Need help on Equivalence principle

1. Dec 5, 2008

JustinLevy

Hello,
I can answer some basic questions about GR, but I don't teach it, and I am not proficient in it. Yesterday a student asked a question that I felt I couldn't do justice to.

Can you help?
I had the student email the question so I could ask around. So far I haven't gotten any useful answers from friends, so I'm trying here.

Another question, which he didn't include in the email was:
Would a charge in gravitational orbit radiate?
If it did, then I would think the equivalence principle would be false (an inertial observer would notice an anomolous force on charged particles), so I told him no. But he followed up by pointing out that even two neutral bodies spiral into each other in finite time. If they can lose energy gravitationally, why not electromagnetically?
I said, (and warned that I was probably incorrect and would double check), that if gravity effectively provided a drag force that sprialled them together, that locally you could not detect any 'anomolous force' since it would act on all particles in proportion to their mass and therefore the equivalence principle does not forbid this while it would for the electromagnetic case.

2. Dec 5, 2008

HallsofIvy

1) The inertial observer sees a force acting on the object and so see a change in its potential energy. It is potential energy changing to kinetic energy that accounts for the increase in kinetic energy.

2) A charge only radiates when its energy changes by a multiple of Plank's constant. It is the fact that electrons in orbit around a nucleus, even though in constant acceleration, do not continually radiate until they lose all energy that led to quantum mechanics.

3. Dec 5, 2008

JustinLevy

I'm sorry, but I don't see how this answers the question, as it seems to only reword it.

For example, a spring pushes a block and moves it. The force was in the direction of motion. It did positive work. The block gained kinetic energy. Because the force is conservative, we can reword this as potential energy changed to kinetic energy. Regardless, the spring is expending energy to accelerate the block. And the spring can provide a constant proper acceleration to the bock only for a finite amount of time.

Now go back to the book on the table example. Is the earth really expending energy, and really doing work on the book just sitting on the table ... and furthermore doing so for an infinite period of time, for the proper acceleration will stay the same potentially forever.

If the Earth really is doing positive work on a book just sitting there, where is this energy coming from?
When a system (such as the spring) gives up energy (goes to a lower potential) it loses mass (think for example bound nuclei weigh less than the nucleons separate), so as the earth gives up energy does it too lose mass?

Quantized for energy states means "discrete", not multiple of a fundemental 'quanta' (which is what the word means when we say charge is quantized). In quantum mechanics the separation between bound energy states can be arbitrarily small (even in your example of the hydrogen atom)... but yes there is still a discrete set of states, and therefore a discrete set of transitions. We don't know how to do quantum-gravity, and the question isn't fundementally a quantum mechanics one, so let's avoid QM for this question.

In this case, the question is one of classical mechanics ... assume GR + classical EM. The charge is in a gravitational orbit about a neutral body. Does it radiate?

I gave my answer and reasoning in the first post, but I am worried it is wrong.

4. Dec 5, 2008

Jonathan Scott

Both good questions.

The first has me somewhat confused, mainly because I'm not sure that you can explain something involving the whole Earth in terms of local inertial frames, but the usual key point here is to note that if an observer is falling in a gravitational field, his clock rate is decreasing with potential, so his view of any static object says that its energy is increasing correspondingly (matching its apparent kinetic energy). I'm not sure if that all works out when you take both SR and GR time dilation into account, but it seems promising.

The question of whether a charge radiates when accelerated by a gravitational field is quite tricky, but the general consensus is now that the answer simply depends on whether the observer is being accelerated by the same field. From the point of view of an observer in free fall next to the charge, there is no overall radiation of energy. From the point of view of an observer who sees the charge being accelerated, there is overall radiation of energy. This is consistent with the Principle of Equivalence and seems quite reasonable to me, as for example if you consider the energy of a charge and its field, then change its velocity, then the kinetic energy of the field changes as well as the kinetic energy of the charge, and that could be interpreted as being due to energy flowing from the charge into the field or vice versa.

5. Dec 5, 2008

Naty1

Seems like the energy comes from the gravitational field. A book doesn't seem held firm to the earth just because of an inertial observer's perspective. Are we mixing relativistic terms with classical reasoning??

I don't understand HallsofIvy answer. Different inertial observers see different KE's according to their relative motion. Sitting with the book in my lap there is no work being done because there is no change in distance...gravity exerts a force and the book remains in its original position; KE is zero, potential energy remains fixed.

