Time dilation on the surface of a planet and escape velocity time dilation

1. Oct 18, 2014

Shaw

If an object falls through a planet's entire grav. field, before impact it's velocity will be the escape velocity. I assume the time dilation for an observer on the object will be the same as for an observer on the planet. They're not added together. Another point - since time dilation on a planet equals the time dilation for an object moving at the escape velocity, and mass changes accompany time changes in space, has anyone checked to see if there is a mass change accompanying the time change on the surface of a planet?

2. Oct 18, 2014

phinds

Not if it hits a terminal velocity

I THINK this is wrong. Time dilation has a gravitational component and a component due to motion. Of course, you have to say time dilation RELATIVE TO WHAT? since time dilation is frame dependant.
Again, I think this is wrong

Invariant mass doesn't change just because an object is moving or gets near a planet. That is, the mass of the object itself doesn't change. Mass calculated by an observer in another frame of reference is frame dependent.

3. Oct 18, 2014

Shaw

Thanks for the reply. Let's assume we reach the escape velocity just before reaching the surface, by any means. Time dilation on the surface of a planet is the same as the time dilation at a velocity equal to the escape velocity. It's a coincidence that many physicists think is quite odd. If the time change is greater on the moving object than on the surface don't we end up with peculiar relativity calculations?

4. Oct 18, 2014

Simon Bridge

No.

To fall through the entire grav field it would have to start an infinite distance away and fall to the center.
In the absence of any other forces, an object falling from infinity to the surface of a planet will arrive with exactly the escape velocity at the surface.
That is what "escape velocity" means.

You mean that however much the object observer discovers the planet observers time is dilated, the same will be true the other way around?

No.
If you travel at high speed your clocks do not slow down ... this is because, with respect to you, you are stationary.
In relativity, movement is what other people have.

Time dilation is always a comparison between two observers - a single object does not have any time dilation.

How do you figure?

5. Oct 18, 2014

Simon Bridge

Fair enough - say there's a hole in the surface there and an observer "Peter" stands on the planet next to the hole and another observer "Roberta" is riding the falling object - when they pass each other, they notice a time dilation effect in each other:

This makes no sense - time dilation is not something that happens to individuals.
P and R will have a time dilation between them which depends on the speed they are passing at that instant and that R is also following a geodesic while P is not so R has an additional time dilation due to that. Neither of them are in an inertial reference frame so it is quite a tricky calculation.

Please provide a reference - who are these "many phycisist" who "think [it's] quite odd"?

These sorts of things are cleared up by being careful about how you word your descriptions.

6. Oct 18, 2014

Simon Bridge

Here's what it sounds like:
Treating an observer in flat space-time as our reference - observer O.
Clocks in curved space-time (observer A - stationary wrt O?) will be time-dilated wrt O, and so will clocks (observer B) moving at a constant speed v in flat space-time.
We can ask: what speed does v have to be, so that the time dilation for B will be the same for some particular observer A.

You assert that v, for this situation, will be equal to the escape velocity from the position of A.

7. Oct 18, 2014

Simon Bridge

Doing the math while waiting for a reply... we all need a hobby:
Assuming A is a distance r from the center of gravity of a non-rotating sphere of mass M, $$\Delta t_A=\Delta t_O\sqrt{1-\frac{R}{r}}$$ ... R is the Schwarzschild radius for mass M.

For B the regular special relativity applies: $$\Delta t_B = \Delta t_O\sqrt{1-v^2}$$

The condition for these to be equal means: $$\Delta t_B = \Delta t_A \implies v^2 = R/r$$ ... which is the escape velocity of A from M.

This what you are talking about?

8. Oct 19, 2014

Shaw

It seems so. A mechanical engineering professor did the Schwarzchild calculation for me which confirmed the fact that gravitational time dilation on the surface equals the time dilation at the escape velocity. A Forum advisor then agreed that it was indeed a recognized odd coincidence. GR predicts a lot of things but it doesn't predict that. Hard to believe it's just a fluke. Consider a thought experiment where an object accelerates itself by ejecting photons in one direction. The speed of the mass increases as the mass diminishes, until it's almost completely depleted. This will occur at a very high velocity common to all mass. The only high speed constant we know about is the speed of light. Time dilation should accompany mass depletion in the same proportion. How much mass does it take to accelerate an object to the escape velocity using this scenario? The mass loss will be in proportion to the time dilation. We see inertial mass increase in a cyclotron, but the accelerated particles are not isolated. The energy for their acceleration is externally applied. An object falling in a gravitational field can be considered an isolated particle if it's small enough. That's why I ask if anyone has checked to see if there are mass changes associated with gravitational time changes. A time change just by itself seems a little odd to me. It looks like there's a missing piece of the puzzle.

9. Oct 19, 2014

A.T.

Gravitational time dilation is due to different gravitational potential.
Kinetic time dilation is due to relative motion.

So when the free falling observer (FF) passes the surface fixed (SF) observer, they will both observe the other's clock going slower, because of the relative motion. They will not observe gravitational time dilation between them, because they are at the same gravitational potential.

A distant observer will observe the same gravitational time dilation for both FF & SF, and depending on his relative motion to FF & SF some kinetic time dilation, which multiplied with the gravitational time dilation.

There are many threads here on the so called "relativistic mass" and why it is a missleading concept.

Last edited: Oct 19, 2014
10. Oct 19, 2014

A.T.

Yes.

There is no additional time dilation due to
not following a geodesic, at the instant when they pass each other

If they would meet twice, the accumulated proper times would depend on the paths they took, but that is not relevant here.

If R is following a geodesic, it is at rest in a local inertial reference frame.

