Gravitational force and acceleration in General Relativity.

[starthaus]

Good to see you have finally come to the conclusion that the equation I gave in #1 for proper acceleration (based mainly on information from the mathpages website) is correct in the context that it was given in. The claim you started in #2 that the equations I gave in #1 are wrong, has been shown by the vast majority of posters in this thread (i.e. everyone but you), to be invalid.

Nope. You got the proper acceleration by using the hack $$a_0=a\gamma^3$$

You have expressed an interest to extend the equations beyond the context of #1, to the more general situation.

This is precisely what the method based on variational mechanics does. Derivation from scratch, not selective "cut and paste" from the mathpages+hacks.

 

[starthaus]

I am, aren't you?]

Edit: All typos and stuff were edited.

Nope, you still have quite a few. Some are typos but the others are serious conceptual errors.

You see, I intentionally can make an error

Intentionally? You still have some more "intentional" errors. You may want to continue copying from post 38, it is all correct there.

 

[yuiop]

Your equation:

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

does not work, because the proper acceleration of a free falling particle is zero. (Attach an accelerometer to a free falling particle and see what it reads.)

...
The math used for deriving the proper and coordinate acceleration in the radial field does not support your claim above. If you think otherwise try deriving the formula for the proper acceleration of a particle dropped from $$r=r_0$$ in a variable field as a function of the radial coordinate $$r$$. Show that your obtain $$a_0=0$$ for any $$r$$

Others have already correctly stated that the proper acceleration of a particle in freefall is zero by definition, but I accept your challenge for the more general case. This is not a formal derivation or proof. Just my best shot based on my physical understanding of the situation and my interpretation of the equations given by references like mathpages. (i.e. using selective "cut and paste" from the mathpages+hacks.)

For easy visualisation, consider the case of a small rocket descending into a gravitational well, with its thrust (Fo) directed in the opposite direction to the local gravitational acceleration $$(a'_g)$$ so that the descent acceleration of the rocket at r, as measured by a stationary observer at r, is $$d^2r'/dt'^2$$ using his local clocks and rulers. (All primed variables indicate measurements made by a stationary local observer at r in the Schwarzschild coordinates). I can now say:

$$F_o = \left( \frac{d^2r'}{dt' ^2} - a'_g \right) m_o$$

where Fo is the proper force directed upwards and m_o is the rest mass of the rocket .

Dividing both sides by m_o gives the proper acceleration $$a_o$$ of the rocket as:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g$$

where positive acceleration is defined here as upwards and negative acceleration means downwards.

A combination of http://www.mathpages.com/rr/s6-07/6-07.htm" gives the velocity dependent Schwarzschild coordinate gravitational acceleration $$a_g$$ as:

$$a_g = - \frac{M}{r^2}(1-2M/r)\left(1-\frac{3(dr/dt)^2}{(1-2M/r)^2}\right)$$

Assuming the following relationships between the local vertical measurements (dr’, dt’, a’) made by a stationary observer at r and the coordinate measurements (dr, dt, a) made by a Schwarzschild observer at infinity using the gravitational boost factor $$\gamma_g = 1/\sqrt{(1-2M/r)}$$:

$$dr' = dr \; \gamma_g$$

$$dt' = dt \; \gamma_g^{-1}$$

$$dr'/dt' = dr/dt \; \gamma_g^{2}$$

$$a' = a \; \gamma_g^{3}$$

it is easy to obtain:

$$a'_g = - \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

where dr’/dt’ is the descent velocity of the rocket as it passes radial coordinate r as measured by a stationary observer located at r.

The more general proper acceleration can now be expressed in the form:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{d^2r'}{dt' ^2} + \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

When the rocket is holding position at r by using its thrust to hover at r, dr’/dt’ = 0 and $$d^2r'/dt' ^2 =0$$ and its proper acceleration is:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = 0 + \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right) = \frac{M}{r^2}\frac{1}{\sqrt{1-2M/r}}$$

which is what most people here expect.

In the specific case of freefall only, dr’/dt’ can be calculated using:

$$\frac{dr'}{dt'} = {\frac{\sqrt{2M (1/r-1/R)}}{\sqrt{1-2M/r}}$$

where R is a constant parameter of the trajectory path representing the height of the apogee and the following relationship is true:

$$\frac{d^2r'}{dt' ^2} = - \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

When the rocket turns off its engines and falls freely in a vacuum, the proper acceleration of the rocket is:

$$a_o = \frac{d^2{r}'}{d{t}' ^2} - a'_g = \frac{d^2{r}'}{d{t}' ^2} + \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right) = 0$$

for all R>2m and r>2m.

