General Relativity Basics: The Principle of Equivalence

[Mentz114]

But basically, the reason we know that space-time is curved is that clocks appear to run at different rates depending on their gravitational potential.

From a post in another topic

https://www.physicsforums.com/showthread.php?t=440657&page=3

 

[granpa]

why didnt you give the link to the post?
https://www.physicsforums.com/showthread.php?p=2951148#post2951148

Gravitatonal time dilation is a REALLY BIG clue that space-time is curved. If you think of gravity as just being a force, there's no reason it should affect how clocks operate. But we observe that it does. Recent experiments can detect the gravitational time dilation for a height difference as small as a foot.

When we draw the resulting space-time diagram, we have diagrams that look like they should be parallelograms, but the sides are of unequal length. (The "sides" of unequal length in this diagram are the clocks which run at different length because it's a space-time diagram).

This isn't possible in an Eulidean geometry, and suggests (as most textbooks will mention) that space-time is curved.

To actually calculate a curvature invariant and show that it is nonzero is a bit more involved, but it's certainly possible in principle. It involves measuring the metric accurately enough to get the second partial derivatives. This will rule out some special cases such as being in an elevator. But basically, the reason we know that space-time is curved is that clocks appear to run at different rates depending on their gravitational potential[/color].

 

[Mentz114]

why didn't you give the link to the post?

Because it's at the top of the page ( in my browser) and you get straight to it.

What are you ? The link police ? If you're going to nit-pick, at least choose something worth the trouble.

 

[granpa]

it wasnt even on my page

it was on page 2

 

[Mentz114]

it wasnt even on my page

it was on page 2

OK, sorry I snapped. But the link I've given is clearly to page 3. Anyway, I don't agree with everything in that post, I just thought the last sentence answered JDoolin's question.

Peace.

 

[granpa]

I know that the link you gave was to page 3. But there are different display modes and for the display mode that I use the post that you intended to link to was on page 2.

 

[PeterDonis]

If I drop something out the window, it falls down, and approaches a point c²/g below, but it will never reach there. The clocks down at that level (x=0) are not ticking at all and nothing can fall past them. If the rocket extended down that far, the gravity there would approach infinity.

I was OK with your post up until this point. It would be OK to say that the rocket cannot extend to x = 0, but it is not OK to say that "clocks at that level are not ticking at all and nothing can fall past them". The reason is simple: if you drop something out the window, it *does* fall past "a point c^2/g below", and its clock continues to tick. However, even this way of stating it is, in my opinion, misleading, which is why I put the phrase "a point c^2/g below" in quotes.

Here's what I think is a better way to describe what's going on: consider a light beam emitted from the event x = 0, t = 0 in the global inertial frame (what I think we were calling the "lab frame" above, as opposed to the "rocket frame"). No observer on the rocket, on any of its floors, will ever see this light beam; it will never catch up to any portion of the rocket (assuming that the rocket, with all its floors, maintains its acceleration forever). This fact is what justifies us referring to the light beam's worldline--the line x = t--as a "horizon" with respect to the rocket. However, a free-falling observer who is dropped off any floor of the rocket *will* see this light beam, after a finite amount of proper time by the free-falling observer's clock. From the rocket's point of view, the free-falling observer's clock will *appear* to move slower and slower as the observer approaches the horizon, but no rocket observer will ever see the free-falling observer cross the horizon (because light rays from events on the horizon will never catch up to the rocket, as above).

(Edit: I should clarify the statement "the free-falling observer's clock will appear to move slower and slower". What I really should have said was that, if we imagine the free-falling observer emitting light beams at values of his proper time closer and closer to the proper time when he crosses the horizon, those light beams will reach the rocket at times, according to the rocket's clock, that increase without bound.)

This way of putting things also puts the "gravity approaches infinity as the horizon approached" statement into proper perspective. The horizon itself is a null surface--only light rays can move "on" the horizon, so there's no such thing as an observer that stays "at" the horizon, not even one with infinite acceleration (or infinite "gravity"). The best you can do is to say that, for hyperbolas x^2 - t^2 = x_0^2, as x_0 gets closer and closer to zero, the acceleration felt by observers moving along the hyperbola gets larger and larger, without bound. (Also, of course, since there's no observer that stays at the horizon, there's no clock that "is not ticking" there.)