Now I change to a free falling frame and pass right nearby the book. I do measure a fixed KE, don't I, dependent on my velocity. The potential energy still appears as it did previously. Seems to me the KE in this example comes from the apparent motion...So I'd think the energy comes from relativistic energy or linear momentum..I don't know how to decide which.

6. Dec 5, 2008

JustinLevy

That sounds impossible to me.
If a charge radiates, there is a back reaction on the charge. This would give the charge a non-zero proper acceleration. Therefore if the charge radiates, everyone knows. The inertial observer, even if he can't detect the radiation for some strange reason because he's travelling with it, would still measure the proper acceleration of the charge to be non-zero. There would appear to be an anomolous force on the charge, and this would break the equivalence principle ... as the laws of physics while free falling would no longer be locally equivalent to an inertial frame. No?

7. Dec 5, 2008

Naty1

Jonathan posted

This free falling frame explanation, I think, matches what I posted....in any case it provides a similar perspective regarding the relativity of KE.

8. Dec 5, 2008

Naty1

Why is it asssumed there is a non zero "proper acceleration"? (I'm ignoring earth's rotational acceleration here.)

If proper acceleration is the rate of change of proper velocity and that's change in position relative to change in time, seems like it should be zero...because position is fixed.

Wiki says " In the absence of any other forces, any object will accelerate in a gravitational field at the same rate, regardless of the mass of the object." Well with a stationary book, there IS a force pressing up against the book radially holding it in place...the force is Mg...so the books sits in place, like a responsible book would be expected to do...proper (physical) acceleration is zero.

Regardless, does this even have any effect on the susbequent questions...I don't think it's relevant...

Last edited: Dec 5, 2008
9. Dec 5, 2008

Naty1

yes, the earth's (magnetic) (edit: This should have said gravitational) field is expending energy to hold everything in place; no it is not doing any WORK because work requires displacement (change of position), and yes the proper acceleration remains the same....as I posted, I believe it to be zero....

The earth will lose an ever so tiny amount of mass over time expending it a gravitational energy....ignoring accumulating space dust...

To say it another way: the earth's gravitational field is expending energy to keep the attractive gravitational force applied to the book...if the book fell off a table, work would be done, it's potential energy would be reduced, its KE would increase during the fall and be dissipated as heat and FT=MV when it hit the floor...then the new equilibrium would be a stationary book in a higher gravitational, lower potential energy location .

Last edited: Dec 5, 2008
10. Dec 5, 2008

Jonathan Scott

That's what I originally learned too, but apparently it doesn't apply in this case. I think one of the important points is that for this equivalence to hold exactly, the acceleration must be constant. I think this was only convincingly clarified fairly recently. Some time ago when I was at an Open University summer school (probably around 1991) it was a current topic of discussion and someone explained to me the latest thinking on it, which I think still holds but has been somewhat refined. Unfortunately, although I remember being convinced at the time, I don't remember the argument!

I suggest Googling for relevant papers; I got some interesting-looking hits using "equivalence principle" "electric charge".

11. Dec 5, 2008

Jonathan Scott

Proper acceleration means the acceleration experienced by an observer at that point. From the GR principle of equivalence point of view, the book is experiencing proper acceleration.

12. Dec 5, 2008

Naty1

posted by Justin,

So an inertial observer sees a charge accelerating, KE energy increasing, as near light speed is approached and no radiation emission? I don't get that.

13. Dec 5, 2008

Jonathan Scott

It's the earth's GRAVITATIONAL field, it's not expending energy, the proper acceleration (at least from the Principle of Equivalence point of view) is non-zero and the Earth isn't losing mass by holding everything in place.

If there's anything else I've missed, you'd better assume it's wrong too.

14. Dec 5, 2008

JustinLevy

I'm not 100% sure what you are referring to here, as the part you quoted from me I was taking a statement and showing why I felt it led to a contradiction... so you could have been disagreeing with something I wrote there and that was exactly the point.

However, since you mistook coordinate acceleration for proper-acceleration earlier, let me point out that something freefalling will often have a non-zero coordinate acceleration (for example according to the rest frame of someone on the earth's surface) but the proper acceleration is zero for a free-falling object. This is because a freefalling frame is locally an inertial frame (by the equivalence principle). So no, the inertial observer following a free-falling charged particle will NOT see the charge accelerate, nor will the KE increase, etc.

Since the inertial observer doesn't see the charge accelerate, the charge should not radiate according to the inertial observer. I argued this means it can't radiate according to any observer since radiation would cause a back-reaction and everyone would see this effect, but apparrently this is disputed.

