# (Alpha) General relativity dissatisfies the equivalence principle

#### zanket

##### Human
Valued Senior Member
(Alpha) General relativity violates the equivalence principle

(The “Alpha” in the title indicates that the Alpha rules apply to this discussion.)

Here is part of the (strong) equivalence principle: “The outcome of any local experiment, whether gravitational or not, in a laboratory moving in an inertial frame of reference is independent of the velocity of the laboratory, or its location in spacetime.” I will show that general relativity (GR) violates this part, thereby violating the principle.

To be strictly correct, a tidal force exists in any frame larger than a point, hence an inertial frame can be no larger than a point and the equivalence principle applies to only a point. However, for the purposes of a given experiment the tidal force can be deemed negligible throughout a frame of any size. This explains how the equivalence principle itself can refer to a lab. The principle allows the following definition for an inertial frame: a frame in free fall and throughout which the tidal force is negligible (not nonexistent).

Let a lab be in free fall in an inertial frame “X”. A thrusting rocket moves freely within the lab, dragging a rope behind itself. The rope straddles the horizon of a black hole. The rocket hovers above the horizon. Assume that the rope will break before the rocket cannot withstand the rope pulling on it. Let “Y” be a copy of X, including its contents, with the sole exception that Y does not straddle a horizon.

GR predicts that anything below a horizon must fall. Then the rope in X must break, whereas the rope in Y need not break. X and Y differ only by the location in spacetime of their respective labs. Then in GR the outcome of a local experiment (e.g. a test of whether or not the rope breaks) in a lab moving in an inertial frame is dependent on the lab’s location in spacetime. And so GR violates the equivalence principle.

One may think that an inertial frame cannot straddle a horizon, due to an infinite tidal force there. But GR predicts that the tidal force is infinite only at the center of a black hole. Straddling a horizon, both X and the lab within it can be larger the larger the black hole is. X need only straddle a horizon temporarily to make the case above.

One may think that the rope in X cannot straddle the horizon. After all, GR predicts that the crew of the rocket cannot observe anything to cross the horizon from their vantage point above the horizon, and GR predicts that the directly measured distance between the rocket and the horizon is infinite. Yet GR allows an object to cross a horizon, and that’s all that’s needed to show that the rope can straddle the horizon. The rocket could have fallen freely until part of the rope crossed the horizon, and then thrust to stay above the horizon.

Wikipedia says, “general relativity ... is thought to be the only theory of gravity that satisfies the strong equivalence principle”. That is refuted above.

Last edited:
Hi zanket,
Great thread, and a worthy 3000th post!

In the interest of unbiased moderation, I wasn't going to participate, but since no one else has responded I can't resist! For this thread, possible violations will still be discussed in the Alpha rules thread, but I will defer to the consensus ruling of CANGAS and BillyT if there is a question of moderator bias.

In your scenario, I'm not sure that you have fulfilled the requirement that tidal forces are negligible.

The rope will break not because it crosses an event horizon, but because it is under more tension than it can take. You can be sure that the tension will be too strong if it the ship is hovering and the rope crosses an event horizon... but is that because tidal forces must be significant in that situation?

Thinking this through, I think the issue is that if the lab is large enough that the rope breakage will definitely be detectable within the lab, then perhaps tidal forces must be significant... because it means that the tension in the rope must be significantly different from one end of the lab to the other.

Last edited:
Ah... I've just had another thought. this one is much simpler:
Is the rope stretchable?
If it is stretchable, then it is not guaranteed that the rope will break if it crosses the EH.
If it is not stretchable, then it must break regardless of whether it crosses an EH or not.

If it is a practical rope (partly but not infinitely stretchable), then high enough acceleration of the rocket will still necessarily break the rope, EH or not, and if there is a difference in the two experiments it is because the lab is large enough that tidal forces must be non-negligible.

(Responding to zanket - Pete)
Claims about tidal forces have been stated with no proof, and, in fact, which seem to be outragious. I am not hitting this TARBABY unless it restated in rational form.

Brer Rabbit.

Last edited by a moderator:
(The “Alpha” in the title indicates that the Alpha rules apply to this discussion.)

the tidal force can be deemed negligible throughout a frame of any size.