 

[yuiop]

Nope. You got the proper acceleration by using the hack $$a_0=a\gamma^3$$

You are missing the point. I am asking you to establish a "baseline of agreement" by conceding that the equations in #1 and the equations given in mathpages are correct, even if you do not agree with my method of using mathpages equations plus my hacks to obtain them.

 

[starthaus]

Assuming the following relationships between the local vertical measurements (dr’, dt’, a’) made by a stationary observer at r and the coordinate measurements (dr, dt, a) made by a Schwarzschild observer at infinity using the gravitational boost factor $$\gamma_g = 1/\sqrt{(1-2M/r)}$$:

$$dr' = dr \; \gamma_g$$

$$dt' = dt \; \gamma_g^{-1}$$

$$dr'/dt' = dr/dt \; \gamma_g^{2}$$

$$a' = a \; \gamma_g^{3}$$

$$a' = a \; \gamma_g^{3}$$? Again?
Try deriving this and you'll see that it isn't correct.

 

[Dale]

No, I was asking this question that if I want to get from

$$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma ^{r}_{tt}U^{t}U^{t}=\frac{c^2 (r-R) R}{2 r^3}$$

to

$$\mathbf A = \left(0,\frac{c^2 R}{2 r^2},0,0\right) = \left(0,\frac{G M}{r^2},0,0\right)$$

it is required to have the term $$\frac{-c^2R^2}{2 r^3}$$ disappear. How does it occur?

AB

Sorry, this is my fault, I wrote the expression down incompletely. This should have been:

$$\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma ^{r}_{tt}U^{t}U^{t}=\frac{c^2 (r-R) R}{2 r^3} \left(1-\frac{R}{r}\right)^{-1/2} \left(1-\frac{R}{r}\right)^{-1/2}$$

The previous expression was only the Christoffel symbol, and did not include the contributions from U.

 

[yuiop]

$$a' = a \; \gamma_g^{3}$$? Again?
Try deriving this and you'll see that it isn't correct.

... and yet unlike you, I obtain the correct results for proper acceleration by using that assumption.

Defend your assertion that it wrong, by showing how you would show:

$$a' \neq a \; \gamma_g^{3}$$

(eg give a counter example)

 

[starthaus]

... and yet unlike you, I obtain the correct results for proper acceleration by using that assumption.

Defend your assertion that it wrong, by showing how would derive:

$$a' \neq a \; \gamma_g^{3}$$

It doesn't work this way, kev. You used the formula, show how you derived it. You showed all the intermediate steps except the last one. Show the last step and I'll show you were your error is.

 

[yuiop]

... and yet unlike you, I obtain the correct results for proper acceleration by using that assumption.

Defend your assertion that it wrong, by showing how (you) would derive:

$$a' \neq a \; \gamma_g^{3}$$

for the colinear vertical (radial) case.

I have already declared that I no mathematician and that my methods are informal. It would take a better mathematician/ relativist like Dalespam, George, rofle2, DrGreg, JesseM or Altabeh to do a more formal proof and determine if I am on the right track or not.

 

[Altabeh]

1. So, why are you wasting time?

Because if no one prevents you from making nonsense, I have to do it, don't I?

2. I must have missed your derivation, where is it?

Takes time to write. It'll come tonight! You’re not the only one who is familiar with the variational methods in Physicsforums, are you!? :-)

3. You editted your post from inadvertently trying to correct my definition of proper "speed" to trying to correct my definition of proper "acceleration". Why are you moving the goalposts? Do you think I missed that?

Nupe, I just meant to write "acceleration" in place of "velocity" in the very first edition of my post but intentionally changed it to a wrong thing to only know if you were going to point it out ironically to me which you unfortunately did! It has nothing to do with the proper velocity in case you see I wrote the proper 4-acceleration’s components. So I’m not going to show any respect to you here and in fact I’ll throw at your face whatever mistake you make like the one you’ve been making us play with inadvertently for weeks now. Yet this whole test that I put you to shows whose tone is condescending. All you’re trying now to do doesn't change your false claim about what proper acceleration is.

4. Your new definition (of proper acceleration) is still wrong, though you copied from post 38.[/QUOTE]

I’m not into copy-paste stuff like you. The definition of proper acceleration can be found in http://en.wikipedia.org/wiki/Proper_acceleration" page.

 

[starthaus]

4. Your new definition (of proper acceleration) is still wrong, though you copied from post 38.

I’m not into copy-paste stuff like you. The definition of proper acceleration can be found in http://en.wikipedia.org/wiki/Proper_acceleration" page.

You still have it wrong. Not one mistake, several.

