Orbital velocities in the Schwartzschild geometry

[starthaus]

Going back to this post, do you disagree that \frac{dt}{d\tau}=\left(\frac{d\tau}{dt}\right)^{-1} ?

Of course not, I am simply saying that both (30) and (37) are wrong, you need to figure out why.

 

[starthaus]

In the interests of intellectual integrity, I have to admit to my own errors and my earlier claim that the third Euler-Lagrange equation quoted by Starthaus:... is not the same as the solution obtained by Espen, is mistaken.

Good , it only took you 150 posts and several hints to understand that.

My earlier calculation contained a sign error that caused the dr/ds term in the Euler-Lagrange equation to differ from Espen's solution by a factor of three. Here is the (hopefully) correct derivation of one from the other:

Given:

-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0

Carrying out the differentiation of the -d/dr(1/\alpha)(dr/ds)^2 term gives 2m/(\alpha^2 r^2)(dr/ds)^2 and the expression becomes:

-d/ds(1/\alpha*2dr/ds)- 2m/ r^2 (dt/ds)^2 - 2m/(\alpha^2 r^2)(dr/ds)^2 + 2r (d\phi/ds)^2) = 0

Mutliply both sides by -\alpha/2

(\alpha/2)* d/ds(1/\alpha*2dr/ds) + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r \alpha(d\phi/ds)^2) = 0

Carrying out the differentiation of the d/ds(1/{\alpha}*2dr/ds) term gives ((2/{\alpha}) (d^2r/ds^2) + d/ds(1/{\alpha})*2(dr/ds)) and the expression becomes:

(\alpha/2)* ( (2/{\alpha}) (d^2r/ds^2) + (d/ds(1/{\alpha})*2(dr/ds) )+ m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0

Now d/ds(1/{\alpha})*2(dr/ds) \Rightarrow d/dr(1/{\alpha})*2(dr/ds)^2 and carrying out the differentiation of this term gives -4m/(\alpha^2 r^2)(dr/ds)^2 and the expression becomes:

\Rightarrow (\alpha/2)* ( (2/\alpha) d^2r/ds^2 -4m/(\alpha^2 r^2)(dr/ds)^2 ) + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0

\Rightarrow d^2r/ds^2 -2m/(\alpha r^2)(dr/ds)^2 + m\alpha/ r^2 (dt/ds)^2 + m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0

\Rightarrow d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0

At this point the messy equation given by Starthaus is now in the cleaner

It isn't messy, I simply left the simple computations to you to finish. It isn't my problem that it took you 160 posts and multiple hints to figure it out. I am glad that you finally figured it out.

can be directly obtained from the Schwarzschild metric and after substituting this expression for (dt/ds)^2 into the Espen/Euler-Lagrange solution above, further simple algebraic manipulations result in this condensed (but still valid for both radial and orbital motion) version:

\frac{d^2r}{ds^2} = -\frac{m}{r^2} +\frac{d\phi^2}{ds^2}(r-3m)

This is the same as the result I obtained more directly by my method

...except that :

-your"method" is an invalid hack that shows crass ignorance of basic calculus

-your "method" is incapable of deriving something as simple s \frac{d\phi}{ds}. Let's see you do it.

 

[starthaus]

You still don't seem to get it. Do we really need to go over all that again? Hint: There is a clue in the phrase "H is a constant".

Since you obviously don't know how H was derived (you don't understand the Euler-Lagrange) formalism), you don't understand that the corret statement is "H does not depend on s but it does depend on r, so kev's attempt of differentiating wrt r by considering H a constant is a mistake"

P.S. Could we stick to the substance of the posts (i.e the physics and maths) rather than personal attacks

Sure we can. If you stopped trying to cover your mistakes by trying silly attacks.

I am quite willing to have a truce if you agree to stick to the spirit of the forum guidelines and stop making statements like "There is already established knowledge that you don't know basic calculus".

Start with understanding what H=r^2\frac{d\phi}{ds}. It doesn't mean "H is constant". In any calculus book. OK?

 

[starthaus]

Rolfle2 (and others) went to great lengths in another thread to explain to you why H is a constant and not a function of r.
.

:LOL: H=r^2\frac{d\phi}{ds}

 

[starthaus]

Your post #53 is very suspect. Let's go through it.