I believe that the fiat statement of arbitrarily claiming that the logic of tidal effect can be revoked at will is unnecessarily obscure and vague and can easily lead to arguements which then have no basis.

If this thread start post is considered acceptable, then I can start a thread in which I state by fiat that a dropped object can fall either up or down at my choice.

If this thread start post is considered acceptable, then I can start a thread in which I state by fiat that momentum in an elastic collision can be transferred in any arbitrary direction according to my own whim.

Is this enough clarification for you, Pete?

By the way, anybody who wants to research the use of Lorentz frames in General Relativity, or who wants to refer to an even more prestigious source than Wikipeda might pick up the book "Gravitation" by Misner/Thorne/Wheeler out of Freeman publishing.

There is extensive description of the manner in which flat Lorentz frames have been adapted to function in curved Geometrodynamic metric space.

Part of my discontent with the thread start post was the ambiguous attitude expressed toward the relationship of flat Lorentz frames to curved General Relativity space. For many decades professional physicists have devised ways in which to seemingly successfully cope with the dichotomy of flat frames in curved space. "You can't argue with success."

And part of my discontent was my understanding that the thread starter stated by fiat that flat Lorentz frames are to be called curved in this thread. Have his cake and eat it too! If a dissenter argues with him one way, he can say "But I stated that you must call the flat frames curved!" And if a dissenter argues with him another way, he can say "But even if they have been making it work for forty years, they are wrong in theory because the flat frames are incompatible with curved space!".

A long time ago I learned to not play cards against a stacked deck.

Great thread, and a worthy 3000th post!
Thanks!

In your scenario, I'm not sure that you have fulfilled the requirement that tidal forces are negligible.
An inertial frame fulfills that requirement by definition. I gave a definition for an inertial frame in the original post (OP). An inertial frame can in principle exist everywhere in spacetime except for a central singularity (where GR itself breaks down). Here is some evidence for that:

From Three Basic Premises of General Relativity:

General relativity postulates that spacetime (the set of all events) is a smooth 4-dimensional Riemannian manifold M, where points are called events, with the properties A1-A3 listed below. A1: Locally, M is Minkowski spacetime (so that special relativity holds locally).

SR requires an inertial frame. It follows that an inertial frame can exist everywhere locally, even temporarily straddling (i.e. crossing) the horizon of a black hole. Lots of books and sites will tell you that you need not notice anything special (like a painful tidal force) when crossing a horizon. That’s because an inertial frame can cross one.

The rope will break not because it crosses an event horizon, but because it is under more tension than it can take. You can be sure that the tension will be too strong if it the ship is hovering and the rope crosses an event horizon...
Wikipedia confirms your thinking (search within for “rope”).

... but is that because tidal forces must be significant in that situation?
It might seem so. But X is an inertial frame, so by definition the tidal force throughout the frame is negligible. GR allows an inertial frame to cross a horizon, and the rocket + rope is an experiment in the lab in X. So the rope does not break due to a tidal force. If there is a contradiction there then that’s GR’s fault not mine. Elsewhere I prove that GR’s violation of the strong equivalence principle (SEP) leads to an internal inconsistency, but it’s out of scope in this thread.

Thinking this through, I think the issue is that if the lab is large enough that the rope breakage will definitely be detectable within the lab, then perhaps tidal forces must be significant... because it means that the tension in the rope must be significantly different from one end of the lab to the other.
Again it might seem so. But to pursue this line of reasoning you would need to show that GR does not allow an inertial frame to cross a horizon.

Ah... I've just had another thought. this one is much simpler:
Is the rope stretchable?
If it is stretchable, then it is not guaranteed that the rope will break if it crosses the EH.
The Wikipedia link above says: “[as the rope is pulled taut by the rocket,] the forces along the rope increase without bound as they approach the event horizon, and at some point the rope must break”. Presumably regardless how (finitely) stretchable it is. There is ambiguity in the OP though as to whether the rope would break in X before the rocket hits the wall of the lab. That can be corrected with a change to the wording; I'll think about the best way to do that.

If it is not stretchable, then it must break regardless of whether it crosses an EH or not.
This is confirmed in the Wikipedia link above, for a rod.