 

[Altabeh]

your moving the goalposts between posts?
4. BTW , your new definition (of proper acceleration) is still wrong, though you copied from post 38.

Don't play with words. That is not the general definition. Here we are just dealing with the radial velocity of a particle near a gravitating body and that definition is to be assigned to this.

All you do is to make people busy on some monkey job you take along yourself to these forums and just give us some garbage and then say that is a mathematically backed physics while still playing down other's arguments over your nonsense. As an example, you repeatedly referred to me copying the definition of proper acceleration from the post 38:

Intentionally? You still have some more "intentional" errors. You may want to continue copying from post 38, it is all correct there.

BTW , your new definition (of proper acceleration) is still wrong, though you copied from post 38.

I simply took my definition from http://en.wikipedia.org/wiki/Proper_acceleration#In_curved_spacetime". If this is called copy-paste, people don't use Einstein's field equations because then will a man with a long torch in his hand come to you while whispering to your ears the phrase All rights reserved. You're so funny, boy!

Nope, you still have quite a few. Some are typos but the others are serious conceptual errors.

AS a way to escape from being criticized about your nonsense, I agree with the second part of your theories "Altabeh has conceptual errors". So I'm waiting to know what they are!

Well, it wasn't that obvious to me when I started developing the alternate approach since I was trying to avoid covariant derivatives. Otherwise the approach would not be original, would it? Besides, I was interested in getting an alternate definition for proper acceleration. To date, none of the definitions (see my exchange with Rolfe2) has been satisfactory. This may mean that the variational mechanics method is good for deriving coordinate acceleration (something that the covariant derivative method can't do) and the covariant derivative method is good at deriving the proper acceleration for the hovering particle only. So, we need both methods.

Okay, I see that you finally admitted to being wrong about what is called a "mistake". I doubt that you even know what the variational methods of general relativity are all about; the essence of those methods in GR is found to be meaningful when one knows that they all have to be consistent with covariant methods. Talking about one being good and the other bad for one special purpose gives me this feeling that you've taken shelter in logics in order to hide your BIG mistake. After going through 21 pages, I have to say "what a mess "!

AB

 

[starthaus]

for the colinear vertical (radial) case.

I have already declared that I no mathematician and that my methods are informal.

There is no such thing as "informal" in math. If you redo the last step in your derivation, you will find out the error, it is quite obvious.

 

[starthaus]

Don't play with words.

I don't. You can't even get the definition right after reading the posts that make use of it.
Your errors are staring you in the face.

 

[Altabeh]

I don't. You can't even get the definition right after reading the posts that make use of it.

Play with words again to maybe have a hope to survive in the zone of physics. What's next? Einstein had mistakes or what?! Lool!

AB

 

[Altabeh]

There is no such thing as "informal" in math. If you redo the last step in your derivation, you will find out the error, it is quite obvious.

This shows how your knowledge of mathematics is limited as well. Not only do we have informal methods in mathematics that are mostly overlooked by mathematical communities, but also there has been a traditional form of mathematics called "Informal mathematics" or "naive mathematics". Consult Google for more info.

AB

 

[starthaus]

for the colinear vertical (radial) case.

Doesn't work for $$\gamma=1/\sqrt{1-2m/r}$$
Only works for $$\gamma=1/\sqrt{1-(v/c)^2}$$

I must have told you this a dozen times. Try showing your last step in your derivation and you'll see the mistake immediately..

 

[Mentz114]

I was pondering this thread Correction term to Newtons gravitation law. when it occurred to me that we should be be able to answer simple questions like "what does General Relativity predict that the weight of a 1kg mass on the surface of a very massive gravitational body to be?" or "what is the initial acceleration of 1kg mass when released from a short distance above a very massive gravitational body?".

I believe these questions are easily answered if you use the method of frames to find the acceleration of a stationary observer. It is worked out here

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

 

[Geigerclick]

Starthaus, Kev, this seems like the time to disengage. I'm not saying you should agree to disagree, just that this has become one hell of a pissing match. This has strayed so far from the previous meaningful discussion that it is annoying as a newer user reading this thread, then hitting what amounts to a clot upstream.

 

[Altabeh]

Others have already correctly stated that the proper acceleration of a particle in freefall is zero by definition, but I accept your challenge for the more general case. This is not a formal derivation or proof. Just my best shot based on my physical understanding of the situation and my interpretation of the equations given by references like mathpages. (i.e. using selective "cut and paste" from the mathpages+hacks.)

