General Relativity Basics: The Principle of Equivalence

[JDoolin]

What I have heard *about* the principle of equivalence is a
great and grave over generalization; primarily that gravity is
equivalent to acceleration.

I would be prepared to acknowledge that it is highly likely that the
behavior of free-falling bodies in the region where F=m*g would be
effectively equivalent to the behavior of inertial bodies as viewed by
an observer on an accelerated platform accelerating at a rate of a=g
on the approach.

Of course, once the accelerating platform PASSES the inertial body,
you have the acceleration going the WRONG WAY, and any resemblance
between acceleration and gravity is now gone. (*Edit: Strike-through of poor reasoning pointed out in first reply.)

Also, if you are in a larger region, such that F= G m1 * m2 / r^2, the
motion of bodies falling in this volume is NOT equivalent to the
motion you would see if you were on a platform accelerating toward
them.

On the other hand, I could imagine trying an approach to determine the
speed of the clock of an observer on an accelerating platform, and
somehow relating this to the speed of a clock of an observer standing
on the ground in a gravitational field. In this one small way, (speed
of clock), gravitational field and acceleration may be equivalent.
But to whom?

You need to produce the inertial observer from whose viewpoint the two
clocks give this ratio. You need to have an initial velocity and a
final velocity of the accelerated clock in the frame of reference of
this inertial observer. You need to have a particular scale of time
between the initial and final velocity.

I've been pondering this for a day or so, but I'm not yet to the point of putting pencil to paper to crank out any math. However, I think I've got a general idea of the problem set-up.

What you need to consider is the moment when an accelerating observer comes to rest with respect to the unaccelerating observer. The way to imagine this is to picture a rocket, pointed upward, but descending, with flames shooting out it's back end. So at first, the fuel is decelerating the rocket, and then it turns around and goes the other direction.

So here is where my confusion lies. We need, at that moment, to compare the speed of the rocket's clock to the speed of my clock. However, the problem is, the rocket is only instantaneously at rest with respect to me, so there is no way to get a reading of the rocket's clock during the time we are at relative rest.

Perhaps there is a way to set up the problem to find \lim_{t\to 0} \frac{t_{rocket}}{t_{inertial}}
where t_{rocket} can be found by the familiar path integral
ds^2 = (cdt)^2 - dx^2 -dy^2 - dz^2
or simply
ds^2 = (cdt)^2 - dx^2
since we are considering acceleration in only one direction.

I still need to determine some details of how to do this. Figuring out (x(t),t(t)) I guess won't be too much of a problem, since x(t)=v0 + a t*. So my big question now is, does the limit come out to be 1, or some other number?

(By the way, if you want to use the site's LaTex feature, I find you must first preview your post, then refresh the page in order to see the preview of your equations.)

Jonathan

*Edit: Wrong equation, sorry: x(t)=\frac {1}{2} a t^2 + v_0 t + x_0

[Mentz114]

I have to say I don't understand what you're saying. The equivalence principle I'm familiar with says that the effect on the observer accelerating at 1 g is indistinguishable in a small region from the effect of gravity on the earth's surface. It is irrelevant where they are or in which direction the accelerating observer is pointing. You don't have to compare the same small region.

I may be misreading your scenario, but at the crucial moment you describe, the accelerating observer actually is not accelerating.

[the Latex thing is a pain, but I'm told if you set your browser not to cache images it won't need to be refreshed]

[JDoolin]

It is irrelevant where they are or in which direction the accelerating observer is pointing.

Okay, yes. I can see your first point. I suppose that an object falling into a hole in the ground would still be equivalent to an object falling through a hole in any accelerating platform. (I remember being bothered by that scenario several years ago, but now I am not recalling exactly what it was that bothered me.)

I still stand by the idea that the "small region" you describe is the same "small region" where you can approximate the force as F=m*g. (But, of course, that region does extends underground.)

But at the crucial moment I describe, the acceleration is NOT zero.

Let v0=9.8m/s, x0=4.9 m, a=-9.8 m/s2

and I believe you'll find that the rocket stops at the origin (x=0) at t=1, but it's acceleration is nonzero.

I also think you introduced a very good term, "crucial moment." If the principle of equivalence is to be applied, it makes sense to make use this crucial event, when the rocket at constant acceleration stops, instantaneously, at the origin.

Jonathan

[JDoolin]

For a "low" acceleration (and by low, I mean one like earth gravity, or Jupiter's gravity, but not like a black hole's gravity),

we can use:
y = \frac{1}{2} a t^2 .

Then
dy=a t dt

To represent the differential time on the rocket, we'll want the path integral

s = \int ds = \int \sqrt{(cdt)^2 -dy^2} = \int ^{t_0}_{-t_0} (\sqrt{c^2 - (at)^2})dt

and our limit becomes
\lim_{t_0 \to 0} \frac{\int ^{t_0}_{-t_0} (\sqrt{c^2 - (at)^2})dt }{2 c t_0}

[JDoolin]

Using Mathematica, I find that the integral \int\sqrt(c^2-a^2 t^2)dt is

\frac{1}{2} t \sqrt(c^2-a^2 t^2)dt+\frac{c^2 \tan ^{-1}\left(\frac{a
t}{\sqrt{c^2-a^2 t^2}}\right)}{2 a}

and the limit does indeed reduce to 1.

This proves that the time experienced by a person standing on an accelerating platform experiences time at the same rate as anyone to whom he is instantaneously, relatively at rest.

So if the principle of equivalence is taken as given, then a person who is standing on the ground in a gravitational field should, equivalently, experience time at the same rate as someone who is relatively stationary, but far from any gravitational bodies.

This is quite contrary to what I have heard (and parroted) in the past, that time should slow down for bodies deep in gravitational wells.

Has anyone any kind of convincing and quantitative argument to support a slowing of time due to gravity? I had always assumed there was one, but putting pencil to paper, and applying the Principle of Equivalence, I'm finding that my own line of reasoning implies there is no such slowing of time.

Jonathan Doolin

[granpa]

imagine a long line of perfecly sychronized stationary clocks and a stationary observer. now imagine that the observer accelerates to velocity v along the line of clocks. since he is now moving with respect to the line of clocks the clocks will by his calculations be out of synch. the further the clocks are away the more out of sych they are.

now imagine that he accelerates continuously. You should be able to see that by his calculations the clocks are running at differnt speeds. some may even be running backwards.

[JDoolin]

imagine a long line of perfecly sychronized stationary clocks and a stationary observer. now imagine that the observer accelerates to velocity v along the line of clocks. since he is now moving with respect to the line of clocks the clocks will by his calculations be out of synch. the further the clocks are away the more out of sych they are.

now imagine that he accelerates continuously. You should be able to see that by his calculations the clocks are running at differnt speeds. some may even be running backwards.

At any given time, your accelerating observer is going to have exactly one relative velocity with all of the clocks. Since it will be the same relative velocity with every clock, it should be the same amount of time dilation for every clock.

Also, the suggestion of a long line of clocks takes us out of the territory of the Principle of Equivalence. Remember, we can only have a small region where the gravitation is constant. By the time someone accelerates to speeds with respect to a long line of clocks, where the time dilation becomes appreciable, he would be well into a region where gravity becomes dominated by the inverse distance squared.

If you want to bring into play a consideration of distance, as well as time, a better thought experiment might involve a comparison of a sky-scraper on the accelerating platform, compared to a sky-scraper floating in space. Maybe some appreciable difference in time-scales would arise.

Naturally, there will be some difference; the top and the bottom of the accelerating sky-scraper will not be in the same reference frame, and naturally wouldn't be, because the particles in the rear must continually push the particles in front of them, so the front lags behind the back in some fashion. The bottom must experience the acceleration first, but does this lead to some slowing of time?

I will have to give this some further thought.

[granpa]

At any given time, your accelerating observer is going to have exactly one relative velocity with all of the clocks. Since it will be the same relative velocity with every clock, it should be the same amount of time dilation for every clock.


the clocks will seem to be running at the same speed but the time on the clocks as calculated by the observer will be different

[Austin0]

the clocks will seem to be running at the same speed but the time on the clocks as calculated by the observer will be different
Do you think that any possible acceleration could lead to the proximate clocks that are passing [or being passed] would have decreasing displays of proper time???
If not what did mean when you said running backwards???
SImply calculaltions for clocks at spatially separated locationa??

[granpa]

clocks far enough behind the accelerating observer will seem to be running backwards (hence the connection with gravitational time dilation)

http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html

[Austin0]

clocks far enough behind the accelerating observer will seem to be running backwards (hence the connection with gravitational time dilation)

http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html
Are you talking about co-accelerating clocks in Rindler coordinates or inertial clocks in the initial rest frame???

[granpa]

I never said anything about the clocks accelerating. acceleration isnt even important. you just need to know that at time t the observer is moving at velocity k*t
the faster he is moving the more out of synch the clocks are (as calculated by the observer) and hence they must be ticking at different rates (as calculated by the accelerating observer and hence the connection with gravitational time dilation)

And its just plain old ordinary Minkowski space
http://en.wikipedia.org/wiki/Minkowski_space

[JDoolin]

clocks far enough behind the accelerating observer will seem to be running backwards (hence the connection with gravitational time dilation)

http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html

Ah, right. As the rocket accelerates, the line of simultaneity comes into line with events further in the past, so technically, objects associated with those events "seem to be running backwards." This is an effect based on the changing of reference frames.

(Austin. This is actually the same phenomenon that Tom Fontenot was talking about in your other thread, when Sue suddenly changed age from 70 to 30.)

Though, I'm sure, experientially a rocketship driver would not so much notice the sudden backward aging (which has to be calculated) as he would notice the sudden lurch into the distance of the image (which would be immediately seen.)

Granpa, One major issue with your argument is that you can't have one without the other. When you have that "backward aging" you also have the aberration effect, where the object of interest leaps into the distance. The backward aging and aberration effects are both more pronounced the further you are from the object of interest.

Standing on earth's surface and looking at a star 4 light years away is in no way equivalent to accelerating toward a star at 9.8 m/s^2. (We certainly do not see an increasing aberration, so we should not expect to measure a backward motion of clocks)

We need to think on a much smaller scale, in the region where gravity is approximately constant--the size of a mountain, perhaps.

The question then is whether this phenomenon of desynchronization would occur on the scale of a mountain, how much it would happen, and how a phenomenon that is based on relative velocity can be resolved with the fact that from any vantage, the tip and base of the mountain are co-moving.

Jonathan

[Al68]

The question then is whether this phenomenon of desynchronization would occur on the scale of a mountain, how much it would happen, and how a phenomenon that is based on relative velocity can be resolved with the fact that from any vantage, the tip and base of the mountain are co-moving.The top and bottom of the mountain are moving at two different velocities relative to an inertial (freefalling) reference frame. In the inertial frame, the distance between the top and bottom of the mountain is shrinking with time.