If you work out the equations of motion for a photon (which by definition always has ds=0) then when you divide both sides by ds^2 you end up with:

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

\frac{0}{0} =\alpha \frac{dt^2}{0}-\frac{1}{\alpha}\frac{dr^2}{0}-r^2d\frac{\phi^2}{0}

which is undefined on the left and a bunch of infinite terms on the right. This is a very shaky foundation for a rigorous derivation.

LOL. You really need to take a class in calculus. If you want the lightlike metric, you only need to set ds=0. If you manage to do this correctly, you will be rewarded by getting:

\alpha dt^2-dr^2/\alpha=0

0-(2m/r^2(dt/ds)^2-0-2r(d\phi/ds)^2)=0

which solves to:

(d\phi/ds)^2 = \frac{m}{R^3} (dt/ds)^2

and not as you claim:

(d\phi/ds)^2 = \frac{m}{R^3}

You can add this to the list of basic algebra blunders you have already made in this thread.

Congratulations for finally managing to substitute correctly \frac{dr}{ds}=0 in the general Euler-Lagrange formula I've shown you. Yes, you managed to find a typo , it is \frac{d\phi}{dt}=\sqrt{m/r^3}, not \frac{d\phi}{ds}=\sqrt{m/r^3}. I made a cut and paste error, it is quite obvious, no need to keep harping on it over 200 posts.

 

[starthaus]

We can however note that he has failed to explicitly state what he thinks d\phi/ds should be, despite being asked to do so.

Why do you have difficulty deriving \frac{d\phi}{ds} when you have been given \frac{d\phi}{dt}=\sqrt{m/r^3} and \frac{ds}{dt}=\sqrt{1-3m/r} in post 53? Do I need to do all the basic calculations for you?We could have saved 200 posts if you made the effort to understand post 53.

 

[matheinste]

Hello Starthaus.

I have been following this thread but most of the content is way above my level but I am learning something from it.

I do have a problem with some of the mathematical wrangling here so I will pose a question to clarify an issue which has been mentioned several times, the answer being for my own benefit and perhaps that of other non-involved spectators.

Even if you see the question as irrelevant could you please still answer it as it only requires a yes or no answer.

Leaving aside mathematical niceties, roughly speaking, talking of the graphical representaion, if the first derivative of a continuous function exists at a point on the graph of the function and is itself continuous and takes the value zero at that point, does this IN GENERAL imply that the second derivative, is zero at that point. More to the physical point if the velocity takes the value zero at some point, does the accelaration necessarily take the value zero at that point.

Matheinste

 

[George Jones]

Leaving aside mathematical niceties, roughly speaking, talking of the graphical representaion, if the first derivative of a continuous function exists at a point on the graph of the function and is itself continuous and takes the value zero at that point, does this IN GENERAL imply that the second derivative, is zero at that point.

No. For example, the parabola y = x^2 has first derivative y' = 2x and second derivative y'' = 2. At x = 0, the first derivative is zero and the second derivative equals two.

 

[matheinste]

No. For example, the parabola y = x^2 has first derivative y' = 2x and second derivative y'' = 2. At x = 0, the first derivative is zero and the second derivative equals two.

Thanks for your reply. This correct answer I am sure most already know. I am absolutely OK with such basic calculus. My point was to try to get Starthaus to say whether he thought the answer was yes or no. Others on this thread have been telling him what you have just said but have, in my impression been fobbed off with non-commital or evasive answers.

Matheinste

 

[yossell]

I haven't followed every twist and turn in this engaging thread, which seems to me to mainly constitute a warning about the dreadful ambiguities of using Leibniz notation in calculus, but starthaus originally pointed out that certain second derivatives were zero all the way back in post 5 (ah, the nostalgia...).

In that context, this was in response to the the OP's wish to analyse a certain kind of circular motion. Here's my understanding, but I wonder if it's correct: in the OP's case, the circular motion is modeled by a FUNCTION with constant radial coordinate r and (since it's circular) constant theta coordinate. When the function is a constant, the first derivative is everywhere zero and the second derivative is everywhere zero. And that's all he's relying on in that post. So, to me at least, at at this early point, it doesn't seem to be a hack, George Jones' point notwithstanding.

 

[starthaus]

Hello Starthaus.
Even if you see the question as irrelevant could you please still answer it as it only requires a yes or no answer.

Leaving aside mathematical niceties, roughly speaking, talking of the graphical representaion, if the first derivative of a continuous function exists at a point on the graph of the function and is itself continuous and takes the value zero at that point, does this IN GENERAL imply that the second derivative, is zero at that point.

