# General Relativity Basics: The Principle of Equivalence

1. Aug 27, 2010

### 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: $$x(t)=\frac {1}{2} a t^2 + v_0 t + x_0$$

Last edited: Aug 27, 2010
2. Aug 27, 2010

### 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]

3. Aug 27, 2010

### JDoolin

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

4. Aug 28, 2010

### 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
$$\lim_{t_0 \to 0} \frac{\int ^{t_0}_{-t_0} (\sqrt{c^2 - (at)^2})dt }{2 c t_0}$$​

Last edited: Aug 28, 2010
5. Aug 30, 2010

### 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

6. Aug 30, 2010

### 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.

7. Aug 30, 2010

### JDoolin

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.

8. Aug 31, 2010

### granpa

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

Last edited: Aug 31, 2010
9. Aug 31, 2010

### Austin0

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??

10. Aug 31, 2010

### 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 [Broken]

Last edited by a moderator: May 4, 2017
11. Aug 31, 2010

### Austin0

Are you talking about co-accelerating clocks in Rindler coordinates or inertial clocks in the initial rest frame???

Last edited by a moderator: May 4, 2017
12. Aug 31, 2010

### 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

Last edited: Aug 31, 2010
13. Aug 31, 2010

### JDoolin

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

Last edited by a moderator: May 4, 2017
14. Aug 31, 2010

### Al68

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.

15. Aug 31, 2010

### 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

Last edited: Aug 31, 2010
16. Aug 31, 2010

### granpa

17. Sep 2, 2010

### JDoolin

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

Last edited by a moderator: Apr 25, 2017
18. Sep 2, 2010

### 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

19. Sep 2, 2010

### 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.

Last edited: Sep 3, 2010
20. Sep 7, 2010

### Austin0

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.