Gravitational force and acceleration in General Relativity.

[starthaus]

You are quite right Al. I think the objection that Starthaus is making is that this implies the next step is differentiating zero with respect to s.

It is much worse than that. Face the music , kev, prof. Kevin Brown's proof is fatally flawed. See post #137. Time for you to go fishing for a different proof on the web.

 

[yuiop]

It is much worse than that. Face the music , kev, prof. Kevin Brown's proof is fatally flawed. See post #137. Time for you to go fishing for another proof on the web.

I don't need to. I have shown in the last post how your main objection is invalid.

I can address your other objections too. There is nothing wrong K. Brown's derivation.

Time for you start quoting from textbooks that you actually understand. The equation you have given for coordinate acceleration is not actually given by Rindler in his book and is just a hash up by you based on a extrapolated misunderstanding of what Rindler actually says .

 

[starthaus]

The clever part of Prof Brown's derivation is that by using the constant k he can get around this and carry out the differentiation.

No, kev, you can't do that , this is basic calculus. You admitted earlier on that you are weak on this subject, you really need to learn it if you want to learn physics. You can't differentiate as if $$k$$ is a constant when k is really a function of $$r$$. Prof. Kevin Brown goofed, what is your excuse?

 

[yuiop]

No, kev, you can't do that , this is basic calculus. You admitted earlier on that you are weak on this subject, you really need to learn it if you want tolearn physics. You can't differentiate as if k is aconstant when k is really a function of r. Prof. Kevin Brown goofed, what is your excuse?

It was you that introduced the Lagrangian and the equations of motion, but it seems you really do not understand the physics behind them. k is a constant for a free falling particle. Read your own post #2 of this thread quoted below:

This is very wrong. You need to start with the metric:


$$ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2$$

$$\alpha=1-\frac{2m}{r}$$

From this you construct the Lagrangian:

$$L=\alpha (\frac{dt}{ds})^2-\frac{1}{\alpha}(\frac{dr}{ds})^2-r^2(\frac{d\phi}{ds})^2$$

From the above Lagrangian, you get immediately the equations of motion:

$$\alpha \frac{dt}{ds}=k$$

$$r^2 \frac{d\phi}{ds}=h$$

whre h,k are constants.

A classic case of being "hung by your own petard" I think. TIme to bow out graciously and admit you are wrong and apologise to K.Brown.

When a particle is in freefall, $$\alpha$$ changes in such a way that exactly cancels out the change in the ratio dt/ds so that k remains constant for any value of r. k is what Prof Brown calls an affine parameter. Read up on Killing vectors.

$$\alpha \frac{dt}{ds}=k$$

 

[starthaus]

It was you that introduced the Lagrangian and the equations of motion, but it seems you really do not understand the physics behind them. k is a constant for a free falling particle. Read your own post here:
When a particle is in freefall, $$\alpha$$ changes in such a way that exactly cancels out the change in the ratio dt/ds so that k remains constant for any value of r. k is what Prof Brown calls an affine parameter. Read up on Killing vectors.

$$\alpha \frac{dt}{ds}=k$$

No , kev, you really don't understand calculus,

The Euler-Lagrange equation is:

$$\frac{d}{ds}(\alpha \frac{dt}{ds})=0$$

This means that :

$$\alpha \frac{dt}{ds}=k$$

where k is NOT a function of s. But, as Kevin Brown does in his failed proof, k can be a function of both t and r. In his case, $$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$.

So, he cannot differentiate wrt $$r$$ as if k were not a function of $$r$$. Because it is.

 

[yuiop]

No , kev, you really don't understand calculus,

The Euler-Lagrange equation is:

$$\frac{d}{ds}(\alpha \frac{dt}{ds})=0$$

This means that :

$$\alpha \frac{dt}{ds}=k$$

where k is NOT a function of s. But, as Kevin Brown does in his failed proof, k can be a function of both t and r. In his case, $$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$.

So, he cannot differentiate wrt $$r$$ as if k were not a function of $$r$$. Because it is.

$$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$

is really:

$$k=\sqrt{1-2m/r_a}$$

where $$r_a$$ is a constant which is equal to the radial coordinate of the particle at its apogee. As the particle falls the value of $$\sqrt{1-2m/r_a}$$ does not change. It is only at the apogee that $$r = r_a$$ and the two values can be used interchangeably and cancel out. It is only at the apogee that you can say $$\sqrt{1-2m/r_a} = \sqrt{1-2m/r} = \sqrt{\alpha}$$

Why did you say in #2 that k is a constant and now you are vehemently arguing that it is not? Do you suffer from multiple personality disorder? Either you are wrong now or you were wrong in #2. Do you retract your statements in #2? You probably shouldn't. I can find plenty of references that show k is a constant. (Not just K.Brown).

