Gravitational force and acceleration in General Relativity.

[starthaus]

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:

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

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

LOL. You are really funny. Too bad that you aren't any good with reading and comprehension.

 

[yuiop]

This is a summary of the equations in your gravitational acceleration paper in your blog:

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

$$k = \alpha \frac{dt}{ds}\;\;(4)$$

Coordinate acceleration:

$$\frac{d^2r}{dt^2} = \frac mr^2 \alpha (2-\frac{3\alpha}{k^2}) \;\; (7)$$

Proper acceleration:

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2}\; \; (8)$$

Equation 7 is exactly the same as the equation I posted in #135 for coordinate acceleration, so finally we are in agreement.

Equation 8 is agreement with the proper acceleration given by #1/Dr Brown/Dalespam/Rindler etc, if we take into account that you are using dr (coordinate distance) rather than the normal local/proper distance.

OK, now at the apogee when the particle is nearly stationary,

$$ds/dt = \sqrt{\alpha}$$

and so the value of the k constant is:

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

Substituting this constant in your equation (7) for coordinate acceleration gives:

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

which is in exact agreement with the equation I gave for coordinate acceleration in #1.


Can we now agree that the equations I gave in #1 are correct and have now been derived in a number of different ways by different people including yourself?

 

[yuiop]

LOL. You are really funny. Too bad that you aren't any good with reading and comprehension.

That was a typo. You replied while I was editing it. The preview for tex does not work on my PC and I have to edit it live for typos after posting. It should have read:

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:

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

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

I have to question your comprehension. We have spent 178 odd posts with you trying to show that my post in #1 was wrong when in fact you have been wrong all along.

Post #1 still stands as correct.

 

[starthaus]

Wrong again. As I said a few posts before, if you start a derivation from an approximation, you have failure built in from the start, if you try to come to any general conclusions outside the scope of the approximation. This is the (long awaited) correct way to derive the coordinate acceleration:

(It is the substance of the derivation I sent to Dalespam and DrGreg, but tidied up and debugged. Thanks are due to Dalespam for looking over it and picking up some of the original typos. )

Derivation of the coordinate acceleration of in Schwarzschild spacetime.

Starting with the radial Schwarzschild metric:

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

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

Solve for dr/dt:

$$\frac{dr}{dt} \;=\; \sqrt{\alpha^2 - \alpha\left( \frac{ds}{dt}\right )^2}$$

From the Lagragian and the equations of motion we have the conserved quantity k defined as:

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

(See #2 by Starthaus https://www.physicsforums.com/showthread.php?p=2710570#post2710570 )

(Note that the equations of motion are only valid for a particle in freefall.)

This value $$(k/\alpha)$$ is substituted for dt/ds in the equation above it.

$$\frac{dr}{dt} \;=\; \sqrt{\alpha^2 - \frac{\alpha^3}{k^2}}$$

Now dr/dt is differentiated again:

$$\frac{d^2r}{dt^2} \;=\; \frac{d(dr/dt)}{dr}\frac{dr}{dt}\;=\; -\frac{M}{r^2}\left(\frac{3\alpha^2}{k^2}-2\alpha \right )$$

You are not allowed to consider $$k$$ as a constant wrt $$t$$ or $$r$$, so you can't differentiate the way you did it. You are not allowed to take $$k$$ when you calculate $$\frac{d}{dr}(\frac{dr}{dt})$$.
You need to differentiate wrt $$s$$ as I showed you in my blog. I explained to you repeatedly, that $$k$$ is independent of $$s$$ not of $$t$$ or $$r$$. This is exactly what the Euler-Lagrange equation teaches you.

 

[starthaus]

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

Umm, no. You are still doing your standard numerology. You calculated the derivatives as if $$k$$ is a constant but now you are setting $$k=\sqrt\alpha=\sqrt{1-2m/r}$$.
If k is constant, it cannot be equal to $$\sqrt{1-2m/r}$$.
If k is not constant, then you can not do the chain derivative the way you are attempting it.

Substituting this constant(!?) in your equation (7) for coordinate acceleration gives:

Wrong again, you claim $$k$$ to be constant but you just set it equal to $$k=\sqrt\alpha=\sqrt{1-2m/r}$$ and last we all checked, $$r$$ was the variable radial coordinate. You are back to doing numerology.

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

which is in exact agreement with the equation I gave for coordinate acceleration in #1.

...only in your hacky derivation. In general, it isn't true. Now, if you want to do things right, you will need to think really hard what is the correct value for $$k$$. It isn't hard to get it correct.

