Is the Equivalence Principle Contradictory in General Relativity?

[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.

I am confused about Einstein's equivalence principle.

Consider a book sitting on the Earth's surface. The book has a non-zero proper acceleration. But I thought it took work to accelerate an object. Specifically, from the point of view of a freefalling observer (an inertial observer according to the equivalence principle) there is a net force on the object and the object is traveling in the direction of the force ... work = force * distance ... and the book is gaining kinetic energy according to the inertial observer. Where is the energy coming from that allows the Earth to continually accelerate the book?

Furthermore, since it is the proper acceleration that is non-zero, all observers and coordinate systems will agree the book is indeed accelerating. I must be misunderstanding the equivalence principle as it seems to lead to contradictions.

Please help,
[ name omitted for privacy ]

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 anomalous 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.


Please help. Since my responses are most likely incorrect.

 

[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.

 

[JustinLevy]

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.

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?

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.

Quantized for energy states means "discrete", not multiple of a fundamental '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.

 

[Jonathan Scott]

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 anomalous 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.


Please help. Since my responses are most likely incorrect.

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.

 

[Naty1]

Consider a book sitting on the Earth's surface. ... Where is the energy coming from that allows the Earth to continually accelerate the book?

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.

 

[JustinLevy]

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.

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 traveling with it, would still measure the proper acceleration of the charge to be non-zero. There would appear to be an anomalous 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?

 

[Naty1]

Jonathan posted

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.

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

 

[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...

 

[Naty1]

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.

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 .

 

[Jonathan Scott]

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 traveling with it, would still measure the proper acceleration of the charge to be non-zero. There would appear to be an anomalous 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?

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".

 

[Jonathan Scott]

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...

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.

 

[Naty1]

posted by Justin,

The inertial observer, even if he can't detect the radiation for some strange reason because he's traveling with it, would still measure the proper acceleration of the charge to be non-zero. There would appear to be an anomalous force on the charge, and this would break the equivalence principle .

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.

 

[Jonathan Scott]

yes, the Earth's magnetic 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...

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.

 

[JustinLevy]

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.

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.

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

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.

That's what I originally learned too, but apparently it doesn't apply in this case.

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?

I think one of the important points is that for this equivalence to hold exactly, the acceleration must be constant.

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.

 

[Naty1]

Jonathan posts

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.

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)...

 

[Jonathan Scott]

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.

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.

 

[Fredrik]

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.

 

[JustinLevy]

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.

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 anyone 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.

 

[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:

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.

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)

There is no difference between an accelerated observer in the absence of a gravitational field and a non accelerated observer with a gravitational field...We can say all observers, no matter what their state of motion, can consider themselves to be at rest as long as they include a suitable gravitational field.

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:

Where is the energy coming from that allows the Earth to continually accelerate the book?

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??

 

[JustinLevy]

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)

There is no difference between an accelerated observer in the absence of a gravitational field and a non accelerated observer with a gravitational field...We can say all observers, no matter what their state of motion, can consider themselves to be at rest as long as they include a suitable gravitational field.

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.

 

[Dale]

Consider a book sitting on the Earth's surface. The book has a non-zero proper acceleration. But I thought it took work to accelerate an object. Specifically, from the point of view of a freefalling observer (an inertial observer according to the equivalence principle) there is a net force on the object and the object is traveling in the direction of the force ... work = force * distance ... and the book is gaining kinetic energy according to the inertial observer. Where is the energy coming from that allows the Earth to continually accelerate the book?

Furthermore, since it is the proper acceleration that is non-zero, all observers and coordinate systems will agree the book is indeed accelerating. I must be misunderstanding the equivalence principle as it seems to lead to contradictions.

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" . 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.

 

[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.

... 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

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?

 

[Dale]

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.

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

As long as the curvature is finite, can't we always talk about a local inertial frame and the equivalence principle?

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

 

[JustinLevy]

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

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.

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

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

 

[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?

 

[atyy]

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.

Doesn't the Killing vector only imply energy conservation for free-falling particles (geodesic trajectories)?

 

[atyy]

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 anomalous 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?

There are some ways of resolving the accelerated charge and equivalence principle question, eg. in Peierls's "Surprises in Theoretical Physics". But I think they are wrong, and the equivalence principle does not apply to accelerated charges. There's a list of references here: https://www.physicsforums.com/showthread.php?p=1921540, to which one may add Eric Poisson's http://relativity.livingreviews.org/Articles/lrr-2004-6/ . The article by Harpaz and Soker has interesting comments on where the energy is coming from, but I have not followed the argument in detail for myself: http://arxiv.org/abs/physics/9910019.

 

[Dale]

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

Yes. That is correct. The equivalence principle is, roughly speaking, that anytime you have a flat spacetime (no tidal effects) you cannot experimentally distinguish between "gravity" and "acceleration". Whenever you see the word "local" in a GR context, it is talking about a region of spacetime that is small enough to be considered flat (tidal effects within the region are too small to be measured).

http://scintilla.nature.com/node/513854": "Locally, the effects of gravitation (motion in a curved space) are the same as that of an accelerated observer in flat space."