 

[starthaus]

$$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$

is really:

$$k=\sqrt{\alpha}=\sqrt{1-2m/r_a}$$

What do you mean "is really"? You are changing your tune from post to post.

 

[starthaus]

$$\alpha=1-\frac{2M}{r}$$

Substituting $$k=\sqrt{\alpha}$$

So, you need to make up your mind, what is the value for $$k$$?

 

[yuiop]

$$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$

is really:

$$k= \sqrt{\alpha} =\sqrt{1-2m/r_a}$$

What do you mean "is really"? You are changing your tune from post to post.

$$\alpha=1-\frac{2M}{r}$$


Substituting $$k=\sqrt{\alpha}$$ gives the coordinate acceleration of a particle at its apogee at r as:

So, you need to make up your mind, what is the value for $$k$$?

I though I made it clear in the last post that:

$$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$

is interchangeable (synonymous) with

$$k=\sqrt{\alpha}=\sqrt{1-2m/r_a}$$

at the apogee, where $$r=r_a$$

It is a bit tricky. You have to pay attention

 

[starthaus]

I though I made it clear in the last post that:

$$k=\sqrt{\alpha}=\sqrt{1-2m/r}$$

is interchangeable (synonymous) with

$$k=\sqrt{\alpha}=\sqrt{1-2m/r_a}$$

at the apogee, where $$r=r_a$$

It is a bit tricky. You have to pay attention

..and its value at infinity is 1. So, varies from 1 to $$\sqrt{1-2m/r_a}$$ as a function of r $$k=\sqrt{1-2m/r}$$. What does all this have to do with your inability to perform a derivative of a function with respect to the r variable?

 

[yuiop]

..and its value at infinity is 1. So, varies from 1 to $$\sqrt{1-2m/r_a}$$ as a function of r $$k=\sqrt{1-2m/r}$$. What does all this have to do with your inability to perform a derivative of a function with respect to the r variable?

No, the value of k is always $$\sqrt{1-2m/r_a}$$.

$$r_a$$ is a constant and it value is fixed at the apogeee of the particles trajectory. For a particle with its apogee at infinity the fixed value of $$r_a$$ is infinity and the fixed value of k for a particle falling from inifinty is:

$$k = \sqrt{1-2m/r_a} = \sqrt{1-2m/\infty} = 1$$

No inconsistency there.

All your objections have been debunked.

 

[yuiop]

... What does all this have to do with your inability to perform a derivative of a function with respect to the r variable?

The derivation is as you say the derivation done by Prof Brown. I made no secret of that. I stated it clearly in #125 when I said:

If the statement's in #1 are correct, it does not really matter how I obtained them, but for the record they are substantially based on the mathpages relativity website which is usually fairly reliable and rigorous.

If the derivatives of the functions are wrong, then it is Prof Brown's inablity to do calculus that is the problem, but I think it is clear that he is an expert in calculus. I did however check his derivatives using algebraic software and they are all correct. That is off course assuming that the Mathematica and Maple software packages do calculus correctly. Seeing as how these software packages are used by serious physicists and engineers all over the world, they probably do.

 

[starthaus]

No, the value of k is always $$\sqrt{1-2m/r_a}$$.

$$r_a$$ is a constant and it value is fixed at the apogeee of the particles trajectory. For a particle with its apogee at infinity the fixed value of $$r_a$$ is infinity and the fixed value of k for a particle falling from inifinty is:

$$k = \sqrt{1-2m/r_a} = \sqrt{1-2m/\infty} = 1$$

No inconsistency there.

All your objections have been debunked.

LOL, I told you that numerology and physics are two different fields. What you are doing is numerology.
Let me try to help you understand why you are doing numerology.

From the metric

$$ds^2=\alpha dt^2- \alpha^{-1} dr^2$$

one derives the Lagrangian:

$$L=\alpha (dt/ds)^2- \alpha^{-1} (dr/ds)^2$$

From the above Lagrangian, one derives two Euler-Lagrange equations, one of them being:

$$\frac{d}{ds}(\alpha*(dt/ds))=0$$

with the solution:

$$\alpha*dt/ds=k$$If you do what K.Brown is doing, dividing the metric by $$dt^2$$:

$$\frac{ds^2}{dt^2}=\alpha-\alpha^{-1} (dr/dt)^2$$

you end up getting:

$$\frac{dr}{dt}=\alpha \sqrt{1-\alpha/k^2}$$ and you reach a dead end.