Can we now agree that the equations I gave in #1 are correct and have now been derived in a number of different ways by different people including yourself?

Nope, your post 1 is incorrect and all the subsequent "derivations" that you attempted on your own in order to justify it are wrong. The only time you did a derivation right was when I guided you through the steps.

 

[starthaus]

I have to question your comprehension. We have spent 178 odd posts with you trying to show that my post in #1 was wrong when in fact you have been wrong all along.

It will take as long until you get it right. Because you still don't have it right.

 

[Al68]

Now, if you want to do things right, you will need to think really hard what is the correct value for $$k$$. It isn't hard to get it correct.

If you are claiming the value of k to be something different, you need to state what the correct value of k is and why.

This isn't the homework forum, so disagreeing with a post while leaving others to guess why "as an exercise" is against forum rules.

It will take as long until you get it right.

Nonsense. This isn't a personal e-mail exchange between a teacher and student, it's a discussion forum with rules that you keep breaking.

 

[starthaus]

If you are claiming the value of k to be something different, you need to state what the correct value of k is and why.

The only way to get kev to realize his mistakes is by guiding him in finding his own mistakes. Even then, he will still claim he was right all along. Anyways, I already gave the correct derivation in my blog, so you can look it up there. Finding out the correct expression for k is trivial.

 

[Al68]

If you are claiming the value of k to be something different, you need to state what the correct value of k is and why.

The only way to get kev to realize his mistakes is by guiding him in finding his own mistakes.

I was asking about your claims.

Forum rules place the burden to substantiate claims on the person making the claim, not on others as an "exercise".

Anyways, I already gave the correct derivation in my blog, so you can look it up there. Finding out the correct value for k is trivial.

The results of your derivations in your blog are identical to kev's. And in your blog, k is a function of r.

I just can't make any sense of your responses in this thread.

 

[starthaus]

I was asking about your claims.

Forum rules place the burden to substantiate claims on the person making the claim, not on others as an "exercise".

Again, can you explain the correct value for k?

Yes, I can. You can find it yourself from the data I supplied in the complete solution.

 

[espen180]

Umm, no. You are still doing your standard numerology. You calculated the derivatives as if $$k$$ is a constant but now you are setting $$k=\sqrt\alpha=\sqrt{1-2m/r}$$.
If k is constant, it cannot be equal to $$\sqrt{1-2m/r}$$.
If k is not constant, then you can not do the chain derivative the way you are attempting it.

I'm not claiming to be an expert, but I really can't follow what you are saying here.

What kev stated is simply that $$k=\sqrt{1-2\frac{m}{r_a}}$$, $$r_a$$ being the apogee radius. Now, when a particle reaches the apogee, $$r=r_a$$, so for a brief moment, $$k=\sqrt{\alpha}$$, because if $$r=r_a$$, $$\sqrt{1-2\frac{m}{r}}=\sqrt{1-2\frac{m}{r_a}}$$. kev seems to observe this and take advantage of it when doing calculations at apogee, yet you seem to imply that kev is stating that $$k=\sqrt{\alpha}$$ holds always.

At least that is how I understood this exchange.

 

[starthaus]

I'm not claiming to be an expert, but I really can't follow what you are saying here.

What kev stated is simply that $$k=\sqrt{1-2\frac{m}{r_a}}$$

kev oscillates between setting k as:

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

and

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

So, the first order item is to get him to commit to one or the other.

$$r_a$$ being the apogee radius. Now, when a particle reaches the apogee, $$r=r_a$$, so for a brief moment, $$k=\sqrt{\alpha}$$, because if $$r=r_a$$, $$\sqrt{1-2\frac{m}{r}}=\sqrt{1-2\frac{m}{r_a}}$$. kev seems to observe this and take advantage of it when doing calculations at apogee, yet you seem to imply that kev is stating that $$k=\sqrt{\alpha}$$ holds always.

At least that is how I understood this exchange.

Yes, you are on the right track. Look at his post 177. Compare against the general formula I provided. It is pretty self explanatory.

 

[espen180]

Look at his post 177. Compare against the general formula I provided. It is pretty self explanatory.

Is this what you are referring to?

OK, now at the apogee when the particle is nearly stationary,

$$ds/dt = \sqrt{\alpha}$$

and so the value of the k constant is:

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

If so, kev clearly states that he is working with a particle at apogee, where the equality holds. I don't see the problem. Please point it out to me.

 

[starthaus]

Is this what you are referring to?
If so, kev clearly states that he is working with a particle at apogee, where the equality holds. I don't see the problem. Please point it out to me.