For easy visualisation, consider the case of a small rocket descending into a gravitational well, with its thrust (Fo) directed in the opposite direction to the local gravitational acceleration $$(a'_g)$$ so that the descent acceleration of the rocket at r, as measured by a stationary observer at r, is $$d^2r'/dt'^2$$ using his local clocks and rulers. (All primed variables indicate measurements made by a stationary local observer at r in the Schwarzschild coordinates). I can now say:

$$F_o = \left( \frac{d^2r'}{dt' ^2} - a'_g \right) m_o$$

where Fo is the proper force directed upwards and m_o is the rest mass of the rocket .

Dividing both sides by m_o gives the proper acceleration $$a_o$$ of the rocket as:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g$$

where positive acceleration is defined here as upwards and negative acceleration means downwards.

A combination of http://www.mathpages.com/rr/s6-07/6-07.htm" gives the velocity dependent Schwarzschild coordinate gravitational acceleration $$a_g$$ as:

$$a_g = - \frac{M}{r^2}(1-2M/r)\left(1-\frac{3(dr/dt)^2}{(1-2M/r)^2}\right)$$

Assuming the following relationships between the local vertical measurements (dr’, dt’, a’) made by a stationary observer at r and the coordinate measurements (dr, dt, a) made by a Schwarzschild observer at infinity using the gravitational boost factor $$\gamma_g = 1/\sqrt{(1-2M/r)}$$:

$$dr' = dr \; \gamma_g$$

$$dt' = dt \; \gamma_g^{-1}$$

$$dr'/dt' = dr/dt \; \gamma_g^{2}$$

$$a' = a \; \gamma_g^{3}$$

it is easy to obtain:

$$a'_g = - \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

where dr’/dt’ is the descent velocity of the rocket as it passes radial coordinate r as measured by a stationary observer located at r.

The more general proper acceleration can now be expressed in the form:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{d^2r'}{dt' ^2} + \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

When the rocket is holding position at r by using its thrust to hover at r, dr’/dt’ = 0 and $$d^2r'/dt' ^2 =0$$ and its proper acceleration is:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = 0 + \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right) = \frac{M}{r^2}\frac{1}{\sqrt{1-2M/r}}$$

which is what most people here expect.

In the specific case of freefall only, dr’/dt’ can be calculated using:

$$\frac{dr'}{dt'} = {\frac{\sqrt{2M (1/r-1/R)}}{\sqrt{1-2M/r}}$$

where R is a constant parameter of the trajectory path representing the height of the apogee and the following relationship is true:

$$\frac{d^2r'}{dt' ^2} = - \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

When the rocket turns off its engines and falls freely in a vacuum, the proper acceleration of the rocket is:

$$a_o = \frac{d^2{r}'}{d{t}' ^2} - a'_g = \frac{d^2{r}'}{d{t}' ^2} + \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right) = 0$$

for all R>2m and r>2m.

All calculations I see above are seamless from a mathematical point of view except that I don't understand the how you manage to get from $$\frac{d^2r'}{dt' ^2} = - \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$ to the last equation. Would you mind explaining it to me?

In fact I don't expect to see anything else but zero in front of the proper acceleration of that rocket in free-fall!

AB

 

[starthaus]

Starthaus, Kev, this seems like the time to disengage. I'm not saying you should agree to disagree, just that this has become one hell of a pissing match. This has strayed so far from the previous meaningful discussion that it is annoying as a newer user reading this thread, then hitting what amounts to a clot upstream.

You are right.

 

[Geigerclick]

You are right.

Thank you, I'm glad you were not insulted, I meant for that to be reasonable, not intrusive. Your discussion is above my pay grade in either case, so I can only comment on the structure of it, not the content at this point.

 

[yuiop]

All calculations I see above are seamless from a mathematical point of view ...

Thanks Altabeh. You are right that is correct mathematical but I have noticed that there is a minor problem with the physics and should taking the conversion from proper time $$(t_o)$$ of the moving particle to the local time (t') of the stationary observer at r into account. It so happens that in the particular case of a hovering particle and a particle at apogee the equation is still correct and the equations in post #1 are still correct. It also happens to give the correct result for when the particle is free falling $$a_o = 0$$ but inbetween the results are not accurate. In effect:

$$a' = a \; \gamma_g^{3}$$

is generally true even when the particle is not stationary at r, but:

$$a_o = a \; \gamma_g^{3}$$

is only true if the particle is stationary at r.

Other than this scaling problem, I believe the general idea I used in #318 is the correct way to o to obtain the general equation for proper acceleration. I will try and fix this problem with scaling factor and repost the general solution for when the particle has arbitrary velocity at a r.