You cannot generate any more information from the above by trying to do a second differentiation wrt t using $$\alpha*dt/ds=k$$ because that was obtained from
$$ds^2=\alpha dt^2- \alpha^{-1} dr^2$$ in first place. You have run out of usable information.
What you have to do is to use the second, independent Euler-Lagrange equation. That equation provides the extra necessary information since it is indeed a function of $$\frac{d^2r}{ds^2}$$, $$\frac{dr}{ds}$$. This is the part I left as an exercise for you but, since you don't want to learn the formalism you are left to attempting all kinds of different hacks that don't really solve anything.

 

[yuiop]

In #2 you said the correct way to do the derivation was like this:

This is very wrong. You need to start with the metric:

$$ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2$$

$$\alpha=1-\frac{2m}{r}$$

From this you construct the Lagrangian:

$$L=\alpha (\frac{dt}{ds})^2-\frac{1}{\alpha}(\frac{dr}{ds})^2-r^2(\frac{d\phi}{ds})^2$$

From the above Lagrangian, you get immediately the equations of motion:

$$\alpha \frac{dt}{ds}=k$$

$$r^2 \frac{d\phi}{ds}=h$$

whre h,k are constants.

and you left the completion as an exercise. I completed that derviation or rather K.Brown did and I carefully explained to you how his derivation works and the physics behind it. Now you are saying that is not the correct way to start the derivation. This makes it clear that when you set the exercise you did not know how to complete your own exercise as I have suspected all along. If you disagree with K.Browns completion of the exercise then you should show how it should correctly be completed. You don't how to? Thought so.

Then you decided that deriving the equations from a field aproximation was the way to go (a bad move in itself) and botched up the job of extrapolating Rindler's equations to obtain the coordinate acceleration and the end result is miles out. You get coordinate accleration greater than proper acceleration and it is blindingly obvious to everyone here that is a wrong conclusion. and even more shockingly, still insist that is correct.

You are now still trying to maintain your self appointed status as my teacher, even though it is now clear that you have a very poor grasp of the physics and mathematics involved. I have shown how at least half a dozen of your statements in recent posts are outright wrong.

Now you want me complete another exercise, presumably because you do not know how to complete it yourself. If you want to save face you probably should show that you can (correctly) complete at least one of the exercises you have set, (You will know if you are on the right track if your results agree with the results I posted in #1) or simply admit you are on a learning curve like the rest of us.

I am pretty sure I do not want to go through another 150 posts of thinly veiled personal attacks from you, trying to guess what you are thinking.

Finish what you started here:

What you have to do is to use the second, independent Euler-Lagrange equation. That equation provides the extra necessary information since it is indeed a function of $$\frac{d^2r}{ds^2}$$, $$\frac{dr}{ds}$$. This is the part I left as an exercise for you but, since you don't want to learn the formalism

 

[Dale]

This is a question for Dalespam.

I know you are an expert in 4 vectors and have already obtained a correct solution for the proper acceleration. Would you be able to obtain a solution for the coordinate gravitational acceleration for us, using those methods?

OK, but the truth is that I am an expert in Mathematica , not 4-vectors. I will use the metric, Christoffel symbols, etc. given https://www.physicsforums.com/showpost.php?p=2712746&postcount=38". The worldline of a particle moving only in the radial direction in these coordinates is given by:
$$\mathbf X = (t,r,0,0)$$

Where r is now a function of t.

Then we can derive the four-velocity as follows:
$$\mathbf U = \frac{d \mathbf X}{d \tau} = c \frac{d \mathbf X}{ds} = c \frac{d \mathbf X}{dt} \frac{dt}{ds} = \left(\frac{c}{\sqrt{-\frac{c^2 R}{r}+c^2+\frac{r \left(r'\right)^2}{R-r}}},\frac{c r'}{\sqrt{-\frac{c^2 R}{r}+c^2+\frac{r \left(r'\right)^2}{R-r}}},0,0\right)$$

Where the ' denotes differentiation wrt the coordinate t.

And we can verify that the norm of the four-velocity is equal to:
$$||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = c^2$$

Now it gets ugly. We can derive the four-acceleration as follows:
$$A^{\mu}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda} =$$
$$\left(\frac{c^2 r' \left(c^2 R (R-r)^2+2 r^3 (r-R) r''-3 r^2 R \left(r'\right)^2\right)}{2 \left(c^2 (R-r)^2-r^2 \left(r'\right)^2\right)^2},\frac{c^4 (R-r)^2 \left(c^2 R (R-r)^2+2 r^3 (r-R) r''-3 r^2 R \left(r'\right)^2\right)}{2 \left(c^2 r (R-r)^2-r^3 \left(r'\right)^2\right)^2},0,0\right)$$

So, we can set all of these equal to 0 to get the geodesic equation describing an object in free-fall along a radial trajectory. Doing so and solving for r'' gives
$$r''=\frac{c^2 R (R-r)^2-3 r^2 R \left(r'\right)^2}{2 r^3 (R-r)}$$

For a momentarily stationary particle r'=0 so
$$r''=\frac{GM}{r^2}\left(\frac{2GM}{c^2r}-1 \right)$$

where I have substituted back in for R.