It's easy to see that a clock at the top runs slightly faster than a clock at the bottom due to velocity based time dilation wrt the inertial frame. Wrt the accelerated frame in which the mountain is at rest, this same effect is called gravitational time dilation, and is, like you suggest, too small to notice on earth without very sensitive equipment.

[granpa]

I will only add that the line of clocks is getting shorter as the observer accelerates and that the clocks seem to the observer to be piling up at a certain point somewhere behind the accelerating observer (the point where time seems to stop). This point would seem to the observer to be a kind of black hole.

Also, time stops at the event horizon of a real black hole due to gravitational time dilation. Inside the event horizon is where one would expect time to be running backward (by analogy with the accelerating platform anyway)

I would also like to point out that general (not special) relativity is really, at heart, a way of explaining why passive gravitational mass is proportional to inertial mass

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

[granpa]

You might find this useful, its an intro to relativity that I wrote:
http://www.physicsforums.com/showthread.php?t=314080

[JDoolin]

If you want to bring into play a consideration of distance, as well as time, a better thought experiment might involve a comparison of a sky-scraper on the accelerating platform, compared to a sky-scraper floating in space. Maybe some appreciable difference in time-scales would arise.

Daryl McCullough posted a related response (http://groups.google.com/group/sci.physics.relativity/msg/b67b2d244f13fb06?hl=en) that approached the problem somewhat along these same lines.

I can summarize it thus:

First of all, assume that the top and the bottom of the rocket have the same acceleration. This acceleration is constant, and has been for some time, so the top and bottom are comoving. (I'm still a little confused about this issue, but I think we can safely say any simultaneity issues will at least be very small, if not absent entirely.)

We have a rocket of length L, accelerating in the y direction, so the front and back of the rocket have the following equations of motion:

y_{back}=\frac{1}{2} a t^2
y_{front}=\frac{1}{2} a t^2 + L

On the back (bottom) of the rocket, we have a clock, and it sends out two signals, at time 0, and time T. The position of the light from those signals is described by:

y_{sig1}(t) = c t ; t>0
y_{sig2}(t)= \frac{1}{2} a T^2 + c (t-T) ; t>T

(The light of signal 1 leaves the rocket at time 0, so it starts at y=0. But the light of signal 2 leaves the rocket at time T, after the light has traveled up a bit.)

Now, we can find the times when the first and second signals reach the top (front) of the rocket, simply by finding t1 and t2, when

y_{sig1}(t_1)=y_{front}(t_1)
y_{sig2}(t_2)=y_{front}(t_2)

Hence
c t_1=\frac{1}{2} a t_1^2 + L
\frac{1}{2} a T^2 + c (t_2-T) =\frac{1}{2} a t_2^2 + L

The ratio of (t_2-t_1)/T should give a fairly precise measure of how much time is slowed in a length L over a region where gravity is approximately constant.

The quadratic equations produce two solutions; as the linear equation passes the quadratic twice. We can eliminate the solution where the quadratic has a slope (velocity) greater than c.

t_1 = \frac{c-\sqrt{c^2-2 a L}}{a}

t_2 = \frac{c-\sqrt{c^2 - 2 a T c + a^2 T^2 - 2 a L}}{a}


(I may edit later and put in explicit solutions for (t2-t1)/T here, if I have time.)

Daryl (using a power-series expansion in a) found this quantity to be (t_2-t_1)/T=1 + (a L)/c^2

In any case, this is the most compelling argument I've seen for the slowing of time in a gravitational field. Curiously, it requires essentially no knowledge of the Special Theory of Relativity to derive.

Jonathan

[granpa]

what you have described is just the expected classical behavior. you are confusing doppler shift with time dilation.

you must introduce special relativity into it to get general relativity

[JDoolin]

After giving some more thought, I think this phenomena is actually very different than the "Doppler effect."

I spent more time thinking about the "comoving" requirement: If you have a rocket accelerating, then the back end has always been accelerating for a slightly longer time than the front end. There will always be impulses from the back that have not yet reached the front. So, essentially the two ends can *never* be co-moving.

As equations for the front and back, then, we should use:

y_back(t)= .5 a t^2
y_front(t)= L + .5 a (t-L/c)^2

But using this equation for the position of the front of the rocket, there was actually no difference in the clocks. I found t_2-t_1=T. So I'm not sure the non-unity clock ratio actually arises with the classical behavior.

On the other hand, if we are talking about gravity, we have to assume the front and the back *are* comoving, and we can use Daryl's equation for the front.

y_back(t)= .5 a t^2
y_front(t)= L + .5 a t^2

So, assuming Daryl's approach is correct, the derivation actually relies on a fundamental difference between acceleration and gravity.

[Austin0]

After giving some more thought, I think this phenomena is actually very different than the "Doppler effect."

I spent more time thinking about the "comoving" requirement: If you have a rocket accelerating, then the back end has always been accelerating for a slightly longer time than the front end. There will always be impulses from the back that have not yet reached the front. So, essentially the two ends can *never* be co-moving.

As equations for the front and back, then, we should use:

y_back(t)= .5 a t^2
y_front(t)= L + .5 a (t-L/c)^2

But using this equation for the position of the front of the rocket, there was actually no difference in the clocks. I found t_2-t_1=T. So I'm not sure the non-unity clock ratio actually arises with the classical behavior.

On the other hand, if we are talking about gravity, we have to assume the front and the back *are* comoving, and we can use Daryl's equation for the front.

y_back(t)= .5 a t^2
y_front(t)= L + .5 a t^2

So, assuming Daryl's approach is correct, the derivation actually relies on a fundamental difference between acceleration and gravity.
Whether the front and the back are co-moving based on the propagation of momentum is a question that it maybe not so simple. It is true that the front can not accelerate until the force reaches it from the point of impulse but whether this means the back is actually moving in the meantime is questionable. Obviously there will be compression but will this be signifcant enough to be actual coordinate motion??
I would think that for this to be true the acceleration factor would have to be so high that actuall structural disruption would occur.
In any case if your assumption is correct shouldn't the ,,,,y_front(t)= L + .5 a (t-L/c)^2 be then (t-L/ speed of sound)^2 .....????

Are you familiar with the Born rigid hypothesis??
In a system accelerating under its assumptions there is a significantly different acceleration in the back I.e. greater magnitude. As far as I know this is the condition for the equivalence principle.

[JDoolin]

Whether the front and the back are co-moving based on the propagation of momentum is a question that it maybe not so simple. It is true that the front can not accelerate until the force reaches it from the point of impulse but whether this means the back is actually moving in the meantime is questionable. Obviously there will be compression but will this be signifcant enough to be actual coordinate motion??
I would think that for this to be true the acceleration factor would have to be so high that actuall structural disruption would occur.
In any case if your assumption is correct shouldn't the ,,,,y_front(t)= L + .5 a (t-L/c)^2 be then (t-L/ speed of sound)^2 .....????

Yes, it should be the speed of sound. I was looking for an upper possible limit of causality between the bottom to the top. The fact is that the speed of sound is almost vertical in the space-time diagram I'm trying to draw. I guess I want to try to imagine a perfect material where the speed of sound through it is equal to the speed of light; and see how it would be malformed by acceleration.

If the back end is sending a signal to the front end to say "speed up: move as I move," then there must be another signal coming from the front end, to the back, saying "No. You slow down. You move as I do." Both signals are delayed, and some compromise is reached. This morning, I was only taking into account the signal from the back end.

I think if you take both into account, then perhaps it makes no difference whether the signal travels at the speed of sound or the speed of light. You will still reach an equilibrium where the front and back are somehow traveling at the same velocity in any co-moving frame, so the equation of motion for the front and back would both be symmetric around t=0.

Are you familiar with the Born rigid hypothesis??
In a system accelerating under its assumptions there is a significantly different acceleration in the back I.e. greater magnitude. As far as I know this is the condition for the equivalence principle.

I said something along that line, here: (http://groups.google.com/group/sci.physics.relativity/msg/72bffe9e7e6d4a56?hl=en)


So what you're saying--(or rather, what will produce the same
mathematical result to what you're saying)--is that the acceleration
of the front of the rocket is somewhat slower than the back--so that
the front and back of the rocket share the very same stationary event
for the origin of their accelerated frame.

This appears to be qualitatively the same prediction as the Born Rigid Hypothesis as you gave it; whether it is quantitatively the same, I have no idea. In any case, if the acceleration in the back is greater, but you think the acceleration is the same, does that mean the clock in the back is going slower, or faster?

You may be better off reading Daryl's answer that led me to think about that. I certainly haven't digested it yet.

[JDoolin]

Whether the front and the back are co-moving based on the propagation of momentum is a question that it maybe not so simple. It is true that the front can not accelerate until the force reaches it from the point of impulse but whether this means the back is actually moving in the meantime is questionable. Obviously there will be compression but will this be signifcant enough to be actual coordinate motion??
I would think that for this to be true the acceleration factor would have to be so high that actuall structural disruption would occur.
In any case if your assumption is correct shouldn't the ,,,,y_front(t)= L + .5 a (t-L/c)^2 be then (t-L/ speed of sound)^2 .....????

Are you familiar with the Born rigid hypothesis??
In a system accelerating under its assumptions there is a significantly different acceleration in the back I.e. greater magnitude. As far as I know this is the condition for the equivalence principle.

I've attached a couple of pictures. The born-rigid picture, I believe, has a certain magic quality to it. If you draw a line from the origin, the slope of the line will be the reciprocal of the slope of the hyperbola. That means that two observers at those two events would regard each other as simultaneous. This structure that has the accelerations set up just right would maintain structural integrity at all times. It wouldn't get smashed by the acceleration.

The question, I guess is whether this has any reason to happen in practice. Is it something that happens in practice as long as the acceleration isn't too fast? I think if the acceleration is much faster than the speed of sound through the object, maybe it will just get crushed anyway.

Edit: What is the property of matter that keeps it from getting crushed when force is applied to it? Not the speed of sound, of course

[JDoolin]

I want to know how to calculate the "radius" or the Rindler Horizon distance from the acceleration.

I can see that these two equations are very close near t=0

x=\frac{1}{2} t^2 +1;
x^2-t^2=1

I want to change them to
x= \frac{a}{2} t^2 + r^2;
x^2 - (ct)^2 = r^2

So I need r(a); r as a function of a.

[Mentz114]

I want to know how to calculate the "radius" or the Rindler Horizon distance from the acceleration.

I can see that these two equations are very close near t=0

x=\frac{1}{2} t^2 +1;
x^2-t^2=1

I want to change them to
x= \frac{a}{2} t^2 + r^2;
x^2 - (ct)^2 = r^2

So I need r(a); r as a function of a.