If it is a practical rope (partly but not infinitely stretchable), then high enough acceleration of the rocket will still necessarily break the rope, EH or not, ...
Agreed, that’s why I say in the OP that the difference between the outcomes in X and Y is the difference between “the rope must break” and “the rope need not break”.

... and if there is a difference in the two experiments it is because the lab is large enough that tidal forces must be non-negligible.
Throughout both labs the tidal force is negligible, given that X and Y are inertial frames and the labs are in free fall in those frames. Because I have validly used inertial frames, and the experiment is wholly within those frames in a lab in free fall, I have eliminated the tidal force as a factor that affects the outcomes of the experiments. About "large enough", think about this: If an inertial frame the size of a breadbox can cross the horizon of black hole J, then what size of an inertial frame can cross the horizon of black hole K that is 10^7000 times larger than J? (I just like that number.)

Last edited:
Zanket said:
the tidal force can be deemed negligible throughout a frame of any size.
I believe that the fiat statement of arbitrarily claiming that the logic of tidal effect can be revoked at will is unnecessarily obscure and vague and can easily lead to arguements which then have no basis.
By that logic I should fully explain every concept of GR that I use. For example, I should explain what a black hole is, what a horizon is, etc. That would be overkill. Just go ahead and dispute whatever you don’t agree with. Or ask me to support it, which I will within reason (e.g. I won’t explain what a black hole is).

I gave the definition for an inertial frame in the original post (OP). This definition agrees with other sources. Nothing in that definition puts an upper limit on the size of the frame. It follows that the tidal force can be deemed negligible (for the purposes of a given experiment) throughout a frame of any size.

And part of my discontent was my understanding that the thread starter stated by fiat that flat Lorentz frames are to be called curved in this thread. Have his cake and eat it too!
Presumably you are talking about my definition for an inertial frame. My definition agrees with that of Taylor, Thorne, and Wheeler, all top relativity physicists.

The resolution to Zanket's thought experiment is that he mixes up the distinctions between "curvature" and "metric". While the "curvature" across a horizon may be negligible, the presence of the horizon is encoded into the metric. The mere presence of a horizon changes the outcome of experiments, which is expected, even though the tidal forces may be negligible. ("Tidal forces" and "curvature" are interchangable, but "metric" and "curvature" most certainly are not!)

An illustration is at my website:

wwwDOTbaylorDOTedu/~Ben_Dundee/horizonDOTpdf

(Sorry for the DOTs, but I don't have enough posts to be able to give links!)

Because the metrics at r1 and r2 are inequivalent (and thus ds1 and ds2), there is no way to define equivalent reference frames. X and Y cannot be compared using the equivalence principle.

Note: The fact that Y is not straddling the horizon is irrelevant, and the logic still holds even if r2 approaches the horizon, which is located at 2Gm.

Awesome. Thanks.

In the interest of transperancy, I followed Zanket here after pointing out the flaws with his logic. Anyone interested in the debate can find it at ilovephysics(DOTcom) forums.

The mere presence of a horizon changes the outcome of experiments, which is expected, even though the tidal forces may be negligible.
Welcome to sciforums, Ben. Lots of questions for you.

Do you mean “presence” literally, like “existence”, or do you mean “proximity”? If the former, then do you agree that an experiment is in the presence of a horizon if one exists anywhere in the observable universe of the experimenter?

The strong equivalence principle says that the outcome of a local experiment in a lab moving in an inertial frame is independent of the lab’s location in spacetime. Here you are effectively saying that the outcome is in fact dependent on that. Then you agree that GR violates the SEP, right? If not, why not?

Because the metrics at r1 and r2 are inequivalent (and thus ds1 and ds2), ...
Your terminology is off there. The two metrics are the same in your illustration; they are both the Schwarzschild metric. A metric is a function. You input different values into the same metric, and got different results (ds1 and ds2).

... there is no way to define equivalent reference frames. X and Y cannot be compared using the equivalence principle.
According to your illustration and your conclusion, you think two inertial frames at different r-coordinates are incomparable. What is your support for this?

In your illustration ds1 and ds2 could be equivalent even when r1 and r2 differ, by adjusting the M’s so that they differ. Would X and Y be comparable then? If not (i.e. if the criterion for comparability is more than just ds1 = ds2), what more precisely are the conditions under which you think the inertial frames are comparable? Like, must the object represented by M be the same object in both frames?