Note that this is the second derivative of the coordinate r wrt the coordinate t, so take it for what it is worth. I suspect this differs from Rindler because he is calculating something other than r'' as defined here.

 

[starthaus]

In #2 you said the correct way to do the derivation was like this:and you left the completion as an exercise.

...which, to date, you couldn't complete. The reason is that you have no clue how to derive the equations of motion from the Lagrangian. Nor are you interested in learning.

I completed that derviation or rather K.Brown did and I carefully explained to you how his derivation works and the physics behind it.

K. Brown did it but his solution has nothing to do with any Lagrangian and nothing to do with any Euler-Lagrange equations, you lifted it from his page (without giving him any credit). I explained that he made a fatal error but, since the error involves calculus, you clearly can't understand it.

If you disagree with K.Browns completion of the exercise then you should show how it should correctly be completed. You don't how to? Thought so.

My solution has nothing to do with the botched solution that you lifted off the mathpages.
It employs the Euler-Lagrange formalism, I even wrote down for you the equations of motion. Once I realized that there is no way you will be able to learn the approach, I gave you the simpler solution based on metrics, potential and gradients. I guided you through your calculus missteps in deriving proper acceleration.

Then you decided that deriving the equations from a field aproximation was the way to go (a bad move in itself) and botched up the job of extrapolating Rindler's equations to obtain the coordinate acceleration and the end result is miles out.

For someone who has no clue how to do the derivation, you have a lot of gall.

You get coordinate accleration greater than proper acceleration and it is blindingly obvious to everyone here that is a wrong conclusion. and even more shockingly, still insist that is correct.

Apparently you have discovered a rule that says that coordinate acceleration needs to be related to proper acceleration via the rule:

$$a_0=a\gamma^3$$

What you refuse to understand, in your blind application of numerology is that the above was derived for the particular case when:

A) $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$
B) $$\vec{a}$$ and $$\vec{v}$$ are co-linearOf course, I have told you this repeatedly but it fell on deaf ears.

Now you want me complete another exercise, presumably because you do not know how to complete it yourself.

No, it is because I wanted to reach you how to derive solutions instead of copying and pasting.

If you want to save face you probably should show that you can (correctly) complete at least one of the exercises you have set, (You will know if you are on the right track if your results agree with the results I posted in #1) or simply admit you are on a learning curve like the rest of us.

What you posted in post #1 is pure junk. I have been showing this to you ever since you posted.

 

[Al68]

Apparently you have discovered a rule that says that coordinate acceleration needs to be related to proper acceleration via the rule:

$$a_0=a\gamma^3$$

What you refuse to understand, in your blind application of numerology is that the above was derived for the particular case when:

A) $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$
B) $$\vec{a}$$ and $$\vec{v}$$ are co-linear

I'm too lazy to sort through this thread again, so is this coordinate acceleration that of a stationary (accelerated) particle wrt an inertial frame, or that of a free falling particle wrt to an accelerated reference frame (origin at constant r)?

And if the latter, is the origin of the non-inertial coordinate system local to the free falling particle's apogee, or at r=0?

 

[yuiop]

...which, to date, you couldn't complete. The reason is that you have no clue how to derive the equations of motion from the Lagrangian. Nor are you interested in learning.

There is no need to derive the equations of motion from the Langrangian, if no one is disputing their validty. I took the equations you gave and went from there. You claimed to have done "the hard work" but the equations you gave are standard fare in any relativity textbook. No original work from you at all. You disgree the derivation based on K.Brown's work, but you have completely failed to state how you think the deriviation should be done from the starting point you gave. Basically you are saying "It's wrong. It's something else!", but you do not know what the something else is!" lol.

...
K. Brown did it but his solution has nothing to do with any Lagrangian and nothing to do with any Euler-Lagrange equations, you lifted it from his page (without giving him any credit). I explained that he made a fatal error but, since the error involves calculus, you clearly can't understand it.

I am sure there other people on this forum who do understand calculus and I do not see any of them backing you up.

I stated in #1 that would provide the references I used if anyone was interested. You never asked. I stated my references later without being asked.

K.Brown does indeed not base his solution on anything to do with the Langragian. I made the connection between the Lagrangian conserved energy quantity you used in your initial approach (that you abandoned) and K.Brown's "affine parameter" and put it all together in an easy to understand package, that even you could understand.

...
My solution has nothing to do with the botched solution that you lifted off the mathpages.
It employs the Euler-Lagrange formalism, ...