... except that I don't understand the how you manage to get from $$\frac{d^2r'}{dt' ^2} = - \frac{M}{r^2}\left(\frac{1-3(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$ to the last equation. Would you mind explaining it to me?

In fact I don't expect to see anything else but zero in front of the proper acceleration of that rocket in free-fall!
AB

This is easy. If I may give a Newtonian analogy. Let us say the result force Fr of two opposing forces Fa and Fb acting on a particle is Fr = Fa-Fb, then when Fa=Fb, then Fr=0 or I can write Fr = Fa - Fb = 0. In effect writing (Fr = Fa - Fb and (Fa = Fb) ) is just a long winded way of writing Fr = 0, but it shows how the zero result is obtained.

In the gravitational acceleration case, proper acceleration can take on any value betwen plus and minus infinity and the case when proper acceleration equals zero is only true when the particle is freefaling making a special case of a more general equation.

Hope that makes some sort of sense.

 

[starthaus]

$$dr' = dr \; \gamma_g$$

$$dt' = dt \; \gamma_g^{-1}$$

$$dr'/dt' = dr/dt \; \gamma_g^{2}$$

OK

$$a' = a \; \gamma_g^{3}$$

No.

$$a'=\frac{d^2r'}{dt'^2}$$

$$a=\frac{d^2r}{dt^2}$$

$$\gamma_g=1/\sqrt{1-2M/r}$$

so you cannot have

$$a' = a \; \gamma_g^{3}$$

Do the calculations and you'll find out why. Here, I'll start it for you:

$$a'=\frac{d^2r'}{dt'^2}=\frac{d}{dt'}(\frac{dr'}{dt'})=\frac{d}{dt}(\frac{dr'}{dt'})\frac{dt}{dt'}=...$$

 

[yuiop]

so you cannot have

$$a' = a \; \gamma_g^{3}$$

Do the calculations and you'll find out why. Here, I'll start it for you:

$$a'=\frac{d^2r'}{dt'^2}=\frac{d}{dt'}(\frac{dr'}{dt'})=\frac{d}{dt}(\frac{dr'}{dt'})\frac{dt}{dt'}=...$$

I have already stated I see there is a problem with my more general solution in #318. You are the calculus expert, so why not make this a cooperative joint effort and post what you know, in the interests of trying to bring this thread to a rapid conclusion and preventing Geigerclick from having a nervous breakdown

 

[starthaus]

OK

No.

$$a'=\frac{d^2r'}{dt'^2}$$

$$a=\frac{d^2r}{dt^2}$$

$$\gamma_g=1/\sqrt{1-2M/r}$$

so you cannot have

$$a' = a \; \gamma_g^{3}$$

Do the calculations and you'll find out why. Here, I'll start it for you:

$$a'=\frac{d^2r'}{dt'^2}=\frac{d}{dt'}(\frac{dr'}{dt'})=\frac{d}{dt}(\frac{dr'}{dt'})\frac{dt}{dt'}=...$$

$$a'=\frac{d^2r'}{dt'^2}=\frac{d}{dt'}(\frac{dr'}{dt'})=\frac{d}{dt}(\frac{dr'}{dt'})\frac{dt}{dt'}= \gamma_g\frac{d}{dt}(\gamma_g^2\frac{dr}{dt})=\gamma_g^3\frac{d^2r}{dt^2}+2(\gamma_g\frac{dr}{dt})^2\frac{d\gamma_g}{dr}$$

Contrary to your beliefs, basic calculus shows that $$a'$$ is not equal to $$a\gamma^3$$. Can we close this thread now?

 

[Altabeh]

Thanks Altabeh. You are right that is correct mathematical but I have noticed that there is a minor problem with the physics and should taking the conversion from proper time $$(t_o)$$ of the moving particle to the local time (t') of the stationary observer at r into account. It so happens that in the particular case of a hovering particle and a particle at apogee the equation is still correct and the equations in post #1 are still correct. It also happens to give the correct result for when the particle is free falling $$a_o = 0$$ but inbetween the results are not accurate. In effect:

$$a' = a \; \gamma_g^{3}$$

is generally true even when the particle is not stationary at r, but:

$$a_o = a \; \gamma_g^{3}$$

is only true if the particle is stationary at r.

Other than this scaling problem, I believe the general idea I used in #318 is the correct way to o to obtain the general equation for proper acceleration. I will try and fix this problem with scaling factor and repost the general solution for when the particle has arbitrary velocity at a r.

This is easy. If I may give a Newtonian analogy. Let us say the result force Fr of two opposing forces Fa and Fb acting on a particle is Fr = Fa-Fb, then when Fa=Fb, then Fr=0 or I can write Fr = Fa - Fb = 0. In effect writing (Fr = Fa - Fb and (Fa = Fb) ) is just a long winded way of writing Fr = 0, but it shows how the zero result is obtained.

