Gravitational force and acceleration in General Relativity.

[starthaus]

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]

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 a_0 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 a_0=\frac{d^2r}{d\tau^2} instead of a_0=2\pi r.

What's relevant here is that in post 1, kev clearly defined a_0 as the proper acceleration measured by an accelerometer by a local stationary observer.

So, what value does GR predict for a_0? You have two choices:

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

2. \frac{m}{r_0^2\sqrt{1-2m/r_0}}?

It obviously depends on whether a_0 is defined as \frac{d^2r}{d\tau^2} or \frac{d^2r'}{d\tau^2}.

 

[Rolfe2]

Hmm, I was under the impression that proper acceleration is related to the four-acceleration \vec{A}=\frac{d^2\vec{R}}{d\tau^2} where R 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]

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.

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

(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 A, nor would we be able to calculate proper acceleration a_0 from the conditon A=(a_0,0) for u=0

 

[starthaus]

It obviously depends on whether a_0 is defined as \frac{d^2r}{d\tau^2} or \frac{d^2r'}{d\tau^2}.

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]

It obviously depends on whether a_0 is defined as \frac{d^2r}{d\tau^2} or \frac{d^2r'}{d\tau^2}.

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 a_0 you want me to base it on.

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

3. a_0=2\pi r, where a_0 is defined as the circumference of the moon.

 

[Rolfe2]

The above is definitely at odds with this.

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

(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.

IF what you were saying were true, the ccordinate acceleration a would not show up in the definition of the four-vector A, nor would we be able to calculate proper acceleration a_0 from the conditon A=(a_0,0) for u=0

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]

There is no :

a_0=a\gamma^3

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 Schwarzschild radial coordinates and Schwarzschild coordinate time. Let's call this the coordinate acceleration of the particle a = (d^2r/dt^2).

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

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

a' = a\gamma^3

where \gamma = 1/\sqrt{(1-2M/r)} 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]

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]

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 Schwarzschild radial coordinates and Schwarzschild coordinate time. Let's call this the coordinate acceleration of the particle a = (d^2r/dt^2).

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

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

a' = a\gamma^3

where \gamma = 1/\sqrt{(1-2M/r)} 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 r_0:

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

a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}

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

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

it still isn't true)

Why does this nonsense about a_0=a\gamma^3 has so much fascination for you?

 

[yuiop]

Why does this nonsense about a_0=a\gamma^3 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 a' =a\gamma^3 where \gamma = 1/\sqrt{(1-v^2/c^2)} 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 a' = a_0

Do you agree that a_0 = d^2x'/dt'^2 and not a_0 = d^2x/dt'^2\;\;? (i.e. the local or proper acceleration is measured using local rulers and clocks and not coordinate distance)

 

[Rolfe2]

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.

 

[starthaus]

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 a' =a\gamma^3 where \gamma = 1/\sqrt{(1-v^2/c^2)} and v is the relative velocity between inertial reference frames S and S'?

Sure, I told you this several times in this thread.
The reason is that in SR \gamma=1/\sqrt{1-(v/c)^2}

Can you take the above in really prove that a_0=a\gamma^3

Once you do that, try the same thing with \gamma=1/\sqrt{1-2m/r}

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

Do you agree that a_0 = d^2x'/dt'^2 and not a_0 = d^2x/dt'^2\;\;? (i.e. the local or proper acceleration is measured using local rulers and clocks and not coordinate distance)

This has nothing to do with the disproof to your claims I just posted. Simple algebra says you're wrong.

 

[yuiop]

But you know that the above can't be true. For a particle dropped from r_0:

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

\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}

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

\frac{d^2r}{d\tau^2}=-\frac{m}{r^2 \sqrt{1-2m/r}}

it still isn't true)

For the stationary observers on the surface, the local clock rate (dt') is not the same as the proper time of the particle (dtau) when the particle has significant velocity relative to the observers.

 

[starthaus]

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.

So, you have no references. Thank you.

 

[starthaus]

For the stationary observers on the surface, the local clock rate (dt') is not the same as the proper time of the particle (dtau) when the particle has significant velocity relative to the observers.

True but this is a total non-sequitur, the first expression is the coordinate acceleration and the second one is the proper acceleration and they are obviously not in the ratio a_0/a=\gamma^3.

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

2a. a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}

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

2b. a_0=\frac{d^2\rho}{d\tau^2}=-\frac{m}{r^2 \sqrt{1-2m/r}}

it still isn't true, there is no way you can ignore the term 3\frac{1-2m/r}{1-2m/r_0}-2}

 

[Al68]

But you know that the above can't be true. For a particle dropped from r_0:

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

a_0=\frac{d^2r}{d\tau^2}=-\frac{m}{r^2}

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

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

it still isn't true)

Why does this nonsense about a_0=a\gamma^3 has so much fascination for you?

A simple rearrangement of the first and last equations above shows that:

\frac{d^2r}{d\tau^2}=\frac{d^2r}{dt^2}(1/\sqrt{1-2m/r})^3 when r=r_0

 

[starthaus]

A simple rearrangement of those two equations shows that:

\frac{d^2r}{d\tau^2}=\frac{d^2r}{dt^2}(1/\sqrt{1-2m/r})^3 when r=r_0

Yes, I can see you can perform simple substitutions. Yet, the above is not true in general. It isn't true for any other value of r. Don't let that trouble you.

 

[Al68]

A simple rearrangement of the first and last equations above shows that:

\frac{d^2r}{d\tau^2}=\frac{d^2r}{dt^2}(1/\sqrt{1-2m/r})^3 when r=r_0

Yes, I can see you can perform simple substitutions. Yet, the above is not true in general. It isn't true for any other value of r. Don't let that trouble you.