Last edited: Oct 19, 2014
11. Oct 19, 2014

harrylin

I'm afraid that a more precise formulation will be necessary to make clear what you are getting at; however the two effects are certainly added together!
Your thought experiment was measured with Gravity probe A . If you can, I advice you to download the paper by Vessot in Phys. Rev. Letters of 1980. See equation 1 there, as well as fig.3.
A good inverse example is clocks on the surface of the Earth: clocks on the equator move fast compared to those near the North pole and they are generally higher up as the Earth is expanded there. For that rather common case, identical clocks stay well in tune.

12. Oct 19, 2014

Simon Bridge

@A.T.
... Fair enough... I was a tad uncertain if the "geodesic = path of longest proper time" did anything special here.

@Shaw
Not exactly verifiable references, but I take it that the calculation I just did is what you were trying to talk about?

... depends what you mean by "a fluke".
I think the idea is that it is not a particularly meaningful concurrence - though it looks suggestive, the result is only for a special case(s?)
Try repeating the calculation for different metrics.

BTW: GR was used to make the prediction ... it was needed to compute the gravitational time dilation.

Anyway: now we know what you are talking about ... what was the question?

As with any rocket ... though there is no need for almost all the rocket to be reaction mass. It sounds like you are thinking of some sort of total conversion - was there a special reason for that.

... well if all the mass were turned into energy, then there would be nothing left to travel. If there were any mass remaining, then it must be travelling less than the speed of light.
We have had rockets which used almost all their mass for reaction-mass but still managed to stay well below light-speed, and different rockets managed to end up with different speeds - not a single common speed.
Why would it be different for total-conversion rockets?

You also forgot to be careful about specifying the observer.

That could be one way of thinking about it - but consider: the amount of reaction mass used (the mass-loss) to achieve a particular delta-vee (say from rest to a speed that happens to be the escape velocity from some particular distance from the center of mass of some non-rotating sphere we happen to be interested in at the time...) will vary with the final mass that has to be moved.
You use the relativistic rocket equation.

... are you referring to the "cyclotron effective mass"?

Yet the energy for the acceleration is provided by an external field too - or, if you like, the intrinsic curvature of space-time? At what education level do you need these answers?

I still cannot tell what that means. Mass changes in what? by "time changes" do you mean the time dilation wrt a flat space-time observer?
It is not clear how the gravitational "time change" asked about here means anything in terms of the rocket before.

Do you mean to compare the mass used up to make a particular delta-vee in some reference frame compares with the time dilation for the final velocity in the same reference frame? (You just happen to be choosing a delta-vee as the difference between some "escape velocity" and rest) Then you need to look at the relativistic rocket equation.

Unfortunately, Nature does not seem to care what you or I find odd. But maybe that's for the best.

I don't know what you mean by a "time change". However - time dilation is not usually "by itself", there is an associated length contraction ... more generally, we talk about the Lorentz transformations and the geometry of space-time rather than just time by itself. Sometimes, though, we just care about comparing the clocks.

Before anyone can hep you, though, you will have to clarify your question, and let us know your education level.
I have a feeling the delta-vee thing is close.

13. Oct 19, 2014

jartsa

Case 1: Two massive clocks are slowly brought together, extracting potential energy from the clocks. Time dilation of clocks is proportional to extracted energy, or extracted mass, or extracted mass-energy. (We know this because we know that redshift is proportional to the change of potential energy, and gravitational time dilation is proportional to redshift.)

Case 2: Two massive clocks are dropped onto each other, the casings of the clocks heat up. Time dilation of clocks is about the same as in case 1. Nothing was extracted, so mass-energy is unchanged.

I don't know what we can conclude from that ... Sometimes there is a mass change, sometimes there is no mass change?

14. Oct 19, 2014

Shaw

Thanks for all the input. I thought my questions were straight forward, but they apparently touch on many difficult concepts.

15. Oct 19, 2014

Simon Bridge

It is not so much that the concepts themselves are difficult as that it is difficult to guess what you are trying to ask us about.

16. Oct 19, 2014

Shaw

Well, to be frank, and I'm loathe to bring it up since it's inappropriate for this forum, I had an idea about mass. I treated it as if it was continually formed by space, and proceeded to think about the consequences. I found out later that this idea has been suggested before, sometime in the 1940' s or 50's I believe. They wanted to bridge the divide between dynamics and electrodynamics. They apparently got some interesting results, but then I guess the trail petered out. Paul Dirac thought was a good idea, if I can believe the information I got off the internet. I'm not going to bother you with a "theory" that is very likely just spurious speculation, but the idea turned out to be predictive. It required that time must slow down on the surface of a planet and the slowing must equal that occurring on an object traveling at the escape velocity. I was not aware at the time that this was correct but not predicted by GR. In addition, the idea absolutely required that both gravitational and inertial mass must also diminish to the same degree. My previous comments need to be viewed in this context. Saying there can't possibly be a mass change along with time change on the surface of a planet, without checking experimentally, doesn't sound like science to me. We can't just rely on theory when there's a perfectly good, Newtonian experiment waiting to put it to the test.

17. Oct 19, 2014

Staff: Mentor

You are thinking about the steady-state universe proposed by Hoyle, Bondi, and Gold as an alternative to the Big Bang hypothesis, I think. It died with the discovery of the cosmic microwave background radiation.
Such an effect would show up pretty clearly in observations of planetary motion when multiple-body effects are included. It's not there.

(and as you mentioned above, we are right at the edge of acceptable topics for PhysicsForums. I'm closing this thread)

Last edited: Oct 19, 2014