No, of course it doesn't but this is not the point of disagreement.
For the case of circular orbits, the radial coordinate r=constant, so
\frac{dr}{ds}=0 everywhere in the domain of definition of r .
Therefore, by virtue of \frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds}) it follows that \frac{d^2r}{ds^2}=0 .

So, when espen180 first set \frac{dr}{ds}=0 into the geodesic equation for circular orbits, that attracted immediately \frac{d^2r}{ds^2}=0 in the same equation.

When one makes \frac{dr}{ds}=0 into the Euler-Lagrange equation I gave eons ago , at post 53 (where this thread should have stopped since it gives the correct general answer to all orbits), this results immediately into \frac{d}{ds}(\frac{dr}{ds})=0, a simple fact that took kev about 200 posts to accept and a fact that Altabeh still has understood.

More to the physical point if the velocity takes the value zero at some point, does the accelaration necessarily take the value zero at that point.

Matheinste

No, of course not, yet, this is not the point (see above)

 

[starthaus]

You are repeating the same errors , you just made L=1=constant, when you differentiate a constant, you get ...zero.

You have added new errors as well. If you want to obtain the lagrangian, then you shout divide by ds, not by dt. If you do this, you get the correct Lagrangian:

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

Once you get the Lagrangian, you can get one of the Euler-Lagrange equations:

r^2d\phi/ds=H_c (compare against your incorrect expression).

The other Euler-Lagrange equation is:

\alpha (dt/ds)=K (compare to your other incorrect expression).

I know that I have written this stuff for you before.

There is a third Euler-Lagrange equation, it is dependent on the other two, so it does not contain any extra information per se:

-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0 (1)

If you make r=R in the above, this means the cancellation of the terms in dr/ds and if you giving you

(d\phi/dt)^2=\frac{m}{R^3} (2) (corrected typo)

i.e.

\omega^2=\frac{m}{R^3}

I am quite sure that I have shown you this before as well.

Substitute d\phi/dt and dr=0 into the metric expression and you get:

ds^2=(1-3m/R)dt^2 (3)

The above makes sense only for R>3m

From (2) and (3) one obtains easily:

\frac{d\phi}{ds}=\sqrt{\frac{m}{R^3}}\frac{1}{\sqrt{1-3m/R}} (4)

It should be noted that (1) is the general Euler-Lagrange equation for planar orbits (d\theta=0) from which all other equations (radial motion, circular orbits, etc) can be easily derived. This thread should have stopped at post 53.

 

[matheinste]

Thankyou Starthaus.

The first part of your reply in post #251 answers my question.

Matheinste.

 

[yuiop]

Since you obviously don't know how H was derived (you don't understand the Euler-Lagrange) formalism), you don't understand that the corret statement is "H does not depend on s but it does depend on r, so kev's attempt of differentiating wrt r by considering H a constant is a mistake"
...
Start with understanding what H=r^2\frac{d\phi}{ds}. It doesn't mean "H is constant". In any calculus book. OK?

You seem to think that a basic level of profiency in calculus (not withstanding some misconceptions you have in this department) is all you need to come in here and tell us how GR works, but you do not understand the physics.

For a free-falling particle that has both radial and orbital motion H is a constant. As it falls r^2 obviously gets smaller but at the same time the angular velocity d\phi/ds increases in such a way that the changes cancel each other out and H has the same value for any value of r. H is therefore independent of r and if you take the derivative of H with respect to r, as if it was a function of r you will get the wrong answers. You would already know this is you really understood the Euler-Lagrange formalism as you claim you do.

 

[starthaus]

Thankyou Starthaus.

The first part of your reply in post #251 answers my question.

Matheinste.

Yes, I knew you would understand.

 

[starthaus]

You seem to think that a basic level of profiency in calculus (not withstanding some misconceptions you have in this department) is all you need to come in here and tell us how GR works, but you do not understand the physics.

For a free-falling particle that has both radial and orbital motion H is a constant.

It is constant wrt s. It is not a constant wrt r.

As it falls r^2 obviously gets smaller but at the same time the angular velocity d\phi/ds increases in such a way that the changes cancel each other out and H has the same value for any value of r.