In the gravitational acceleration case, proper acceleration can take on any value betwen plus and minus infinity and the case when proper acceleration equals zero is only true when the particle is freefaling making a special case of a more general equation.

Hope that makes some sort of sense.

That is not that much easy. Actually I noticed that only at apogee the formula

$$a' = a \; \gamma_g^{3}$$

works. If you expand $$a'$$ in a differential form, you'll figure that

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

So we have to either abandon this approach or modify it in such a way that the proper acceleration of rocket in free-fall would be zero. I'm still thinking about it...

AB

 

[Geigerclick]

I just want to say, since I took the liberty of criticizing this discourse, that it seems to be back in line with what one would hope from intelligent people debating math and science. As a casual observer, thanks to all of you.

 

[yuiop]

That is not that much easy. Actually I noticed that only at apogee the formula

$$a' = a \; \gamma_g^{3}$$

works. If you expand $$a'$$ in a differential form, you'll figure that

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

So we have to either abandon this approach or modify it in such a way that the proper acceleration of rocket in free-fall would be zero. I'm still thinking about it...

AB

Hi Altabeh,

Could you show me some of the steps you took to obtain this result?

Could you also confirm that you are using the notation that I was using, i.e a' = dr'/dt' for the acceleration measured by a local observer that is stationary at r and $$a_o$$ for the proper acceleration of the test particle?

 

[yuiop]

$$a'=\frac{d^2r'}{dt'^2}=\frac{d}{dt'}(\frac{dr'}{dt'})=\frac{d}{dt}(\frac{dr'}{dt'})\frac{dt}{dt'}= \gamma_g\frac{d}{dt}(\gamma_g^2\frac{dr}{dt})=\gamma_g^3\frac{d^2r}{dt^2}+2(\gamma_g\frac{dr}{dt})^2\frac{d\gamma_g}{dr}$$

OK, thanks for that. It agrees with the expansion given by Altabeh although I am not sure if Altabeh just expanded on your differentiation. Anyway, I will expand on your result in a different way to obtain a result in terms of radial velocity.

$$a' = a\:\gamma_g^3+2\left(\gamma_g\frac{dr}{dt}\right)^2\frac{d\gamma_g}{dr} = a\:\gamma_g^3+2 \left(\gamma_g\frac{dr}{dt}\right)^2\left( \frac{-M }{c^2r^2}\right)\gamma_g^3 = a\:\gamma_g^3\: - \:\frac{2M}{r^2} \left(\frac{dr}{dt}\right)^2 \gamma_g^5 = a\:\gamma_g^3 -\frac{2M}{r^2} \left(\frac{dr'}{dt'}\right)^2 \gamma_g$$

Now we should be in a position to correct my previous derivation of the general expression for proper radial acceleration.
The following is a rehash of my first attempt in #318 (using units of G=c=1):

=================================================


For easy visualisation, consider the case of a small rocket descending into a gravitational well, with its thrust (Fo) directed in the opposite direction to the local gravitational acceleration $$(a'_g)$$ so that the descent acceleration of the rocket at r, as measured by a stationary observer at r, is $$d^2r'/dt'^2$$ using his local clocks and rulers. (All primed variables indicate measurements made by a stationary local observer at r in the Schwarzschild coordinates). I can now say:

$$F_o = \left( \frac{d^2r'}{dt' ^2} - a'_g \right) m_o$$

where Fo is the proper force directed upwards and m_o is the rest mass of the rocket .

Dividing both sides by m_o gives the proper acceleration $$a_o$$ of the rocket as:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g$$

where positive acceleration is defined here as upwards and negative acceleration means downwards.

A combination of http://www.mathpages.com/rr/s6-07/6-07.htm" gives the velocity dependent Schwarzschild coordinate gravitational acceleration $$a_g$$ as:

$$a_g = - \frac{M}{r^2}(1-2M/r)\left(1-\frac{3(dr/dt)^2}{(1-2M/r)^2}\right) = - \frac{M}{r^2}(1-2M/r)(1-3(dr'/dt')^2)$$

Assuming the following relationships between the local vertical measurements (dr’, dt’, a’) made by a stationary observer at r and the coordinate measurements (dr, dt, a) made by a Schwarzschild observer at infinity using the gravitational boost factor $\gamma_g = 1/\sqrt{(1-2M/r)}$:

$$dr' = dr \; \gamma_g$$

$$dt' = dt \; \gamma_g^{-1}$$

$$dr'/dt' = dr/dt \; \gamma_g^{2}$$

$$a' = a\: \gamma_g^3 -\frac{2M}{r^2} \left(\frac{dr'}{dt'}\right)^2 \gamma_g$$

it is fairly easy to obtain:

$$a'_g = - \frac{M}{r^2}\left(\frac{1-(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

where dr’/dt’ is the descent velocity of the rocket as it passes radial coordinate r, as measured by a stationary observer located at r.