 

[PeterDonis]

One interesting aspect of this is that the speed of time approaches infinity as x \to \infty. So as an ambiguous answer to my ambiguous question, the infinite speed of time would mean there is no natural zero for a corresponding gravitational potential.

I'm curious: why are you so insistent on finding an interpretation of what's going on in terms of "gravitational potential"? As far as relativity is concerned, "gravitational potential" is not a fundamental concept; it's an approximation that works well in a certain domain (basically, under conditions such that gravity can be usefully treated as a Newtonian force, determined by the gradient of a potential) but does not extend well beyond that domain.

 

[granpa]

If I drop something out the window, it falls down, and approaches a point c²/g below, but it will never reach there. The clocks down at that level (x=0) are not ticking at all and nothing can fall past them. If the rocket extended down that far, the gravity there would approach infinity.

from the point of view of someone on the rocket[/color] that is exactly what will happen and the clocks there are indeed not ticking at all but gravity at that point will not be infinite.

it is a common fallacy to assume that infinite time dilation equals infinite gravity. (time dilation is proportional to gravitational potential not gravitational field strength)

beyond that point Δt continues to increase and the clocks will be ticking backwards. (again from the point of view of someone on the rocket)

this can easily be seen with a long line of sychronized clocks using nothing more than special relativity.

an accelerating rocket can be thought of as creating a uniform gravitational field and the Δt can be shown to be proportional to the distance from the rocket so the time dilation is proportional to the potential.

 

[JDoolin]

I was OK with your post up until this point. It would be OK to say that the rocket cannot extend to x = 0, but it is not OK to say that "clocks at that level are not ticking at all and nothing can fall past them". The reason is simple: if you drop something out the window, it *does* fall past "a point c^2/g below", and its clock continues to tick. However, even this way of stating it is, in my opinion, misleading, which is why I put the phrase "a point c^2/g below" in quotes.

Here's what I think is a better way to describe what's going on: consider a light beam emitted from the event x = 0, t = 0 in the global inertial frame (what I think we were calling the "lab frame" above, as opposed to the "rocket frame"). No observer on the rocket, on any of its floors, will ever see this light beam; it will never catch up to any portion of the rocket (assuming that the rocket, with all its floors, maintains its acceleration forever). This fact is what justifies us referring to the light beam's worldline--the line x = t--as a "horizon" with respect to the rocket. However, a free-falling observer who is dropped off any floor of the rocket *will* see this light beam, after a finite amount of proper time by the free-falling observer's clock. From the rocket's point of view, the free-falling observer's clock will *appear* to move slower and slower as the observer approaches the horizon, but no rocket observer will ever see the free-falling observer cross the horizon (because light rays from events on the horizon will never catch up to the rocket, as above).

(Edit: I should clarify the statement "the free-falling observer's clock will appear to move slower and slower". What I really should have said was that, if we imagine the free-falling observer emitting light beams at values of his proper time closer and closer to the proper time when he crosses the horizon, those light beams will reach the rocket at times, according to the rocket's clock, that increase without bound.)

This way of putting things also puts the "gravity approaches infinity as the horizon approached" statement into proper perspective. The horizon itself is a null surface--only light rays can move "on" the horizon, so there's no such thing as an observer that stays "at" the horizon, not even one with infinite acceleration (or infinite "gravity"). The best you can do is to say that, for hyperbolas x^2 - t^2 = x_0^2, as x_0 gets closer and closer to zero, the acceleration felt by observers moving along the hyperbola gets larger and larger, without bound. (Also, of course, since there's no observer that stays at the horizon, there's no clock that "is not ticking" there.)

I realize this is probably not the way its usually described, but I'm basing this on a simulation that I made myself a few years ago--(well before I had any idea that there should be a Rindler Horizon.)

I decided to make a video of this, this morning and upload it to youtube.


I described the Rindler horizon as the spot that everything falls toward, but never reaches. You describe the Rindler horizon as the spot from which a light beam cannot reach us. In either case, the "spot" we are referring to is actually an "event."

They are both legitimate ways of explaining what is going on. The light coming from the Rindler Horizon can't ascend from the horizon, and the light and objects falling toward the rindler horizon approach it asymptotically, but never reach it. Objects beyond the rindler horizon, mathematically also move toward the rindler horizon, but their clocks are mathematically going backwards.