Well, most things that came up was the opposite situation. i.e. Does a charge sitting on the table radiate?

Regardless, there doesn't seem to be a consensus. There is a lot in the literature written on this, and I don't know enough (hence the openning post) to parse all this information and pare out the 'reliable' stuff.

I'm not sure what "it" is in that sentence. Are you saying there is no back-reaction due to radiation in this case for some reason? Or are you saying the equivalence principle doesn't hold for some reason here?

But we're primarily talking about freefalling observers and relating this to what they would see with a charge in an actual inertial frame. Here the proper-acceleration is indeed a constant, it is zero. On the other hand if we refer to an observer on the earth's surface, his proper-acceleration is also constant (although non-zero).

If a charge at rest in an inertial frame doesn't radiate, then by the equivalence principle a particle at rest in a free falling frame doesn't radiate either. Therefore a charge in a gravitational orbit should not radiate. I don't see what is wrong with that sequence of logic.

Last edited: Dec 5, 2008
15. Dec 5, 2008

Naty1

Jonathan posts
I was agreeing with your post ...that it seems like a contradiction....

I am unsure about how far the equivalence principle goes. I am comfortble with an acceleration of 1g in flat space creating a force equivalent to sitting in gravitational field of 1g. I also know that the equivalence principle has limits.

"the proper acceleration (at least from the Principle of Equivalence point of view) is non-zero..."
I agree the book experiences a force equal to 1g, but I don't understand how the local/proper acceleration is non zero...What is the proper velocity? looks to me like dx/dt is zero, as I posted previously...time changes but position doesn't...

And if the source of energy in a gravitational field is not mass, what is the source?? I'll see if I can find a source for that statement...we do agree the gravitational field has energy,right?

Bravo to whoever asked the original question(s)...

Last edited: Dec 5, 2008
16. Dec 5, 2008

Jonathan Scott

You may be right; at least some of the papers say that an electric charge falling or remaining at rest under gravity does not radiate at all as observed locally basically because its field is accelerated with it, whereas if you push the charged object (for example via an electromagnetic field), that causes changes to propagate through the field.

However, I don't think you can deduce that there would be no radiation from a charge in orbit just from the fact that a charge in free fall doesn't radiate as observed in the free fall frame. For the principle of equivalence to apply, the observer who tests for the presence of radiation has to be in the same frame. To extend this to a general free fall case, where the radiation is being measured by a static observer rather than a free fall one you would need to rely on the additional assumption that the existence of radiation can somehow be measured in an invariant way, and I'm not sure this is true, even though the quantum picture of radiation as discrete photons would seem to suggest it.

At least one paper that I've read suggested that an observer would calculate that energy was flowing in or out if the observer was in a frame where the charge appeared to be accelerating, and I think this is plausible. I also suspect that once you get to a scale where a single local inertial frame cannot apply, such as an orbit, then the acceleration is non-uniform and the charge MUST radiate. Consider a highly charged object orbiting a dense neutral mass. I would find it difficult to believe that it would not create an electromagnetic wave.

17. Dec 5, 2008

Fredrik

Staff Emeritus
I didn't read all the replies, so maybe this has been mentioned already. The answer to the student's question is surprisingly simple: There's no motion in the direction of the acceleration, so no work is performed. Proper acceleration is the coordinate acceleration in a co-moving local inertial frame, and the object's velocity in a co-moving frame is zero.

18. Dec 6, 2008

JustinLevy

That last statement seems vacuously true.
In the instantaneously co-moving inertial frame, the object's velocity is zero. Yes... that is always true.

But if you think that allows you to argue no work is done on the particle here, then that would be like taking the position that no work can ever be done on an object.

I don't believe that is a valid argument, since an instant later in any one of those inertial frames, work will be done on the object. To claim there is zero work you need to add the work as seen from a different frame for each instant, and it is not valid to just add coordinate dependent values from different frames.

Pick one freefalling frame and stick to it for a finite amount of time. The object will accelerate, and in the direction of motion. It appears work was done.

19. Dec 6, 2008

Naty1

This is an especially frustrating thread...A difficulty,I think, is that too many questions and concepts are involved...In hopes of resolving my own uncertainties and providng more of an asnwer to Justin, I'd like to discuss further.

I'll be the first to admit, I am still confused by equivalence as was posted by Justin...maybe we can discuss equivalence further??? Wikipedia has a barely decent discussion, but I did not find concrete application to this thread.

