# B Gravity vs Acceleration

1. Sep 8, 2016

### calinvass

My question is, if gravity and acceleration are equivalent the what happens to a clock c1 , after a for example 1 year, if it is being accelerated by a force F, at g relative to a reference frame R, and what happens to another clock c2 , if it at rest relative to the same reference frame R, in a gravity field g=10 [mps/s] producing a force G=F?

What I understand is as the moving clock c1 time starts to slow down, the acceleration gets smaller and it an increased force is needed to maintain the acceleration g relative to the frame R. The same happens to the clock under G force. If we increase the force clock 1 time will go slower, if not they will show the same time. Is that correct ?

2. Sep 8, 2016

### Grinkle

I was / am hoping to see someone who better knows what they are talking about respond, but in the meantime ...

I think if your question includes a third clock and you say that this third clock is in a reference frame that observes clock 1 accelerating and also observes clock 2 at rest but in a gravity well, then I think this third clock says that clocks 1 and 2 are ticking at the same rate.

I could be wrong.

3. Sep 8, 2016

4. Sep 8, 2016

### Stephanus

I think as @Grinkle has pointed out, you need the third clock.
To calculate the accelerated force, you use SR. While to calculate the clock in gravity field you would use GR. I don't know GR at all, but that's how to calculate it.

Increased force? Ye-e-ess, but... Wouldn't it be faster than the speed of light if c1 is accelerated in someway that with respect to R frame it keeps constant acceleration?
Supposed c1 is accelerated relative to R 1 m /s2. The speed of light is: 299,792,458m/second
So, after c1 speed is 299,792,456, (99.9999993328718% c) than 1 second later (according to whom? R?) then it will be 299,792,487 m/s. (99.9999996664359% c).
Let my try it with relativity addition formula.
So the speed difference wrt C1 would be ... 0.333 times the speed of light or 99,930,817 meter/seconds! I don't know if my calculator is accurate.
So c1 will be jerked (not accelerated) 99 thousands km/second wrt C1 to reach 299,792,487 m/s from 299,792,486m/s wrt R. I have no idea how much force to propel 1 kg mass that way. Of course you might want to calculate it yourself.

5. Sep 8, 2016

### Staff: Mentor

They are only equivalent locally, i.e., in a small patch of spacetime.

This thought experiment is not limited to a small patch of spacetime, so the results in the two cases don't have to be the same. And in this case you can't even really compare them. See below.

You are mixing up frame-dependent and frame-independent quantities. The frame-independent quantity is the acceleration that is felt by each clock; in order for the equivalence principle to apply at all, even in a small patch of spacetime, both have to be the same. So that is the condition of the problem: that the acceleration felt by both clocks, the one accelerating in free space and the one hovering at rest in a gravitational field. This acceleration--the precise term for it is proper acceleration--will not change at all throughout the problem--at least it shouldn't for the comparison you are trying to make.

For the clock accelerating in free space, if the acceleration it feels is constant, the acceleration it will appear to have relative to a fixed inertial frame--the precise term for this is "coordinate acceleration"--will decrease with time.

Not really, because there is no "the same" in this case, because there is no fixed inertial frame we can use globally in the presence of gravity. The only thing we can compare between the two cases is the acceleration felt by the clocks--their proper acceleration.

We can't even compare "how much time elapses" for the two clocks, because they are in different spacetimes and there is no way to pick out "corresponding" events on their worldlines. This is just part of the fact I mentioned above, that the equivalence principle only covers small patches of spacetime--"small" meaning small enough that all these other issues don't arise.

6. Sep 8, 2016

### calinvass

Possibly.
From the conditions of the problem the proper acceleration for both clocks was supposed to be the same. An example in the real world would be a clock1 in a spacecraft, clock2 on earth and reference frame at say in space at 10 light seconds from Earth and in the Sun reference frame . Relative speed between the sun an object on earth is negligible.

I also said that in order to maintain proper acceleration on clock1 you need to increase the force, perhaps it wasn't clear or I didn't understand your answer.
If the "coordinate acceleration " decreases it means it is similar to what happens on the clock2 reference frame.

7. Sep 8, 2016

### Ibix

The problem as you described it holds coordinate acceleration constant while the proper acceleration (the acceleration felt by passengers in the rocket) grows without bound. It reaches infinity (obviously impossible to do) in under a year of the original rest frame's time. You want to hold proper acceleration constant.

However, what do you mean by time going slower? What are you comparing it to? You need a clock outside your rocket and far away from your planet to compare to. As soon as you do that, though, you run into obvious differences between the gravity and the rocket - for example the rocket observers will notice the external clock starting to move past them, while the planet observers will see it stationary at all times. This means that they aren't making a fair comparison - which is the start of the problems Peter mentioned.