If that was the case, then $$\alpha$$ is a constant and he cannot differentiate wrt it. Yet he does differentiate...this is precisely how eq (7) has been derived from (6).

 

[espen180]

If that was the case, then $$\alpha$$ is a constant and he cannot differentiate wrt it. Yet he does differentiate...this is precisely how eq (7) has been derived.

I don't see this. What I think kev is doing it something analogous to (or equivalent to) saying two functions are tangent at a certain point. It is like saying that $$e^{x-1}=x$$ when $$x=1$$. That isn't saying $$e^{x-1}=x$$ always.

Can you show me an instance where kev uses $$k=\sqrt{\alpha}$$ except at apogee?

As for the derivation of (7), I probably wouldn't be able to follow it.

 

[Al68]

I was asking about your claims.

Forum rules place the burden to substantiate claims on the person making the claim, not on others as an "exercise".

Again, can you explain the correct value for k?

Yes, I can. You can find it yourself from the data I supplied in the complete solution.

All I see in your solution are derivations that completely agree with kev's, including k being a function of r.

Can you clearly explain your disagreement with kev's results in a post in this thread?

I'd say it just a tiny bit (188 posts) overdue. :uhh:

 

[starthaus]

As for the derivation of (7), I probably wouldn't be able to follow it.

well, you should in order to carry this discussion.

 

[starthaus]

All I see in your solution are derivations that completely agree with kev's, including k being a function of r.

Wrong. I made a very clear point in my derivation to never differentiate wrt $$r$$. The only differentiation going on is wrt $$s$$. This is due to the fact that $$\frac{dk}{ds}=0$$ as explained in the beginning of the derivation.

Can you clearly explain your disagreement with kev's results in a post in this thread?

I'd say it just a tiny bit (188 posts) overdue. :uhh:

1. kev's results are not derived, they are put in by hand (see post 1)
2. when kev attempts a derivation, it is always flawed, most of the time due to his limitations in terms of proficiency in elementary calculus (to his credit, kev has admitted to this limitation)
3. since there are no derivations per se, it is very difficult to convince him that he's made errors. This is why I got fed up and I did the whole derivation, in a rigorous manner, from scratch.
4. the most offensive part in kev's post 1 is the $$a_0=a\gamma^3$$ . kev simply "borrowed" it from the case when $$\gamma=1/\sqrt{1-(v/c)^2}$$ . Simply re-labelling $$1/\sqrt{1-2GM/(rc^2)}$$ as "gamma" does not mean that the formula can be applied blindly in the case of motion in a gravitational field. We clash often on this issue , of mis-application of formulas outside their domain of applicability.
5. There is the flipping and the flopping between $$k=\sqrt{1-2m/r}$$ and $$k=\sqrt{1-2m/r_a}$$. The first expression is a function, the second is a constant, kev needs to make up his mind.
6. Probably the best way is to have kev restate his post 1 and to include a derivation of his claims, the way I did it. He can feel free to borrow from my derivation. Once he does that , I will respond to each claim and to each derivation of such claim.

 

[espen180]

well, you should in order to carry this discussion.

I'm not here to debate about the validity of eqn #7 or the derivation thereof. I just pointed out that I think you are falsely accusing kev, maybe due to a misunderstanding or because you seem to really have it out for the guy.

5. There is the flipping and the flopping between $$k=\sqrt{1-2m/r^2}$$ and $$k=\sqrt{1-2m/r_a^2}$$. The first expression is a function, the second is a constant, kev needs to make up his mind.

Please point to an instance where kev uses $$k=\sqrt{\alpha}$$ except at apogee.

 

[starthaus]

I'm not here to debate about the validity of eqn #7 or the derivation thereof. .

Post 177. Since he's putting in results in by hand (using the detailed derivation of eq(7) from eq(6)) and since you admit that you can't follow the mentioned derivation, the point is moot.

 

[yuiop]

kev oscillates between setting k as:

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

and

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

So, the first order item is to get him to commit to one or the other.

Espen is right. I am using $$r_a = r$$ for the specific instance when a particle is at apogee when the statement is true. This is context of post #1 about particles that are stationary, (or so nearly stationary that any slight movement causes a insignificant error). For example something like 10m/s is insignificant compared to the speed of light squared. If you catch me using $$r_a = r$$ or $$k = \sqrt{\alpha}$$ when I am not talking about a particle at its apogee then you have the right to pull me up on it.

