Gravitational force and acceleration in General Relativity.


by yuiop
Tags: acceleration, force, gravitational, relativity
starthaus
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#235
May28-10, 01:14 PM
P: 1,568
Quote Quote by Al68 View Post
I merely assumed that kev was using [tex]a_0=\frac{d^2r'}{d\tau^2}[/tex], where r' is proper distance, because that's the most commonly used definition of proper acceleration.
Hmm, I was under the impression that proper acceleration is related to the four-acceleration [tex]\vec{A}=\frac{d^2\vec{R}}{d\tau^2}[/tex] where [tex]R[/tex] is clearly the coordinate , not proper ,distance (see here). Has the definition of proper acceleration changed lately? Dalespam and I went over this issue in another thread.
Al68
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#236
May28-10, 01:26 PM
P: 801
Quote Quote by starthaus View Post
Hmm, I was under the impression that proper acceleration is related to the four-acceleration [tex]\vec{A}=\frac{d^2\vec{R}}{d\tau^2}[/tex] where [tex]R[/tex] is clearly the coordinate , not proper ,distance (see here). Has the definition of proper acceleration changed lately? Dalespam and I went over this issue in another thread.
The issue of how different people define proper acceleration is a semantical issue, not worthy of argument.

What's relevant here is that in post 1, kev clearly defined [tex]a_0[/tex] as the proper acceleration measured by an accelerometer by a local stationary observer.
yuiop
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#237
May28-10, 01:36 PM
P: 3,967
Quote Quote by kev View Post
At some lower radius the proper time (s) of the particle has increased and k holds the same value because it is independent of the proper time. Now it is obvious that at the lower radius the variable r has changed and so has the coordinate time (t), so if k is independent of s, then it is also independent of t and r.
Quote Quote by starthaus View Post
Can you prove this ? By using math?
It seems so obvious, I am not sure why a proof is needed. I can not prove it any more than I can rigously prove 2+1=3. It seems obvious to any normal person, but actually proving it formally and rigorously would probably take a mathematician a whole book to do.

The best I can do is demonstrate that using k as a constant that is indepent of r produces reasonable results and using k as a function of r produces incorrect results.

Let us initially assume the value of k is [tex]\sqrt{1-2M/R}[/tex] where M and R are constants of a given trajectory and R is the apogee.

This gives:

[tex]\frac{dt}{ds} = \frac{\sqrt{1-2M/R}}{(1-2M/r)} [/tex]

where r is a variable of the trajectory. Now if the location of the particle (r) coincides with the apogee (R) this gives

[tex]\frac{dt}{ds} = \frac{1}{\sqrt{1-2M/r}} [/tex]

in this specific instance and is what we would expect for the difference between coordinate time and proper time for a stationary particle at R.

If the apogee of the particle is at infinity the equation reduces to:

[tex]\frac{dt}{ds} = \frac{1}{(1-2M/r)} [/tex]

which is also what we would expect the time dilation ratio to be for a particle falling from infinity to r and takes both the gravitational and velocity time dilation components into account. This result has been obtained in many other threads.

Now let us assume k is a function of (r) such that [tex]k=\sqrt{1-2M/r}[/tex] where r is a variable.

This gives:

[tex]\frac{dt}{ds} = \frac{\sqrt{1-2M/r}}{(1-2M/r)} = \frac{1}{\sqrt{1-2M/r}} [/tex]

This is an incorrect result because it says the time dilation ratio is independent of the height of the apogee and of the instantaneous velocity of the falling particle and so is not generally true.

It is easy to see that time dilation ratio of the falling particle does depend on the velocity of the falling particle by taking the radial Schwarzschild solution:

[tex]ds^2 = \alpha dt^2 - \alpha^{-1} dr^2 [/tex]

and solving for ds/dt:

[tex]\frac{ds}{dt} = \sqrt{\alpha - \frac{1}{\alpha} \frac{dr^2}{dt^2}} [/tex]

From the above it obvious that ds/dt is only independent from dr/dt when [itex]dr^2/(\alpha dt ^2)= 0[/itex] which is not generally true.

That should be enough to convince any reasonable person that k is not a function of the variable r.

Treating k as constant is consistent with what we know about the Schwarzschild metric.

If the constant k is inserted into the Schwarzschild metric it is easy to obtain the falling velocity in terms of proper time as:

[tex]\frac{dr}{ds} = \sqrt{k^2 -\alpha} = \sqrt{(1-2M/R) -(1-2M/r)} = \sqrt{2M/r - 2M/R}[/tex]

This allows us to calculate the terminal velocity of a falling particle at r when released from a height R. Setting the apogee R to infinite gives [tex]dr/ds = \sqrt{2M/r}[/tex] which is the Newtonian escape velocity at r and is related to the conversion of potential energy to kinetic energy. I guess a formal proof of the constancy of k would be based on energy considerations. I leave that to a better person.
starthaus
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#238
May28-10, 01:38 PM
P: 1,568
Quote Quote by Al68 View Post
The issue of how different people define proper acceleration is a semantical issue, not worthy of argument.
But it is very worthy. You need to be able to choose a set of rules and stick to it.