If X and Y approach each other, at what point do they suddenly become comparable? Like, how many significant digits must their r-coordinates share?

Also you said at the top “The mere presence of a horizon changes the outcome of experiments”. But your subsequent text declared only that different r-coordinates mean incomparable inertial frames. How does a horizon fit into that?

According to GR:

From Three Basic Premises of General Relativity:

General relativity postulates that spacetime (the set of all events) is a smooth 4-dimensional Riemannian manifold M, where points are called events, with the properties A1-A3 listed below. A1: Locally, M is Minkowski spacetime (so that special relativity holds locally).

And according to SR:

From Special relativity – Wikipedia:

First postulate - Special principle of relativity - The laws of physics are the same in all inertial frames of reference. In other words, there are no privileged inertial frames of reference.

So SR holds locally in GR, and in SR the laws of physics are the same in all inertial frames. Then the laws of physics must be the same in all inertial frames in GR too. How do you explain that the laws of physics are the same in two inertial frames but the outcomes of experiments in those frames can be affected by the frame’s location in spacetime or the presence a horizon?

zanket said:
Let a lab be in free fall in an inertial frame “X”. A thrusting rocket moves freely within the lab, dragging a rope behind itself. The rope straddles the horizon of a black hole. The rocket hovers above the horizon.

Then the rocket must be accelerating constantly. The rope is attached to the rocket. The rest frames of both are non-inertial.

Assume that the rope will break before the rocket cannot withstand the rope pulling on it. Let “Y” be a copy of X, including its contents, with the sole exception that Y does not straddle a horizon.

So, Y is at a different distance from the horizon of a black hole than X is. Therefore, the acceleration of the rocket will be different to keep it "hovering".

On the other hand, perhaps Y has no horizon at all, in which case the rocket is not stationary but is still presumably accelerating.

In either case, if the rocket is accelerating at the same rate, that will create the same stress on the rope, and the rope will break if the forces on it are too great.

Then the rocket must be accelerating constantly. The rope is attached to the rocket. The rest frames of both are non-inertial.
Yes to all. The rocket + rope is an experiment in the lab. The lab is in free fall in an inertial frame (X or Y). The lab’s frame is inertial. The rocket’s noninertial frame is irrelevant.

So, Y is at a different distance from the horizon of a black hole than X is. Therefore, the acceleration of the rocket will be different to keep it "hovering".
Y is defined as a copy of X including its contents, with the sole exception that Y does not straddle a horizon. Then the rocket accelerates the same in Y as it does in X. My intent with the wording was that Y doesn’t necessarily have something to do with a black hole; the rocket in Y doesn’t necessarily hover above anything. But I see that the wording can mislead. I’ll think of the best way to clarify it.

In either case, if the rocket is accelerating at the same rate, that will create the same stress on the rope, and the rope will break if the forces on it are too great.
The stress on the rope wouldn’t be the same.

In X the rope must break regardless of its (finite) strength. (Search here for “rope must break”. The rocket will pull the rope taut—and keep pulling—simply by hovering.) In the lab’s frame, the end of the rope below the horizon is falling relative to it (GR’s prediction), and the other end is at rest with respect to it. Then the rope is being stretched that way; its proper length is increasing. No matter how hard the rocket pulls the rope, and no matter the rope’s strength, the rope could never be fully reeled into the rocket.

Whereas in Y the rope need not break if it is strong enough. The only stretching force on the rope is due to the rocket alone. In Y the proper length of the rope need not increase. The rope could possibly be fully reeled into the rocket. (The ability to fully reel in the rope would also be a different outcome.)

Last edited:
Hi Zanket---you have frustrated me once, I won't let it happen again Last time I made my posts too long and you didn't read them I think. This time I will make my responses more succinct. That being said, I have just realized that this is a long post. Sorry, but I have responded specifically to your questions, so I think it is ok.

In order to simplify things, we should take an empty universe accept for a Schwarzchild black hole located at r = 0. Also look at the picture in the link. Please.

First to James:
In either case, if the rocket is accelerating at the same rate, that will create the same stress on the rope, and the rope will break if the forces on it are too great.