In the view of the rocket, the event at (x=0,t=0) is an event that cannot be crossed because nothing can happen there.

 

[PeterDonis]

They are both legitimate ways of explaining what is going on. The light coming from the Rindler Horizon can't ascend from the horizon, and the light and objects falling toward the rindler horizon approach it asymptotically, but never reach it. Objects beyond the rindler horizon, mathematically also move toward the rindler horizon, but their clocks are mathematically going backwards.

In the view of the rocket, the event at (x=0,t=0) is an event that cannot be crossed because nothing can happen there.

If you use the term "horizon" to refer only to the single *event* at x = 0, t = 0, then I can see why you're describing things this way. Obviously a free-falling observer who drops off the rocket at event x = s_0, t = 0 will never pass through the event x = 0, t = 0. However, that observer *will* cross the line x = t (at x = s_0, t = s_0), so if the term "horizon" refers to that entire line (which is the standard usage, and also makes more sense--see below), it is simply false to say that "objects falling toward the rindler horizon approach it asymptotically, but never reach it". The objects' worldlines *do* reach the horizon, and pass beyond it. That's the fact. It is also a fact that observers in the rocket will never *see* that portion of a free-falling object's worldline, but that doesn't mean that portion of the worldline doesn't exist.

Why does it make more sense to use the term "horizon" to refer to the *line* x = t, instead of just the single *event* x = 0, t = 0? (Actually, if we include the other two space dimensions, the horizon is a null *surface* which includes the line x = t.) Because the hyperbolas along which the Rindler observers move, x^{2} - t^{2} = s_{0}^{2}, asymptote to the line, *not* the single point. Put another way, it's not just the single event x = 0, t = 0 that can't send light signals to any of the Rindler observers; it's *any* event on the line x = t (or in the region "above" it--i.e., with t > x--so the line x = t is the *boundary* of the region, which is what "horizon" means in standard usage). The only reason to single out the particular event x = 0, t = 0 is that it is the "pivot point", so to speak, of all the lines of simultaneity for the Rindler observers as they move along their hyperbolas. But since, as you note, nothing actually *happens* at that event, singling it out tempts you to ignore real stuff that *does* happen, like free-falling observers crossing the line x = t.

 

[Mentz114]

What PeterDonis says above is correct. At the risk of repeating this, have a look at Greg Egan's artricle, which has useful diagrams.

http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

 

[JDoolin]

If you use the term "horizon" to refer only to the single *event* at x = 0, t = 0, then I can see why you're describing things this way. Obviously a free-falling observer who drops off the rocket at event x = s_0, t = 0 will never pass through the event x = 0, t = 0. However, that observer *will* cross the line x = t (at x = s_0, t = s_0), so if the term "horizon" refers to that entire line (which is the standard usage, and also makes more sense--see below), it is simply false to say that "objects falling toward the rindler horizon approach it asymptotically, but never reach it". The objects' worldlines *do* reach the horizon, and pass beyond it. That's the fact. It is also a fact that observers in the rocket will never *see* that portion of a free-falling object's worldline, but that doesn't mean that portion of the worldline doesn't exist.

Why does it make more sense to use the term "horizon" to refer to the *line* x = t, instead of just the single *event* x = 0, t = 0? (Actually, if we include the other two space dimensions, the horizon is a null *surface* which includes the line x = t.) Because the hyperbolas along which the Rindler observers move, x^{2} - t^{2} = s_{0}^{2}, asymptote to the line, *not* the single point. Put another way, it's not just the single event x = 0, t = 0 that can't send light signals to any of the Rindler observers; it's *any* event on the line x = t (or in the region "above" it--i.e., with t > x--so the line x = t is the *boundary* of the region, which is what "horizon" means in standard usage). The only reason to single out the particular event x = 0, t = 0 is that it is the "pivot point", so to speak, of all the lines of simultaneity for the Rindler observers as they move along their hyperbolas. But since, as you note, nothing actually *happens* at that event, singling it out tempts you to ignore real stuff that *does* happen, like free-falling observers crossing the line x = t.

There are two different things we're talking about here. The events, which never cross the line x=ct (or x=-ct for that matter), and the objects, which never cross the point x=0.

 

[PeterDonis]

There are two different things we're talking about here. The events, which never cross the line x=ct (or x=-ct for that matter), and the objects, which never cross the point x=0.