You still don't see the connection.. even though I have spoon fed it to you. Sigh.

...
Apparently you have discovered a rule that says that coordinate acceleration needs to be related to proper acceleration via the rule:

$$a_0=a\gamma^3$$

What you refuse to understand, in your blind application of numerology is that the above was derived for the particular case when:

A) $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$
B) $$\vec{a}$$ and $$\vec{v}$$ are co-linear


Of course, I have told you this repeatedly but it fell on deaf ears.

No. After stating the values for proper and coordinate radial gravitational acceleration, I simply made the observation that the ratio between those two values is $$a_0=a\gamma^3$$ when $\gamma=1/\sqrt{(1-2M/r)}$ which is similar to the transformation of parallel acceleration in SR where $\gamma=1/\sqrt{(1-(v/c)^2)}$ which I think is kinda cool. There is nothing wrong with making observations. Kepler made observations about orbits, even though he did not understand the physics behind it and he was not castigated for making them. It was Newton who later filled in the physics.

In fact, I can make the further observation that as far as I can tell, all vertical and horizontal measurements (with the trivial exception of velocity) made by a stationary local observer in the Schwarzschild metric and measurements made by an observer at infinity, transform in the same way as longitudinal and transverse measurements respectively in SR, if you use $\gamma=1/\sqrt{(1-(v/c)^2)}$ in SR and $\gamma=1/\sqrt{(1-2M/r)}$ in the Schwarzschild metric. If this turns out to be true, it makes a huge simplification in understanding the Schwarzschild metric.

...
What you posted in post #1 is pure junk. I have been showing this to you ever since you posted.

The equations posted in #1 are supported by:

The calculations of Prof.Brown on his mathpages website.
The calculations quoted in the Hartle textbook. (https://www.physicsforums.com/showpost.php?p=2711553&postcount=10")
The calculations by Dalespam using four vectors. (Just awesome. See #155)

and I have even shown they are they are compatible with the Lagrangian equations of motion introduced by you.

Your comment here is just abusive.

 

[yuiop]

I'm too lazy to sort through this thread again, so is this coordinate acceleration that of a stationary (accelerated) particle wrt an inertial frame, or that of a free falling particle wrt to an accelerated reference frame (origin at constant r)?

And if the latter, is the origin of the non-inertial coordinate system local to the free falling particle's apogee, or at r=0?

Have a look at #158 that I posted just after you posted this and see if it answers your question. If not, feel free to ask again.

 

[yuiop]

I'm too lazy to sort through this thread again, so is this coordinate acceleration that of a stationary (accelerated) particle wrt an inertial frame, or that of a free falling particle wrt to an accelerated reference frame (origin at constant r)?

The coordinate acceleration is the initial acceleration measured by an observer at infinity, when a particle is released from height r.

It is also the acceleration of a particle in free fall with its apogee at r, when the particle is at its apogee, as measured by an observer at infinity.

It is stationary in the sense that it is momentarily stationary.

And if the latter, is the origin of the non-inertial coordinate system local to the free falling particle's apogee, or at r=0?

It is not the latter. The acceleration of a free falling particle (at its apogee) wrt to an accelerated local stationary observer at r is its proper acceleration. It has the same magnitude at the acceleration measured by an a stationary accelerometer at r.

 

[yuiop]

... so is this coordinate acceleration that of a stationary (accelerated) particle wrt an inertial frame,

The acceleration of a stationary (accelerated) particle as measured by an inertial observer that is (almost) stationary (i.e. the observer is at his apogee at r) has the same magnitude as the proper acceleration of the particle. This is closely related to the clock hypothosis and the concept of the Momentarily Comoving Inertial Frame (MCIF).

or that of a free falling particle wrt to an accelerated reference frame (origin at constant r)?

The magnitude of this measurement is also the same as the proper acceleration. The observer in this case is stationary at r and the particle is at its apogee at r.

And if the latter, is the origin of the non-inertial coordinate system local to the free falling particle's apogee, or at r=0?

Yes, at its apogee. Proper acceleration is a local measurement.

P.S. I was answering at the same time as you were editing your post and we might have got our wires crossed. Just ask for any clarifications.

 

[Al68]

The acceleration of a stationary (accelerated) particle as measured by an inertial observer that is (almost) stationary (i.e. the observer is at his apogee at r) has the same magnitude as the proper acceleration of the particle. This is closely related to the clock hypothosis and the concept of the Momentarily Comoving Inertial Frame (MCIF).

The magnitude of this measurement is also the same as the proper acceleration. The observer in this case is stationary at r and the particle is at its apogee at r.
Yes, at its apogee. Proper acceleration is a local measurement.