This was one of my first thoughts, too, but I don't think it has to be the case, for example, one can have a sufficiently large radius, and thus sufficiently small accelerations.

Do you mean “presence” literally, like “existence”, or do you mean “proximity”? If the former, then do you agree that an experiment is in the presence of a horizon if one exists anywhere in the observable universe of the experimenter?

You should take presence to be proximity.

Your terminology is off there. The two metrics are the same in your illustration; they are both the Schwarzschild metric. A metric is a function. You input different values into the same metric, and got different results (ds1 and ds2).

Your terminology is off! (also thanks for looking at the picture.) The metrics are both the Schwarzchild metric, but if one is to do a calculation, the metric must be evaluated at a specific point in space-time. I input two different values into the metric because that's what it wanted.

According to your illustration and your conclusion, you think two inertial frames at different r-coordinates are incomparable. What is your support for this?

The support is that the metric wants two different coordinates. The metric tells you the global structure of space-time, but it is a function of the space-time coordinates. This means that the metric is also a local function of space-time.

In your illustration ds1 and ds2 could be equivalent even when r1 and r2 differ, by adjusting the M’s so that they differ. Would X and Y be comparable then?

Yes, one could adjust m, but then we are changing the initial assumption. The black hole cannot change its mass. Let us assume that we have a Schwarzchild black hole of constant mass. Otherwise we have to worry about all kinds of new things.

If not (i.e. if the criterion for comparability is more than just ds1 = ds2), what more precisely are the conditions under which you think the inertial frames are comparable? Like, must the object represented by M be the same object in both frames?

I think that ds1=ds2 is not quitesufficient for comparison---one must also have the same causal structure. One cannot compare an experiment done across the horizon of a black hole to some other experiment preformed somewhere else because of the causal structure of the horizon. (I'm affraid that you will ask me to define "causal structure" in more detail. I will have to think about a sufficiently non-technical definition!) One can transform one frame to the other, this I will talk about in more detail below.

If X and Y approach each other, at what point do they suddenly become comparable? Like, how many significant digits must their r-coordinates share?

Let us not worry about significant digits, this is a complication. But, needless to say, you can formulate these claims using "to first order" or "to second order", and talk about verification or violation of the SEP to arbitrary accuracy.

Also you said at the top “The mere presence of a horizon changes the outcome of experiments”. But your subsequent text declared only that different r-coordinates mean incomparable inertial frames. How does a horizon fit into that?

The horizon is located at a specific r. Here's an example which may (or may not) illuminate the matter. One could compare the metrics of a very large star and a very massive black hole. If one chose numbers judiciously enough, they could compare the curvature across the horizon of the black hole and somewhere in the space-time of the star and find the same values. The curvatures are the same, but the fact that there is a horizon sitting across your reference frame will interfere with world lines and Penrose diagrams and such. The causal structure of the two space-times is different. Search for "Penrose diagrams" on wikipedia. (Your moderators, in their infinite wisdom, will not afford me the privlidge of linking you directly there just yet No offense--I know the reasons.)

So SR holds locally in GR, and in SR the laws of physics are the same in all inertial frames.

Hmmm. Ok. This sounds reasonable. But one has to be careful that he's not sitting at a singularity. I think that the way these arguments work is that the metric and its first derivative must both be continuous so that one may define a good tangent space. But ok, in most cases I'll agree with this.

Then the laws of physics must be the same in all inertial frames in GR too. How do you explain that the laws of physics are the same in two inertial frames but the outcomes of experiments in those frames can be affected by the frame’s location in spacetime or the presence a horizon?

Perhaps you are missing a step of logic here. Let me think.
1)The laws of physics are good in all inertial reference frames in SR.
2)SR holds locally in GR.
3)Therefore, the laws of physics must be good in all inertial reference frames.

Ahh yes here it is. From your same link "Three basic principles"...

A3 (Strong Equivalence Principle) All physical laws that hold in flat Minkowski space (ie. "special relativity") are expressible in terms of vectors and tensors, and are meaningful in the manifold M, continue to hold in every frame (provided we replace derivatives by covariant derivatives).

I added the emphasis.