For a constantly accelerating observer, I would write


a^2x^2-a^2t^2-1=0


in which case the horizon is 1/a behind the observer. 'a' being the acceleration.

[PeterDonis]

I want to know how to calculate the "radius" or the Rindler Horizon distance from the acceleration.

PMFJI, but I saw that this thread had become active again after being dormant, which prompted me to read it. The answer to the question you posed just above is simple: the "radius" r(a) is

r = \frac{c^{2}}{a}

in conventional units, or \frac{1}{a} in units where c = 1, as Mentz117 has just pointed out as I type this.

However, I also wanted to comment on what you said regarding Born rigid acceleration:

The question, I guess is whether this has any reason to happen in practice.

Austin0 may remember that some time ago we had a rather long discussion about Born rigid acceleration, among other things; the thread is at

http://www.physicsforums.com/showthread.php?t=334460

One of the points that I think we established is that, although it is extremely difficult to *arrange* for an object to undergo Born rigid acceleration from the start, as it were (since it would require a very precise application of very precisely calculated amounts of thrust to each small piece of the object), in practice, one actually would expect a fairly "normal" object, such as a rocket undergoing thrust from an engine at the rear, to eventually "settle" into a state of Born rigid acceleration--in other words, Born rigid acceleration is a sort of "equilibrium state" for an accelerating object. This is because, as you note, an object in a state of Born rigid acceleration maintains constant, unchanging proper distances between all of its parts, and so such an object will maintain constant internal stresses (assuming that internal stresses are basically a function of proper distances between the parts--which is pretty much what we have observed experimentally). Of course, an object that "settles" into such a state, such as a rocket with an engine providing thrust at the rear, would have a quite different "profile" of internal stresses while under acceleration than it would have if it were moving inertially (in the latter state the internal stresses would, presumably, be zero in the typical case, whereas under acceleration they would have to be nonzero and varying from the front to the back of the object in a definite way--the thread linked to above goes into this).

One other note: if "Born rigid acceleration" is a good "buzzword" for the second of your two diagrams in post #22, "Bell's spaceship paradox" would be a good buzzword for the first one.

[JDoolin]

I'm trying to get a sense of the scale of the Rindler coordinates.

Let's say that we have a rocket accelerating so that it's the same as the gravitational constant on earth is 9.80665 m/s^2 (N/kg)

And we are able to detect a difference in gravity in the 6th figure. How far would we have to climb along our rocket before we notice that the gravity has dropped to 9.80664 N/kg?

(3*10^8)^2 \left(\frac {1}{9.80664} - \frac{1}{9.80665}\right)=9.36*10^9 m = 31.2 light seconds

Let me also calculate how far you would have to go to find the same amount of change in the gravitation, if you were to climb a building on earth.


\begin{matrix}
r = \sqrt {\frac {(G)(M_e)}{g}}
\\
\Delta r = \sqrt{\frac{(6.673*10^{-11})(5.9742*10^{24})}{9.80664}}-\sqrt{\frac{(6.673*10^{-11})(5.9742*10^{24})}{9.80665}}
\\
=63758820.792-6375879.541 \\
=3.25 m

\end{matrix}

[JDoolin]

r = \frac{c^{2}}{a}


Thanks for the equation for r. and the link and the summary. Post 13 (http://www.physicsforums.com/showthread.php?p=2334164#post2334164) also had a good diagram showing the lines of simultaneity.

[PeterDonis]

I'm trying to get a sense of the scale of the Rindler coordinates.


These calculations look right to me. If you want an intuitive way to guess the orders of magnitude, observe that the "radius" corresponding to a 1-g acceleration is about 1 light-year, which is about 30 *million* light-seconds; your value of 31.2 light seconds is 6 orders of magnitude smaller, which makes sense if you're looking for a difference in the 6th figure. For the case of the Earth, since the radius is about 6 million meters, your figure of 3.25 meters is again 6 orders of magnitude smaller. (The extra factor of 2 or so--3.25 is about half of 6--comes from the square root in the formula for the Earth's radius as a function of acceleration.)

Of course the obvious question is why the "Rindler radius" is about 9 orders of magnitude larger. One could argue that the comparison isn't really a valid one: the "Rindler radius" is the distance to the Rindler horizon, whereas the Earth's radius isn't the distance to any kind of horizon, since the Earth isn't a black hole. However, we can imagine the Earth replaced by a black hole of the same mass, and the acceleration due to gravity at a radius of 6.37 million meters would *still* be 1 g--and that radius *could* be viewed as a "distance to the horizon" (the actual Schwarzschild radius of the Earth-mass black hole would only be a few millimeters, which is too small an "error" to worry about), as long as we're willing to overlook some technicalities in how to define the "distance" to a black hole horizon.

In my opinion, the 9 orders of magnitude difference is simply a (rather striking) illustration of the difference between flat and curved spacetime. I'd be interested to know what other regulars on these forums think, though.

[JDoolin]

In my opinion, the 9 orders of magnitude difference is simply a (rather striking) illustration of the difference between flat and curved spacetime. I'd be interested to know what other regulars on these forums think, though.

Is there a definitive quantitative measurement for the curvature of space? Could it be said that g/c2 could be a measure of the curvature of space? I think I've heard that the curvature may be related to the gravitational potential, rather than the gravitational pull.

[Mentz114]

Is there a definitive quantitative measurement for the curvature of space? Could it be said that g/c2 could be a measure of the curvature of space? I think I've heard that the curvature may be related to the gravitational potential, rather than the gravitational pull.

Spatial curvature alone is not enough to give a semblance of Newtonian gravity. It is in fact the 'time curvature' -(1+ c2g00) which gives a Newtonian potential from the Schwarzschild metric. I don't think 'time curvature' is a proper term, which is why I've quoted it.

Full space-time curvature is encoded in the Riemann tensor, which has 20 independent components.

[PeterDonis]

Is there a definitive quantitative measurement for the curvature of space?

The usual quantitative measure of spacetime curvature is tidal gravity--the magnitude of tidal accelerations. For the spacetime outside a gravitating body like the Earth, tidal accelerations go like

\frac{2GM}{r^{3}} \Delta r

where r is the distance from the center of the Earth and \Delta r is the size of the body being subjected to the tidal accelerations.

(Technically, according to GR, the quantitative measure of spacetime curvature is the Riemann curvature tensor, whose components are sort of a "limit" of tidal acceleration as the size of the object goes to zero--I say "sort of" because the limit of the above formula as \Delta r goes to zero is zero, so obviously that isn't quite what the curvature tensor is. But its "radial" components outside a gravitating body like the Earth look just like the above formula, but with the \Delta r taken out--i.e., just \frac{2GM}{r^{3}}. There's been plenty of discussion on these forums of the Riemann curvature tensor; I can try and find some good threads and post links to them.)

Could it be said that g/c2 could be a measure of the curvature of space?

Actually, in a sense, yes, despite what I just said above--but it's still curvature of *spacetime*, not just space. There's an interesting discussion of this point in Misner, Thorne, and Wheeler, in Box 1.6 (pp. 32-33 in my edition). They compare the trajectories of a baseball, thrown over a distance of 10 meters, and a bullet fired from a rifle over the same distance. Obviously the spatial curvature of these two trajectories will be very different--they assume a slow thrown baseball, not a Nolan Ryan fastball, so the baseball's trajectory will be visibly curved in space, unlike the bullet's, whose spatial curvature is microscopic (but it still is there, as we'll see). But if we take time into account, we find that the curvatures of these two paths *are* the same (actually, we find that the tracks themselves are not "curved", being geodesics--but they reveal the curvature of the spacetime through which they travel, which has to be the same for both). Here's a quick back of the envelope calculation of that using their numbers:

Horizontal distance: 10 m for both baseball and bullet.

Vertical "height" of trajectory (highest point relative to start and end points): 5 m for baseball, 5 x 10^4 m (half a millimeter) for bullet.

Speed: 5 m/sec for baseball (about 10 mph, so a *very* slow pitch, hence the high trajectory), 500 m/sec for bullet.

Travel time: 2 sec for baseball, 2 x 10^-2 sec for bullet.

Spacetime interval: here MTW use the special relativistic formula s^{2} = {ct}^{2} - x^{2}; yes, this formula is only valid in an inertial frame, but since all the distances and times are very small compared to the radius of curvature we'll be calculating, we can treat all the numbers above as being measurements with respect to a momentarily comoving inertial frame. The intervals work out to be 6 x 10^8 meters for the baseball, and 6 x 10^6 meters for the bullet. (Note that the horizontal distance itself, 10 m, is negligible here, so the interval is basically ct; the horizontal distance only comes into play indirectly in determining t from the speed of each object.)

Curvature: here MTW use the formula (radius of curvature) = (spacetime interval)^2 / (8 * vertical rise). The derivation of this formula from the components of the Riemann curvature tensor is in the text; I won't go into it here. If you plug in the numbers above for the baseball and the bullet, you'll see that both of them result in a radius of curvature of about 1 light-year, which, as we've already seen, is indeed c^2 / g for a 1-g acceleration. And in fact, we can see that their formula for radius of curvature *must* give this result; if we use h for the vertical rise and t for the travel time, we have:

h = \frac{1}{2} g \left( \frac{1}{2} t \right)^{2} = \frac{1}{8} g t^{2}

r = \frac{\left( ct \right)^{2}}{8 h} = \frac{c^{2}}{g}

Note that we have to square 1/2 t in the formula for h because each object goes up and then comes down again, so the height h must be calculated using the time for half of the trajectory.

So this means that the actual "radius of curvature" of spacetime around the Earth *is* 1 light-year, even though the spatial radius of the Earth itself is only 6.3 million meters or so, or 9 orders of magnitude smaller.

[PeterDonis]

So this means that the actual "radius of curvature" of spacetime around the Earth *is* 1 light-year, even though the spatial radius of the Earth itself is only 6.3 million meters or so, or 9 orders of magnitude smaller.

Thinking more about this, I realized that the 9 orders of magnitude difference illustrates a relationship that must always hold between the "geometric mass" of an object (its mass in geometric units) and the "radius of curvature" of spacetime around the object. Using r_{c} for the radius of curvature, r_{m} for the geometric mass, and g for the gravitational acceleration at radius r, we have:

g = \frac{GM}{r^{2}}

r_{c} = \frac{c^{2}}{g} = \frac{c^{2} r^{2}}{GM}

r_{m} = \frac{GM}{c^{2}}

Putting these equations together and rearranging slightly, we get:

\frac{r_{c}}{r} = \frac{r}{r_{m}}

The Earth's geometric mass is about 4.4 millimeters, so this relationship does indeed appear to hold given the numbers calculated in this thread.

(Edit: I should note that the "geometric mass" for a given mass is *half* the Schwarzschild radius of a black hole with that mass.)