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html" : "So modern usage demotes the uniform "gravitational" field back to its old status as a pseudo-field. And the hallmark of a truly GR problem (i.e. not SR) is that spacetime is not flat. "

http://en.wikipedia.org/wiki/Equivalence_principle" : "If an observer measures a patch of space to be flat, then the strong equivalence principle suggests that it is absolutely equivalent to any other patch of flat space elsewhere in the universe."

 

[Dale]

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?

It is a little different in Newtonian gravity where gravity is considered a real force and therefore a free-falling frame is non-inertial, but I think the final conclusion is similar.

In a Newtonian free-falling frame there is a fictitious force that counteracts gravity so that the free-falling object does not accelerate and no net work is done on a free-falling object. So a book on a table in a free-falling frame experiences 3 forces, gravity, the normal force, and the fictitious force. These result in a net force on the book, which accelerates, and gains KE.

 

[Fredrik]

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.

No it's not. If a rocket in free fall switches on its engines, the velocity and the acceleration will be in the same direction. The engine is clearly doing work in this case.

I don't believe that is a valid argument, since an instant later in anyone 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.

Any metric in GR has the property of local Lorentz invariance. That means that the components of the metric in a local inertial frame are approximately $\eta_{\mu\nu}$ in a small region of spacetime that contains the origin. The approximation becomes exact in the limit where the size of that region goes to zero. When you use the local inertial frame to describe a finite region of spacetime (i.e. when you don't take that limit), things won't look the same as in SR. It will be more like the free-falling frame in Newtonian mechanics that DaleSpam and atyy discussed. The work is being done by a fictitious force.

 

[Jonathan Scott]

In my experience, the principle of equivalence is very robust, and only fails when the zone in which the observation is being made is large enough to detect significant tidal effects or variations in the direction of the acceleration.

In the case of the book on the table, this is locally just like a book on a table in a spaceship accelerating upwards relative to the table at 1g. A free-falling observer in that spaceship will see the kinetic energy of the book increasing simply because it is being pushed by the spaceship engines, but an observer sitting beside the book will see the falling observer being accelerated by a fictitious force which looks just like gravity.

When this is converted to the Earth gravity picture (for example by landing the spaceship), locally there will be no difference (and I think this applies even if electric charges are involved). From the falling observer's point of view, the book will appear to acquire kinetic energy. My opinion is that the question about the ultimate source of the apparent energy in that case cannot be answered under the Principle of Equivalence because we cannot extend our local inertial frame to include the whole Earth.

 

[JustinLevy]

atyy,
Thanks for the links, but when I search published literature there appear to be contradictory views on the charged particle problem. Even you disagree with some of the things you linked.

As stated I do not have the GR skills to pare good published resources vs. probably incorrect publications (yes, incorrect papers do get published). So links to archiv papers just makes me even more wary. I don't know enough to figure out what 'authority' is correct. That is the problem.

Is the following two sentences a fair summary of a response to the charged particle question?

There is still scientific debate on these topics, so the question cannot be definitively answered.
Current consensus is that the orbitting electric charge will spiral inwards, but the concept of radiation is not agreed on with an inertial observer claiming only charges with non-zero proper acceleration radiate and observer's having non-zero proper acceleration can claim otherwise.

Yes. That is correct. The equivalence principle is, roughly speaking, that anytime you have a flat spacetime (no tidal effects) you cannot experimentally distinguish between "gravity" and "acceleration". Whenever you see the word "local" in a GR context, it is talking about a region of spacetime that is small enough to be considered flat (tidal effects within the region are too small to be measured).

Then the equivalence principle is not a princple of GR, but a principle of SR. Since while a spacetime point can always be given the -1,1,1,1 metric, the curvature is intrinsic and even at zero size the point region isn't equivalent to the flat spacetime of a Rindler observer in an accelerating rocket.

That doesn't sound right to me. I thought the equivalence principle referring to acceleration is a GR principle.

Can someone state mathematically what the principle is?
I think this would cut through any mis-communication that may be due to word choice issues.
This is also probably the easiest way for me to correct my understanding.

No it's not. If a rocket in free fall switches on its engines, the velocity and the acceleration will be in the same direction. The engine is clearly doing work in this case.

The velocity and acceleration according to who?
The velocity of the rocket in an instantaneously co-moving inertial frame is still zero. Hence why it is a vacuous truth. If you are going to argue there isn't work being done on the book because of this arguement, then the engine is not doing work on the rocket by the same argument. You can't have it both ways.

Different frames are not going to agree on energy. But they will still agree energy is conserved. According to a freefalling frame there is work being done on the book. Where is this energy coming from?