The long and the short of it is that the equivalence principle is approximately true in some fairly limited circumstances. You are trying to operate outside those circumstances, so you need more sophisticated tools.

Last edited: Sep 8, 2016
8. Sep 9, 2016

### Staff: Mentor

And how long will clock1 stay within this reference frame? Certainly not for a year. And 10 light seconds is much too large for the equivalence principle to apply; differences in the strength of the Earth's and probably even the Sun's gravity are observable over that range. In other words, your thought experiment is not limited to a small enough patch of spacetime to apply the EP.

Not as measured by the person in the spacecraft. The "force" as calculated in a fixed inertial frame will increase, yes.

I don't understand what you mean. In a reference frame fixed to the Earth, clock2 has zero coordinate acceleration; it is at rest and stays at rest. It has nonzero (and constant) proper acceleration.

9. Sep 9, 2016

### calinvass

I meant to say that acceleration and gravity both produce time dilation.
For example, time for an object on earth over a period of time of 1 year will be 0.0219 s less than time for an distant object, without taking into consideration Earth rotation.

Last edited: Sep 9, 2016
10. Sep 9, 2016

### Stephanus

Constant velocity and gravity both produce time dilation. If you accelerate, then you'll change frame. And time difference between velocity and gravity is very significant.

11. Sep 9, 2016

### calinvass

Thanks, I get it now. It is also consistent with the way light travels as a wave. Under a gravity field similar to that produces by Earth EM waves are slightly affected. When a light source travels towards an object when reaching a certain speed, It gets blue-shifted, so the effect is significant and with acceleration it is progressive.

Last edited: Sep 9, 2016
12. Sep 9, 2016

### Staff: Mentor

That's not a very precise way of putting it. The precise way of putting it is this: if we have two observers inside a rocket accelerating in flat spacetime ("accelerating" meaning "experiencing proper acceleration"), at rest relative to each other (as measured by them exchanging round-trip light signals and seeing that the round-trip travel time by each of their clocks remains constant), the observer at the bottom of the rocket will be time dilated (clock running slower) relative to the observer at the top (as measured by noting the elapsed time on both of their clocks between two successive round-trip light signals). Similarly, if we have two observers at rest in a gravitational field (and experiencing proper acceleration as required to hold them at rest), at rest relative to each other (same measurement method as above), the one lower down will be time dilated (clock running slower) relative to the one higher up (same measurement method as above).

Note that what causes the time dilation is not "acceleration" or "gravity" but difference in height between observers experiencing proper acceleration (where "height" just means "distance along the direction of proper acceleration"). Note also that the observers have to be at rest relative to each other; if they are moving relative to each other, things get more complicated (though in many cases there will still be a good approximation in which we can separate time dilation due to height from time dilation due to motion).

Note also that the above has a wider application than just a small patch of spacetime; that is, the above is not just a restatement of the equivalence principle. The EP makes a stronger claim: it says that locally--in a patch of spacetime small enough that both observers, to within the accuracy of measurement, experience the same proper acceleration--the same height difference produces the same time dilation in both cases (accelerating rocket in flat spacetime, at rest in a gravity field). But that claim no longer holds once we consider regions of spacetime large enough for differences in "field strength" (differences in the proper acceleration experienced by the two observers) to be measurable. In that case the same height difference no longer produces the same time dilation in both cases.

13. Sep 10, 2016

### calinvass

Thanks, things are now very clear for me.

14. Sep 14, 2016

### David Lewis

If my understanding is correct, time appears to pass more slowly on Earth from the viewpoint of the space object because gravitational field strength is weaker far from the Earth. And conversely, clocks seem to run fast for space objects from the perspective of Earth observers. Both clocks, if they are functioning correctly, actually tick at the same rate in reality. It's the difference in viewpoint that creates the illusion.

15. Sep 14, 2016

### Staff: Mentor

This is not a good way to put it, because "in reality" implies that there is some "real" clock rate and the others are all "illusions" (your word). There is no such thing as the "real" clock rate. There is proper time along a particular world line, and there are measurements you can make to try to compare clock rates along different worldlines--for example, two observers both at rest relative to a large mass, but at different altitudes, can exchange round-trip light signals and measure the elapsed proper times on each of their clocks between two successive signals, and thereby determine that the clock at lower altitude is "running slow" compared to the higher one--because it shows less elapsed time between two successive signals. But all of these various observations and measurements are equally "real"; none of them has any privileged position, and none of them corresponds to how time flows "in reality".

