[jeffreyH (OP)]

For a mass m we can define a function v_e = f(r) for escape velocity. We should then also be able to define a function t_d = f(r). Can we then define a function t_d = f(v_e)? What would this relate to? Everything is relative so there is no universal frame of reference.

[jeffreyH (OP)]

I found these two consecutive posts on physicsforums.

1) The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.

2) Meaning, I assume, in free space, far away from all gravitating bodies?

Is this correct?

[Ethos_]

I found these two consecutive posts on physicsforums.

1) The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.

2) Meaning, I assume, in free space, far away from all gravitating bodies?

Is this correct?

It's really very simple Jeff. Think of this in terms of the black hole. We are all familiar with the escape velocity of a black hole as being equal to c.

Understanding this, the fact that escape velocity for any mass in question is caused by the gravitational influence, and as such is directly proportional  to the time dilation associated with that mass. That fact should come as no surprise.

[jeffreyH (OP)]

I found these two consecutive posts on physicsforums.

1) The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.

2) Meaning, I assume, in free space, far away from all gravitating bodies?

Is this correct?

It's really very simple Jeff. Think of this in terms of the black hole. We are all familiar with the escape velocity of a black hole as being equal to c.

Understanding this, the fact that escape velocity for any mass in question is caused by the gravitational influence, and as such is directly proportional  to the time dilation associated with that mass. That fact should come as no surprise.

Well, that being said, I am asking if the same escape velocity for any size of mass will give the same time dilation. This requires a variation in density.

EDIT: This should relate to black hole entropy.

http://en.wikipedia.org/wiki/Black_hole_thermodynamics

This also involves the Bekenstein Bound

http://en.wikipedia.org/wiki/Bekenstein_bound

[Ethos_]

 This requires a variation in density.

That's very true Jeff but as densities are different for rocky planets like earth, and gas giants like Jupiter, these differences in density also effect where the surface of each is found. And consequently, the radius of each which is used to calculate the escape velocities of those bodies.

[jeffreyH (OP)]

Distance traveled in free fall is:

s = (1/2)gt^2

Since g = GM/r^2

We can say that

S = (1/2)(GM/r^2)t^2

Then 2Sr/t^2 = GM/r

Ve = SQRT(2GM/r)

So:

4Sr/t^2 = 2GM/r

The factor of 4 is now present

SQRT(4Sr/t^2) = SQRT(2GM/r)

Since g = GM/r^2

Then

Ve = SQRT(2gr) for any mass.

[alancalverd]

But that all assumes point masses.

[jeffreyH (OP)]

But that all assumes point masses.

Yes and that is a major problem. Which highlights just why GR isn't a piece of cake.

[Bill S]

We are all familiar with the escape velocity of a black hole as being equal to c.

If that were the case, wouldn't light escape?

[Ethos_]

We are all familiar with the escape velocity of a black hole as being equal to c.

If that were the case, wouldn't light escape?

Yes of course...........I should have said "escape velocity greater than c." But greater by only the smallest degree.

[jeffreyH (OP)]

With Ve = SQRT(2gr) if we hold g constant and vary r as in a change in density then we have a differing value for Ve. So gravitational acceleration and escape velocity are not proportional. This means that while for the earth the value of g at the surface is much less than Ve there can be situations in which g is greater than Ve near very dense objects. In theory this can result in superluminal acceleration.

[Ethos_]

With Ve = SQRT(2gr) if we hold g constant and vary r as in a change in density then we have a differing value for Ve. So gravitational acceleration and escape velocity are not proportional. This means that while for the earth the value of g at the surface is much less than Ve there can be situations in which g is greater than Ve near very dense objects. In theory this can result in superluminal acceleration.

I think you've misunderstood my use of the word proportional Jeff. I was not saying they were equal, Webster's defines proportional as:

"to arrange the parts of (a whole) so as to be harmonious. a ratio"

[jeffreyH (OP)]

With Ve = SQRT(2gr) if we hold g constant and vary r as in a change in density then we have a differing value for Ve. So gravitational acceleration and escape velocity are not proportional. This means that while for the earth the value of g at the surface is much less than Ve there can be situations in which g is greater than Ve near very dense objects. In theory this can result in superluminal acceleration.

I think you've misunderstood my use of the word proportional Jeff. I was not saying they were equal, Webster's defines proportional as:

"to arrange the parts of (a whole) so as to be harmonious. a ratio"

Well there is a proportionality and it is balanced but not like the inverse square nature of the gravitational field. Add time dilation to the mix and stir. Can we say time dilation is a function of Ve, g or both?

[JohnDuffield]

It isn't a function of g, because that denotes the "local slope" of gravitational potential. Gravitational time dilation denotes the depth of gravitational potential. And to escape it, you need to "take a run at it", wherein your escape velocity is related to the gravitational time dilation. 

The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.

It's no coincidence. The thing is, GR doesn't actually explain why matter falls down. It doesn't actually say why an object acquires some given speed, which is escape velocity when you flip things round. However it's very easy to understand if you think about the wave nature of matter and stuff electron diffraction and spin. Just simplify the electron to light going round and round, then simplify it further to light going round a square path. The horizontals bend in the gravitational gradient, and the electron falls down. It's similar if you accelerate the electron in gravity-free space. Sadly you never seem to see any texts which combine relativity with the wave nature of matter.