Sure, I knew this was what you were referring to. But from reading this thread, I thought maybe starthaus was referring to a different situation, such as a free falling particle distant from an accelerating observer in such a way that the coordinate acceleration of the particle could be greater than the proper acceleration of the observer.

I can't make any other sense of this thread.

 

[yuiop]

Sure, I knew this was what you were referring to. But from reading this thread, I thought maybe starthaus was referring to a different situation, such as a free falling particle distant from an accelerating observer in such a way that the coordinate acceleration of the particle could be greater than the proper acceleration of the observer.

I can't make any other sense of this thread.

Well I don't think you will get any sense by asking Starthaus. He will probably give you the the two postulates of relativity and leave the rest as "an exercise".

 

[starthaus]

There is no need to derive the equations of motion from the Langrangian, if no one is disputing their validty.

Of course it is, this is the stuff I tried to teach you. Since you don't know anything about the subject and since you don't want to learn, I showed you the correct derivation for proper acceleration straight from the Schwarzschild metric, something that you could follow with great difficulty.

I took the equations you gave and went from there. You claimed to have done "the hard work" but the equations you gave are standard fare in any relativity textbook. No original work from you at all.

I gave you both Euler-Lagrange equations, there is a small step from there in order to determine the acceleration. Since you can't learn the formalism, I gave up and tried to teach you the simpler one, based on the determination of the potential from the Schwarzscild metric. It took you about 10 posts and a lot of proding to get the proper acceleration right.

You disgree the derivation based on K.Brown's work, but you have completely failed to state how you think the deriviation should be done from the starting point you gave.

I explained to you repeatedly the failure of the K.Brown method. The method that you copied dutifully. Now, if you want to learn the Euler-Lagrange method and act in a respectful way, I am willing to teach you.

I am sure there other people on this forum who do understand calculus and I do not see any of them backing you up.

This is basic stuff (maybe not for you), I don't need anyone elses "backup".

K.Brown does indeed not base his solution on anything to do with the Langragian. I made the connection between the Lagrangian conserved energy quantity you used in your initial approach (that you abandoned) and K.Brown's "affine parameter" and put it all together in an easy to understand package, that even you could understand.

Yes, even I could understand. I understood that it is based on an embarasiing error. So, you cited a method that is a failure.

No. After stating the values for proper and coordinate radial gravitational acceleration, I simply made the observation that the ratio between those two values is $$a_0=a\gamma^3$$ when $\gamma=1/\sqrt{(1-2M/r)}$ which is similar to the transformation of parallel acceleration in SR where $\gamma=1/\sqrt{(1-(v/c)^2)}$ which I think is kinda cool.

Nope, you didn't. Look at post 1. I am glad that you are finally backpeddaling on your silly claim.
Do you even know how to derive $$a_0=a\gamma^3$$?

In fact, I can make the further observation that as far as I can tell, all vertical and horizontal measurements (with the trivial exception of velocity) made by a stationary local observer in the Schwarzschild metric and measurements made by an observer at infinity, transform in the same way as longitudinal and transverse measurements respectively in SR, if you use $\gamma=1/\sqrt{(1-(v/c)^2)}$ in SR and $\gamma=1/\sqrt{(1-2M/r)}$ in the Schwarzschild metric. If this turns out to be true, it makes a huge simplification in understanding the Schwarzschild metric.

Physics and numerology are different disciplines. What you are trying above is numerology.

The equations posted in #1 are supported by:

The calculations of Prof.Brown on his mathpages website.

...that contain an embarassing error

The calculations quoted in the Hartle textbook. (https://www.physicsforums.com/showpost.php?p=2711553&postcount=10")

I sincerely doubt Hartle ever wrote such nonsense as $$a_0=a\gamma^3$$

The calculations by Dalespam using four vectors. (Just awesome. See #155)

The difference between I, Dalespam on one side and you, on the other side is that we can derive the results. You can't.

and I have even shown they are they are compatible with the Lagrangian equations of motion introduced by you.

You don't even know what Lagrangian mechanics is. You don't even want to learn.

 

[Al68]

Well I don't think you will get any sense by asking Starthaus. He will probably give you the the two postulates of relativity and leave the rest as "an exercise".

Why would he do that? I was only asking for clarification about his post, I never asked to be his student.

I still don't get the disagreement about the relationship between coordinate acceleration and proper acceleration. In the scenario you referred to, the relative velocity is zero, which results in a pretty simple and obvious relationship. What am I missing?

 

[yuiop]

Do you even know how to derive $$a_0=a\gamma^3$$?

If you mean in SR, then the answer is yes.