16. Sep 14, 2016

### David Lewis

Peter, thank you for correcting me. Would it be equally valid to say the the clock at the lower altitude is not running slow, but rather less time has passed?

17. Sep 14, 2016

### Staff: Mentor

The question is, less time has passed between what two events on the clock's worldline?

This is always the issue when trying to compare "clock rates" between spatially separated objects. What are the "corresponding" events on the two worldlines? In the standard twin paradox there is an easy answer: the two worldlines intersect at the start and end of the scenario, so those events are obvious choices for "corresponding". In other cases the choice can be more problematic; in the general case there is no single choice that is picked out by the scenario, all of them are arbitrary, which means that in the general case there is no way of comparing "clock rates" between spatially separated objects.

In the case under discussion, the fact that both objects are at rest, relative to each other and to the large mass, picks out a particular choice of "corresponding" events: basically they are the ones that are picked out by the natural simultaneity convention of the large mass's rest frame. With that choice, yes, we can say that the clock at the lower altitude has less time pass between a pair of corresponding events. But it's still good to be aware that that statement depends on a particular choice of simultaneity convention, i.e., of which pairs of events on the two worldlines are "corresponding".

18. Sep 14, 2016

### A.T.

Gravitational time dilation depends on the potential, not the field strength. In the center of the Earth the field strength is zero, but the time dilation is maximal (lowest potential).

It's not an illusion, and it's not symmetrical like kinetic time dilation.

19. Sep 14, 2016

### A.T.

Time passed is what clocks measure.

20. Nov 22, 2016

### calinvass

A clock in free fall under a gravitational field vs one hovering above the earth is equivalent to one clock accelerating in respect to another.

21. Nov 22, 2016

### Ibix

If the two clocks are passing "close" to each other in the gravitational field and you mean to compare this to two clocks in flat spacetime, one moving inertially and one undergoing proper acceleration, then these two cases are the same, yes. Literally so, by the equivalence principle, although scare quotes are needed because "close" here depends on your measurement precision.

If that is the case, proper acceleration is a frame independent thing and you do not need to specify "in respect to". It's worth specifying the "proper" though, unless you are very sure that your audience will both be able to understand it from context and trust you to have used it correctly.

If you are meaning something else then it depends on what you actually meant.

22. Nov 22, 2016

### calinvass

I meant to say a hypothetical constant gravitational field,so there is no change in gravitational filed during descent , but only gravity potential difference. This case works for a limited period of time, otherwise the object falling exceeds the speed of light. However, I don't understand acceleration as frame independent. It is something I've missed. Is it because no matter the current velocity (in one frame or another) , the acceleration remains constant by definition?

Last edited: Nov 22, 2016
23. Nov 22, 2016

### Staff: Mentor

There is no such thing. More precisely, there is no solution of the Einstein Field Equation that has this property. The closest you can get is Rindler coordinates in flat Minkowski spacetime; but the "gravitational field" in that case is still not constant, it varies with "height" (but it doesn't vary "horizontally", i.e., its magnitude changes but not its direction). Also, the "field" only covers a portion of spacetime--there is a Rindler horizon beyond which observers that are "at rest" at a constant height in the "gravitational field" can never see.

In the case of Rindler coordinates, above, this is not a correct description: what happens is that a free-falling object falls below the Rindler horizon and can't be seen any longer by the static observers. But from the viewpoint of a global inertial frame, the free-falling object is just sitting at rest in flat Minkowski spacetime; it certainly isn't "exceeding the speed of light" or traveling on a spacelike worldline or anything like that.

Proper acceleration--the acceleration that is measured by an accelerometer and that you feel as weight--is frame independent. But coordinate acceleration--the second derivative of position with respect to time--is not. Carefully distinguishing between these two concepts is critical in understanding the topic under discussion.

24. Nov 22, 2016

### calinvass

That is why I said hypothetical and between some time limits. But, I, understand it is not possible to use it because you can't write any equations. It is possible to imagine it though.

25. Nov 22, 2016

### Staff: Mentor

The inability to write an equation is a very strong hint that the hypothetical you're imagining is not internally consistent. Starting with an internally inconsistent hypothesis pretty much guarantees confusing and illogical conclusions.

However, if by "within limits" you mean that you want to study the behavior of an object across a small enough region of spacetime that we can ignore any variations in the gravitational field and treat it as uniform.... There's no difficulty writing equations that cover that situation, so we're good. (for example, it takes some seriously sensitive instruments to detect any non-uniformity in the earth's gravitational field within a volume 100 meters on the side at the earth's surface). A free-falling object within that volume will experience zero proper acceleration and coordinate acceleration determined by whatever coordinates you use.