Huh? Events don't "cross" anything--they're single points in spacetime. Objects travel on worldlines, which are ordered sequences of events (the usual ordering/parametrization of the worldline is given by the "proper time" according to the object traveling on that worldline; each event on the worldline has its own unique proper time). The point where a free-falling object crosses the horizon, the line x = ct, is an event just like any other on that object's worldline.

What I've just given are the standard definitions of "event" and the other terms. If you're using different definitions in the statement I just quoted above, please state them explicitly, because I don't understand the distinction you're making.

 

[JDoolin]

Huh? Events don't "cross" anything--they're single points in spacetime. Objects travel on worldlines, which are ordered sequences of events (the usual ordering/parametrization of the worldline is given by the "proper time" according to the object traveling on that worldline; each event on the worldline has its own unique proper time). The point where a free-falling object crosses the horizon, the line x = ct, is an event just like any other on that object's worldline.

What I've just given are the standard definitions of "event" and the other terms. If you're using different definitions in the statement I just quoted above, please state them explicitly, because I don't understand the distinction you're making.

We may be talking in different contexts, but I want to make sure you realize: no event has a unique proper time. As you say, "proper time" is determined by an object traveling on "that world line" but since there is no unique worldline through any event, there can be no unique proper time at that event.. Under the context we have been discussing there are two proper times assigned to each event. We can assign a one proper time to each event according to the reading on the rocket's clocks. We could also assign a proper time by using clocks stationary in the inertial reference frame. But the proper time of an event is not uniquely, since it depends on the prior motion of the clocks arriving at that event.

I see how our concepts differ about events. You think of events remaining in place, while you progress forward in time. I think of events drifting from the future into the present toward the past, while I remain in the present.

The events also move according to the rules of Special Relativity--Lorentz Transformations. It is under this context, the people aboard the rocket would model events approaching the line x=c t and x=-c t from either side, but not crossing.

The objects' worldlines *do* reach the horizon, and pass beyond it. That's the fact. It is also a fact that observers in the rocket will never *see* that portion of a free-falling object's worldline, but that doesn't mean that portion of the worldline doesn't exist.

True, the worldlines of the falling objects do cross the line x = c t, but always at a point t>0. That event of crossing will remain forever in the future for the observer on the rocket. It will always be something that has not happened yet. (Whether something that happens in the future actually "exists" is a metaphysics question, but in this case of perfectly predictable motion, the worldlines can be modeled, of course.)

 

[PeterDonis]

We may be talking in different contexts, but I want to make sure you realize: no event has a unique proper time.

I agree.

I see how our concepts differ about events. You think of events remaining in place, while you progress forward in time. I think of events drifting from the future into the present toward the past, while I remain in the present.

I don't have a problem with either of these points of view; however, I'm not sure the first one accurately captures the one I've been implicitly using. The point of view I've been implicitly using is that *nothing moves*: spacetime, all of it, just *is*, as a four-dimensional geometric object. When we make statements about events, worldlines, etc., we're making geometric statements about geometric objects within this overall geometric object.

True, the worldlines of the falling objects do cross the line x = c t, but always at a point t>0. That event of crossing will remain forever in the future for the observer on the rocket. It will always be something that has not happened yet.

Yes, that's one way of putting it. However, that way of putting it tempts you, as I said before, to say things like...

Whether something that happens in the future actually "exists" is a metaphysics question

...which is *not* justified, in my opinion. The event of crossing is only "forever in the future" for observers on the rocket; it does *not* remain forever in the future for inertial observers. That's not a metaphysical question; it's a direct logical consequence of the construction of the spacetime, as a geometric object.

Furthermore, it's a logical consequence that is accessible to the observers on the rocket; even though they can't themselves "see" the event of the free-falling observer crossing the horizon, they can tell that there *must* be such an event (and further events after it along the free-falling worldline). How? By integrating the proper time along the free-falling worldline (using *their* metric), and realizing that, even as their "accelerated" coordinate time goes to infinity, the worldline of the free-falling observer only contains a finite amount of proper time (because the integral converges to a finite value as Rindler coordinate time t goes to infinity). But worldlines can't just stop at a finite proper time; more precisely, the Rindler observers can find no physical *reason*, even in their frame, why the free-falling worldline would just stop at a finite proper time. It's not just that there's no catastrophic event there, no explosion, no laser blast blowing the free-falling observer to bits, no "wall" for the observer to run into. Even if there were such an event, the debris from it would have to go *somewhere*--there would be other worldlines to the future of the catastrophic event. In other words, physically, the free-falling worldline *has to have a future* past the last point the rocket observers can see; there must be a further portion of the free-falling worldline that they can't observe.