The more general proper acceleration can now be expressed in the form:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{d^2r'}{dt' ^2} + \frac{M}{r^2}\left(\frac{1-(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

When the rocket is holding position at r by using its thrust to hover at r, dr’/dt’ = 0 and $$d^2r'/dt' ^2 =0$$ and its proper acceleration is:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{M}{r^2}\frac{1}{\sqrt{1-2M/r}}$$

which is the expected result.

When the rocket turns off its engines and falls freely in a vacuum, the proper acceleration of the rocket is:

$$a_o = \frac{d^2{r}'}{d{t}' ^2} - a'_g = 0$$

for all R>2m and r>2m.

That, I hope is now a correct result, but I can not find any references to check it against for the case where the particle is neither free falling nor stationary.

 

[Altabeh]

Hi Altabeh,

1- Could you show me some of the steps you took to obtain this result?

2- Could you also confirm that you are using the notation that I was using, i.e a' = dr'/dt' for the acceleration measured by a local observer that is stationary at r and $$a_o$$ for the proper acceleration of the test particle?

Hi and sorry for being late to reply to your post. I've had a really busy week.

(I just assigned a number to each of these questions so that I can answer them in order.) In the following I will, for simplicity, drop the index g of Lorentz factors.

1- First we expand $$a'$$ in the following form:

$$a'=\frac{d^2r'}{dt'^2}=\frac{d}{dt'} \frac{dr'}{dt'}=\frac{dt}{dt'}\frac{d}{dt}(\frac{dr'}{dt'})=\gamma\frac{d}{dt}(\frac{dr}{dt}\gamma^{2})=\frac{d^2r}{dt^2}\gamma^3+\gamma\frac{dr}{dt}\frac{d}{dt}(\gamma^{2}).$$

Now from the fact that the gravitational Lorentz factor is written by

$$\gamma=\frac{1}{\sqrt{1-2m/r(t)}},$$

one would get

$$\frac{d}{dt}(\gamma^2)=2\gamma\frac{d}{dt}(\gamma)=2\gamma\frac{-m}{r^2}\frac{1}{\frac{1}{\sqrt{1-2m/r(t)}}(1-2m/r)^2}\frac{dr}{dt}=-\frac{2m}{r^2}\frac{dr}{dt }\gamma^4.$$

Introducing this into the very first equation yields

$$a'=\gamma^3(\frac{d^2r}{dt^2}+\frac{-2m}{r^2}[\frac{dr}{dt}]^2\gamma^2).$$

Since the radial velocity of a rocket starting at $$r_0=r(t_0)$$ to fall with a zero initial velocity $$(dr/dt)_0=0$$ near a gravitating body is given by (c=1)

$$\frac{dr}{dt}=\frac{r-2m}{r}\sqrt{1-\frac{1-2m/r}{1-2m/r_0}},$$

therefore the penult equation must be

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

which is the equation from an early post by me.

2- By looking at the coordinate velocity I'm using above you can see that all measurements related to the coordinate pair $$(r,t)$$ are done by a Schwarzschild observer at infinity; and that an observer being locally stationary at r measures the coordinate pair $$(r',t')$$. About the proper acceleration, since I've not yet made use of any symbol for it, I'll use the same notation as yours.

AB

 

[starthaus]

OK, thanks for that.

Yeah, took only about 200 posts to convince you of your error.

When the rocket turns off its engines and falls freely in a vacuum, the proper acceleration of the rocket is:

$$a_o = \frac{d^2{r}'}{d{t}' ^2} - a'_g = 0$$

You were supposed to derive the above, not to put in the answer by hand. How do you derive it?