I sincerely doubt Hartle ever wrote such nonsense as $$a_0=a\gamma^3$$

If Hartle gives:

$$a_0 = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}}$$

and:

$$a= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)$$

(In agreement with equations in #1) and if we define:

$$\gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}$$

then I am sure that even you can see that it logically follows that:

$$a_0=a\gamma^3$$

It is just simple algebra.

The difference between I, Dalespam on one side and you, on the other side is that we can derive the results. You can't.

But Dalespam has derived the correct coordinate acceleration and it agrees with my conclusion and disagrees with yours.

You have not yet correctly derived anything in this thread. Your derivation of the proper acceleration is flawed because it has the fatal error of being derived from an incorrect equation for coordinate acceleration.

It also has the problem of being derived from a strong field aproximation, which is dangerous in the hands of someone who can not clearly quantify where the aproximation breaks down (someone like you).

The weak field aproximation is OK when used in the correct context, because the error introduced by the aproximation can be quantified precisely.

 

[starthaus]

If you mean in SR, then the answer is yes.

Based on you repeated use of numerology, I really doubt you can derive it.

If Hartle gives:

$$a_0 = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}}$$

and:

$$a= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)$$

(In agreement with equations in #1) and if we define:

$$\gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}$$

then I am sure that even you can see that it logically follows that:

$$a_0=a\gamma^3$$

It is just simple algebra.

It is just a coincidence. You used it in your posting results by hand in post 1. This is one of the reasons post 1 is just junk.

You have not yet correctly derived anything in this thread.

Coming from someone who didn't derive anything this is rich.
Coming from you who took forever to derive the proper acceleration after I held your hand at each step, this is super-rich.

Your derivation of the proper acceleration is flawed because it has the fatal error of being derived from an incorrect equation for coordinate acceleration.

Nope. It derived through the proper application of gradients. Actually, it is exactly the derivation used by Rindler. I even gave you the exact reference several times. (11.15).

It also has the problem of being derived from a strong field aproximation, which is dangerous in the hands of someone who can not clearly quantify where the aproximation breaks down (someone like you).

Talk with Rindler LOL

 

[yuiop]

I still don't get the disagreement about the relationship between coordinate acceleration and proper acceleration. In the scenario you referred to, the relative velocity is zero, which results in a pretty simple and obvious relationship. What am I missing?

We are not talking about relativistic effects due to relative velocity, but due to relative graviational potential.

The proper acceleration is measured by a stationary local observer and the coordinate acceleration is measured by an observer at infinity. The relationship between what they measure is not one to one.

 

[starthaus]

OK, but the truth is that I am an expert in Mathematica , not 4-vectors. I will use the metric, Christoffel symbols, etc. given https://www.physicsforums.com/showpost.php?p=2712746&postcount=38". The worldline of a particle moving only in the radial direction in these coordinates is given by:
$$\mathbf X = (t,r,0,0)$$

Where r is now a function of t.

Then we can derive the four-velocity as follows:
$$\mathbf U = \frac{d \mathbf X}{d \tau} = c \frac{d \mathbf X}{ds} = c \frac{d \mathbf X}{dt} \frac{dt}{ds} = \left(\frac{c}{\sqrt{-\frac{c^2 R}{r}+c^2+\frac{r \left(r'\right)^2}{R-r}}},\frac{c r'}{\sqrt{-\frac{c^2 R}{r}+c^2+\frac{r \left(r'\right)^2}{R-r}}},0,0\right)$$

Where the ' denotes differentiation wrt the coordinate t.

And we can verify that the norm of the four-velocity is equal to:
$$||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = c^2$$

Now it gets ugly. We can derive the four-acceleration as follows:
$$A^{\mu}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda} =$$
$$\left(\frac{c^2 r' \left(c^2 R (R-r)^2+2 r^3 (r-R) r''-3 r^2 R \left(r'\right)^2\right)}{2 \left(c^2 (R-r)^2-r^2 \left(r'\right)^2\right)^2},\frac{c^4 (R-r)^2 \left(c^2 R (R-r)^2+2 r^3 (r-R) r''-3 r^2 R \left(r'\right)^2\right)}{2 \left(c^2 r (R-r)^2-r^3 \left(r'\right)^2\right)^2},0,0\right)$$

So, we can set all of these equal to 0 to get the geodesic equation describing an object in free-fall along a radial trajectory. Doing so and solving for r'' gives
$$r''=\frac{c^2 R (R-r)^2-3 r^2 R \left(r'\right)^2}{2 r^3 (R-r)}$$

For a momentarily stationary particle r'=0 so
$$r''=\frac{GM}{r^2}\left(\frac{2GM}{c^2r}-1 \right)$$

where I have substituted back in for R.

Note that this is the second derivative of the coordinate r wrt the coordinate t, so take it for what it is worth. I suspect this differs from Rindler because he is calculating something other than r'' as defined here.

