Gravity's effects

Passionflower

Actually that is an interesting proposal.
Can I challenge you demonstrate it with numbers and formulas? I think this would be of great didactic value.

Indeed two stationary test observers at different r values in a Schwarzschild solution show different time flow as you call it but can you demonstrate that is not due to the difference in proper acceleration?

Similarly with various test observers radially free falling at different relative velocities, their proper time derivative is all different when they at an instant all fly by at the same location, but can you demonstrate that is not due to the Lorentz factor?

For instance:

A stationary test observer at a given r coordinate value undergoes constant acceleration to resist the escape velocity at that location.

The escape velocity at a given r is (rs is the Schwarzschild radius):
[tex]\Large v_{{{\it escape}}}=\sqrt {{\frac {{\it rs}}{r}}}[/tex]

While the time dilation is:
[tex]\Large d\tau_{{r}}=\sqrt {1-{\frac {{\it rs}}{r}}}[/tex]

However if we simply take the escape velocity and apply the Lorentz factor we get the same time dilation.
Thus how would you want to prove any gravitational time dilation through this method?
And the same situation arises for the various free falling test observers.

PeterDonis

Observers at varying radii on disks rotating at varying angular velocities. By varying the radius and angular velocities appropriately you can pretty much achieve any combination of proper acceleration and linear velocity relative to the center of the disk that you want, subject of course to restrictions on velocities not reaching the speed of light.

Sure, because the different time flow is due to g_00, while the acceleration is due to the radial rate of change of g_00. By varying the mass M of the central body and the radius r, you can achieve pretty much any combination of rate of time flow and acceleration you like.

If observers are in relative motion, the relative motion always contributes to their comparative rates of time flow, but in general it won't be the only contribution. If all the observers are also at exactly the same radial coordinate r above the same central mass M, obviously their relative velocities are the only respect in which they differ, so in that particular case, that's the only thing that *can* contribute to a difference in rate of time flow. That's not the type of case we've been talking about, or at least I don't think it is; but I agree that it's always good to clarify exactly what can affect what.

Passionflower

Peter I mean by demonstrating using a test situation and formulas to prove it. It is so because it is so, is just a tautology.

I just added some formulas to my prior posting to show you what I am getting at.

PeterDonis

So far in this thread we've been talking about static observers only, or at least I think we have. Saying that a static observer's time dilation arises from "resisting escape velocity" is fuzzy reasoning; I don't understand how it validates your derivation of the time dilation factor, since a static observer is not moving at escape velocity in the coordinates in question.

You are correct that freely falling observers are different; in fact, since they *are* moving inward at escape velocity, you could argue that your time dilation formula should apply to them, *not* to static observers. Unfortunately, you left out an important piece of information: a freely falling observer is changing his r coordinate as well as his t coordinate, so to calculate his rate of time flow you have to integrate dtau along his worldline using both the dt^2 and the dr^2 terms in the line element; you can't just look at g_00 like you can for static observers. This is easier to do in Painleve coordinates, particularly if you want to carry the computation inside the horizon. You've posted such calculations in these forums before, so I know you know how to do them; such a calculation is also given on the Wikipedia page on Painleve coordinates:

http://en.wikipedia.org/wiki/Gullstr...A9_coordinates

Passionflower

They would be moving wrt to a rain frame in GP coordinates. It is simply a matter of perspective and coordinates.

Passionflower

Again I wait for you to show a case with formulas to prove what you claim.

How about this:
Let's take three observers:

rs=1

O1 = Stationary at R1
O2 = Free falling at escape velocity at R1
O2 = Free falling at 0.5* escape velocity at R1

If we consider their resp tau differentials at that location can we prove their differences are gravitational instead of velocity based?

Can you show it with math?

PeterDonis

I described how to set up the "test situations" you speak of, but sure, I'll post a few formulas as well to give more detail on how it goes.

(1) Observers at radius r on a disk rotating with angular velocity [itex]\omega[/itex], in flat spacetime, have a linear velocity v and proper acceleration a of:

[tex]v = \omega r[/tex]

[tex]a = \frac{\omega^{2} r}{1 - \omega^{2} r^{2}} = \frac{\omega v}{1 - \omega^{2} r^{2}} = \omega \frac{v}{1 - v^{2}}[/tex]

It should be obvious from the above that I can first pick whatever v I like (subject to the constraint 0 <= v < 1), and then adjust [itex]\omega[/itex] appropriately so as to make a assume any value > 0 that I want, by adjusting r in concert with [itex]\omega[/itex] so as to keep their product constant and equal to v. So I can choose v and a independently and achieve any combination of the two within the constraints, as I said.

(2) Observers "hovering" at a constant radial coordinate r above a gravitating body experience time dilation and proper acceleration of:

[tex]\frac{d\tau}{dt} = \gamma = \sqrt{1 - \frac{2 M}{r}} = \sqrt{1 - 2 U}[/tex]

[tex]a = \frac{M}{r^{2} \sqrt{1 - \frac{2 M}{r}}} = \frac{U}{r \gamma} = \frac{1}{r} \frac{U}{\sqrt{1 - 2 U}}[/tex]

So again, it should be obvious that I can pick any value for [itex]\gamma[/itex] that I like (subject to the constraint that [itex]\gamma > 1[/itex]), which fixes the ratio M/r = U, and then use the second formula to adjust r appropriately, adjusting M in concert with r to keep U constant, to make a assume any value > 0 I want. So again I can choose [itex]\gamma[/itex] and a independently and achieve any combination of the two within the constraints, as I said.