(Edit #2: I should also note that the above formulas are only valid when the radius r is much greater than the "geometric mass". For small radius, the acceleration g is no longer given by the above formula; it has an extra factor in the denominator that diverges as r approaches the Schwarzschild radius, 2 r_{m}.)

[JDoolin]

The usual quantitative measure of spacetime curvature is tidal gravity--the magnitude of tidal accelerations. For the spacetime outside a gravitating body like the Earth, tidal accelerations go like

\frac{2GM}{r^{3}} \Delta r

where r is the distance from the center of the Earth and \Delta r is the size of the body being subjected to the tidal accelerations.

(Technically, according to GR, the quantitative measure of spacetime curvature is the Riemann curvature tensor, whose components are sort of a "limit" of tidal acceleration as the size of the object goes to zero--I say "sort of" because the limit of the above formula as \Delta r goes to zero is zero, so obviously that isn't quite what the curvature tensor is. But its "radial" components outside a gravitating body like the Earth look just like the above formula, but with the \Delta r taken out--i.e., just \frac{2GM}{r^{3}}.

More simply,

\begin{matrix} g=-G M r^{-2} \\ \frac {dg}{dr}=2 G M r^{-3} \end{matrix}

[Passionflower]

Let me also calculate how far you would have to go to find the same amount of change in the gravitation, if you were to climb a building on earth.


\begin{matrix}
r = \sqrt {\frac {(G)(M_e)}{g}}
\\
\Delta r = \sqrt{\frac{(6.673*10^{-11})(5.9742*10^{24})}{9.80664}}-\sqrt{\frac{(6.673*10^{-11})(5.9742*10^{24})}{9.80665}}
\\
=63758820.792-6375879.541 \\
=3.25 m

\end{matrix}


I do not believe the formula you use is correct.

First you should calculate the Schwarzschild radius rs using G and the mass of the Earth then solve the following equation for r :


{\it r_s \over 2r^2}(1-\it r_s}/r)^{-1/2} = 9.80664_{r2}, \, 9.80665_{r1}


Then if you want to be exact you can calculate the distance using:


\sqrt {{\it r2}\, \left( {\it r2}-{\it r_s} \right) }-\sqrt {{\it r1}\,
\left( {\it r1}-{\it r_s} \right) }+{\it r_s}\,\ln \left( {\frac {
\sqrt {{\it r2}}+\sqrt {{\it r2}-{\it r_s}}}{\sqrt {{\it r1}}+\sqrt {{
\it r1}-{\it r_s}}}} \right)


This step is optional though as the difference is extremely small.

[PeterDonis]

More simply,

\begin{matrix} g=-G M r^{-2} \\ \frac {dg}{dr}=2 G M r^{-3} \end{matrix}

Yes. I was groping towards that. :-)

[PeterDonis]

I do not believe the formula you use is correct.

First you should calculate the Schwarzschild radius rs using G and the mass of the Earth then solve the following equation for r ...

Technically this is correct, but since JDoolin was only considering things to the sixth significant figure, and the Schwarzschild radius is eight or nine orders of magnitude smaller than the radius of the Earth, the approximation he used was fine. Here are the numbers I come up with using your formula (ignoring the ln term since, as you say, it's way way smaller even than the small corrections we'll see below):

For g = 9.80664 m/s^2, r = 6375882.769 meters.

For g = 9.80665 m/s^2, r = 6375879.518 meters.

As you see, the approximation JDoolin used is fine to six figures; the differences with the more exact formula you gave only show up in the eighth figure.

(Edit: Actually it's the ninth figure.)

[JDoolin]

The usual quantitative measure of spacetime curvature is tidal gravity--the magnitude of tidal accelerations. For the spacetime outside a gravitating body like the Earth, tidal accelerations go like

\frac{2GM}{r^{3}} \Delta r

where r is the distance from the center of the Earth and \Delta r is the size of the body being subjected to the tidal accelerations.

(Technically, according to GR, the quantitative measure of spacetime curvature is the Riemann curvature tensor, whose components are sort of a "limit" of tidal acceleration as the size of the object goes to zero--I say "sort of" because the limit of the above formula as \Delta r goes to zero is zero, so obviously that isn't quite what the curvature tensor is. But its "radial" components outside a gravitating body like the Earth look just like the above formula, but with the \Delta r taken out--i.e., just \frac{2GM}{r^{3}}. There's been plenty of discussion on these forums of the Riemann curvature tensor; I can try and find some good threads and post links to them.)



Actually, in a sense, yes, despite what I just said above--but it's still curvature of *spacetime*, not just space. There's an interesting discussion of this point in Misner, Thorne, and Wheeler, in Box 1.6 (pp. 32-33 in my edition). They compare the trajectories of a baseball, thrown over a distance of 10 meters, and a bullet fired from a rifle over the same distance. Obviously the spatial curvature of these two trajectories will be very different--they assume a slow thrown baseball, not a Nolan Ryan fastball, so the baseball's trajectory will be visibly curved in space, unlike the bullet's, whose spatial curvature is microscopic (but it still is there, as we'll see). But if we take time into account, we find that the curvatures of these two paths *are* the same (actually, we find that the tracks themselves are not "curved", being geodesics--but they reveal the curvature of the spacetime through which they travel, which has to be the same for both). Here's a quick back of the envelope calculation of that using their numbers:

Horizontal distance: 10 m for both baseball and bullet.

Vertical "height" of trajectory (highest point relative to start and end points): 5 m for baseball, 5 x 10^4 m (half a millimeter) for bullet.

Speed: 5 m/sec for baseball (about 10 mph, so a *very* slow pitch, hence the high trajectory), 500 m/sec for bullet.

Travel time: 2 sec for baseball, 2 x 10^-2 sec for bullet.

Spacetime interval: here MTW use the special relativistic formula s^{2} = {ct}^{2} - x^{2}; yes, this formula is only valid in an inertial frame, but since all the distances and times are very small compared to the radius of curvature we'll be calculating, we can treat all the numbers above as being measurements with respect to a momentarily comoving inertial frame. The intervals work out to be 6 x 10^8 meters for the baseball, and 6 x 10^6 meters for the bullet. (Note that the horizontal distance itself, 10 m, is negligible here, so the interval is basically ct; the horizontal distance only comes into play indirectly in determining t from the speed of each object.)

Curvature: here MTW use the formula (radius of curvature) = (spacetime interval)^2 / (8 * vertical rise). The derivation of this formula from the components of the Riemann curvature tensor is in the text; I won't go into it here. If you plug in the numbers above for the baseball and the bullet, you'll see that both of them result in a radius of curvature of about 1 light-year, which, as we've already seen, is indeed c^2 / g for a 1-g acceleration. And in fact, we can see that their formula for radius of curvature *must* give this result; if we use h for the vertical rise and t for the travel time, we have:

h = \frac{1}{2} g \left( \frac{1}{2} t \right)^{2} = \frac{1}{8} g t^{2}

r = \frac{\left( ct \right)^{2}}{8 h} = \frac{c^{2}}{g}

Note that we have to square 1/2 t in the formula for h because each object goes up and then comes down again, so the height h must be calculated using the time for half of the trajectory.

So this means that the actual "radius of curvature" of spacetime around the Earth *is* 1 light-year, even though the spatial radius of the Earth itself is only 6.3 million meters or so, or 9 orders of magnitude smaller.

It sounds like Misner Thorne Wheeler (MTW) might be an text on General Relativity, with comprehendible examples. I haven't seen it, so I'l see if I can locate it through the library system.


I'm curious/confused about the fomula
h = \frac{1}{2} g \left( \frac{1}{2} t \right)^{2} = \frac{1}{8} g t^{2}

How is this time calculated; is it the time the ball take to go 10 meters? t=x/v ? Is this all done in a region of constant g?

I spent some time trying to collect some of the concepts in your posts--I haven't got to Passionflower's just yet.



\begin{center}
\begin{tabular}{| c | c | c | c| }
\hline
Formula &
ind. vars &
name &
derivation \\ \hline

r_m=\frac {GM}{c^2} &
M &
Geometric Mass &
2*Schwartzchild Radius \\ \hline


r_c=\frac{c^2}{a} &
a &
Rindler Horizon &
position of stationary event in accelerating frame \\ \hline

r_c= \frac {c^2 r^2} {G M} &
M, r &
spatial curvature &
Rindler Horizon with Newtonian Gravity \\ \hline

r_c=\frac{s^2}{8 h} &
s,h &
radius of curvature &
Components of Riemann Curvature Tensor \\ \hline

-(1+ c^2g00) &
? &
time curvature &
? \\ \hline

\end{tabular}
\end{center}

[PeterDonis]

It sounds like Misner Thorne Wheeler (MTW) might be an text on General Relativity, with comprehendible examples.

It is one of the classic GR texts--some would say *the* classic GR text, although, as the Usenet Physics FAQ says (I link to the page below), it "may not the best first book on GR for most students, in part because by offering so much it is liable to overwhelm a newcomer." Also, since it was published in 1973, it doesn't cover some important subject areas that were not really understood then (e.g., black hole evaporation). It's a weighty tome, so be advised that if you get it from the library, you may need a wheelbarrow to get it home. :-)

A more recent book is Wald's 1984 text "General Relativity", which covers a lot of the same ground as MTW but includes topics like black hole evaporation that are left out of MTW.

This page on the Usenet Physics FAQ is a pretty comprehensive guide to relativity texts:

http://www.edu-observatory.org/physics-faq/Administrivia/rel_booklist.html

There are texts listed there that may be better introductory ones; I'm not familiar enough with any of them to give an opinion (I have both MTW and Wald so I tend to draw examples from them).

I'm curious/confused about the fomula
h = \frac{1}{2} g \left( \frac{1}{2} t \right)^{2} = \frac{1}{8} g t^{2}

How is this time calculated; is it the time the ball take to go 10 meters? t=x/v ? Is this all done in a region of constant g?

I just used the Newtonian formula for distance as a function of time, d = \frac{1}{2} g t^{2}, noting that the "t" in this formula will be the time for the ball or bullet to travel half the 10 meters, or 5 meters, since the formula applies to half the trajectory only (either half, since they're symmetric); since we were using the symbol "t" for the time to travel the whole 10 meters, I inserted "1/2 t" in the Newtonian formula.

Obviously this is an approximation, which assumes that g is constant in the spacetime region of interest, that the Earth's surface is flat to within the desired accuracy, and that the trajectories are parabolic. All of these assumptions are valid for the level of accuracy MTW were using (only 1 or 2 significant figures); they would *not* all be valid for the six-figure accuracy you were using, since, as you saw, g varies in the sixth figure over a height of 3 meters or so, less than the 5 meter height of the ball's trajectory. (I haven't calculated the error in the other assumptions, but I'm pretty sure the spatial curvature of the Earth, at least, shows up sooner than the sixth figure.)

