Force due to gravity vs force due to accelratio

1. Oct 17, 2010

Reuben Smith

Lets start over and let me rephrase my question, a force due to gravity permeates the area around it with time dilation an object in free fall in a room on the earth would be affected by time dilation because it is accelerating at 32 ft per s per s,while an object in free fall in a room that was accelerating at 32 ft per s per s would not because it would be still and not accelerating.My experiment would be to make two identical oscillating circuits one in the barrel of a machine gun that is rapidly firing ,one in a vacuum levitated by a magnetic field would either circuit show signs of time dilation?

2. Oct 17, 2010

Staff: Mentor

I am not sure that I agree with this. Time dilation is essentially a coordinate effect. If you are using an accelerating coordinate system, e.g. Rindler coordinates, then an object in free fall in a room would indeed experience the same time dilation as an equivalent object in a gravitational field. If you are in a gravitational field then the equivalent coordinate system would be some sort of free-fall coordinates, e.g. the rain frame. In those coordinates the free falling object would not be subject to time dilation even in a gravitational field.

3. Oct 17, 2010

PAllen

Wrong, on several counts. It is true that a massive body influences the region around it. It is false that this is a 'time dilation field'. The rate of a time is always a function of the path of given 'clock' through spacetime. A clock sitting on the earth is following a non-inertial path through space time (in its own frame of reference, one may describe its path as (0,0,0,t), but it is still a path through spacetime). A clock free falling toward the ground would advance at a faster rate than one sitting on the ground (because it is following an inertial path through spacetime), and the free falling clock would approximate the behavior of a clock in deep space.

Basically, your two clocks would behave the same, and must, within the approximate limits described in my earlier post. If this were not true, the whole framework of general relativity would be wrong.

4. Oct 17, 2010

Reuben Smith

Let me see if Iv got this straight a clock on earth would run slower than one in deep space away from all other bodies.

5. Oct 17, 2010

inflector

That doesn't seem to agree with what was discussed in this thread:

Where it was described that there would be gravitational time dilation inside a massive sphere even though there would be no acceleration because the gravitational potential would be non-zero inside the sphere.

I think that means there would be as you say: "free-fall coordinates." Am I missing a subtle (or even obvious) distinction?

6. Oct 17, 2010

PAllen

Yes.

7. Oct 17, 2010

inflector

8. Oct 17, 2010

Reuben Smith

okay, now here is my question,three identical oscillating circuits one in the barrel of a machine gun monted on a rocket that compinsates for the recoil that is rapidly firing in deep space away from the earth and all bodies ,one on earth in a vacuum levitated by a magnetic field ,and one in deep space by its self ,would they oscillated at the same frequecy and if not what which ones would be slower in coparison ?

9. Oct 17, 2010

Staff: Mentor

I am not sure where you think the conflict is between the two threads. This thread and the previous thread that you linked to were both dealing with a misapplication of the equivalence principle. Here the equivalence principle was misapplied locally by comparing accelerating coordinates in curved spacetime (Schwarzschild coordinates) with inertial coordinates in flat spacetime. In the other thread the equivalence principle was misapplied by considering a large enough region where the spacetime curvature could not be neglected.

10. Oct 17, 2010

inflector

Earlier, you stated:
In the other thread, it was stated that:

1) Time dilation was due to the gravitational potential and not due to acceleration.

2) An object that was not accelerating could be experiencing time dilation because the acceleration is the derivative of the gravitational potential. You experience zero acceleration even when the gravitational potential is non-zero as long as the potential is the same as adjacent space so that the derivative is zero. So an object in an area of constant but non-zero potential would have time dilation.

3) An object in the center of a massive sphere would have non-zero but constant gravitational potential so there would be no acceleration but there would be time dilation.

This appears to conflict with your statement above that I have bolded because if one considers the free-fall coordinates of an object in the center of the massive sphere, it would be experiencing no acceleration but time dilation according to the other thread.

I certainly could have got this wrong (again) but it seemed to conflict to me so that is why I asked. I want to get this stuff right. Thanks for taking the time to look into this.

11. Oct 17, 2010

Staff: Mentor

An clock at rest in the center of a massive hollow sphere would experience gravitational time dilation relative to a clock at the surface. This is a large region of spacetime, not a small one, therefore the equivalence principle does not hold. As I said:
Is that clear now? Two local clocks at rest in the center of the hollow sphere experience no time dilation relative to each other and the equivalence principle applies. A clock at the center experiences time dilation relative to a clock at the surface and the equivalence principle does not apply.

Two free falling clocks which are at rest near enough to each other for the equivalence principle to hold would not experience any time dilation regardless of if they were in deep space, in the center of a hollow sphere, or even falling through the event horizon of a large black hole. There may be time dilation between two free falling clocks which are at rest wrt each other provided they are far enough apart that the equivalence principle does not hold.

Last edited: Oct 17, 2010
12. Oct 17, 2010

inflector

Okay, it appears that I my understanding was correct but that I was reading too much into your statement. Thanks for the clarification.

13. Oct 17, 2010

Staff: Mentor

No problem. Sorry about any confusion I caused.

14. Oct 17, 2010

PAllen

I am not sure I agree with the analysis of the center of hollow gravitating sphere. Consider the following fundamental statement from:

http://relativity.livingreviews.org/Articles/lrr-2006-3/ [Broken]

"The outcome of any local non-gravitational experiment is independent of the velocity of the freely-falling reference frame in which it is performed."

"The outcome of any local non-gravitational experiment is independent of where and when in the universe it is performed."