Maybe we can clear up some misconceptions...some may well be mine...I'll just pick two issues:

Jonathan posted:
I disagree: that does not, to me, represent the "equivalence principle". Here is, I believe, a correct formulation via Stephen Webb, OUT OF THIS WORLD, PAGE 35, which reflects my understanding: (and this appears consistent with that from other sources Smolin, Greene, for example)

This is in part why I posted earlier that I believed proper velocity and acceleration of the book were zero...also because there was no displacement. Hence no work is done since there is no displacement.
//////////////////////////
Original post: Justin:
I see three questions here: If the book is accelerated, does it take energy? Does the gravitational field have energy? If so, where does it come from. Yes,yes,unsure:

F = -dU/dr = -GMm/r^2 (Halliday and Resnick) so I say yes, acceleration requires a force and energy .

The gravitational field clearly has potential energy .

I previously posted the energy comes from earth mass...I read, but can't find, that the sun and earth come closer by (don't quote me on the exact figure here) 1 cm per hundred years due to gravitational radiation energy losses....don't know if it's true, of course, but I did not make up that answer on pure personal speculation. It's incomplete and even if "accurate" not very satisfactory for obvious reasons...I personally wonder (and this IS purely speculation) if the real source is the Higgs ocean (and/or vacuum energy) which imparts mass. I did not see an alternative explanation posted...anybody seen anything??

20. Dec 6, 2008

JustinLevy

You seem to be misunderstanding those quotes.
If you don't understand it intuitively, let me try just brute forcing the point using the logic from that quote:
Would you agree that an accelerated observer in the absence of a gravitational field has a non-zero proper acceleration? If so, then so too must an a non accelerated (non coordinate acceleration) observer in a gravitational field have a non-zero proper acceleration.

The terms in that quote are referring to coordinate quantities ("at rest", "accelerated", "non-accelerated"), and the part about 'absense of a gravitational field' has an implied inertial coordinate system.
Hopefully that helps some.

Last edited: Dec 6, 2008
21. Dec 6, 2008

Staff: Mentor

Hi Justin,

I think your student has hit on one of the big limitations of GR. It isn't the equivalence principle, per se, but rather how GR treats energy. It is pretty well-known that http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html" [Broken]. So basically if we look at a small scenario energy will be conserved, but once we get to the point that the curvature of spacetime becomes significant then energy becomes hard even to define.

So back to the student's scenario. If we look at the book on the table in a free-falling frame then the work is obviously done by the normal force of the table. No problem, work is done on the book whose KE increases, energy conserved. Now, where did the table get that energy? There is no chemical reaction, no electrical energy, no potential energy, etc., instead the energy comes from work being done on the table by the normal force from the floor. No problem, work is done on the table whose KE increases and which does work on the book whose KE increases, energy conserved. Now, where did the floor get that energy? ... This goes on until you get to the point where you are no longer able to neglect the curvature of spacetime and you can no longer speak about inertial frames and the equivalence principle at all and you can only talk about energy with great dificulty and with careful definitions.

I am not sure that will satisfy the student, but I think it is important to let them know the difficulties with talking about energy on a large scale in GR. Particularly if this particular student is perceptive enough to come up with this objection.

Last edited by a moderator: May 3, 2017
22. Dec 6, 2008

JustinLevy

While I don't understand the full subtly of energy conservation issues in GR, I do understand the basics and am aware of the issue. I feel you are using the 'energy issue' as a scape goat here. We have a static spacetime, and even a timelike Killing vector. Energy conservation is not an issue here.

I think I'm misunderstanding you here. As long as the curvature is finite, can't we always talk about a local inertial frame and the equivalence principle?

Last edited: Dec 6, 2008
23. Dec 6, 2008

Staff: Mentor

Huh? Energy conservation is obviously an issue here since it is the subject of the student's question.

No, the equivalence principle holds only when the curvature is negligible, i.e. ~0.

24. Dec 6, 2008

JustinLevy

But defining energy is not a problem here, because not only do we have a static spacetime, but we have a timelike Killing vector. So the problematic issue of defining energy in GR under general conditions is not a relevant issue here.

What!? Are you saying the equivalence principle only holds exactly in flat space-time?
That makes absolutely no sense to me. Please do explain.

25. Dec 6, 2008

atyy

Doesn't the question make sense in Newtonian gravity (Isn't the book gaining energy in a free-falling frame)? Or is there an aspect that is specific to GR?