General Relativity Basics: The Principle of Equivalence

[PeterDonis]

It comes out so whether you do the calculation by (X and T) or by (x and $\tau$), you get the very same result for the acceleration.

This looks OK to me as a calculation of the acceleration (or rather a verification that the calculation gives the same results in the inertial frame and the accelerated frame). I assume you would then just take the derivative with respect to x to get dg/dx?

 

[JDoolin]

This looks OK to me as a calculation of the acceleration (or rather a verification that the calculation gives the same results in the inertial frame and the accelerated frame). I assume you would then just take the derivative with respect to x to get dg/dx?

Yes. I think that comes out to be

$$\frac{dg}{dx}=-\frac{c^2}{x^2}$$.​

http://groups.google.com/group/sci.physics.relativity/msg/6ff91383c387f7a1?hl=en" tidal gravity as a "divergence of free-fall paths." So is dg/dx the definition of tidal gravity? Could it be that what we have here is the divergence of free-fall paths caused by Lorentz contraction alone?

 

[PeterDonis]

URL="http://groups.google.com/group/sci.physics.relativity/msg/6ff91383c387f7a1?hl=en"]Daryl McCullough describes[/URL] tidal gravity as a "divergence of free-fall paths." So is dg/dx the definition of tidal gravity? Could it be that what we have here is the divergence of free-fall paths caused by Lorentz contraction alone?

No. The *free-fall paths* are not diverging; the *accelerated* paths are (actually, they're "converging" from the point of view of the global inertial frame, but converging is just "negative diverging", so to speak). The free-fall paths remain parallel (this is obvious in the global inertial frame); they don't converge or diverge. Hence, the tidal gravity is zero.

In the spacetime around the gravitating body, on the other hand, the actual free-fall paths converge or diverge; which they do depends on their relative orientation. For example, two free-fall paths separated radially will diverge (their separation will increase with time), while two free-fall paths separated tangentially will converge (their separation will decrease with time).

I realize that, as it stands, the above may not be very enlightening, because "separation", as it stands, is frame-dependent. After all, the accelerated observers see their own separations as constant, whereas inertial observers (in the flat-spacetime scenario, at least) see their separations as decreasing with time (due to Lorentz contraction). However, it turns out that there *is* an invariant way to define "separation" for *nearby* geodesics (free-falling paths): pick one free-falling path as the "origin" or "reference worldline", and construct a local inertial frame around it. Then the "separation" between the two free-falling paths is just the spatial distance between them in the local inertial frame, and the *change* in the separation is the derivative of that spatial distance with respect to proper time along the reference worldline. (Note that in curved spacetime, a "local" inertial frame must be "local" in time as well as in space, so actually, we have to construct a separate local inertial frame and go through the above procedure at each *event* on the reference worldline.)

If we do this for the accelerated paths in flat spacetime (the "Rindler worldlines"), we find that the "separation" x in Rindler coordinates does *not* meet the above requirements, so we can't use it to measure the "separation" we need to evaluate tidal gravity. Instead, we have to look at the free-falling worldlines themselves and construct local inertial frames around them. Of course in this particular case that's easy, because all of them are just the worldlines of objects at rest in the global inertial frame, which therefore qualifies as the "local inertial frame" for all of them (with the origin shifted appropriately, if you want to be precise), and all of them obviously maintain constant separation in that frame. So there's no tidal gravity.

In the curved spacetime around a gravitating body, on the other hand, we have to go through the full procedure I described above, and when we do, we find that the separations of nearby free-falling worldlines *do* converge or diverge as I stated above.

The technical term for the math behind the procedure I described above is "equation of geodesic deviation"; the sections on that subject in MTW or Wald are decent places to start if you want to study it further. However, their treatment requires full-blown tensor algebra. If I can find a discussion that's more oriented towards the physical aspects I'll post a link.

 

[JDoolin]

The *free-fall paths* are not diverging; the *accelerated* paths are (actually, they're "converging" from the point of view of the global inertial frame, but converging is just "negative diverging", so to speak).

I can't tell whether you are disagreeing with the physics, or just the words I am using.