[PeterDonis]

I spent some time trying to collect some of the concepts in your posts

A few comments/corrections:

(1) What I've been calling the "geometric mass" is usually just called "mass" or "mass in geometric units" and is usually just denoted M in the textbooks, which usually use geometric units (MTW does, for example; what you denote "M" in your table, MTW would denote M_{conv} to make clear that it's in "conventional" rather than geometric units). It is *half* the Schwarzschild radius (which is usually written 2M).

(2) What I've been calling the "radius of curvature" r_{c} is not one of the components of the Riemann curvature tensor; it's a scalar which can be calculated from those components, but it's an invariant (the same in all coordinate systems), which the curvature tensor components, individually, are not. (Actually, in 4-dimensional spacetime, there is in general not a single "radius of curvature", since that radius can be different along different dimensions--which is why we need a tensor to completely describe the curvature. For the specific case we've been considering, what we've been calling the "radius of curvature" should more properly be called something like the "radius of curvature in the r-t plane".)

(Edit: I should also note that the two items you've labeled as "spatial curvature" and "radius of curvature" are the same thing; the formula s^{2} / 8 h from MTW for calculating the latter is just a shortcut way of calculating the former for that specific problem. Obviously you could set up problems with baseballs or bullets or other objects traveling at different speeds on different trajectories, but what you'll find is that no matter how you set the problems up, the interval s, the height h, and any other variables, will all vary together in such a way that the "radius of curvature" at a given radius r from the central mass is always the same, and is given by the formula you gave for "spatial curvature".)

(3) I haven't really discussed what you've labeled as "time curvature" yet, but for the case we've been considering, the metric component g_{00}, in Schwarzschild coordinates, is

g_{00} = g_{tt} = 1 - \frac{2 G M}{c^{2} r}

where I've used conventional units. The "time dilation factor", which is the amount of proper time elapsed for an observer hovering at radius r (and therefore accelerating to hold station), for a given amount of coordinate time t, is the square root of this. The expression Mentz114 gave in post #30 comes from the desire to show the correspondence between GR and Newtonian gravity; if we write the Newtonian "potential" as

U = - \frac{G M}{r}

we can see that

g_{tt} = 1 + \frac{2 U}{c^{2}}

which can be rearranged to look pretty much like what Mentz114 wrote down. (Note that I've used a "timelike" sign convention in the above, where the sign of the tt term in the metric is positive and the signs of the "spacelike" terms, rr and so forth, are negative; Mentz114's expression implicitly used the opposite convention, where the timelike terms are negative and the spacelike terms are positive, so his signs are switched around from mine.)

[PeterDonis]

Curvature: here MTW use the formula (radius of curvature) = (spacetime interval)^2 / (8 * vertical rise). The derivation of this formula from the components of the Riemann curvature tensor is in the text; I won't go into it here.

This statement has been nagging at me, so I went back and looked through that section of MTW more carefully (I was more or less going on memory before--I just briefly glanced at MTW to verify that the formula in Box 1.6 was the one I remembered). The statement I made in the quote above, as it stands, is false, as far as I can tell: I can't find any specific derivation of the "radius of curvature" formula from the Riemann tensor in MTW. In fact, no explicit derivation is given: there is simply a note in Box 1.6 of MTW about comparing the track to the "arc of a circle". That made me think of the following: suppose, in Euclidean geometry, you have a circle cut by a chord of length s, forming an arc of "height" (i.e., the length of the largest perpendicular from the arc to the chord) h. If r is the radius of the circle, the Pythagorean theorem gives:

\left( r - h \right)^{2} + \left( \frac{s}{2} \right)^{2} = r^{2}

We can rearrange this to (assuming I've done the algebra correctly):

r = \frac{s^{2}}{8 h} \left( 1 + 4 \frac{h^{2}}{s^{2}} \right)

which, assuming h << s, gives the formula MTW used.

Taken at face value, this means that MTW's "radius of curvature" calculation is basically hand-waving; they're just assuming that a formula from Euclidean geometry can be carried over to a calculation in a local inertial frame in GR. However, I don't think it's a coincidence that the result of the formula *exactly* equals the "Rindler horizon" value c^{2} / g. There probably is some connection with the curvature tensor that I haven't yet spotted that makes the "radius of curvature" formula make sense in this situation. However, it certainly isn't as simple as I had implied in my earlier post.

[JDoolin]

\begin{center}
\begin{tabular}{| c | c | c | c| }
\hline
Formula &
ind. vars &
name &
derivation \\ \hline

r_m=\frac {GM_{conv}}{c^2} &
M_{conv} &
Geometric Mass &
.5*Schwartzchild Radius \\ \hline


r_c=\frac{c^2}{a} &
a &
Rindler Horizon &
position of stationary event in accelerating frame \\ \hline

r_c= \frac {c^2 r^2} {G M} &
M, r &
spatial curvature &
Rindler Horizon with Newtonian Gravity \\ \hline

r_c=\frac{s^2}{8 h} &
s,h &
radius of curvature &
See attached diagram \\ \hline

-(1+ c^2g00) &
? &
time curvature &
? \\ \hline

g_{tt}=1-\frac{2 G M_{conv}} {c^2 r} & M_{conv}, r & (grav time dilation)^2 & ? \\ \hline
g_{tt}=1-\frac{2 U} {c^2} & U \in \lbrace - \infty , 0 \rbrace & (grav time dilation)^2 & ? \\ \hline


\end{tabular}
\end{center}


I've ordered both Wald, and MTW through the library, so I'll get out the wheelbarrow soon.

I am just applying some corrections and additions to the table based on your responses. I fixed the Sch. radius and added equations for gtt.

I also attached a diagram related to your calculation of s2/8h. I was thinking that the s was the space-time-interval, but that is definitely not the case in this diagram. Or if it is the case, it is only because (ct)^2 - h^2 \approx (ct)^2+h^2 for this low-velocity case.

[PeterDonis]

I...added equations for gtt.

The second equation for g_{tt} should have a plus sign in front of the U term; the term only becomes negative because the sign of U itself is negative (as you note in the table).

I also attached a diagram related to your calculation of s2/8h. I was thinking that the s was the space-time-interval, but that is definitely not the case in this diagram. Or if it is the case, it is only because (ct)^2 - h^2 \approx (ct)^2+h^2 for this low-velocity case.

The diagram as you've drawn it is backwards; the "radius" lines should go to the center of the circle.

In the diagram, of course, "s" is just a distance, as you note; but in MTW, "s" does refer to a spacetime interval (which is basically just the time interval, since it's so much larger than the space interval--here both "intervals" refer to the horizontal travel of the ball or the bullet, so the "time interval" is c times the travel time, 2 sec or 0.02 sec, and the "space interval" is just 10 meters in both cases; h only comes in when we apply the formula for radius of curvature). That's why I said the MTW argument was basically hand-waving.

[JDoolin]

The second equation for g_{tt} should have a plus sign in front of the U term; the term only becomes negative because the sign of U itself is negative (as you note in the table).



Are you sure? The equation now has it so the gravitational time dilation is 1 when you are far away from any masses, and it increases toward infinity as you go into the black hole. The way you're saying, it would go to zero fairly quickly, (so time would be speed up in the gravitational well) and then tend toward negative infinity.



The diagram as you've drawn it is backwards; the "radius" lines should go to the center of the circle.

In the diagram, of course, "s" is just a distance, as you note; but in MTW, "s" does refer to a spacetime interval (which is basically just the time interval, since it's so much larger than the space interval--here both "intervals" refer to the horizontal travel of the ball or the bullet, so the "time interval" is c times the travel time, 2 sec or 0.02 sec, and the "space interval" is just 10 meters in both cases; h only comes in when we apply the formula for radius of curvature). That's why I said the MTW argument was basically hand-waving.

I made a new diagram.

c = 3*10^8; a = 9.8;
xmin = -c^2/a;
xmax = 5 c^2/a;
tbound = 3 c/a;
p1 = ContourPlot[(x + c^2/a)^2 - (c*t)^2 == c^4/a^2, {x, xmin,
xmax}, {t, -tbound, tbound}, Axes -> True];(*hyperbolic*)
p2 = ContourPlot[x == a/2 t^2, {x, xmin, xmax}, {t, -tbound, tbound},
Axes -> True];(*quadratic*)
p3 = ContourPlot[
x + c^2/a == c t, {x, xmin, xmax}, {t, -tbound, tbound}];(*linear*)
p4 = ContourPlot[
x + c^2/a == -c t, {x, xmin, xmax}, {t, -tbound, tbound}];(*linear*)
p5 = ContourPlot[(x - c^2/a)^2 + (c t)^2 == (c^2/a)^2, {x, xmin,
xmax}, {t, -tbound, tbound}];(*circular*)
p6 = Graphics[{Thick, Blue, Line[{{-c^2/a, 0}, {c^2/a, 0}}]}];(*radius lines*)
Show[p1, p2, p3, p4, p5, p6]

[PeterDonis]

Are you sure? The equation now has it so the gravitational time dilation is 1 when you are far away from any masses, and it increases toward infinity as you go into the black hole. The way you're saying, it would go to zero fairly quickly, (so time would be speed up in the gravitational well) and then tend toward negative infinity.

Look again at post #39; the two formulas for g_{tt} that I gave there have to agree, given the formula for U that I gave there (which agrees with the range you give for U). That means there needs to be a plus sign in front of the U term in the second formula, since U is negative.

You may be interpreting "time dilation" backwards from the way I was using the term. (It's kind of a difficult term since it can reasonably be interpreted either way; unfortunately, I'm not aware of any other handy colloquial term for what we're talking about.) The metric coefficient g_{tt} (or more precisely its square root) tells you how much proper time elapses for each unit of coordinate time. So if you're deep in a gravity well, your g_{tt} will be *less* than 1, meaning that, for example, *less* than 1 second of your proper time elapses for each second of Schwarzschild coordinate time t (the time that is experienced by observers very far away from the gravity well). (As I said, "time dilation" is probably not the best name for this effect, but I don't know of any other handy name for it.) This corresponds to U getting more and more *negative* as you get deeper into the gravity well; but again, the sign in front of U in the formula for g_{tt} needs to be a plus sign for g_{tt} to get smaller as U gets more and more negative.

It is true, by the way, that g_{tt} goes to zero as the radius r goes to the Schwarzschild radius, and becomes negative (with the sign convention we're using) inside that radius, ultimately going to negative infinity as the radius goes to zero. This is one of the reasons that Schwarzschild coordinates are not good coordinates for doing physics close to or inside a black hole horizon. We've only been dealing with situations where the radius is much larger than the Schwarzschild radius, so we haven't had to deal with that problem.

I made a new diagram.

The circle part looks good.