\frac{d}{ds}(r^2\frac{d\phi}{ds})=0

so

r^2\frac{d\phi}{ds}=H

where H is not a function of s

Please plug in the value for \frac{d\phi}{ds} in the above expression for H. If you have difficulties, you can look at post 252.

if you take the derivative of H with respect to r, as if it was a function of r you will get the wrong answers.

I am not the one attempting such silly hacks in my derivation. In fact, my derivation uses only the third Euler-Lagrange equation. If you still have difficulties following, I rewrote post 53 a little into post 252. Try reading it.

 

[Altabeh]

It is your problem that you don't know that \frac{d}{ds}(\frac{dr}{ds})=0 for \frac{dr}{ds}=0. I do not know why you have so much difficulty with this elementary subject.

I can give you millions of counter-examples to this "elementary" result which would smash it in the face and turn it into a "nonsense". So if you keep using such hacks in your derivations this means you're doing a double hack: first supporting a nonsense and second insisting on it being "correct". Use my interpretation and stand corrected. Such silly hacks are to only blame for the lack of capability to deral with elementary calculus unless you start getting on the right track.

AB

 

[Altabeh]

Thanks for your reply. This correct answer I am sure most already know. I am absolutely OK with such basic calculus. My point was to try to get Starthaus to say whether he thought the answer was yes or no. Others on this thread have been telling him what you have just said but have, in my impression been fobbed off with non-commital or evasive answers.

Matheinste

Yeah, most of students learning basic calculus in high school know of such thing but unfortunately starthaus doesn't! If be lucky, more than 200 posts might be spent on making him understand that supporting the nonsense "dr/ds=0\Rightarrow d^2r/ds^2=0" would not probably save his hacks from getting smashed.

AB

 

[yuiop]

When the function is a constant, the first derivative is everywhere zero and the second derivative is everywhere zero. And that's all he's relying on in that post.

Yes, for purely circular motion the Starthaus fallacy dr/sd=0 \Rightarrow d^2r/ds^2 =0 happens to be true by luck in that special case, but Starthaus believes it applies to radial motion too. See for example post #209 quoted below where I am clearly talking about purely radial motion:

\frac{\text{d}^2r}{\text{d}\tau^2} = -\frac{c^2 r_s}{2r^2} = -\frac{GM}{r^2}

This result means that the acceleration using these coordinates is independent of the falling velocity dr/ds because we have made no assumption of dr/ds=0. It also means that the acceleration is non-zero for any value dr/ds and so it is proved that the claim by Starthaus that the acceleration is zero when dr/ds=0, is false.

And the response by Starthaus was:

If you insist on hacking the metric by putting \frac{dr}{ds}=0 by hand as you've been doing, then, by virtue of elementary calculus, you'd get \frac{d^2r}{ds^2}=0.

Clearly Starthaus thinks dr/ds=0 \Rightarrow d^2r/ds^2 =0 is a universal truth and that it also applies to purely radial motion. Since Starthaus claims to be a calculus expert any newcomers might think that he is correct and share his misconception. Starthaus needs to make it clear that he made a elementary calculus blunder here. It is not just a typo. He has been defending this claim for many posts. See all of page 4 of this thread for example (https://www.physicsforums.com/showthread.php?t=409241&page=4) (and surrounding pages) to see the strength of his conviction that the Starthaus fallacy dr/ds=0 \Rightarrow d^2r/ds^2 =0 is a universal truth.

The above quote by Starthaus is also an example of his hypocrisy because he has been inserting \frac{dr}{ds}=0 by hand in plenty of his derivations.

 

[yuiop]

From (2) and (3) one obtains easily:

\frac{d\phi}{ds}=\sqrt{\frac{m}{R^3}}\frac{1}{\sqrt{1-3m/R}} (4)

It should be noted that (1) is the general Euler-Lagrange equation for planar orbits (d\theta=0) from which all other equations (radial motion, circular orbits, etc) can be easily derived. This thread should have stopped at post 53.

Actually the correct solution for circular motion (based on Espen's derivation in his document) was given by me 26 posts earlier:

I have had another look at your revised document and checked all the calculations from (9) onwards and they all seem to be correct. Your final result can be simplified to:

\frac{rd\phi}{ds} = c \sqrt{\frac{GM}{rc^2 - 3GM}

Divide both sides of the above equation by r and simplify and you obtain:

\frac{d\phi}{ds} = \sqrt{\frac{GM/r^3}{1 - 3GM/(rc^2)}

Your solution for circular motion in #53 came later and contained a major error (which you have now corrected.) If you had checked your solution against my (and Espen's) solution in #27 you might have noticed your error earlier.