Two stationary particles in the inertial frame will appear, in the rocket frame, to rise up from below, decelerating, and spreading out until they stop. Then they will appear to fall down, accelerating and contracting. Both the apparent spreading out, and the contracting are mathematically identical to what one would expect from Lorentz Contraction of a ruler placed between the two objects.

Do you disagree with this?

Edit: With one extra caveat--the top and bottom of any free-falling object will appear to be moving at different velocities. In general Lorentz Transformation alone means the front and back are moving at the same velocity

 

[JDoolin]

Then the "separation" between the two free-falling paths is just the spatial distance between them in the local inertial frame, and the *change* in the separation is the derivative of that spatial distance with respect to proper time along the reference worldline. (Note that in curved spacetime, a "local" inertial frame must be "local" in time as well as in space, so actually, we have to construct a separate local inertial frame and go through the above procedure at each *event* on the reference worldline.)

If we do this for the accelerated paths in flat spacetime (the "Rindler worldlines"), we find that the "separation" x in Rindler coordinates does *not* meet the above requirements, so we can't use it to measure the "separation" we need to evaluate tidal gravity.

I didn't understand the above requirements. When you say "the local inertial reference frame" are you referring to the momentarily comoving reference frame of the accelerating body? Of course that quantity doesn't change. That's kind of the point of Lorentz Contraction.

Possibly, we are talking about two different x's. You are talking about x's being drawn up on the side of the rocket-ship. I'm talking about the x coordinate of a free-falling body, as measured by the ruler drawn up on the side of the rocket ship.

 

[PeterDonis]

I can't tell whether you are disagreeing with the physics, or just the words I am using.

Two stationary particles in the inertial frame will appear, in the rocket frame, to rise up from below, decelerating, and spreading out until they stop. Then they will appear to fall down, accelerating and contracting. Both the apparent spreading out, and the contracting are mathematically identical to what one would expect from Lorentz Contraction of a ruler placed between the two objects.

Do you disagree with this?

Edit: With one extra caveat--the top and bottom of any free-falling object will appear to be moving at different velocities. In general Lorentz Transformation alone means the front and back are moving at the same velocity

I don't disagree with the physics, but I do say that what you've described above is not "tidal gravity". The apparent spreading out and contracting in the rocket frame are, as you note, kinematic effects; we can transform them away by switching to the global inertial frame. Tidal gravity can't be transformed away by changing reference frames. See my next post responding to your next post.

 

[PeterDonis]

I didn't understand the above requirements. When you say "the local inertial reference frame" are you referring to the momentarily comoving reference frame of the accelerating body?

Sort of. The key thing is that, to measure geodesic deviation, you have to compare two nearby geodesics, and the worldline of an accelerating observer is not a geodesic. In the MCIF anywhere along the accelerating observer's worldline, you can pick two nearby geodesics and look at whether they converge or diverge as judged from the MCIF; but in doing that, you will have to look beyond the "momentarily comoving" part of the MCIF, because if you restrict yourself to the "momentarily comoving" part only, it by definition is too small to see tidal effects!

In the particular case we're discussing, suppose we pick the MCIF of the rocket at the instant when one free-falling object is momentarily at rest relative to the rocket (and at the same spatial point). That MCIF will then be the same as the rest frame of the free-falling object (which, since spacetime is flat in this scenario, is basically the global inertial frame), but it will only be "momentarily comoving" with the rocket in a very small patch surrounding the instant when the two objects are momentarily at rest relative to each other. Within that small patch, we can't tell if the spacetime is flat or curved (i.e., whether there are or are not tidal effects); but outside that patch, we can see that, in the inertial frame, the two free-falling objects maintain constant separation for all time, so there's no tidal gravity. We can't use their changing separation in the rocket frame for this purpose because the rocket frame doesn't "stay inertial" outside the small patch.

