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.

[STRIKE]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.[/STRIKE] (*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 [STRIKE]x(t)=v0 + a t[/STRIKE]*. 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 isn't 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:
https://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 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, 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 [strike]speed of sound[/strike] 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

https://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. https://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. :-)