This is clarified in the text to mean any frame over which tidal effects can be ignored. So, the question is, are there tidal effects inside the sphere or are these cancelled? Is there curvature inside the sphere? I have assumed that tidal effects are cancelled, and spacetime is flat inside. If they are cancelled, then it clearly follows that either an atom decays at the same rate in the sphere (assuming sphere and reference frame at its center are inertial) as far away from it, or the Einstein equivalence principle is violated.

A final comment, again assuming flatness inside the sphere: then the metric is purely locally minkowski. The proper time computed for an inertial clock in the center is pure minkowski. How can it come out different from a clock in deep space?

[EDIT] OK, I researched this a bit and see the error of my ways. One can say that the inertial frame inside the sphere acts like an inertial frame moving at constant speed away from a distant inertial frame at rest relative to the sphere. This is basically due to the red shift and time slowing that would happen climbing throught the space between the sphere and the distant observer.

Last edited by a moderator: May 5, 2017
15. Oct 17, 2010

pervect

Staff Emeritus
The key in all cases is how you compare the clocks.

If you have a NIST_ standard clock, and some sort of oscillating machine gun platform or electronic circuit, both located at the center of a hollow planet, and you compare the two while they're side by side, they'll agree.

They'll also agree if you're in deep space (or anywhere else for that matter).

The first issue comes up when you compare the clock at the center of the hollow planet to the one in deep space. How do you compare it? Well, you probably use the obvious way - you send out a radio signal. And the propagation time for the radio signal is constant for a round trip.

Using this method, you find then that both the NIST clock and the machine gun contraption out in space seem to run faster than the one on the planet, when compared by radio signals.

So, case closed? Not quite.

Suppose now, we put our two clocks (and our oscillating contraption, which doesn't add much to the story since it always reads the same as a NIST standard clock, but is perhaps not quite as precise), in an accelerating rocketship.

Now, our rocketship is accelerating, but if you work out the round-trip time for a light signal, the round trip time is constant. But while you can see from the inertial frame analysis that the round-trip time is constant, you disagree about the rate at which the clocks are running.

What happens from the point of view of the inertial frame is that one clock accelerates and experiences a doppler shift - but there is no "gravitational field" in the inertial frame, and there is no "gravitational time dilation" in the inertial frame either.

So what's going on? It's just another example of how simultaneity is relative. One method of comparing the rate of the two clocks gives a different result than another method. This is due to the counter-intuitive fact that there isn't any absolute way to compare clocks at all, unless they are side by side. The notion of simultaneity in the acclerating frame is a different notion than the one in the non-accelerating frame - which shouldn't be too surprising, every inertial frame has a self-consistent but _different_ notion of simultaneity, and the accelerating's frame is sort of "stitched together" from different notions in different inertial frames.

BTW, the inertial frame analysis isn't any more or less "correct" than the non-inertial analysis in GR - because GR is designed to handle arbitrary coordinates. The idea that you do everything in inertial frames is fundamentally due to the Newtonian analysis.

16. Oct 17, 2010

PAllen

Let me pose the following question about these issues. My understanding of the moment is that a free fall clock near the earth's surface would go faster than one on the ground. Locally, one is inertial, the other isn't, the inertial clock goes faster if they are 'close enough'.

However, I now suspect that the falling clock still run slower than a distant inertial clock because of the gravity well effect. Neither can really be said to inertial relative to the other because there are no inertial frames over this large distance. All comparisons between them are affected by the curvature over a large region.

Yet another way I can think about this is as follows. There is no 'equivalence' between two widely separated inertial observers not moving relative to each other, in a pure SR universe, with one of them surrounded by an accelerating rocket (soon to have floor hit them, but not yet); and two widely separated inertial observers in GR, one 'far from everything' the other free falling near a massive body. To compare such distant observers requires accounting for space between them, which is just not the same.

17. Oct 18, 2010

Passionflower

This is quoted ad infinitum at various places but never have I seen a formula demonstrating that.

It would be nice if we could have the formula so we have numerical proof that GR indeed predicts this.

Say we have a sphere of mass X and a cavity of size N (you could use the Schwarzschild R for simplicity). Then what is the time dilation wrt an observer at infinity?

Also a clock can free fall at various velocities.

One approach is the calculate the time dilation(r) for a stationary observer and then calculate the dilation(r) for free falling observers at various proper velocities wrt the stationary observer.

If we take rs=1 and radial motion for simplicity then for the stationary observer we have:

$${t}^{2}={\tau}^{2} \left( 1-{r}^{-1} \right) ^{-1}$$

For free falling observers with an initial velocity v0 the proper velocity v(r) is:

$$v(r) = \sqrt {1- \left( 1-{v_0}^{2} \right) \left( 1-{r}^{-1} \right) }$$

Applying the Lorentz factor would give us the time dilation wrt an observer at infinity.

Last edited: Oct 18, 2010
18. Oct 18, 2010

PAllen

Actually, the theory on this that I found in reputable sources states that the clock inside the sphere would run slow compared to a distant observer with minimal motion relative to the sphere. Compared to one on ther surface, the whole derivation I saw does not apply. My guess would be that the rate on the surface would generally be the *same* as inside the sphere, but it might depend a little on the assumed thickness of the sphere.

Someone should double check this.

19. Oct 18, 2010

Staff: Mentor

Yes, this is correct.

Can you link to the derivation you are referring to? I am sure I can extend it to include a surface clock.

20. Oct 18, 2010

PAllen

I'll look for it, but that derivation obviously cannot be so generalized. It explained the whole effect as due to the integrated curvature or total redshift from the surface to the disant observer, and showed why the same would apply to a source inside the hollow sphere. In fact, this derivation strongly implies (but doesn't state) that there would be little or no difference from the internal clock compared to one on the surface of the hollow sphere.