 

[Altabeh]

OK, thanks for that. It agrees with the expansion given by Altabeh although I am not sure if Altabeh just expanded on your differentiation. Anyway, I will expand on your result in a different way to obtain a result in terms of radial velocity.

$$a' = a\:\gamma_g^3+2\left(\gamma_g\frac{dr}{dt}\right)^2\frac{d\gamma_g}{dr} = a\:\gamma_g^3+2 \left(\gamma_g\frac{dr}{dt}\right)^2\left( \frac{-M }{c^2r^2}\right)\gamma_g^3 = a\:\gamma_g^3\: - \:\frac{2M}{r^2} \left(\frac{dr}{dt}\right)^2 \gamma_g^5 = a\:\gamma_g^3 -\frac{2M}{r^2} \left(\frac{dr'}{dt'}\right)^2 \gamma_g$$

Now we should be in a position to correct my previous derivation of the general expression for proper radial acceleration.
The following is a rehash of my first attempt in #318 (using units of G=c=1):

=================================================


For easy visualisation, consider the case of a small rocket descending into a gravitational well, with its thrust (Fo) directed in the opposite direction to the local gravitational acceleration $$(a'_g)$$ so that the descent acceleration of the rocket at r, as measured by a stationary observer at r, is $$d^2r'/dt'^2$$ using his local clocks and rulers. (All primed variables indicate measurements made by a stationary local observer at r in the Schwarzschild coordinates). I can now say:

$$F_o = \left( \frac{d^2r'}{dt' ^2} - a'_g \right) m_o$$

where Fo is the proper force directed upwards and m_o is the rest mass of the rocket .

Dividing both sides by m_o gives the proper acceleration $$a_o$$ of the rocket as:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g$$

where positive acceleration is defined here as upwards and negative acceleration means downwards.

A combination of http://www.mathpages.com/rr/s6-07/6-07.htm" gives the velocity dependent Schwarzschild coordinate gravitational acceleration $$a_g$$ as:

$$a_g = - \frac{M}{r^2}(1-2M/r)\left(1-\frac{3(dr/dt)^2}{(1-2M/r)^2}\right) = - \frac{M}{r^2}(1-2M/r)(1-3(dr'/dt')^2)$$

Assuming the following relationships between the local vertical measurements (dr’, dt’, a’) made by a stationary observer at r and the coordinate measurements (dr, dt, a) made by a Schwarzschild observer at infinity using the gravitational boost factor $\gamma_g = 1/\sqrt{(1-2M/r)}$:

$$dr' = dr \; \gamma_g$$

$$dt' = dt \; \gamma_g^{-1}$$

$$dr'/dt' = dr/dt \; \gamma_g^{2}$$

$$a' = a\: \gamma_g^3 -\frac{2M}{r^2} \left(\frac{dr'}{dt'}\right)^2 \gamma_g$$

it is fairly easy to obtain:

$$a'_g = - \frac{M}{r^2}\left(\frac{1-(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

where dr’/dt’ is the descent velocity of the rocket as it passes radial coordinate r, as measured by a stationary observer located at r.

The more general proper acceleration can now be expressed in the form:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{d^2r'}{dt' ^2} + \frac{M}{r^2}\left(\frac{1-(dr'/dt')^2}{\sqrt{1-2M/r}}\right)$$

When the rocket is holding position at r by using its thrust to hover at r, dr’/dt’ = 0 and $$d^2r'/dt' ^2 =0$$ and its proper acceleration is:

$$a_o = \frac{d^2r'}{dt' ^2} - a'_g = \frac{M}{r^2}\frac{1}{\sqrt{1-2M/r}}$$

which is the expected result.

When the rocket turns off its engines and falls freely in a vacuum, the proper acceleration of the rocket is:

$$a_o = \frac{d^2{r}'}{d{t}' ^2} - a'_g = 0$$

for all R>2m and r>2m.

That, I hope is now a correct result, but I can not find any references to check it against for the case where the particle is neither free falling nor stationary.

Sounds flawless. If the particle is neither freely falling nor stationary, then there must be an external force other than gravitational applied to it to be able to move that way which is the case when the right-hand side of the geodesic equations is non-zero. Until we don't know what this force is, there is not going to be any known formula for the related equations of motion.

By the way, I'm preparing a similar proof for the coordinate radial acceleration given in one of the starthaus's blog entries.

AB

 

[Altabeh]

You were supposed to derive the above, not to put in the answer by hand. How do you derive it?

No derivation is needed unless knowing the fact that the acceleration measured by the stationary observer at r in free-fall is just locally the acceleration caused by gravitational force.

Yeah, took only about 200 posts to convince you of your error.

Now what about you to correct this big error after spending a couple of weeks? LOL

AB

 

[Geigerclick]

No derivation is needed unless knowing the fact that the acceleration measured by the stationary observer at r in free-fall is just locally the acceleration caused by gravitational force.



Now what about you to correct this big error after spending a couple of weeks? LOL

AB

I think Altabeh is the winner in this game.

I realize from reading this thread, how infuriating this has been for kev and starthaus, but I for one learned a great watching this back and forth. 200 interesting posts from the standpoint of this hobbyist. :)