This is interesting, I got tired of waiting for kev to ever learn how to use the Euler-Lagrange formalism so I wrote the derivation for both proper and coordinate acceleration. I attached to the blog. Note that proper acceleration in my derivation is defined as $$\frac{d^2r}{ds^2}$$. You can easily get $$\frac{d^2r}{d\tau^2}$$ using covariant derivatives since the method derives the coordinate speed trivially. The coordinate acceleration is parametrized by the parameter k.

 

[yuiop]

Nope. It derived through the proper application of gradients. Actually, it is exactly the derivation used by Rindler. I even gave you the exact reference several times. (11.15).

I have looked at the Rindler book http://books.google.com/books?id=Mu...a&dq=rindler 9.6&pg=PA183#v=onepage&q&f=false Page 230 Section 11.15 and I see that although he gives the correct proper acceleration (the one no one is disputing) he does not actually give an equation for the coordinate acceleration. The coordinate acceleration you give, is not due to Rindler but is is your own botched extrapolation of what Rindler says, pretty much as I suspected all along.

 

[starthaus]

I have looked at the Rindler book http://books.google.com/books?id=Mu...a&dq=rindler 9.6&pg=PA183#v=onepage&q&f=false Page 230 Section 11.15 and I see that although he gives the correct proper acceleration

Wow! Rindler meets with your approval. Talking about grandomania!

(the one no one is disputing) he does not actually give an equation for the coordinate acceleration.

He doesn't deal with fools.

The coordinate acceleration you give, is not due to Rindler but is is your own botched extrapolation of what Rindler says, pretty much as I suspected all along.

So, you are still unable to figure out the gradient with respect to the radial coordinate , eh? So, when Rindler explains to anybody that has basic grasp of physics and calculus that $$\vec{a}=-grad (\Phi)$$ you still struggle to figure out what that means, even after you looked up the wiki definition? Way to go, kev!

 

[Al68]

We are not talking about relativistic effects due to relative velocity, but due to relative graviational potential.

The proper acceleration is measured by a stationary local observer and the coordinate acceleration is measured by an observer at infinity. The relationship between what they measure is not one to one.

So, essentially, you're referring to the coordinate acceleration of a free falling particle relative to two different stationary observers, one local and one at infinity, and the relationship between those two different values for coordinate acceleration of the free falling particle?

I know that's a cumbersome way of stating it, but it also seems more descriptive of the two quantities being compared, given the context. And it would seem to also imply a simple and obvious relationship between those quantities, the relationship you already provided. That's why the disagreement seems so bizarre.

 

[yuiop]

So, essentially, you're referring to the coordinate acceleration of a free falling particle relative to two different stationary observers, one local and one at infinity, and the relationship between those two different values for coordinate acceleration of the free falling particle?

I know that's a cumbersome way of stating it, but it also seems more descriptive of the two quantities being compared, if I got it right.

Yes. I think that is it in a nutshell. #1 is about the case of a particle with very low velocity so that relativistic effects due to velocity are insignificant. However, if you look at the equations that include the k constant, the appropriate value of k can be used that also takes the velocity into account. The value of the k constant is set to $$sqrt{(1-2M/r_a)}$$ where $$r_a$$ is the height of the apogee (so k=1 if the particle is dropped from infinity).

P.S. I bet someone, somewhere grits their teeth every time I use that inifity word. It makes me feel uncomfortable too sometimes lol.

 

[yuiop]

So, you are still unable to figure out the gradient with respect to the radial coordinate , eh? So, when Rindler explains to anybody that has basic grasp of physics and calculus that $$\vec{a}=-grad (\Phi)$$ you still struggle to figure out what that means, even after you looked up the wiki definition? Way to go, kev!

I have no trouble at all with understanding that acceleration is the gradient of the potential in Newtonian physics and it works for proper acceleration in Schwarzschild coordinates too. The problem is how it should be treated under coordinate transformations and it clear that the value you got from carrying out the transformation is wrong.

I have just looked at your derivation in your blog and I see that you (finally) get the correct the correct equation for coordinate acceleration i.e:

$$a= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)$$

which shows that the coordinate acceleration you obtained from the potential gradient method:

$$a= \frac{GM/r^2}{(1-2GM/(rc^2))}$$

is wrong. Clearly you too are struggling with the gradient concept too.

 

[starthaus]

I have no trouble at all with understanding that acceleration is the gradient of the potential in Newtonian physics and it works for proper acceleration in Schwarzschild coordinates too. The problem is how it should be treated under coordinate transformations and it clear that the value you got from carrying out the transformation is wrong.

I didn't "carry any coordinate transformation". There is no such animal in GR.