Any metric in GR has the property of local Lorentz invariance. That means that the components of the metric in a local inertial frame are approximately $\eta_{\mu\nu}$ in a small region of spacetime that contains the origin. The approximation becomes exact in the limit where the size of that region goes to zero. When you use the local inertial frame to describe a finite region of spacetime (i.e. when you don't take that limit), things won't look the same as in SR.

Consider a coordinate system with the metric diagonal -1,1,1,1 right at the point of the book (make the book a point particle if you want) and in which the book has a non-zero velocity. Sure, I realize that due to spacetime curvature the metric can't be exactly that of flat spacetime in a finite region, but I only need to talk about a region of size limiting to zero. What is the velocity and acceleration of the book right at that point where the metric is -1,1,1,1 ? It can, according to an inertial observer, give results showing that work is being done on the book. Where is this energy coming from?

 

[Dale]

Then the equivalence principle is not a princple of GR, but a principle of SR.

That's fine. Classify it however you will, any principle of SR is also a principle of GR.

Since while a spacetime point can always be given the -1,1,1,1 metric, the curvature is intrinsic and even at zero size the point region isn't equivalent to the flat spacetime of a Rindler observer in an accelerating rocket.

Let's say you have a measurement technique that is sensitive to curvature (tidal effects) greater than a finite magnitude k, then it is always possible to define a region of finite size s such that the tidal effects within that hypervolume are less than k. Such a region can be treated as flat.

 

[atyy]

atyy,
Thanks for the links, but when I search published literature there appear to be contradictory views on the charged particle problem. Even you disagree with some of the things you linked.

As stated I do not have the GR skills to pare good published resources vs. probably incorrect publications (yes, incorrect papers do get published). So links to archiv papers just makes me even more wary. I don't know enough to figure out what 'authority' is correct. That is the problem.

All the papers I referenced were from excellent sources, not merely posted on arXiv, but in peer-reviewed journals with good reputations (eg. Foundations of Physics, American Journal of Physics, Physical Review; except the Chiao article, which is published in a crackpot volume edited by Physics Nobel laureates). So I agree there is no consensus, at least not at the level for Maxwell's equations or Taylor and Hulse's observations and analyses.

Is the following two sentences a fair summary of a response to the charged particle question?

There is still scientific debate on these topics, so the question cannot be definitively answered.
Current consensus is that the orbitting electric charge will spiral inwards, but the concept of radiation is not agreed on with an inertial observer claiming only charges with non-zero proper acceleration radiate and observer's having non-zero proper acceleration can claim otherwise.

I believe there is little disagreement remaining over the physics (predictions of experimental outcomes), but there is substantial disagreement over when the "equivalence principle" applies (because the predictions can be made using Einstein's and Maxwell's field equations, without the "equivalence principle"). I would like to suggest this modification of your summary:

"There is still scientific debate on these topics, so the question cannot be definitively answered. There is consensus that: the orbitting electric charge will spiral inwards; the ability to detect radiation is observer dependent; a comoving observer will not detect radiation in at least some situations; some non-comoving observers will detect radiation. There is no consensus on whether this may be derived from or is consistent with various forms of the equivalence principle".

OK, now some points which I don't expect agreement on. The orbitting charge is not falling freely, hence the comoving observer is not inertial. There is a loose, "heuristic", "historical" form of the equivalence principle stated from outside GR. There is a "mathematical" equivalence principle stated from within GR that (i) in curved Lorentzian spacetime at any point (or sufficiently close) one may set up a coordinate system in which the metric is exactly flat Lorentzian at the point, and deviates from flatness away from that point only at or above second order in Taylor series (ii) the known fundamental laws of physics (Maxwell's equations), but possibly not the derived laws involving second derivatives, in curved Lorentzian spacetime at that point have the same form as those in flat Lorentzian spacetime.

 

[Fredrik]

The velocity and acceleration according to who?
The velocity of the rocket in an instantaneously co-moving inertial frame is still zero. Hence why it is a vacuous truth. If you are going to argue there isn't work being done on the book because of this arguement, then the engine is not doing work on the rocket by the same argument. You can't have it both ways.

You're definitely right about this. I don't know how I was able to confuse myself enough to make such a claim.

Consider a coordinate system with the metric diagonal -1,1,1,1 right at the point of the book (make the book a point particle if you want) and in which the book has a non-zero velocity. Sure, I realize that due to spacetime curvature the metric can't be exactly that of flat spacetime in a finite region, but I only need to talk about a region of size limiting to zero. What is the velocity and acceleration of the book right at that point where the metric is -1,1,1,1 ? It can, according to an inertial observer, give results showing that work is being done on the book. Where is this energy coming from?

Man, I feel dumb right now. The equivalence principle, as I stated it in my previous post, applies to any local inertial frame, not just the ones that have a time axis that's tangent to the world line of the object that concerns us. And, there's obviously motion in the direction of the acceleration in some of them. D'oh.