 

[yossell]

Kev,

thanks for your helpful post and for taking the time to explain things to me.

To be fair, it seems to me that there's unclarity in the notation and it's not clear to me that starthaus is necessarily making that mistake, even in the posts you point to - though I admit, I may be misunderstanding things.

In some contexts, it's natural to read dr/ds as referring to a function rather than the value of the function at a point. Then dr/ds = k, or dr/ds = 0 can be read as saying that the *function* is k everywhere or 0 everywhere. Understood in this way, the inference from dr/ds = 0 to d^2 r/ds^2 = 0 is correct. So what (I take) starthaus to be getting at in, say, post 57 seems, at least as it stands, fine.

What would be mistaken is inferring anything about d^2 r/dx^2 from the value of dr/ds at a certain point. In this thread, when people have written dx/dt = 0, they've not always made it clear whether they're talking about the value of this function at some point, as in an initial condition, or whether they're making the whole function this value.

I think this is the source of a lot of confusion here.

Now, in, say, espen180's post 56, to which starthaus is replying, espen180 does say he'll `impose dr/d\tau = d\phi /d\tau = d\theta/d\tau = 0'

Now (and here I admit I am at my least certain) I can't see how to understand this condition in the context of this derivation other than as a condition on the *functions* - i.e. in a way which validates his argument at 57. This is not an initial condition-style problem, where we are given the values of the functions at a certain point, which we can then substitute in somewhere.

 

[yuiop]

I think we could spend another 100 posts discussing the misconceptions and errors in calculus, algebra and physics of Starthaus, but maybe it is time to move on and summarize what the vast majority of contributers in this thread have concluded about orbital motion in Schwarzschild geometry in terms of the proper time of the particle and then move on.

The following 3 general equations for radial and orbital motion are all equivalent:

(Eq1) Quoted from a textbook by Starhaus.

-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0

http://5554229043997450163-a-1802744773732722657-s-sites.googlegroups.com/site/espen180files/Schwartzschild.pdf?attachauth=ANoY7cq5YcjBEfWs9NLohb9pW4ZSy8PeA1mye4fi5O5a1gEGkYzDwlDazDnETq05RiOCkTZT8VXKhWuZqMuPeHbOXoRfYD9KsLXbZrCGovBLGPZZtajp8B-RE2DI945uVj6eaA_GW7DZajPzhP90N63csOTzgkYM4mbE0PVDyXqcOwZ2L40PYADGrKjBHcbmOFjqsVwwmfy7mUPChM0x_BVp5G_Fr_KFmg%3D%3D&attredirects=0" .

d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0

https://www.physicsforums.com/showpost.php?p=2781228&postcount=211".

d^2r/ds^2 + m/r^2 - (r-3m)(d\phi/ds)^2 = 0

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

From any of the above, it is easy to obtain the special case for a circular orbit:

\frac{d\phi}{ds} = \sqrt{\frac{ (m/r^2)}{(r-3m)}}

and the special case for purely radial motion:

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

(Note: The above expression for acceleration in terms of proper time is not the proper acceleration as some people seem to think.)

To the above list we can add this gem obtained from Altabeh's calculations:

... the local velocity V_R of a particle in a circular orbit (radius =R) according to a stationary/ hovering observer at R is:

V_R/c = \sqrt{1-\alpha (1-3M/R)}

which correctly gives the expected result that the local velocity is c when R=3M.

with the warning that the above equation is not in terms of proper time of the particle, but in terms of the proper time of the stationary clock used by the hovering observer at r.

Now the OP (Espen) has expressed a desire to analyse the motion of the particle in terms of coordinate time and I will try and focus my efforts in assisting him with his quest in subsequent posts.

 

[starthaus]

I can give you millions of counter-examples to this "elementary" result which would smash it in the face and turn it into a "nonsense".

Your counter-examples are based on the same elementary misunderstanding, so what is the point? You don't understand the difference between a function and the value of a function in a point.

 

[starthaus]

Clearly Starthaus thinks dr/ds=0 \Rightarrow d^2r/ds^2 =0 is a universal truth

Of course it is, it is a fundamental property of functions. Like Altabeh, you are unable to tell the difference between a function and the value of a function in a point.