What's relevant here is that in post 1, kev clearly defined [tex]a_0[/tex] as the proper acceleration measured by an accelerometer by a local stationary observer.
So, what value does GR predict for [tex]a_0[/tex]? You have two choices:

1. [tex]\frac{m}{r_0^2}[/tex] (my derivation based on lagrangian mechanics and K.Brown's)

2. [tex]\frac{m}{r_0^2\sqrt{1-2m/r_0}}[/tex]?
yuiop
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#239
May28-10, 01:54 PM
P: 3,967
Quote Quote by starthaus View Post

[tex]\frac{d^2r}{dt^2}=-\frac{m}{r^2}(1-2m/r)(3\frac{1-2m/r}{1-2m/r_0}-2})[/tex]

so, post 1 is in error. This is what we spent close to 200 posts to prove.
When the particle is at apogee and [tex]r=r_0[/tex] then the equation resolves to:

[tex]\frac{d^2r}{dt^2}=-\frac{m}{r^2}(1-2m/r)[/tex]

or:

[tex]\frac{d^2r}{dt^2}=-\frac{m}{r_0^2}(1-2m/r_0)[/tex]

(the two equations are synonymous at that instant).

This agrees with the equation given for coordinate acceleration in post #1 for the specific instance of a particle at its apogee and with the equation beautifully derived by Dalespam for the same specific instance.
yuiop
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#240
May28-10, 02:03 PM
P: 3,967
Quote Quote by starthaus View Post
So, what value does GR predict for [tex]a_0[/tex]? You have two choices:

1. [tex]\frac{m}{r_0^2}[/tex] (my derivation based on lagrangian mechanics and K.Brown's)

2. [tex]\frac{m}{r_0^2\sqrt{1-2m/r_0}}[/tex]?
K.Brown does not call equation (1.) the "proper acceleration. He calls it "the second derivative of r with respect to the proper time t of a radially moving particle" which is exactly what it is. Interesting that you are now using K.Brown in your defence when very recently you were calling him a blundering fool.

Quote Quote by Rolfe2 View Post
No, that's not the proper acceleration. What you've written there is the second derivative of the Schwarzschild radial coordinate r with respect to the proper time tau. The proper acceleration is actually the second derivative of what you might call the proper radial coordinate rho with respect to the proper time, and it is given by -m/r^2 [1/sqrt(1-2m/r)].
Agree.
Quote Quote by Al68 View Post
This is the type of non-answer that has led to hundreds of fruitless posts in this thread.

So, did it really just take over 200 posts to realize that the real issue is that you are using [tex]a_0=\frac{d^2r}{d\tau^2}[/tex] while kev was using [tex]a_0=\frac{d^2r'}{d\tau^2}[/tex], where r' is proper distance?
Agree.
starthaus
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#241
May28-10, 02:15 PM
P: 1,568
Quote Quote by kev View Post
K.Brown does not call equation (1.) the "proper acceleration. He calls it "the second derivative of r with respect to the proper time t of a radially moving particle" which is exactly what it is. Interesting that you are now using K.Brown in your defence when very recently you were calling him a blundering fool.
I am not "using" him as any defense. I am just pointing out that his solution coincides with mine. How about answering the question I asked Al68 instead of trolling?
Al68
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#242
May28-10, 02:26 PM
P: 801
Quote Quote by starthaus View Post
Quote Quote by Al68 View Post
The issue of how different people define proper acceleration is a semantical issue, not worthy of argument.
But it is very worthy.
It's not worthy of my time and attention. If someone chose to define [tex]a_0[/tex] as the circumference of the moon for the purpose of a post, I would accept their choice and move on instead of trying to convince him to use a different symbol. And I wouldn't try to convince him that [tex]a_0=\frac{d^2r}{d\tau^2}[/tex] instead of [tex]a_0=2\pi r[/tex].
Quote Quote by Al68 View Post
What's relevant here is that in post 1, kev clearly defined [tex]a_0[/tex] as the proper acceleration measured by an accelerometer by a local stationary observer.
So, what value does GR predict for [tex]a_0[/tex]? You have two choices:

1. [tex]\frac{m}{r_0^2}[/tex] (my derivation based on lagrangian mechanics and K.Brown's)