Has anyone in this thread claimed that a_0=a\gamma^3 was true in any case other than when r=r_0 as specified in post 1?

If anyone neglected to specify "when r=r_0 ", I'm sure it's because (almost) everyone in this thread was talking about the case specified in post 1, where it was made perfectly clear that a referred to "the initial coordinate acceleration of a test mass released at r".

Was it unclear that "the initial coordinate acceleration of a test mass released at r" means r=r_0 ?

 

[Rolfe2]

So, you have no references. Thank you.

You mis-read my message. I provided the references you requested, and I also explained your errors. You're welcome.

 

[starthaus]

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)].

I don't think so. For the correct definition, see http://wapedia.mobi/en/Proper_acceleration .
Using the correct definition, proper acceleration for the general case is:

a_0=-\frac{m}{r^2}\frac{\sqrt{1-2m/r_0}}{1-2m/r}

 

[Rolfe2]

I don't think so. For the correct definition, see http://wapedia.mobi/en/Proper_acceleration .
Using the correct definition, proper acceleration for the general case is:

a_0=-\frac{m}{r^2}\frac{\sqrt{1-2m/r_0}}{1-2m/r}

Since you're unable or unwilling to follow rational explanations, and will only accept things on "authority", please check page 152 of Wald (for just one example), where he says "It is easy to check that static observers in the Schwarzschild spacetime must undergo a proper acceleration (in order to "stand still" in the "gravitational field") given by a = (1-2M/r)^-1/2 M/r^2..." This of course is perfectly consistent with the correct definition of proper acceleration, as has been explained to you already.

 

[starthaus]

Since you're unable or unwilling to follow rational explanations, and will only accept things on "authority", please check page 152 of Wald (for just one example), where he says "It is easy to check that static observers in the Schwarzschild spacetime must undergo a proper acceleration (in order to "stand still" in the "gravitational field") given by a = (1-2M/r)^-1/2 M/r^2..." This of course is perfectly consistent with the correct definition of proper acceleration, as has been explained to you already.

You gave a bogus definition of proper acceleration. So I pointed you towards the correct one (derivative of proper speed wrt coordinate time).

PS; if you made a little effort you would have recovered the formula in Wald by making r=r_0 in my formula . I gave you the general formula, derived from scratch, not gleaned from a book. <shrug>

Hey, thanks for playing anyway, better luck next time.

 

[Shackleford]

Hey, who's right? It seems like that would be easy to determine on here. lol.

 

[Al68]

I don't think so. For the correct definition, see http://wapedia.mobi/en/Proper_acceleration .
Using the correct definition, proper acceleration for the general case is:

a_0=-\frac{m}{r^2}\frac{\sqrt{1-2m/r_0}}{1-2m/r}

What do you mean by "general case"? Doesn't proper acceleration by definition mean the acceleration relative to a local (r=r_0) inertial path?

 

[starthaus]

What do you mean by "general case"? Doesn't proper acceleration by definition mean the acceleration relative to a local (r=r_0) inertial path?

If you drop a particle from r_0 what will be its acceleration at r?
If a particle hovers at r_0 what is its acceleration?

 

[Rolfe2]

You gave a bogus definition of proper acceleration. So I pointed you towards the correct one (derivative of proper speed wrt coordinate time).

Nope. See the definition of proper acceleration in any reputable reference, such as page 67 of Rindler's Essential Relativity. Again, it is the second derivative of the local co-moving inertial coordinates (including the proper space coordinates) with respect to proper time. This isn't controversial, it is simply what the term "proper acceleration" means and has always meant.

PS; if you made a little effort you would have recovered the formula in Wald by making r=r_0 in my formula .

First you insisted that the proper acceleration is -m/r^2, and you insisted that I was wrong for giving the correct proper acceleration of a stationary particle at a radial position r, along with the derivation. Then you switch to a different expression, one which you proudly announce reduces to my formula (the one you had been insisting was wrong for the past several posts) for a stationary particle, which is what my formula was stated to be in the first place. Weird.

I gave you the general formula, derived from scratch, not gleaned from a book.

You refused to accept the correct derivation provided to you, back when you were insisting the proper acceleration was -m/r^2, and you demanded that I provide a reference for the correct expression. Now when I provide a reference, you snidely acuse me of "gleaning it from a book". Very odd.

 

[Al68]

What do you mean by "general case"? Doesn't proper acceleration by definition mean the acceleration relative to a local (r=r_0) inertial path?

If you drop a particle from r_0 what will be its acceleration at r?
If a particle hovers at r_0 what is its acceleration?

LOL. The first question refers to coordinate acceleration. The second question refers to proper acceleration (r=r_0).

So, again, what do you mean by proper acceleration for the "general case"?

 

[Cyosis]

So, what value does GR predict for a_0 ? You have two choices:

1. \frac{m}{r^2} (my derivation based on lagrangian mechanics and K.Brown's)

2. \frac{m}{r^2\sqrt{1-2m/r}} ?

The above is definitely at odds with this.

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

(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 A, nor would we be able to calculate proper acceleration a_0 from the conditon A=(a_0,0) for u=0

The expression you obtained for the four-acceleration, using your convention, is A=(-m/r^2,0,0,0). Its magnitude given by

\sqrt{-g_{\mu \nu}A^\mu A^\nu}=\sqrt{-g_{00}A^0 A^0}=\frac{m}{r^2\sqrt{1-\frac{2m}{r}}}<br />.

 

[starthaus]

LOL. The first question refers to coordinate acceleration. The second question refers to proper acceleration (r=r_0).

So, again, what do you mean by proper acceleration for the "general case"?

You are trolling again. What is the proper acceleration of a particle dropped from r_0 when the particle arrives at location r? The observer "rides" on the particle.