 

[starthaus]

The following 3 general equations for radial and orbital motion are all equivalent:

(Eq1) Quoted from a textbook by Starhaus.

-d/ds(1/\alpha*2dr/ds)-(2m/r^2(dt/ds)^2-d/dr(1/\alpha)(dr/ds)^2-2r(d\phi/ds)^2)=0

http://5554229043997450163-a-1802744773732722657-s-sites.googlegroups.com/site/espen180files/Schwartzschild.pdf?attachauth=ANoY7cq5YcjBEfWs9NLohb9pW4ZSy8PeA1mye4fi5O5a1gEGkYzDwlDazDnETq05RiOCkTZT8VXKhWuZqMuPeHbOXoRfYD9KsLXbZrCGovBLGPZZtajp8B-RE2DI945uVj6eaA_GW7DZajPzhP90N63csOTzgkYM4mbE0PVDyXqcOwZ2L40PYADGrKjBHcbmOFjqsVwwmfy7mUPChM0x_BVp5G_Fr_KFmg%3D%3D&attredirects=0" .

d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0

Good , you finally realized your errors of misdirection, you could have accepted that my equation was correct as shown to you 200 posts ago, at post 53.
BTW, it isn't "quoted from a textbook", it is derived from scratch from the Euler-Lagrange equations. You should make the effort to learn the formalism sometimes.

https://www.physicsforums.com/showpost.php?p=2781228&postcount=211".

d^2r/ds^2 + m/r^2 - (r-3m)(d\phi/ds)^2 = 0

Except that the equivalent derivations based on the geodesic or on the Euler-Lagrange formalism are correct whereas yours is based on a series of hacks. As to your post 26, you show no derivation whatsoever, you simply agree with espen180's result based on the geodesic formalism.

 

[starthaus]

The above quote by Starthaus is also an example of his hypocracy because he has been inserting \frac{dr}{ds}=0 by hand in plenty of his derivations.

LOL , "hypocracy"?

...except that when I do correctly insert \frac{dr}{ds}=0 in the general Euler-Lagrange equation, this attracts immediately \frac{d}{ds}(\frac{dr}{ds})=0 in the same equation. And it isn't "by hand", it is a consequence of the very basic conditionr=constant
My method is in direct contradiction to what espen180 was doing to his geodesic (he was retaining the term in \frac{d^2r}{ds^2}. This is why I called him on it earlier. To this day, you still don't understand the difference. He understood. So do matheinste and yossell, after only one explanation.

 

[starthaus]

Kev,

thanks for your helpful post and for taking the time to explain things to me.

To be fair, it seems to me that there's unclarity in the notation and it's not clear to me that starthaus is necessarily making that mistake, even in the posts you point to - though I admit, I may be misunderstanding things.

In some contexts, it's natural to read dr/ds as referring to a function rather than the value of the function at a point. Then dr/ds = k, or dr/ds = 0 can be read as saying that the *function* is k everywhere or 0 everywhere. Understood in this way, the inference from dr/ds = 0 to d^2 r/ds^2 = 0 is correct. So what (I take) starthaus to be getting at in, say, post 57 seems, at least as it stands, fine.

Great, you understood it as well. Now, if only kev and Altabeh ever understood it, that would be a miracle.

 

[yuiop]

Kev,

thanks for your helpful post and for taking the time to explain things to me.

To be fair, it seems to me that there's unclarity in the notation and it's not clear to me that starthaus is necessarily making that mistake, even in the posts you point to - though I admit, I may be misunderstanding things.

Your welcome You are right that there is some ambiguity in the notation. It becomes tedious thinking thinking of new symbols (GR is already full of subscripts and superscripts) for acceleration at the apogee, acceleration in radial freefall, acceleration when orbiting, etc. so we sometimes use the same symbols to mean different things and clarify what we mean in the surrounding text. Starthaus is a mathematician and only focuses on the symbols and ignores the surrounding text and this causes a lot of the problems.

In some contexts, it's natural to read dr/ds as referring to a function rather than the value of the function at a point.

This is also true. Sometimes we mean dr/ds=0 at a point and sometimes we mean dr/ds=0 means a function which is true for all s. We (Espen, Altabeh and myself) usually make it clear which usage we mean in the surrounding text, but Starthaus seems to be unable to grasp this. You seem to uderstand this because you have ackowledged that there are two interpretations. Starthaus claims there is "no such thing as when dr/ds=0" when you using dr/ds as a function, but all we mean is that we are considering the special case and we are careful to be aware that any conculsions we draw from making that assumption, are only valid at that point in time.