 

[Al68]

All I see in your solution are derivations that completely agree with kev's, including k being a function of r.

Wrong. I made a very clear point in my derivation to never differentiate wrt $$r$$. The only differentiation going on is wrt $$s$$. This is due to the fact that $$\frac{dk}{ds}=0$$ as explained in the beginning of the derivation.

Your resulting equations are identical to kev's. I don't care why. My question was about kev's results, not their derivation, so this is a moot point, anyway.

Can you clearly explain your disagreement with kev's results in a post in this thread?

I'd say it just a tiny bit (188 posts) overdue. :uhh:

1. kev's results are not derive, they are put in by hand (see post 1)
2. when kev attempts a derivation, it is always flawed, most of the time due to his limitations in terms of proficiency in elementary calculus (to his credit, kev has admitted to this limitation)
3. since there are no derivations per se, it is very difficult to convince him that he's made errors. This is why I got fed up and I did the whole derivation, in a rigorous manner, from scratch.
4. the most offensive part in kev's post 1 is the $$a_0=a\gamma^3$$ . kev simply "borrowed" it from the case when $$\gamma=1/\sqrt{1-(v/c)^2}$$ . Simply re-labelling $$1/\sqrt{1-2GM/(rc^2)}$$ as "gamma" does not mean that the formula can be applied blindly in the case of motion in a gravitational field. We clash often on this issue , of mis-application of formulas outside their domain of applicability.
5. There is the flipping and the flopping between $$k=\sqrt{1-2m/r^2}$$ and $$k=\sqrt{1-2m/r_a^2}$$. The first expression is a function, the second is a constant, kev needs to make up his mind.
6. Probably the best way is to have kev restate his post 1 and to include a derivation of his claims, the way I did it. He can feel free to borrow from my derivation. Once he does that , I will respond to each claim and to each derivation of such claim.

Items 1-6 above are all irrelevant to my question, which referred to kev's results, not their derivation.

Again, can you clearly explain your disagreement with kev's results in a post in this thread?

 

[starthaus]

You asked to be "pulled up, here it is from your post 177

This is a summary of the equations in your gravitational acceleration paper in your blog:
Equation 7 is exactly the same as the equation I posted in #135 for coordinate acceleration, so finally we are in agreement.

Equation 8 is agreement with the proper acceleration given by #1/Dr Brown/Dalespam/Rindler etc, if we take into account that you are using dr (coordinate distance) rather than the normal local/proper distance.

OK, now at the apogee when the particle is nearly stationary,

$$ds/dt = \sqrt{\alpha}$$

and so the value of the k constant is:

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

So, in this case $$k=\sqrt{1-2m/r_a}$$

Substituting this constant in your equation (7) for coordinate acceleration gives:

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

But , by borrowing my derivation, you forgot the fact that $$\alpha$$ is a variable (since you are differentiating wrt it), i.e. $$\alpha=1-2m/r$$.
So, you cannot cancel out terms in $$\alpha$$ with terms in $$k^2$$. The first one is a variable (a function), the second one is a constant. I have explained this to you several different ways.

 

[starthaus]

Your resulting equations are identical to kev's.

Wrong, they aren't. You need to pay attention.

Items 1-6 above are all irrelevant to my question, which referred to kev's results, not their derivation.

Physics is about the ability to derive correct results, not about cutting and pasting. He has no valid derivation to speak off and our answers are not the same (unless he copies mine).

 

[yuiop]

As for the derivation of (7), I probably wouldn't be able to follow it.

The derivation of (7) by Starthaus is a little tricky and involves implicit differtiation and it took me a while to figure it out (not being an expert in calculus). This is how it works for equation (8) which is a bit shorter, but uses the same methods as used for (7).

Given:

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

$$k = \alpha \frac{dt}{ds}\;\;(4)$$

$$\frac{dr}{ds} = \sqrt{k^2-\alpha}\:\: (5)$$

$$\frac{dr}{dt} = \sqrt{\alpha^2 - \frac{\alpha^3}{k^2}} \;\; (6)$$

Define a funtion for alpha as f(s) meaning alpha is a a function of s, so that the dr/ds (Equation 5) can be expressed as:

$$\frac{dr}{ds} = \sqrt{k^2-f(s)}$$

Note that k has been left as a constant.