2. [tex]\frac{m}{r_0^2\sqrt{1-2m/r_0}}[/tex]?
It obviously depends on whether [tex]a_0[/tex] is defined as [tex]\frac{d^2r}{d\tau^2}[/tex] or [tex]\frac{d^2r'}{d\tau^2}[/tex].
Rolfe2
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#243
May28-10, 02:34 PM
P: 18
Quote Quote by starthaus View Post
Hmm, I was under the impression that proper acceleration is related to the four-acceleration [tex]\vec{A}=\frac{d^2\vec{R}}{d\tau^2}[/tex] where [tex]R[/tex] is clearly the coordinate , not proper ,distance (see here). Has the definition of proper acceleration changed lately? Dalespam and I went over this issue in another thread.
No, the definition of proper acceleration hasn't changed. The space-time coordinates comprising the four-vector R are (and always have been) local co-moving inertial coordinates, which implies that the spatial components are the "proper" space coordinates.

By the way, it isn't a good idea to talk about "the coordinate" variables, because there are infinitely many coordinate systems, and even the so-called "proper" variables are equal to coordinate variables for some suitable choice of coordinates. To avoid confusion, it's best to just say exactly what you mean (assuming you know what you mean).
starthaus
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#244
May28-10, 04:32 PM
P: 1,568
Quote Quote by Rolfe2 View Post
No, the definition of proper acceleration hasn't changed. The space-time coordinates comprising the four-vector R are (and always have been) local co-moving inertial coordinates, which implies that the spatial components are the "proper" space coordinates.
The above is definitely at odds with this.

[tex]U=(dx/d\tau,dy,d\tau,dz/d\tau,d(ct)/d\tau)[/tex]

(see also Rindler, p.99, Moller p.288))


IF what you were saying were true, the coordinate acceleration a would not show up in the definition of the four-vector [tex]A[/tex], nor would we be able to calculate proper acceleration [tex]a_0[/tex] from the conditon [tex]A=(a_0,0)[/tex] for [tex]u=0[/tex]
starthaus
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#245
May28-10, 04:34 PM
P: 1,568
Quote Quote by Al68 View Post
It obviously depends on whether [tex]a_0[/tex] is defined as [tex]\frac{d^2r}{d\tau^2}[/tex] or [tex]\frac{d^2r'}{d\tau^2}[/tex].
LOL. You need to pick the one that matches the measurement. Which of the two matches "what a comoving observer dynamometer will measure"? This is not up to debate based on definition.
Al68
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#246
May28-10, 04:53 PM
P: 801
Quote Quote by starthaus View Post
Quote Quote by Al68 View Post
It obviously depends on whether [tex]a_0[/tex] is defined as [tex]\frac{d^2r}{d\tau^2}[/tex] or [tex]\frac{d^2r'}{d\tau^2}[/tex].
LOL. You need to pick the one that matches the measurement. This is not up to debate based on definition.
If you want me to pick between your choices, I would be glad to if you specify which definition of [tex]a_0[/tex] you want me to base it on.

Otherwise, I will choose a third option allowed by GR:

3. [tex]a_0=2\pi r[/tex], where [tex]a_0[/tex] is defined as the circumference of the moon.
Rolfe2
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#247
May28-10, 05:23 PM
P: 18
Quote Quote by starthaus View Post
The above is definitely at odds with this.

[tex]U=(dx/d\tau,dy,d\tau,dz/d\tau,d(ct)/d\tau)[/tex]

(see also Rindler, p.99)
No, as always, the references you've cited confirm what I said and refute what you are saying. The x, y, and z coordinates appearing in those expressions are local co-moving inertial coordinates, which in the context of a gravitational field are the proper distance coordinates.

Quote Quote by starthaus View Post
IF what you were saying were true, the ccordinate acceleration a would not show up in the definition of the four-vector [tex]A[/tex], nor would we be able to calculate proper acceleration [tex]a_0[/tex] from the conditon [tex]A=(a_0,0)[/tex] for [tex]u=0[/tex]
What you typed there is gibberish. You need to learn the meanings of various systems of coordinates, both in flat spacetime, and in the presence of a gravitational field. Again, the proper acceleration is, by definition, expressed in terms of the local co-moving inertial coordinates. You don't appear to understand what that means, so I'd suggest you start by going back to basics and trying to learn the meanings of coordinate systems in general relativity.
yuiop
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#248
May28-10, 05:33 PM
P: 3,967
Quote Quote by starthaus View Post
There is no :

[tex]a_0=a\gamma^3[/tex]
Imagine we have some observers on the surface of a massive body of Schwarzschild radius r. They have a set of ideal clocks and a set of vertical rulers that are calibrated using radar devices, so that the average vertical speed of light is c by their measurements. A test particle is dropped from higher up. (It does not have to be from infinity.)

The acceleration of the particle as it passes the surface observers is measured by the Schwarzschild observer at infinity using Schwarschild radial coordinates and Schwarzschild coordinate time. Let's call this the coordinate acceleration of the particle [tex]a = (d^2r/dt^2)[/tex].