In the corresponding case in curved spacetime, if we again chose our MCIF the same way, the inertial frame of the free-falling object, *extended outside the small momentarily comoving patch*, would be our "local inertial frame" for measuring geodesic deviation. Then we would see that, *outside* the small momentarily comoving patch, the separation of the other nearby free-falling object would *not* remain constant--it would converge or diverge (depending on the direction of the separation), even as judged from the inertial frame. This is not a kinematic effect; it's a real, invariant convergence or divergence of the geodesics and can't be transformed away by changing reference frames.

 

[yuiop]

In the spacetime around the gravitating body, on the other hand, the actual free-fall paths converge or diverge; which they do depends on their relative orientation. For example, two free-fall paths separated radially will diverge (their separation will increase with time), while two free-fall paths separated tangentially will converge (their separation will decrease with time).

In the attached diagram I have plotted some worldlines of free falling particles (blue lines) near the event horizon in Schwarzschild coordinates. In these coordinates the particles seem to be initially diverging radially but as they approach the event horizon (vertical green line at r = 2m =1) they converge radially. (The red lines are the Fermi normal lines of the primary free falling observer which is indicated by the black boxes at the intersections.)

Are you saying that in the free falling frame the particles will appear to continue diverging according to the free falling observer as they approach and pass through the event horizon? It would be nice is someone could plot the free falling wordlines from the point of view of the free falling observer to prove that is the case.

 

[yuiop]

Are you saying that in the free falling frame the particles will appear to continue diverging according to the free falling observer as they approach and pass through the event horizon?

O.K. I think I have answered my own question. I have managed to plot the the worldlines of free falling observers (blue curves) in Gullstrand-Painleve coordinates in the attached diagram and they do indeed appear to diverge continuously as the event horizon is approached and continue to diverge after passing through the event horizon from the point of view of the free falling GP observers.

The green curves are ingoing null paths and the red curves are outgoing null paths. The event horizon is at x=2 in the diagram.
It can also be seen from the way the null paths have been plotted, that the radar distance between neighbouring free falling observers is continuously increasing as they fall.

If anyone wants the equations to plot these curves, just ask

 

[Passionflower]

O.K. I think I have answered my own question. I have managed to plot the the worldlines of free falling observers (blue curves) in Gullstrand-Painleve coordinates in the attached diagram and they do indeed appear to diverge continuously as the event horizon is approached and continue to diverge after passing through the event horizon from the point of view of the free falling GP observers.

The green curves are ingoing null paths and the red curves are outgoing null paths. The event horizon is at x=2 in the diagram.
It can also be seen from the way the null paths have been plotted, that the radar distance between neighbouring free falling observers is continuously increasing as they fall.

If anyone wants the equations to plot these curves, just ask

Yes, the equations would be nice. Plots (and even better animated plots) are extremely useful to get a quick overview of what an equation 'does'. In this day and age we have wonderful tools to visualize information and demonstrating something with a plot is an excellent means.

 

[Mentz114]

Are you saying that in the free falling frame the particles will appear to continue diverging according to the free falling observer as they approach and pass through the event horizon? It would be nice is someone could plot the free falling wordlines from the point of view of the free falling observer to prove that is the case.

There is no horizon at r=2m in the Painleve chart.
The metric is

$$\begin{align*} \left[ \begin{array}{cccc} \frac{2\,M}{r}-1 & -\sqrt{\frac{2M}{r}} & 0 & 0\\ -\sqrt{\frac{2M}{r}} & 1 & 0 & 0\\ 0 & 0 & {r}^{2} & 0\\ 0 & 0 & 0 & {r}^{2}\,{sin\left( \theta\right) }^{2} \end{array} \right] \end{align*}$$
Note the flat spatial sections.

An observer at rest wrt to the 'rain', will see tidal tensor with $T_{rr}=2m/r^3,T_{\theta\theta}=T_{\phi\phi}=-m/r^3$

 

[PeterDonis]

There is no horizon at r=2m in the Painleve chart.

This is not correct. The horizon is a frame-independent, invariant feature of the geometry, and it's there regardless of what coordinate chart you use. Yes, the spatial sections in the Painleve chart are flat, but that doesn't mean an observer at r < 2m can send signals to r > 2m. The curvature of the spacetime that makes that impossible is all in the "time dimension" (so to speak) in the Painleve chart, but it's still there.