Now carry out the implicit differentiation. I used http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=calculus&s2=differentiate&s3=advanced and typed in sqrt(k^2 -f) in the expression box, s in the variable box and f(s) in the functions box.)

$$\frac{d^2r}{ds^2} = \frac{d(dr/ds)}{ds}= \frac{d(\sqrt{k^2-f(s)})}{ds} = \frac{-f'(s)}{2\sqrt{k^2-f(s)}} = \frac{-1}{2\sqrt{k^2-\alpha}} \frac{d\alpha}{ds}$$

Note that k is not defined as a function of anything and the derivative works just as well with K being treated as a constant.

In the next step implicit differentiation of alpha with respect to s is carried out so that:

$$\frac{d\alpha}{ds} = \frac{d(1-2m/r)}{ds} = 0 -\frac{d(2m/r)}{ds} = -\frac{d(2mr^{-1})}{ds} = -\frac{(-2mr^{-2})dr}{ds} = \frac{2mr}{r^2} \frac{dr}{ds}$$

The rest is simple algebra.

None of the derivatives carried out by Starthaus depend on k being a function of r or t, so he has not demonstrated that k is not a constant. Prof Brown also consistently treats k as constant in his mathpages and other references refer to k as the conserved energy of the free falling particle. i.e k does not change with r and does not change over time.

That is why when Prof Brown treats k as independent of r when carrying out differentiations with respect to r, he gets the correct results and despite the fact that Starthaus insists Prof. Brown is wrong he gets the same results as Starhaus.

 

[Al68]

Wrong, they aren't. You need to pay attention.

Physics is about the ability to derive correct results, not about cutting and pasting. He has no valid derivation to speak off and our answers are not the same (unless he copies mine).

You are simply not making any logical sense. First you say he doesn't have a derivation, then you say the non-existent derivation is wrong without explaining why. And then you say I'm wrong about your equations being identical with ones used by kev, then you say that his are the same as yours because he copied them.

Are your posts all part of some kind of bizarre and twisted joke?

Again, can you clearly explain your disagreement with kev's results in a post in this thread?

 

[starthaus]

The derivation of (7) by Starthaus is a little tricky and involves implicit differtiation and it took me a while to figure it out (not being an expert in calculus). This is how it works for equation (8) which is a bit shorter, but uses the same methods as used for (7).

Given:
Define a funtion for alpha as f(s) meaning alpha is a a function of s, so that the dr/ds (Equation 5) can be expressed as:

$$\frac{dr}{ds} = \sqrt{k^2-f(s)}$$

Note that k has been left as a constant.

Now carry out the implicit differentiation. I used http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=calculus&s2=differentiate&s3=advanced . Type in sqrt(k^2 -a) in the expression box, s in the variable box and a(s) in the functions box.)

$$\frac{d^2r}{ds^2} = \frac{-f'(s)}{2\sqrt{k^2-f(s)}} = \frac{-1}{2\sqrt{k^2-\alpha}} \frac{d\alpha}{ds}$$

Note that k is not defined as a function of anything and the derivative works just as well with K being treated as a constant.

I did not ask you to explain why my derivation is correct. I asked you to explain your errors.
The reason there is no derivative of k wrt s is that the first Euler-Lagrange equation I derived for you tells us that $$\frac{dk}{ds}=0$$

In the next step implicit differentiation of alpha with respect to s is carried out so that:

$${d\alpha}{ds} = \frac{d(1-2m/r)}{ds} = 0 -\frac{d(2m/r)}{ds} = -\frac{d(2mr^{-1})}{ds} = -\frac{(-2mr^{-2})dr}{ds} = \frac{2mr}{r^2} \frac{dr}{ds}$$

The rest is simple algebra.

But in post 177 (where the errors are) you aren't doing differentiation wrt $$s$$, you are attempting differentiation wrt $$r$$. I did not ask you to explain why my solution is correct, I asked you to admit the errors in yours.

None of the derivatives carried out by Starthaus depend on k being a function of r or t, so he has not demonstrated that k is not a constant.

LOL. This is not the point. The point is that the only thing you know from the Euler-Lagrange equation is that $$k$$ is not a function of $$s$$. This is the only thing you are allowed to use. We don't know (and we don't care) if $$k$$ is a function of $$r$$ (turns out it is).

Leave prof. Brown out of this, address your own errors.

 

[yuiop]

I did not ask you to explain why my solution is correct, I asked you to admit the errors in yours.

You didn't ask, but espen180 expressed an interest. This is an open forum and all are allowed to take part you know.

LOL. This is not the point. The point is that the only thing you know from the Euler-Lagrange equation is that $$k$$ is not a function of $$s$$. This is the only thing you are allowed to use. We don't know (and we don't care) if $$k$$ is a function of $$r$$ (turns out it is).