Edit: OK, I see where the rest of the diagram is coming from. Basically, we have three curves that all look the same in the vicinity of the events in question: the hyperbola, the parabola, and the circle:

\left( x + r \right)^{2} - \left( c t \right)^{2} = r^{2}

x = \frac{ \left( c t \right)^{2} }{2 r}

\left( x - r \right)^{2} + \left( c t \right)^{2} = r^{2}

where I've written r, the "radius of curvature", in place of c^2 / a. These equations can be rearranged as follows:

r = \frac{ \left( c t \right)^{2} - x^{2} }{2 x}

r = \frac{ \left( c t \right)^{2} }{2 x}

r = \frac{ \left( c t \right)^{2} + x^{2} }{2 x}

When x is small enough, all three of these equations reduce to the second, which is basically MTW's formula (the only difference being, as I noted in a previous post, that MTW's time value "t" refers to a time twice as long as the "t" that appears in the above equations, so a factor of 8 appears in their denominator instead of 2).

[pervect]

What page of MTW is the bit about "radius of curvature"? If it is in conjunction with the Rindler space-time, I'm sure they weren't talking about the Riemann - because the Riemann tensor of that space-time is zero.

[PeterDonis]

What page of MTW is the bit about "radius of curvature"? If it is in conjunction with the Rindler space-time, I'm sure they weren't talking about the Riemann - because the Riemann tensor of that space-time is zero.

It's pp. 32-33 in my edition, Box 1.6, where they talk about the "radius of curvature" of the path of a ball and a bullet in the gravitational field of the Earth. I thought at first, based on my (obviously imperfect) memory of that section, that there was at least something there about relating the radius of curvature to the Riemann tensor, but as I noted in post #40, when I checked my memory by re-reading the section in detail, I couldn't find anything.

[JDoolin]

Look again at post #39; the two formulas for g_{tt} that I gave there have to agree, given the formula for U that I gave there (which agrees with the range you give for U). That means there needs to be a plus sign in front of the U term in the second formula, since U is negative.

You may be interpreting "time dilation" backwards from the way I was using the term. (It's kind of a difficult term since it can reasonably be interpreted either way; unfortunately, I'm not aware of any other handy colloquial term for what we're talking about.)

Indeed I was interpreting "time dilation" backwards from the way you were using it. I think in Special Relativity, the time-dilation-factor is usually:

\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}= \frac {\Delta t_{observer(local)}}{\Delta t_{observed(distant)}}

The passage of time for the observer is faster than the passage of time for the observed. The input variable is the relative velocity v, and the output has a range from (1,\infty)

while the gravitational based time dilation you are using, if you consider both observers, is

\frac {\Delta t_{observed(distant)}}{\Delta t_{observer(local)}}=\frac{c^2+2U_{observed}}{c^2+ 2U_{observer}}

(This simplifies to the equation we were using before if you plug in U_{observer} = 0)

The input variables are U_{observer} \in (-\infty,0) and U_{observed} \in (-\infty,0) and the output has a range from (-\infty,\infty ).

[JDoolin]

1-\frac{2 G M_{conv}} {c^2 r}

G in N m^2 / kg^2 = m^3 / (s^2 kg)
M in kg
c in m/s
r in m

U=- G M / r
G in m^3/(s^2 kg)
M in kg
r in m
U in m^2/s^2 or Joules/kg

U is in Joule's per kilogram, which simplifies to m^2/s^2. Analtogous to voltage, V, in Joules per Coulomb, which doesn't simplify.

[PeterDonis]

I guess we're technically using potential energy per unit mass; sort of an equivalent to Voltage, but for mass.

Yes, the units of the "potential" U are velocity squared; that's true in Newtonian physics as well as relativity. The reason we can get away with this is the equivalence principle--more precisely, the version of it that says (in Newtonian terms) that all objects fall with the same acceleration in a gravitational field. That allows us to factor out the "test mass", the mass of the object we are using to probe the field, and do everything in units "per unit mass"--not just energy, but also, in more complicated problems, other quantities like angular momentum.

As you say, this is analogous to how voltage is used in problems dealing with electricity. However, there is no analogue of the equivalence principle for electricity; there are objects with different charge to mass ratios, which therefore "fall" with different accelerations in an electric field. It just happens that for many problems, we can ignore this because all the moving charge carriers have the same charge/mass ratio (typically they are all electrons), so we can factor it out and use voltage instead of the actual electric field.

[JDoolin]

Yes, the units of the "potential" U are velocity squared; that's true in Newtonian physics as well as relativity. The reason we can get away with this is the equivalence principle--more precisely, the version of it that says (in Newtonian terms) that all objects fall with the same acceleration in a gravitational field. That allows us to factor out the "test mass", the mass of the object we are using to probe the field, and do everything in units "per unit mass"--not just energy, but also, in more complicated problems, other quantities like angular momentum.

As you say, this is analogous to how voltage is used in problems dealing with electricity. However, there is no analogue of the equivalence principle for electricity; there are objects with different charge to mass ratios, which therefore "fall" with different accelerations in an electric field. It just happens that for many problems, we can ignore this because all the moving charge carriers have the same charge/mass ratio (typically they are all electrons), so we can factor it out and use voltage instead of the actual electric field.

I guess I've never been quite comfortable with the difference between "potential" and "potential energy"

"potential energy" has units of kg \frac{m^2}{s^2} and, if I'm not mistaken, "potential" doesn't get introduced until it is introduced as "electric potential" with Coulomb's Law in most introductory physics books.

In any case, it's beginning to dawn on me how acceleration-based time dilation occurs in Rindler coordinates.

In the attached diagram, you see one observer traveling along arc AC, another along BD. Both observers view Events A&B to be simultaneous, and events C&D to be simultaneous. However, both also agree that BD takes much longer than AC.

Now I need to find the quantitative relationship between those two times.

[PeterDonis]

I guess I've never been quite comfortable with the difference between "potential" and "potential energy"

"potential energy" has units of kg \frac{m^2}{s^2} and, if I'm not mistaken, "potential" doesn't get introduced until it is introduced as "electric potential" with Coulomb's Law in most introductory physics books.

I agree "potential" by itself is kind of a vague term, it gets used in a number of different ways. It would probably be better if relativity texts stuck to the term "energy per unit mass", which is what "potential" usually means when gravity is the only force under consideration, but which generalizes nicely to situations where other forces are in play without changing its meaning. Energy per unit mass also has the nice property that in "relativistic units", where the speed of light = 1, it is dimensionless.

Now I need to find the quantitative relationship between those two times.

One fairly easy way to do it is to transform to Rindler coordinates, as described in the Wikipedia page:

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

In these coordinates, the hyperbolas you've drawn become lines of constant x, and the lines AB and CD become lines of constant t. Once you write the metric in those coordinates, it's easy to calculate the times AC and BD, since each of them are measured along lines where only one coordinate changes (the Rindler t, in this case), so only one term in the metric is relevant.

[JDoolin]

I agree "potential" by itself is kind of a vague term, it gets used in a number of different ways. It would probably be better if relativity texts stuck to the term "energy per unit mass", which is what "potential" usually means when gravity is the only force under consideration, but which generalizes nicely to situations where other forces are in play without changing its meaning. Energy per unit mass also has the nice property that in "relativistic units", where the speed of light = 1, it is dimensionless.





Forgive me for nitpicking, but resolving to 1 does not mean "dimensionless." It just means that the correlation coefficient comes out to be 1.

You might have c=1 ly/y, or c= 1ls/s. The unfortunate fact is that we don't have any word that sounds intuitively like a distance that makes the speed of light = 1. The closest I've been able to come is 1 foot/nanosecond, which is close to the speed of light.

I do believe that it is nice when quantitities come out to be 1, but even when this happens, the units are still important.

Little things you notice after teaching introductory physics for a couple semesters: For instance, Kepler's Third Law is

\frac{T^2}{r^3}=\frac{4 \pi^2}{G M_{sun}} which becomes

\frac{T^2}{r^3}=1 year^2 / AU^3
if you use units of Astronomical Units for r and Years for T.

This is getting off topic, but I don't really "believe" in dimensionless quantities, even when they come out dimensionless. The coefficient of friction is not unitless, but Newtons/Newton. i.e. newtons of frictional force per newton of parallel force.

On the other and if you take Torque, and divide by Work, you get a unitless quantity, but as far as I know, there's really no physical application of this at all. Work involves a dot-product of F*d and torque involves a cross-product of F*d. The directions of the force and distance make these two quantities completely conceptually different, even though they happen to have the same units.
Then there's the whole discussion of radians and degrees, which in some way measure the portion of a circle. They should have some kind of exact analogy in rapidities. If anything could convince me to "believe" (so to speak) in dimensionless quantities, it would be angles and rapidities.


One fairly easy way to do it is to transform to Rindler coordinates, as described in the Wikipedia page:

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

In these coordinates, the hyperbolas you've drawn become lines of constant x, and the lines AB and CD become lines of constant t. Once you write the metric in those coordinates, it's easy to calculate the times AC and BD, since each of them are measured along lines where only one coordinate changes (the Rindler t, in this case), so only one term in the metric is relevant.

I'm looking over the Wikipedia article. Now a couple of vaguely coherent thoughts are coming to mind.

they use t to represent the time according to one of the accelerated observers whose acceleration g=1. Off the top of my head, I'm thinking 1 ly/year^2 and 1 ls/s^2 are different quantities; but I'm not sure. I am calling this observer whose acceleration is g=1 the "reference observer"

I would prefer to see the equation relating tanh(t)=T/X with all of the variables, (a, c or r=c2/a) still present.

The value for t does not represent what any clocks would say, (except for the reference observer whose g=1). I'm not sure what notation, exactly, I'm looking for to find the relative speed of the clocks. (the issue might be entirely resolved if we have the equation with a and c or r=c2/a)

Finally, it may be interesting to note, precisely, the relationship between the reference-observer's passage of time, and his rapidity.

[PeterDonis]

they use t to represent the time according to one of the accelerated observers whose acceleration g=1. Off the top of my head, I'm thinking 1 ly/year^2 and 1 ls/s^2 are different quantities; but I'm not sure. I am calling this observer whose acceleration is g=1 the "reference observer"

Yes, 1 ly/y^2 and 1 ls/s^2 are different quantities; 1 ly/y and 1 ls/s are the same velocity, but when you divide that by 1 y you get a different acceleration than when you divide it by 1 s--the latter is a much larger acceleration (about 30 million times larger, since that's the number of seconds in a year).

I would prefer to see the equation relating tanh(t)=T/X with all of the variables, (a, c or r=c2/a) still present.