Edit: Maybe what you meant to say is that there is no *coordinate singularity* at r = 2m in the Painleve chart?

 

[Mentz114]

This is not correct. The horizon is a frame-independent, invariant feature of the geometry, and it's there regardless of what coordinate chart you use. Yes, the spatial sections in the Painleve chart are flat, but that doesn't mean an observer at r < 2m can send signals to r > 2m. The curvature of the spacetime that makes that impossible is all in the "time dimension" (so to speak) in the Painleve chart, but it's still there.

Edit: Maybe what you meant to say is that there is no *coordinate singularity* at r = 2m in the Painleve chart?

Yes. Thanks for the lecture, anyway.

 

[yuiop]

Yes, the equations would be nice. Plots (and even better animated plots) are extremely useful to get a quick overview of what an equation 'does'. In this day and age we have wonderful tools to visualize information and demonstrating something with a plot is an excellent means.

OK, here is how the equations to plot the paths in the https://www.physicsforums.com/showpost.php?p=2949294&postcount=79" were obtained.

The GP coordinates are given by Wikpedia as:

$$dt_{GP} = dt\, +(1-2m/r)^{-1} \sqrt{2m/r}\, dr$$

$$dr_{GP} = dr$$

Where dt and dr are the familiar Schwarzschild coordinates.

Now we know that the equation for a falling particle with apogee at infinity in Schwarzschild coordinates is $dr/dt = -(1-2m/r)^{-1} (2m/r)^{-1/2}$ and we can substitute this equation into the definition of the GP time coordinate to elliminte dt and obtain:

$$dt_{GP} = - \sqrt{ \frac{r}{2m}}\, dr$$

(This turns out to be the same as the velocity measured by a local stationary Schwarzschild observer at r). Integrating the above gives the equation for the path of the falling particle in GP coordinates:

$$t_{GP} = -\frac{2r}{3}\sqrt{\frac{r}{2m}} +C$$

where C is a constant of integration.

Substitution of the GP time coordinate into the radial Schwarzschild metric gives the GP metric:

$$dtau^2 = (1-2m/r) dt_{GP}^2\, - 2\sqrt{2m/r}\, dr\, dt_{GP} \, - dr^2$$

Setting dtau = 0 gives the null path of a photon and solving the resulting quadratic gives the differential equation:

$$dr/dt_{GP} = (\pm 1)\, - \sqrt{2m/r}$$

where the sign is determined by whether the photon is ingoing or outgoing. For an ingoing photon the speed at the EH is -2c and the outgoing photon is zero.

Integrating gives the equations for the null paths in GP coordinates:

$$t_{GP} = r(2\sqrt{2m/r} \, \pm 1) - 4m\, \text{artanh}\left(1/\sqrt{2m/r}\right)\, \pm 2m \ln(r-2m) \, + C$$

or alternatively for the outgoing photon:

$$t_{GP} = r(2\sqrt{2m/r} \, +1) \, + 2m \ln\left((r-2m)\frac{1-\sqrt{2m/r}}{1+\sqrt{2m/r}} \right) \, + C$$

and for the ingoing photon:

$$t_{GP} = r(2\sqrt{2m/r} \, -1)\, - 2m \ln\left((r-2m)\frac{1+\sqrt{2m/r}}{1-\sqrt{2m/r}} \right)\, + C$$

The alternative form allows the null paths to be plotted across the event horizon without problems.

 

[JDoolin]

There is no horizon at r=2m in the Painleve chart.
The metric is

$$\begin{align*} \left[ \begin{array}{cccc} \frac{2\,M}{r}-1 & -\sqrt{\frac{2M}{r}} & 0 & 0\\ -\sqrt{\frac{2M}{r}} & 1 & 0 & 0\\ 0 & 0 & {r}^{2} & 0\\ 0 & 0 & 0 & {r}^{2}\,{sin\left( \theta\right) }^{2} \end{array} \right] \end{align*}$$
Note the flat spatial sections.