PeterDonis

No, static observers are not moving relative to GP coordinates. The GP radial coordinate r is defined the same as it is for Schwarzschild coordinates, so the orbits of static observers, which are lines of constant r, theta, phi in Schwarzschild coordinates, are also lines of constant r, theta, phi in Painleve coordinates. The only thing that changes between Schwarzschild and Painleve coordinates is the surfaces of simultaneity; in Schwarzschild they are orthogonal to the static observers' worldlines, but in Painleve they are orthogonal to the worldlines of observers freely falling inward "from infinity".

PeterDonis

Obviously not, since the case you posed has all three observers at the *same* radial coordinate, and I already said that in that special case, the relative velocity *is* the only contribution. I was merely saying that if we have observers at *different* radial coordinates, who also happen to be in relative motion, there will be a contribution from their relative velocity *and* a contribution from the difference in heights. The GPS satellites are an example; the adjustment to their clock frequencies to make GPS time run at the same rate as UTC includes an adjustment for height *and* an adjustment for relative velocity.

Passionflower

It depends on the chosen coordinates if the difference in heights constitutes a velocity or not.

Passionflower

Rain frame Peter, I am talking about a rain frame in GP coordinates.

I am fully aware of your point I merely wanted to demonstrate that from another perspective it is velocity based instead of gravitational based.

PeterDonis

Huh? I'm not sure what you're getting at here. If you mean that the term "relative velocity" does not have a unique well-defined meaning in curved spacetime for observers that are spatially separated, yes, you're correct; one would have to settle on a specific meaning for the term. The one I had in mind was velocity relative to the Earth-Centered Inertial frame, since that's the relative velocity definition that was used, for example, to analyze the results of the Hafele-Keating experiment, but I should have made that explicit.

Passionflower

I think there is no need to settle anything. I think that considering multiple viewspoints of things tend to enrich understanding.
That was all I wanted to show. :)

PeterDonis

You mean rain frame as defined on the Wiki page?

http://en.wikipedia.org/wiki/Gullstr...A9_coordinates

As that line element is written, it mixes two different radial coordinates, so I'm not sure what it's supposed to mean physically. I don't believe such a frame is valid globally. I see the Wiki page references Taylor-Wheeler, I'll have to break open my copy and re-read their discussion of it.

Edit: Oops, read "Taylor-Wheeler" and immediately thought "Spacetime Physics" without reading the actual book title on the Wiki page. I don't have Exploring Black Holes but it's on my list to get.

Passionflower

You might also want to search for Lemaître observers.

Sulu: Captain! The stars. they are gone! We can't navigate!
Spock: Captain, I am locating an unidentified large object rapidly approaching us, with increasing speed. (Lemaitre)
Captain: Scottie, full speed reverse.
Scottie: Aye captain! Captain, we keep the engines at full speed and we are now just staying ahead of this 'thing'. Soon we will deplete our dilithium crystals. (Static)
Spock: Captain, I have a theory, it might work.
Scottie: We are just about to burn all of them out captain!
Spock: We should stop the engines and give some small lateral impulse power.
Captain: Sulu do it now!
Sulu: Captain it works, but now this thing is following us at a constant distance. (Hagihara)
Captain: Following us? Is it intelligent Spock?

It all depends on the point of view!

jk22

This is what I meant, using the equivalence principle to compute the speed of the local inertial frame.


This velocity of escape is calculated with a classical kinetic energy : U = 1/2mv^2

However shouldn't we use a relativistic expression : U = mc2*(1/Sqrt(1-v2/c2) - 1) ?

Then we find the velocity throuh equating with the potential energy : GMm/r2

Passionflower

The energy of a test observer in a Schwarzschild solution is constant as long as it travels on a geodesic.

You can use the following equality:
[tex]\Large E=\sqrt {1-{\frac {{\it r_s}}{r}}}{\sqrt {1-{v}^{2}}} ^{-1}[/tex]
So if you know the energy at one location it is very easy to calculate either the velocity or position at another location or velocity.

For instance it is easy to see that the energy of an observer free falling at escape velocity is exactly 1, just plugin v_escape in the equation above.

From the energy we can get the apogee of a test radial observer, because when the velocity is zero we get:
[tex]\Large E=\sqrt {1-{\frac {r_{{s}}}{r_{{{\it ap}}}}}}
[/tex]
Or alternatively we can get the apogee from the velocity at a given r value:
[tex]\Large {\frac {r_{{s}}}{r_{{{\it ap}}}}}=- \left( {\frac {r_{{s}
}}{r}}-{v}^{2} \right) \left( {v}^{2}-1 \right) ^{-1}
[/tex]