The "relativistic units" they're using express everything in terms of length, so converting back to conventional units just means replacing every time t with ct and every acceleration a with a / c^{2}, then canceling c's where possible (and rearranging so the c's don't appear in front of the t's, if desired). I come up with:

t = \frac{c}{g} arctanh \left( \frac{cT}{X} \right), x = \sqrt{X^{2} - c^{2}T^{2}

T = \frac{x}{c} sinh \left( \frac{gt}{c} \right), X = x cosh \left( \frac{gt}{c} \right)

ds^{2} = - \frac{g^{2} x^{2}}{c^{2}} dt^{2} + dx^{2} + dy^{2} + dz^{2}

The value for t does not represent what any clocks would say, (except for the reference observer whose g=1). I'm not sure what notation, exactly, I'm looking for to find the relative speed of the clocks. (the issue might be entirely resolved if we have the equation with a and c or r=c2/a)

That's what the metric, ds^{2}, tells you. The observers in question all have constant x, y, z, so the only non-zero coordinate differential along their paths is dt. That means their proper time, which is just \sqrt{- ds^{2} / c^{2} }, is given by

\tau = \frac{gx}{c^{2}} t

[JDoolin]

Yes, 1 ly/y^2 and 1 ls/s^2 are different quantities; 1 ly/y and 1 ls/s are the same velocity, but when you divide that by 1 y you get a different acceleration than when you divide it by 1 s--the latter is a much larger acceleration (about 30 million times larger, since that's the number of seconds in a year).



The "relativistic units" they're using express everything in terms of length, so converting back to conventional units just means replacing every time t with ct and every acceleration a with a / c^{2}, then canceling c's where possible (and rearranging so the c's don't appear in front of the t's, if desired). I come up with:

t = \frac{c}{g} arctanh \left( \frac{cT}{X} \right), x = \sqrt{X^{2} - c^{2}T^{2}

T = \frac{x}{c} sinh \left( \frac{gt}{c} \right), X = x cosh \left( \frac{gt}{c} \right)

ds^{2} = - \frac{g^{2} x^{2}}{c^{2}} dt^{2} + dx^{2} + dy^{2} + dz^{2}



That's what the metric, ds^{2}, tells you. The observers in question all have constant x, y, z, so the only non-zero coordinate differential along their paths is dt. That means their proper time, which is just \sqrt{- ds^{2} / c^{2} }, is given by


\tau = \frac{gx}{c^{2}} t


I feel compelled to make up another table:


\begin{center} \begin{tabular}{| c | c | c | }\hline
Quantity & Circ & Hyperbola \\ \hline
Unit Circle/Hyperbola & X^2+Y^2=1 & X^2-(cT)^2=\left(\frac{c^2}{a} \right)^2 \\ \hline
Circle & X^2+Y^2=r^2 & X^2-(cT)^2=x^2 \\ \hline
Vertical Component & Y=r sin(\theta) & c T=x sinh \left(\frac{a t}{c}\right) \\ \hline
Horizontal Component & X=r cos(\theta) & X=x cosh \left(\frac{a t}{c}\right) \\ \hline
Arc Length/Proper Time & \Delta s = r \theta & c \tau = x \left(\frac{a t}{c }\right) \\ \hline

\end{tabular}\end{center}


In this, it appears that t takes on the role of an angle (the time passed by the accelerated observer at x=c^2/a) , while x is analogous to the role of a radius (the distances from the rindler horizon as observed by the accelerated observers) A major difference, I think, is that while arc-length increases happily with both X and Y, proper-time goes up by cT and down by X.

I know \theta represents the arc-length divided by the radius; which seems a perfect geometric description of the angle. However, I don't know if at/c has such an easy relationship*. I know of a geometric analogy between polar angle and hyperbolic angle which is shown here (http://en.wikipedia.org/wiki/File:Funhipgeom.png) and I went ahead and attached it. (*Never mind, Actually, it has basically the same relationship. at/c = c tau / x. theta = Delta s / r.)

[PeterDonis]

I know \theta represents the arc-length divided by the radius; which seems a perfect geometric description of the angle. However, I don't know if at/c has such an easy relationship.

Obviously it does, since your last formula says that at / c is the arc length c \tau (which is the proper time converted to units of length) divided by the radius x.

Also, the slope of the line of simultaneity through a given point on the hyperbola (meaning the line through the origin that also passes through that point), which is just the boost velocity at that point, relative to the global inertial frame, is given by

\frac{c T}{X} = \beta = tanh \left( \frac{a t}{c} \right)

So I think we can equate at / c with the rapidity.

[JDoolin]

\begin{center} \begin{tabular}{| c | c | c | }\hline Quantity & Circ & Hyperbola \\ \hline

radius, proper distance & X^2+Y^2=r^2 & X^2-(cT)^2=x^2 \\ \hline

Unit Circle/Hyperbola & X^2+Y^2=(l_u)^2 & X^2-(cT)^2=\left(\frac{c^2}{a_u} \right)^2 \\ \hline

Arc Length/Proper Time & \Delta s = r \theta & \Delta\tau = \frac{x\phi}{c} =\frac{x a_u t}{c^2}\\ \hline


Angle/Rapidity & \theta= \frac{\Delta s}{r}=arctan(Y/X) & \phi = \frac{a_u t}{c}=\frac{c\Delta\tau}{x} =arctanh(cT/X)\\ \hline

Vertical Component & Y= r sin(\theta) & c T=x sinh (\phi) \\ \hline

Horizontal Component & X= r cos(\theta) & X=x cosh (\phi) \\ \hline

\end{tabular}\end{center}Some further observations:


The passage of proper time is affected signifcantly, as a function of proper distance, x and rapidity acceleration.
The distances x are exactly the same as in the inertial frame. i.e. the rindler coordinate system does not have any effect on the measurement of proper length.
I have replaced 1 with lu, and a with au to denote a unit length and a unit acceleration. The unit circle is a different entity depending on whether you use feet, meters, etc. The unit hyperbola will also be different depending on what units you use.


I think I'm now ready to apply the equivalence principle to this system. From the perspective of the accelerated body, if I'm not mistaken, it would look like g->infinity as x->0, and g->0 as x->infinity. Potential energy would be determined as Work=Force*distance; "Plain Potential" would be Work/mass=acceleration * distance. Especially of interest is to determine g(x) , U(x), and d \tau(x)/d T.

Once I understand how the relationship between time-dilation and potential energy is found in this system, I may begin to understand the relationship between time-dilation and real gravitational potential energy.

[PeterDonis]

The distances x are exactly the same as in the inertial frame. i.e. the rindler coordinate system does not have any effect on the measurement of proper length.

I'm not sure this is completely correct. I agree that in the metric for Rindler coordinates that I quoted above, the metric coefficient for all the space coordinates is 1, meaning that observers at constant x, y, z coordinates will observe a constant spacelike interval ds between each other along curves of constant Rindler time t. However, the physical meaning of this is a little more complicated. The Wikipedia page I linked to in an earlier post discusses various physical notions of "distance" that do not all give the same answers in Rindler coordinates, whereas they do in, for example, global inertial coordinates such as the T, X we are using for the global inertial frame.

I think I'm now ready to apply the equivalence principle to this system. From the perspective of the accelerated body, if I'm not mistaken, it would look like g->infinity as x->0, and g->0 as x->infinity. Potential energy would be determined as Work=Force*distance; "Plain Potential" would be Work/mass=acceleration * distance. Especially of interest is to determine g(x) , U(x), and d \tau(x)/d T.

I haven't gone back and checked earlier posts in this thread, so I don't know if this has been brought up before, but what you're contemplating here goes beyond the equivalence principle. (Which doesn't mean that what you're contemplating is not of interest or worth doing.) The equivalence principle only applies locally; more precisely, it only applies over a small enough region of spacetime that tidal gravity can be neglected. Of course in Rindler coordinates, which are coordinates on flat spacetime, tidal gravity is zero everywhere; but the spacetime you'll be comparing them to, presumably Schwarzschild coordinates outside a gravitating body, is *not* flat, so you'll have to restrict your comparisons to a region of that spacetime that's small enough to set up a comparison with a region of Rindler coordinates. The exact size of that region will vary with the specific case; but certainly looking at what happens as any quantity of interest goes to infinity will *not* fall within a small region.

One example of how global comparisons break down between these two systems is the following: consider two observers accelerating at 1 g, one on a rocket out in deep space, the other standing on the surface of the Earth. Both observers launch an object straight up (i.e., in the direction they are accelerating) at "escape velocity", 11 km/s. The observer in the rocket ship will eventually catch up with the object and pass it. The observer on the Earth will see the object recede forever. Yet these observers are equivalent as far as the equivalence principle is concerned.

[JDoolin]

I'm not sure this is completely correct. I agree that in the metric for Rindler coordinates that I quoted above, the metric coefficient for all the space coordinates is 1, meaning that observers at constant x, y, z coordinates will observe a constant spacelike interval ds between each other along curves of constant Rindler time t. However, the physical meaning of this is a little more complicated. The Wikipedia page I linked to in an earlier post discusses various physical notions of "distance" that do not all give the same answers in Rindler coordinates, whereas they do in, for example, global inertial coordinates such as the T, X we are using for the global inertial frame.



The distances in x, and the values of X match at time T=t=tau=0. I think the wikipedia article refers to this as the "ruler distance"

These other notions of distance are interesting, too. I strongly suspect that if you were looking up, the same ruler-distances would look further than the same ruler-distances looking down. However, if you consider the radar-distance; timing a signal as it takes to reflect back. The radar distance would appear further going down then up than going up then down. Since the further apparent distance results in a shorter radar-distance, that seems a bit counter-intuitive; I may have to do some math to verify that.


I haven't gone back and checked earlier posts in this thread, so I don't know if this has been brought up before, but what you're contemplating here goes beyond the equivalence principle. (Which doesn't mean that what you're contemplating is not of interest or worth doing.) The equivalence principle only applies locally; more precisely, it only applies over a small enough region of spacetime that tidal gravity can be neglected.

Of course in Rindler coordinates, which are coordinates on flat spacetime, tidal gravity is zero everywhere; but the spacetime you'll be comparing them to, presumably Schwarzschild coordinates outside a gravitating body, is *not* flat, so you'll have to restrict your comparisons to a region of that spacetime that's small enough to set up a comparison with a region of Rindler coordinates. The exact size of that region will vary with the specific case; but certainly looking at what happens as any quantity of interest goes to infinity will *not* fall within a small region.

One example of how global comparisons break down between these two systems is the following: consider two observers accelerating at 1 g, one on a rocket out in deep space, the other standing on the surface of the Earth. Both observers launch an object straight up (i.e., in the direction they are accelerating) at "escape velocity", 11 km/s. The observer in the rocket ship will eventually catch up with the object and pass it. The observer on the Earth will see the object recede forever. Yet these observers are equivalent as far as the equivalence principle is concerned.