Leave prof. Brown out of this, address your own errors.

You say you don't care, but you keep banging on about k not being a constant. I can find plenty of references that say it is.

 

[starthaus]

You didn't ask, but espen180 expressed an interest. This is an open forum and are allowed to take part you know.

Fine, thank you for explaining why my solution is correct.
Now, can you address the errors in your approach? Can you write a coherent derivation from end to end, now that you have had plenty of opportunity to examine the one I posted?

 

[yuiop]

Fine, thank you for explaining why my solution is correct.
Now, can you address the errors in your approach? Can you write a coherent derivation from end to end, now that you have had plenty of opportunity to examine the one I posted?

I have done a full and correct deriviation in https://www.physicsforums.com/showpost.php?p=2733502&postcount=135" based on K.Brown's methods.
Dalespam has done an awesome derivation based on four vectors.
You have done a derivation based on Euler-lagrange formamalism.

They all agree. Why do we need another one?

You seem to object to the fact that I used k as constant that is independent or in my derivation. Since I took derivatives with respect to r while assuming k is independent of r, I would not have got the the same results as you and Dalespam, if k was not a constant, would I?

Why do you want me to do another one based on your method? I have already expanded on your method to make it clearer to anyone that is not familiar with implicit differentiation. What more do you want me to do?

I have checked your latest derivation to make sure it is correct, because it agrees with my results and if yours was wrong it would indicate mine was wrong too. Despite the fact I gave you plenty of time to post a derivation before I did, the fact is I posted mine first and your derivation is substantially based on mine.

The only wrong derivation in this thread is your derivation of coordinate acceleration from the strong field aproximation, which gave embarrassingly different results to all the other derivations, including your latest derivation. Perhaps you should look into where you goofed up on that one, eh?

 

[yuiop]

But , by borrowing my derivation, you forgot the fact that $$\alpha$$ is a variable (since you are differentiating wrt it), i.e. $$\alpha=1-2m/r$$.
So, you cannot cancel out terms in $$\alpha$$ with terms in $$k^2$$. The first one is a variable (a function), the second one is a constant. I have explained this to you several different ways.

I was using the general formauls to work out the specific case of a particle on a specific trajectory and at a specific place in time and space (at its apogee). At that precise point it is valid to equate $$\alpha$$ with $$k^2$$.

 

[Al68]

I was pondering this thread https://www.physicsforums.com/showthread.php?t=401713" 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?".

Not being able to find a clear, simple definitive answer to that question on the internet, I decided to do some "back of the envelope" calculations and this is what I came up with (using a combination of hunches, intuition and guesses)...

...Does that seem about right?

I'd say that the fact that, after almost 3 weeks and over 200 posts, your equations in post #1 have not been shown to be incorrect is a pretty good sign.

Those must have been some pretty good "hunches, intuition and guesses".

 

[starthaus]

I have done a full and correct deriviation in https://www.physicsforums.com/showpost.php?p=2733502&postcount=135" based on K.Brown's methods.

I explained to you repeatedly that the derivation you copied from K. Brown is wrong. I explained why you can't do the derivatives wrt $$r$$ while considering $$k$$ constant wrt $$r$$.

I have spent enough time on this nonsense.

 

[Al68]

Can you clearly explain your disagreement with kev's results in a post in this thread?

1. kev's results are not derive, they are put in by hand (see post 1)
2. when kev attempts a derivation, it is always flawed, most of the time due to his limitations in terms of proficiency in elementary calculus (to his credit, kev has admitted to this limitation)
3. since there are no derivations per se, it is very difficult to convince him that he's made errors. This is why I got fed up and I did the whole derivation, in a rigorous manner, from scratch.

I can't help but point out the obvious logical fallacy here, and persisting in your posts. The fallacy is claiming that a result is wrong because you disagree with the way it was derived.

For example, consider this simple derivation:

Given: x = 16/64.

Canceling the 6 on the top and bottom, we obtain: x = 1/4.

Claiming that this result is incorrect would be a logical fallacy.

 

[starthaus]

I can't help but point out the obvious logical fallacy here, and persisting in your posts. The fallacy is claiming that a result is wrong because you disagree with the way it was derived.

For example, consider this simple derivation:

Given: x = 16/64.

Canceling the 6 on the top and bottom, we obtain: x = 1/4.

Claiming that this result is incorrect would be a logical fallacy.

It's not my fault that you don't understand basic calculus. Based on your logic someone copying a correct result and producing a bogus derivation has produced valuable work instead of junk.