The observers on the shell measure the acceleration using their local measuring devices. Let's call this the local acceleration [tex]a' = (d^2r'/dt' ^2)[/tex]

The ratio between [tex](d^2r/dt^2)[/tex] and [tex](d^2r'/dt' ^2)[/tex] will always be:

[tex]a' = a\gamma^3[/tex]

where [itex]\gamma = 1/\sqrt{(1-2M/r)} [/itex] and r is the radius of the massive body, to first order and for a sufficiently local measurement, for a particle dropped from any height greater than r. (as far I can tell)
starthaus
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#249
May28-10, 05:45 PM
P: 1,568
Quote Quote by Rolfe2 View Post
No, as always, the references you've cited confirm what I said and refute what you are saying. The x, y, and z coordinates appearing in those expressions are local co-moving inertial coordinates, which in the context of a gravitational field are the proper distance coordinates.
I gave you three references that disagree with what you are saying, how about you gave me a couple of references that agree with what you are saying.


What you typed there is gibberish. You need to learn the meanings of various systems of coordinates, both in flat spacetime, and in the presence of a gravitational field. Again, the proper acceleration is, by definition, expressed in terms of the local co-moving inertial coordinates. You don't appear to understand what that means, so I'd suggest you start by going back to basics and trying to learn the meanings of coordinate systems in general relativity.
How about you get off your high horse?
starthaus
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#250
May28-10, 05:56 PM
P: 1,568
Quote Quote by kev View Post
Imagine we have some observers on the surface of a massive body of Schwarzschild radius r. They have a set of ideal clocks and a set of vertical rulers that are calibrated using radar devices, so that the average vertical speed of light is c by their measurements. A test particle is dropped from higher up. (It does not have to be from infinity.)

The acceleration of the particle as it passes the surface observers is measured by the Schwarzschild observer at infinity using Schwarschild radial coordinates and Schwarzschild coordinate time. Let's call this the coordinate acceleration of the particle [tex]a = (d^2r/dt^2)[/tex].

The observers on the shell measure the acceleration using their local measuring devices. Let's call this the local acceleration [tex]a' = (d^2r'/dt' ^2)[/tex]

The ratio between [tex](d^2r/dt^2)[/tex] and [tex](d^2r'/dt' ^2)[/tex] will always be:

[tex]a' = a\gamma^3[/tex]

where [itex]\gamma = 1/\sqrt{(1-2M/r)} [/itex] and r is the radius of the massive body, to first order and for a sufficiently local measurement, for a particle dropped from any height greater than r. (as far I can tell)
But you know that the above can't be true. For a particle dropped from [tex]r_0[/tex]:

[tex]a=\frac{d^2r}{dt^2}=-\frac{m}{r^2}(1-2m/r)(3\frac{1-2m/r}{1-2m/r_0}-2})[/tex]

[tex]a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}[/tex]

(Even if we accepted that the above might be of the form:

[tex]a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2 \sqrt{1-2m/r}}[/tex]

it still isn't true)

Why does this nonsense about [tex]a_0=a\gamma^3[/tex] has so much fascination for you?
yuiop
yuiop is offline
#251
May28-10, 06:13 PM
P: 3,967
Quote Quote by starthaus View Post
Why does this nonsense about [tex]a_0=a\gamma^3[/tex] has so much fascination for you?
Do you agree that in the colinear case in SR, the relationship between the proper acceleration of a particle as measured in the MCIF (S') and the acceleration in inertial reference frame S is [tex]a' =a\gamma^3[/tex] where [tex]\gamma = 1/\sqrt{(1-v^2/c^2)}[/tex] and v is the relative velocity between inertial reference frames S and S'?

Do you agree that a' as measured in the MCIF is equivalent in magnitude to the proper acceleration of the particle? i.e [tex]a' = a_0[/tex]

Do you agree that [tex]a_0 = d^2x'/dt'^2[/tex] and not [tex]a_0 = d^2x/dt'^2\;\; [/tex]? (i.e. the local or proper acceleration is measured using local rulers and clocks and not coordinate distance)
Rolfe2
Rolfe2 is offline
#252
May28-10, 06:28 PM
P: 18
Quote Quote by starthaus View Post
I gave you three references that disagree with what you are saying, how about you gave me a couple of references that agree with what you are saying.
Again, the references that have been cited all agree with what I'm saying, and they all disagree with what you are saying. The x, y, z, and t coordinates appearing in the four-vector expression for proper acceleration are local co-moving inertial coordinates, which signifies that x,y,z are the proper space coordinates (by definition). If you think those coordinates represent something else (Schwarzschild coordinates?? Starthaus Normal Coordinates??) then feel free to say so.


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