In "comparing" an infinite uniform gravitational field to a real gravitational field, I meant to highlight the differences; not just the similarities. You make an excellent point, bringing up tidal gravitation; if I'm not mistaken, this is dg/dr. However, notice that we have established that the rindler coordinates also has a nonzero dg/dx, since the born rigid system must have different accelerations at different ruler distances.

[JDoolin]

I think I can establish that in the inertial frame, the structure's acceleration at any given point X, at T=0 is given as:

g(X)=\frac{c^2}{X}

I think this is the value of
g(X)=\frac{d^2 X}{dT^2}
Where (X,T) represents the path of a particular accelerated object.

However, what is the appropriate value of the gravity g(x) in the accelerating frame? We want to use the local time tau, instead of the inertial coordinate time T.

g(x)=\frac{d^2 x_{obj}}{d \tau^2}

We need to consider not dx, because in the accelerated frame, x is the ruler distance, which is not moving. Instead we need to consider the position, xobj, of something that is stationary in the inertial frame, but it's position will change relative to x over time.

I'm kind of speculating that once we account for this, we'll find that the measured gravity will be the same for everybody on board the rocket, even though it looks like different accelerations to the inertial observer.

However, since tau gets faster and acceleration gets slower as you go up, I think it won't come out so nice. Or maybe it will. I'm not working through it real clearly just yet.

[PeterDonis]

You make an excellent point, bringing up tidal gravitation; if I'm not mistaken, this is dg/dr. However, notice that we have established that the rindler coordinates also has a nonzero dg/dx, since the born rigid system must have different accelerations at different ruler distances.

Not quite. It's true that, for the particular case we're considering, the Schwarzschild spacetime outside a gravitating body, the tidal gravity happens to equal dg/dr; but that's not true in general, and the fact that dg/dx is also nonzero in Rindler coordinates does not mean that there is nonzero tidal gravity in flat spacetime (which is what Rindler coordinates apply to).

The general rule is that "tidal gravity" is just another name for "spacetime curvature"; so tidal gravity is only nonzero if spacetime is curved, which it is outside a gravitating body, but is not (obviously) in flat spacetime, even for accelerating observers. Mathematically, tidal gravity/spacetime curvature is quantified by the Riemann tensor and various other tensors formed from it. In flat spacetime the Riemann tensor is zero everywhere; if we were to calculate its components in Rindler coordinates that's what we would find. In Schwarzschild spacetime, the particular component of the Riemann tensor that gives what we've been calling dg/dr is R^{r}_{trt}, which, when you work through the math, turns out to equal dg/dr. (Actually, it may only equal dg/dr for r much greater than the Schwarzschild radius, which of course it is for the cases we've been considering; for values of r close to the horizon, there is an extra factor in the denominator in the formula for g, which makes it diverge to infinity at the horizon, but all the Riemann tensor components are finite and well-behaved at the horizon. I'll have to go back and check the exact formulas.)

[JDoolin]

However, notice that we have established that the rindler coordinates also has a nonzero dg/dx, since the born rigid system must have different accelerations at different ruler distances.

Actually, what we established was that there was a nonzero dg/dX when g was based on the acceleration in the inertial frame. But because the rate of proper time \tau is different at those different x values, this means that in the accelerated frame, the local value of g is also different at those different x values.

Instead of
g(x)=\frac{d^2 X_{rocketfloor}}{d T_{inertialclock}^2}

we want:

g(x)=\frac{d^2 x_{inertialobject}}{d \tau_{rocketclock}^2}


Not quite. It's true that, for the particular case we're considering, the Schwarzschild spacetime outside a gravitating body, the tidal gravity happens to equal dg/dr; but that's not true in general, and the fact that dg/dx is also nonzero in Rindler coordinates does not mean that there is nonzero tidal gravity in flat spacetime (which is what Rindler coordinates apply to).

Whether we claim zero tidal gravity, or not, I think it is important to explicitly define our variables. In the rindler coordinates, neither dg/dr, nor dg/dx are sufficient, because g is a function of the local rate of time.

The general rule is that "tidal gravity" is just another name for "spacetime curvature"; so tidal gravity is only nonzero if spacetime is curved, which it is outside a gravitating body, but is not (obviously) in flat spacetime, even for accelerating observers.

I thought it was obvous that spacetime was flat, even for accelerating observers, but now we have this nonzero dg/dX in the inertial frame which means it is no longer obvious that dg/dx is zero in the accelerated frame.

[PeterDonis]

I thought it was obvous that spacetime was flat, even for accelerating observers, but now we have this nonzero dg/dX in the inertial frame which means it is no longer obvious that dg/dx is zero in the accelerated frame.

I think it will indeed turn out when you do the computation you describe in post #59 that dg/dx will *not* be zero in the accelerated frame. The reason is simple: observers at different Rindler coordinates x certainly *feel* different accelerations, and that's what the function g(x) is supposed to capture--the acceleration that an observer at a given x *feels* (would measure with an accelerometer, etc.). So I expect g to vary with x.

But even assuming this is true, as I said before, it does *not* mean that the spacetime is curved. You can't change flat spacetime to curved spacetime by changing your coordinate system (or your state of motion); whether spacetime is flat or curved is a geometric invariant, just like the curvature of the Earth's surface. If you want, we can verify this by explicitly computing the components of the Riemann tensor in Rindler coordinates and verifying that they are all zero (or maybe a kind soul will point us to someplace where that computation has already been done).

[JDoolin]

I think it will indeed turn out when you do the computation you describe in post #59 that dg/dx will *not* be zero in the accelerated frame. The reason is simple: observers at different Rindler coordinates x certainly *feel* different accelerations, and that's what the function g(x) is supposed to capture--the acceleration that an observer at a given x *feels* (would measure with an accelerometer, etc.). So I expect g to vary with x.

I think I have a possible approach to the problem, somewhat inspired by your earlier reference to Misner Thorne Wheeler. To find the "gravity" measured by occupants of the rocket at any given x, we can use a diagram similar to the one attached, and use

(x_2 - x_1) = \frac{1}{2} a \tau ^2

I think that \tau is fully determined by x1 and \Delta x, so we should be able to find an explicit value of a for any given x.

But even assuming this is true, as I said before, it does *not* mean that the spacetime is curved. You can't change flat spacetime to curved spacetime by changing your coordinate system (or your state of motion); whether spacetime is flat or curved is a geometric invariant, just like the curvature of the Earth's surface. If you want, we can verify this by explicitly computing the components of the Riemann tensor in Rindler coordinates and verifying that they are all zero (or maybe a kind soul will point us to someplace where that computation has already been done).

That would be fine, I suppose, with the caveat that I'm effectively tensor illiterate. I got an MS degree in math and physics, but none of my professors ever talked about tensors. PS, I have looked at a few of the Susskind lectures on YouTube, and I am at least skimming MTW, and Wald now, but so far, it's all a little too abstract for me.

[SinghRP]

Newtonian equation motion in a gravitational field, written in full:

(Inertial mass) * (Acceleration) = (Intensity of gravitational field) * (Gravitational mass).

It is only when there is numerical equality between the inertial and gravitational mass that the acceleration is independent of the nature of the body. This is the Principle of Equivalence, which has double meanings.
Reference: The Meaning of Relativity, Albert Einstein.

[Mentz114]

That would be fine, I suppose, with the caveat that I'm effectively tensor illiterate.

You don't have to be. You understand the concept of spacetime and the geometric invariant ds2. All that's needed is to make the leap to curved spacetime, where one *must* use covariant *and* contravariant components to define a geometric invariant.

I tried to explain this in post#9 and #11 in this thread http://www.physicsforums.com/showthread.php?t=431843

Do dimensional indexes throw you, i.e. writing t as x0, x as x1, y as x2 and so on ? Or the summation convention xaxa=x0x0+x1x1+x2x2+x3x3 ?

If you hate all the indices try the diff. geom. approach.

[SinghRP]

I read discussions under General Physics, Classical Physics, and Specia Relativityl & General Relativity. I THINK there are several slightly varying interpretations of mass, force, accelaration, gravity, etc.

Gamow's One, Two, Three, Infinity raised my curiosity well enough to help me decide to become a physicist. Universities turned me into a marching physicist. Wigner's Symmetries and Reflections made me a humble physicist. I strongly recommend Wigner's essays. What an insigthful book!

[JDoolin]

I think it will indeed turn out when you do the computation you describe in post #59 that dg/dx will *not* be zero in the accelerated frame. The reason is simple: observers at different Rindler coordinates x certainly *feel* different accelerations, and that's what the function g(x) is supposed to capture--the acceleration that an observer at a given x *feels* (would measure with an accelerometer, etc.). So I expect g to vary with x.

Well, I think I've come to a conclusion; right now, the result seems surprising to me, but probably won't to you.

\begin{matrix}
x cosh(\phi)=x+h \\
\phi=arccosh\left ( \frac{x+h}{x}\right )\\
\Delta \tau = \frac{x \phi}{c}\\
a = \frac{2 h}{\Delta \tau^2}\\
a=\lim_{h \to 0}\frac{2 h c^2}{x^2 ArcCosh^2(\frac{x+h}{x})}=c^2/x
\end{matrix}

(The attached diagram might help.)

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.

[SinghRP]

General relativity has it that the spacetime continuum is curved. The physics of continuum is dealt with [stress] tensors.
My questions:
(1) The presence of a mass creates the curvature. By how?
(2) If the curvature due to matter is positive, is the sign due to antimatter negative?
(3) Do other fundamental interactions (the strong, the weak, and electromagnetic) create curvatures in their respective fields?

[Mentz114]

General relativity has it that the spacetime continuum is curved. The physics of continuum is dealt with [stress] tensors.
My questions:
(1) The presence of a mass creates the curvature. By how?
(2) If the curvature due to matter is positive, is the sign due to antimatter negative?
(3) Do other fundamental interactions (the strong, the weak, and electromagnetic) create curvatures in their respective fields?

Do you want to start another topic with these questions - you are hijacking someone else's topic.

[JDoolin]

You don't have to be. You understand the concept of spacetime and the geometric invariant ds2. All that's needed is to make the leap to curved spacetime, where one *must* use covariant *and* contravariant components to define a geometric invariant.

I tried to explain this in post#9 and #11 in this thread http://www.physicsforums.com/showthread.php?t=431843

Do dimensional indexes throw you, i.e. writing t as x0, x as x1, y as x2 and so on ? Or the summation convention xaxa=x0x0+x1x1+x2x2+x3x3 ?

If you hate all the indices try the diff. geom. approach.

I posted a reply (post 13) there in the other thread; yes I still want to understand tensors.

[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}.
Daryl McCullough describes (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 :smile:

[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 :smile:
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 Gullstrand-Painleve chart posted in #79 (http://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 Schwarszchild 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

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

[granpa]

why didnt you give the link to the post?
http://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.
http://www.youtube.com/watch?v=2LLAwJW3GLk

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