Then dr/ds = k, or dr/ds = 0 can be read as saying that the *function* is k everywhere or 0 everywhere. Understood in this way, the inference from dr/ds = 0 to d^2 r/ds^2 = 0 is correct. So what (I take) starthaus to be getting at in, say, post 57 seems, at least as it stands, fine.

In post #57 Starhaus is responding to an equation by Espen about purely radial motion (where the Starthaus fallacy dr/ds=0 \Rightarrow d^2r/ds^2 is false. Espen was in turn responding to a quote by me where I specified "the initial coordinate acceleration of a test mass released at r" where the word "initial" was the indication that we were talking about the motion at a point and not implying that dr/ds is 0 everywhere.

Take this series of exchanges between Espen and Starhaus:

I corrected the error I made in my previous attempt and made a new derivation from scratch.

Please see section 4 (Pages 5&6) in this document for the derivation and result.
Download

(30) is wrong.
(37) is as wrong as before.

Going back to this post, do you disagree that \frac{dt}{d\tau}=\left(\frac{d\tau}{dt}\right)^{-1} ?

Of course not, I am simply saying that both (30) and (37) are wrong, you need to figure out why.

If we look at equation (30) in Espen's document we see that the equation given by Espen is:

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

Now in the surrounding text Espen says "To simplify the case, we study the situation where we drop a test particle from rest at r and study it's acceleration immediately after dropping it "
The word immediately is Espen's indication that he talking about the motion at a point rather than at any arbitary time after releasing the particle. i.e. he is talking about the motion in the limit that s goes to zero if the particle is released at time s=0.

Espen further clarifies equation (30) by stating it is the "acceleration measured by a stationary observer at infinity" making it clear that he talking about coordinate acceleration. In the context given by Espen, equation (30) is correct. Starthaus says it wrong because he has not taken the time to read the surrounding text carefully.

The equation given by Espen is the same as the one given by myself in the first post of this old thread https://www.physicsforums.com/showpost.php?p=2710548&postcount=1 and Starthaus spent nearly 400 posts trying to prove it wrong and by common consensus he failed. The equation is correct in the context it was given in.

Although analysing the acceleration at a given point might not seem very useful, it is a very good starting point for determining the proper acceleration of a stationary particle at rest at r as measured by an accelerometer. The proper acceleration measured by an accelerometer of a stationary particle is equal in magnitude (and opposite in direction) to the initial acceleration of a particle released from r as measured by a local stationary observer at r, in the limit that s goes to zero, if the particle is released at time s=0.

The acceleration of a particle with purely radial motion is not zero at the apogee when dr/ds=0 and this is the important point Starthaus does not seem to get. Check post #264 and you will see he is still defending the Starthaus fallacy, despite counterproofs by myself, Altabeh and George.

 

[yossell]

Kev,

thanks again.

I've looked at espen180's paper, your reference. In his opening paragraph, of that section, he says that, in the case of radial motion, d\theta/d \tau = d \phi /d tau = 0. I understand this a constraint on the functions. But then, when he says,

`we study the situation where we drop a test particle from rest at r and study it's (sic) acceleration immediately after dropping it relative to a stationary observer at r. Therefore dr/d\tau = 0'

that second equation is in fact to be understood as not about the function, but as true only at a particular coordinate, (t, x, y z)? (I recognise this may be too restrictive, that it may just one coordinate which is fixed - but the point is, it is not the general function that is being talked about here). So (??) we can only infer the truth of the equations that he goes on to derive in this section as being true *at* a particular point (t, x, y z) (or set of points) - it's just that mention of this point or points is implicit?

That makes sense to me - but again, the notation in the paper seems inconsistent. In section 4, on pure radial motion, similar equations are written, but here the equations can be interpreted as referring to the functions, rather than being implicitly restricted.

Is this correct?

 

[Altabeh]

Your counter-examples are based on the same elementary misunderstanding, so what is the point? You don't understand the difference between a function and the value of a function in a point.

Your nonsense here means that the fallacy dr/ds=0\Rightarrow d^2r/ds^2=0 refers to the values of the second derivative of r in every point, or in your sense to a function, while you refuse to believe that this function may not follow the fallacy at some point(s) contained in its domain. What a mess! LOL.

AB