One must use the covariant derivative, which depends on the metric. I have forgotten this before (oops). Sorry but I need a few equations, if for no other reason than to establish my LaTeX proficiency...

$$\mathcal{D}_{a} A^b = \partial_{a}A^b + \Gamma^b_{ac} A^c$$

where A is some four vector and the capital Gamma is called a Christoffel symbol, the "backwards 6" is a partial derivative (i.e. derivative in flat space) and for the sake of completeness...

$$\Gamma^i_{kl} \sim \frac{1}{2} g^{im}\partial_l g_{mk} + ...$$

Here, g is the metric. What does all of this mean? You have to replace the derivative in SR with the covariant derivative in GR. The covariant derivative tells you how to get from one frame to another (it encodes the holonomy of your Rimannian manifold $$\mathcal{M}$$) via the Lorentz Transformations. The Lorentz transformation of a four vector is a funtion of the derivatives of that four-vector. In GR, those derivatives depend on the metric.

So what does that mean for our logic above? I think we should replace (3) with "Therefore, the laws of physics are good locally in GR." What are the implications of this? Suppose I do an experiment in X, and you do the same experiment in Y. Locally, supposing we have inertial frames, we don't notice anything out of the ordinary. I see the rocket flying around no problem, and you see the rocket's rope break (perhaps). We call each other and you conclude---ahh, obviously the SEP is violated. I contest, saying, well, first we should connect your frame to mine with a Lorentz Transformation, and then see what happens. Because we will have to take a derivative in doing this, we will immediately recognize that the covariant derivative is now a function of the metric, which is a local function of space-time. The metric encodes the information about the horizon, and we will notice this in transforming from my frame to yours.

It is hard to see how a simple Lorentz transformation will repair the rope! But now exchange "broken rope" for "measurement of quantum process", and it is less hard to swallow.

So, to recap. Because the metric in each frame is different (i.e. the Schwarzchild metric evaluated locally at different space-time coordinates), we expect that experiments will yield different results. That's ok, because in order to make the results of the two frames consistent, one only need to apply a Lorentz transformation from X to Y, which depends on the covariant derivatives, and thus the metric of the space.

The situation is completely analagous to measuring, say, the lifetime of a pion. In the lab, one measures a pion at rest to have a lifetime of something on the order of nanaoseconds. But pions are produced in the upper atmosphere with large momentum, and can be observed decaying at the surface of the earth, giving them a lifetime of seconds. The solution is to connect the two frames with a suitable Lorentz Transformation. After accounting for the time dialation experienced by the relativistic pion, one concludes that the two sets of results are consistent.

Sorry for the long post. Hopefully anyone who was interested has followed this. I will strive in the future to make my replies shorter.

Thanks Ben. Before I respond in full, what about this question I asked above?:

Zanket said:
The mere presence of a horizon changes the outcome of experiments, which is expected, even though the tidal forces may be negligible.
The strong equivalence principle says that the outcome of a local experiment in a lab moving in an inertial frame is independent of the lab’s location in spacetime. Here you are effectively saying that the outcome is in fact dependent on that. Then you agree that GR violates the SEP, right? If not, why not?

I think, if you take another statement of the SEP (from the website you referenced), namely

A3 (Strong Equivalence Principle) All physical laws that hold in flat Minkowski space (ie. "special relativity") are expressible in terms of vectors and tensors, and are meaningful in the manifold M, continue to hold in every frame (provided we replace derivatives by covariant derivatives).

You're forgetting the imporatant caveat (that I also forgot) that we replace derivatives with covariant derivatives. The covariant derivative tells us how to get from one frame to another, and how to transform measurements between frames (the Lorentz Transformations). An example is the pion lifetime measurements, or any of Einstein's original thought experiments that were resolved by simply taking the Lorentz transform between frames.

Ben, your argument seems to be illogical. There’s no need to continue down your path if I can prove that, so first I want to see if you can thwart me.

You say that X is incomparable to Y. But X is comparable to X. Wholly within the lab in X, let there be X2 and Y2, just like X and Y respectively, only smaller. Then X2 must be comparable to Y2, or else X cannot be comparable to X. But according to your criterion (different r-coordinates) they are still incomparable. By the rules of logic, they cannot be both comparable and incomparable.