An observer at rest wrt to the 'rain', will see tidal tensor with $T_{rr}=2m/r^3,T_{\theta\theta}=T_{\phi\phi}=-m/r^3$

So the proper time of a particle in this region could be measured by:

$$ds^2= (dr^2 -dt^2 + r^2 d\theta ^2 +r^2 \sin ^2 \theta d\phi ^2) +(\frac{ 2 M}{r}dt^2 -2 \sqrt{\frac{2 M}{r}} dr dt)$$​

This is the same as flat space except for the two terms at the end. Is this Painleve chart the same as the Schwarzschild metric?

This is not correct. The horizon is a frame-independent, invariant feature of the geometry, and it's there regardless of what coordinate chart you use. Yes, the spatial sections in the Painleve chart are flat, but that doesn't mean an observer at r < 2m can send signals to r > 2m. The curvature of the spacetime that makes that impossible is all in the "time dimension" (so to speak) in the Painleve chart, but it's still there.

Edit: Maybe what you meant to say is that there is no *coordinate singularity* at r = 2m in the Painleve chart?

What makes a heavy mass into a black hole is that it's own physical radius is smaller than its own horizon, right?

 

[Passionflower]

What makes a heavy mass into a black hole is that it's own physical radius is smaller than its own horizon, right?

I think a little more accurately stated:

An object becomes (or is already) a black hole if the ratio between the area it occupies and the area that represents its mass is smaller than 4. As soon as this happens the object's occupied area will shrink to zero.

E.g.:

$${A_{occupied} \over A_{mass} } < 4$$

You notice that mass by itself is not in indication of whether something is a black hole or not, it is only the ratio that matters.

In general relativity mass is represented by a physical area.

Because spacetime is curved when we have mass it is no longer a trivial matter to derive a radius from those areas.

 

[JDoolin]

I live on a rocket that is under constant acceleration.

The rocket has many floors. My floor is labeled $x=9.18 \times 10^{ 15} m$, and on my floor, the gravity is $g=c^2/x$.

If I climb up, the gravity gets smaller. If I look up, the clocks above me are ticking faster.
If I climb down, the gravity gets greater. If I look down, the clocks below me are ticking slower.

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 still need to determine the speed of the clocks as a function of either g or x. Then answer the more ambiguous question, is the speed of time determined by gravitational potential?

 

[granpa]

does this help?
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

http://en.wikipedia.org/wiki/Rapidity

 

[JDoolin]

I still need to determine the speed of the clocks as a function of either g or x. Then answer the more ambiguous question, is the speed of time determined by gravitational potential?

An approximation follows:

We adopt a standard unit of time being the time it takes for an object to fall 1 meter. The following can only be used where x>>1, because when x is near or less than 1, the acceleration is not constant.

$$\begin{matrix} \Delta x = \frac{1}{2}g \Delta t^2 \\ \Delta t = \sqrt{\frac{2 \Delta x}{g}}= \sqrt{\frac{2}{g}}=\sqrt{\frac{2 x}{c^2}} \\ \frac{\Delta t(x)}{\Delta t(x_0)}=\sqrt{\frac{x}{x_0}} \end{matrix}$$​

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.

 

[Passionflower]

If I drop something out the window, it falls down, and approaches a point c²/g below, but it will never reach there.

As soon as you drop something out of the window it is no longer in the accelerating frame you are describing but instead it is free falling and it keeps free falling forever.

The clocks down at that level (x=0) are not ticking at all and nothing can fall past them.

Close to this region clocks run slower wrt clocks higher up but locally clocks just run as normal. Free falling objects would pass by without hindrance.

I still need to determine the speed of the clocks as a function of either g or x. Then answer the more ambiguous question, is the speed of time determined by gravitational potential?

If I am not mistaken the relationship between proper time and coordinate time for a given acceleration is:

$$\tau={c\, \over \alpha} {\it arcsinh} \left( {\frac {\alpha t}{c}} \right)$$

While the relationship between acceleration and position is:

$$\alpha={\frac {{c}^{2}}{x}}$$

 

[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.

 

[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 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.)