 

[Passionflower]

We may be talking in different contexts, but I want to make sure you realize: no event has a unique proper time.

I agree.

Peter I am not sure what you agree on. As far as I understand it an event is a point in spacetime, points don't have proper time or any other time for that matter. Time is a timelike interval between two events. Proper time is such an interval for a real clock traversing a path in spacetime.

 

[PeterDonis]

Peter I am not sure what you agree on. As far as I understand it an event is a point in spacetime, points don't have proper time or any other time for that matter. Time is a timelike interval between two events. Proper time is such an interval for a real clock traversing a path in spacetime.

The term "proper time" can be used that way, yes, but it can also be used to denote the single *value* of the "proper time" used to parametrize a worldline, at a given event on that worldline. But since any event will lie on multiple worldlines, each of which could assign a different value of their "proper time" parameter to the event, no event can have a single, unique value of the "proper time" parameter. That's what I was agreeing to, and I'm pretty sure that's what JDoolin meant (he'll correct me, I'm sure, if he meant something else).

 

[yuiop]

But worldlines can't just stop at a finite proper time; more precisely, the Rindler observers can find no physical *reason*, even in their frame, why the free-falling worldline would just stop at a finite proper time. It's not just that there's no catastrophic event there, no explosion, no laser blast blowing the free-falling observer to bits, no "wall" for the observer to run into.

Would you agree that this exposes a limitation of using Rindler coordinates as an analogy of what happens in the case of real black hole, because we know in that case, there is a real singularity at the centre of the black hole (where the worldline of a free falling observer does stop at a finite proper time) and yet there is no indication of this real singularity in Rindler coordinates. In Rindler coordinates the free falling observer continues to fall for infinite proper time after crossing the horizon so if we use Rindler coordinates to analyse what happens in the case of a real black hole we would be forced to conclude that the free falling observer never arrives at the real central singularity in finite coordinate or proper time.

 

[PeterDonis]

Would you agree that this exposes a limitation of using Rindler coordinates as an analogy of what happens in the case of real black hole, because we know in that case, there is a real singularity at the centre of the black hole (where the worldline of a free falling observer does stop at a finite proper time) and yet there is no indication of this real singularity in Rindler coordinates.

Yes; in fact, the case is even stronger than you suggest, because this...

In Rindler coordinates the free falling observer continues to fall for infinite proper time after crossing the horizon so if we use Rindler coordinates to analyse what happens in the case of a real black hole we would be forced to conclude that the free falling observer never arrives at the real central singularity in finite coordinate or proper time.

is false as it stands; the correct way to state it is that Rindler coordinates do not even *cover* the portion of spacetime "below" the horizon--in other words, we can't say that "In Rindler coordinates the free falling observer continues to fall...after crossing the horizon", because we can't even *assign* coordinates at all, not even infinite ones, to any events below the horizon in Rindler coordinates, even though we can show that the events themselves must be there.

Also, strictly speaking, we wouldn't use "Rindler coordinates" in the case of a real black hole; we'd use Schwarzschild coordinates, which are analogous to Rindler coordinates. But those coordinates have exactly the same limitation: they don't cover the portion of spacetime at or below the horizon, so they can't be used to analyze what happens there.

(Technically, the "exterior Schwarzschild coordinates", with r > 2M, are the ones analogous to Rindler coordinates; there are also "interior Schwarzschild coordinates", with r < 2M, which *do* cover the portion below the horizon, but do *not* cover the portion above. Popular books often obscure this by using the term "Schwarzschild coordinates", without qualification, to refer to both sets of coordinates as though they were one single coordinate system, but they're not; they're two separate ones, which can't be connected into one because they are both singular at the horizon, r = 2M, which acts as an impassable "barrier", so to speak, between them.)

 

[Passionflower]

In this context for an interesting lecture about 'Rindler and Schwarzschild' see:

Lecture 12 of Leonard Susskind's Modern Physics concentrating on General Relativity. Recorded December 9, 2008 at Stanford University: http://academicearth.org/lectures/modern-physics-general-relativity-12