnetic field) are the same phenomenon. In a particular frame, they are each concretely defined. But the variations of each field (in time and space) depend to some extent on the other. This is codified in Maxwell’s equations.
In addition, in going from one inertial frame to another (in special relativity), E and B get ‘mixed up’ somewhat, much like position and time coordinates do.
flex,
Good catch! It was a typo that I have corrected.
As an example of what Rob says in #2, suppose in some frame there is a charge q at rest (v=0) in a region of space where the electric field E is zero and there is only a magnetic field B. The force F on the charge will be zero, since F=q(E +vxB). Hence it will have no acceleration.
But now if we observe this in a frame that is moving constant velocity with respect to the original frame, the charge will have a velocity but we find that both E and B transform to new values in this frame so that the force on the charge will still be zero. The acceleration will still be zero.
Thanks Rob, I was worried that I missed something.
Thanks for the refresher as well on Maxwell’s equations (although I always understood they were really the Maxwell-Heaviside equations). As a EE, I took a lot more courses in electrodynamics using partial differentials then theoretical physics using the Maxwell-Heaviside equations.
You know the ones: a round wire of diameter a and resistance r, is formed into a square shape of where each side is of length b, and in fixed rotation around an axis through the center of the square and intersecting two sides at the center of the sides (see accompanying illustration). The wire is rotating at a speed of v, and sitting in a non-linear magnetic field with equation T(x). Generate the equation for the current generated in the wire.
I couldn’t do it today, and I’m not even certain how I passed the courses then.
Thanks Mano #3,
I think I’ve got it. In relationship to a frame moving at constant velocity to the original frame, the E field may no longer be zero, and the B field will take a different value from what was observed in the original frame in order to keep the acceleration of the charge at zero.
And rather than just playing with numbers to make the accelerations appear the same, the measurements of both the E and B field from the second frame (if they are possible to be measured) would be the values generated by the mathematical transformation.
Thanks!
Doh! bolding failure in my comment. But I think you know what I meant.
[I fixed it.-Mano]
flex @4:
Well, Heaviside prettied them up with vector notation. That’s no small thing; four equations instead of twenty! But Heaviside’s contribution was mathematical. Maxwell had already done the physics. That’s why most physicists just say “Maxwell’s equations”.
Mano @3:
Ah, I hate to ask questions that turn out to be naive, but here I go again. You said, when E and v are zero:
But then you said:
The only way that could happen, if E and B (and q) have positive non-zero values, is if E=-vB, correct? So why is v always negative? I realise that velocity is directional, and you said the frame is moving relative to the original, but even if you choose a negative value for v you could envisage moving it in the opposite direction, making its value positive, resulting in an unsatisfiable equation. What did I miss this time?
file thirteen @8: Suppose you look at a frame moving with velocity v relative to the rest frame. Then the velocity of the particle as seen in this frame is −v.
For nonrelativistic speeds, the fields transform like;
E —-> E + vxB
B —-> B
Since E is zero in the original frame, we have just
E —-> vxB
So the force is, remembering that the particle’s velocity is −v;
q(vxB + (−v)xB) = 0
I should have had
B —-> B − (vxE)/c²
But since E is zero in the rest frame, the second term on the right is zero.
file thirteen @#8,
To add to what Rob says, since we are dealing with three vectors, it is not just the direction of the velocity that matters. Whatever the direction of the velocity, the vectors E and B adjust accordingly to give a net force of zero.
E and B are vectors! I knew my question had to be naive. Thanks again. 🙂
And thanks Rob, that makes perfect sense.
Me @12:
Just discovered that it’s convention to express vectors in bold. So E and B are vectors by definition. I thought it was just highlighting. 🙂