[jeffreyH (OP)]

It isn't a function of g, because that denotes the "local slope" of gravitational potential. Gravitational time dilation denotes the depth of gravitational potential. And to escape it, you need to "take a run at it", wherein your escape velocity is related to the gravitational time dilation. 

The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.

It's no coincidence. The thing is, GR doesn't actually explain why matter falls down. It doesn't actually say why an object acquires some given speed, which is escape velocity when you flip things round. However it's very easy to understand if you think about the wave nature of matter and stuff electron diffraction and spin. Just simplify the electron to light going round and round, then simplify it further to light going round a square path. The horizontals bend in the gravitational gradient, and the electron falls down. It's similar if you accelerate the electron in gravity-free space. Sadly you never seem to see any texts which combine relativity with the wave nature of matter.

Of course it has to be a function of g. If you are stationary on the earth you are experiencing time dilation. You are not accelerating away from it. The velocity involved in escaping the field is due to kinetic energy. If the gravitational field were absent the time dilation would be due to the velocity. At a stationary position on the earth there is still a potential for acceleration. However there is NO potential for escape unless there is an impetus away from the surface. This will INDUCE a time dilation. Without the impetus there is no dilation due to Ve. Can't you see that? You only have the value of g operating.

[jeffreyH (OP)]

What are the consequences of reformulating the Ve equation this way?

c = SQRT(2gr)

This is then simplified to:

g = c^2/2r

EDIT: Of course this should be:

g = c^2/2r_s

This is likely the reason for the survival of the G2 gas cloud.

[jeffreyH (OP)]

In the proper Schwarzschild form we have:

a = (GM)/(r^2*SQRT[1-r_s/r])

Where the above is:

g = c^2/2r_s

In the Schwarzschild form the acceleration goes infinite at the event horizon.

[jeffreyH (OP)]

Considering a point an infinitesimal distance from r_s denoted by n we can then produce the following,

g = c^2/(2[r_s+n]*SQRT(1-r_s/[r_s+n]))

The value of n can then be increased as needed to plot the gradient of g away from the event horizon.

[JohnDuffield]

Of course it has to be a function of g.

No it isn't. See this depiction of gravitational potential:


CCASA image by AllenMcC, see Wikipedia.

The time dilation is represented by the depth of potential, how low in the plot you are, whilst g is the slope of the plot. Note that there's an inflection, so g at one elevation is the same as g at another. You can find two places on the plot where you can draw the same tangent.   

If you are stationary on the earth you are experiencing time dilation.

Yes.

You are not accelerating away from it.

Yes.

The velocity involved in escaping the field is due to kinetic energy.

Yes, and in your ascending rocket going faster and faster you are swapping gravitational time dilation for special-relativistic time dilation. 

If the gravitational field were absent the time dilation would be due to the velocity.

Yes.

At a stationary position on the earth there is still a potential for acceleration. However there is NO potential for escape unless there is an impetus away from the surface. This will INDUCE a time dilation.

Yes. And as you ascend away from the surface you reduce the gravitational time dilation. 

Without the impetus there is no dilation due to Ve. Can't you see that?

Yes.

You only have the value of g operating.

The acceleration due to gravity is g, and it is due to the local slope of the potential. That's like the local slope of the gravitational time dilation. If your clock at the ceiling runs at the same rate as your clock at the floor, your pencil doesn't fall down. If your clock at the ceiling runs faster than your clock at the floor, your pencil falls down. If your clock at the ceiling runs much faster than your clock at the floor, your pencil falls down much faster. Note that your clocks might be light clocks, and that g can be expressed as the local slope of your plot of the coordinate speed of light.

[jeffreyH (OP)]

Of course it has to be a function of g.

No it isn't. See this depiction of gravitational potential:


CCASA image by AllenMcC, see Wikipedia.

The time dilation is represented by the depth of potential, how low in the plot you are, whilst g is the slope of the plot. Note that there's an inflection, so g at one elevation is the same as g at another. You can find two places on the plot where you can draw the same tangent.   

If you are stationary on the earth you are experiencing time dilation.

Yes.

You are not accelerating away from it.

Yes.

The velocity involved in escaping the field is due to kinetic energy.

Yes, and in your ascending rocket going faster and faster you are swapping gravitational time dilation for special-relativistic time dilation. 

If the gravitational field were absent the time dilation would be due to the velocity.

Yes.

At a stationary position on the earth there is still a potential for acceleration. However there is NO potential for escape unless there is an impetus away from the surface. This will INDUCE a time dilation.

Yes. And as you ascend away from the surface you reduce the gravitational time dilation. 

Without the impetus there is no dilation due to Ve. Can't you see that?

Yes.

You only have the value of g operating.

The acceleration due to gravity is g, and it is due to the local slope of the potential. That's like the local slope of the gravitational time dilation. If your clock at the ceiling runs at the same rate as your clock at the floor, your pencil doesn't fall down. If your clock at the ceiling runs faster than your clock at the floor, your pencil falls down. If your clock at the ceiling runs much faster than your clock at the floor, your pencil falls down much faster. Note that your clocks might be light clocks, and that g can be expressed as the local slope of your plot of the coordinate speed of light.

You have just contradicted yourself and described time dilation as a function of g.