Orbital velocities in the Schwartzschild geometry

[Altabeh]

Thanks Altabeh,

I'm sorry by I'm not seeing the lack of sense in your quote from my post.

The argument that, if the first derivative of a function vanishes then so does the second, is this:

...

(3) The derivative of any constant function is zero.

...

Yes but only with the parameter chosen to be differentiated with respect to. Here since the metric can be re-scaled through choosing $$s=a\tau+b$$ where a (nonzero) $$a$$ and $$b$$ are arbitrary constants, thus the the ODE $$dr/ds=0$$ would not mean $$r$$ is constant anywhere since we can give a solution like $$r=C(\tau)$$ to this equation where $$s=a\tau+b$$ and $$ds=ad\tau$$ so that the metric is re-scaled by a constant which has no impact on the form of equations if not say the terms appearing in the metric are "dilated" or "compressed" by $$a^2$$ or $$a^{-2}.$$ So that his claims are nonsense.

I'm afraid I don't really understand what you're doing when you say `take y' to be zero which is a constant function.' As you've shown, y' = x^3 - x^2, and its values vary as x varies. You can't now just stipulate that it's a constant function. Of course, y' equals zero when x equals zero and when x = 1 - so we can talk of its values at various points. But it's not a constant function.

I've not shown $$y' = x^3 - x^2$$ but rather I assumed $$y= x^3 - x^2$$ and said that following y'=0=constant would only lead to your results if one is afraid to spacify the "nature" of function $$y$$. Talking of "y'=0=constant so that y''=0" requires one to always look at y itself to end up with the desired result and here this has nothing to do with whether we are to consider all points or only a special set of them. In general $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ is nothing but a famous fallacy created first by starthaus.

AB

 

[yuiop]

Thanks Altabeh,

I'm sorry by I'm not seeing the lack of sense in your quote from my post.

The argument that, if the first derivative of a function vanishes then so does the second, is this:

(1) Suppose the derivative of a function f vanishes (assumption)

(2) So f' is the constant function zero.

(3) The derivative of any constant function is zero.

(4) So f'', the derivative of f', is zero.

(5) So the second derivative of f vanishes.

Again, to stress, this argument works only if we're both clear that 'function f vanishes' means 'f(x) = 0 for all x' - that is, that we're talking about *functions*. I think that's what he's trying to stress with his quote. None of this goes if 'f vanishes' means 'f vanishes at point p' where point p is some fixed point we've implicitly agreed to focus on. As has already been shown, there are examples where f at p can be zero while f' at p is not zero.

I'm afraid I don't really understand what you're doing when you say `take y' to be zero which is a constant function.' As you've shown, y' = x^3 - x^2, and its values vary as x varies. You can't now just stipulate that it's a constant function. Of course, y' equals zero when x equals zero and when x = 1 - so we can talk of its values at various points. But it's not a constant function.

You are quite right if by dr/dt=0 we mean dr/dt =0 for all time, but in some contexts (like at the apogee) we mean dr/dt=0 at that instant and at at no other time. It is all a matter of context and the likes of Espen, Altabeh, Al68, myself etc. seem to be able to communicate with each other and understand what we mean by using words instead of symbols. In some ways I think using words is helpful for newcomers, because it can be a bit daunting to see equations for the first time that contain symbols that you have never seen in your life before.

 

[yuiop]

EDIT: It seems my conclusion is in conflict with kev's by a factor. I will have to inspect my derivation.

I will inspect your paper (and my derivation too) to see if I can locate where one of us has gone wrong. I think we are close to agreeing a solution.

 

[matheinste]

Hello Starthaus.

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

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 $$\frac{d^2r}{ds^2}=0$$ .

I think this question and answer shows that Starthaus understands the point that the implication is not true in general.

Matheinste

 

[espen180]

EDIT: It seems my conclusion is in conflict with kev's by a factor. I will have to inspect my derivation.

I will inspect your paper (and my derivation too) to see if I can locate where one of us has gone wrong. I think we are close to agreeing a solution.

I managed to locate the error and have updated my paper with the corrected derivation.

http://5554229043997450163-a-1802744773732722657-s-sites.googlegroups.com/site/espen180files/Schwartzschild.pdf?attachauth=ANoY7crvgkJaf2YSb9wipuKnvEoacyFDok-MWepVPK4tZrlLjK8If_HxjS_lzaZVPS-0StjbvJLQU3Bvx-t_oZcneyRG3Orj_V5UXTUFarMpmWYhgtEJNgxcRpypcyted9W0Qa7Re2gn-2ib6A0f-WJV5jukKhp_IfJhNoCnhscy-m2uBhjTO2cHlLX45JEfEI8kNFatSDfwlnBKGeus9Ufx5OdK13A-sw%3D%3D&attredirects=1"

The error consisted of me forgetting the factor 2 when writing out the schwartzschild radius in the final steps of the calculation.

 

[Altabeh]

I think this question and answer shows that Starthaus understands the point that the implication is not true in general.

Matheinste

Actually that is a clear shot at withdrawal from his early posts where he only insisted on the old fallacy. You know why? Because he knows that others and I in this thread are all aware of the "crystal obvious" result of basic calculus he is giving in this post and this is just my interpretation in post #235. All we attepted to do here was to convince him that if he only uses the old fallacy of his, he couldn't get r=const. by any means. So he came to this conclusion that it's better to admit to "mistakes" sometimes and do add $$r=const$$ to the fallacy so as to make it meaningful.

AB

 

[matheinste]

Actually that is a clear shot at withdrawal from his early posts where he only insisted on the old fallacy. You know why? Because he knows that others and I in this thread are all aware of the "crystal obvious" result of basic calculus he is giving in this post and this is just my interpretation in post #235. All we attepted to do here was to convince him that if he only uses the old fallacy of his, he couldn't get r=const. by any means. So he came to this conclusion that it's better to admit to "mistakes" sometimes and do add $$r=const$$ to the fallacy so as to make it meaningful.

AB

Thanks for pointing that out.

Matheinste.

 

[yuiop]

I managed to locate the error and have updated my paper with the corrected derivation.

http://5554229043997450163-a-1802744773732722657-s-sites.googlegroups.com/site/espen180files/Schwartzschild.pdf?attachauth=ANoY7crvgkJaf2YSb9wipuKnvEoacyFDok-MWepVPK4tZrlLjK8If_HxjS_lzaZVPS-0StjbvJLQU3Bvx-t_oZcneyRG3Orj_V5UXTUFarMpmWYhgtEJNgxcRpypcyted9W0Qa7Re2gn-2ib6A0f-WJV5jukKhp_IfJhNoCnhscy-m2uBhjTO2cHlLX45JEfEI8kNFatSDfwlnBKGeus9Ufx5OdK13A-sw%3D%3D&attredirects=1"

The error consisted of me forgetting the factor 2 when writing out the schwartzschild radius in the final steps of the calculation.

Yep, I found it too, but you were just ahead of me.

Amazing how we solved for the solution using coordinate time and came to a mutually agreeable result in just a couple of posts, when we ignore Starhaus isn't it?

 

[yuiop]

I'm afraid I don't really understand what you're doing when you say `take y' to be zero which is a constant function.' As you've shown, y' = x^3 - x^2, and its values vary as x varies. You can't now just stipulate that it's a constant function. Of course, y' equals zero when x equals zero and when x = 1 - so we can talk of its values at various points. But it's not a constant function.

Perhaps this might help. When we say dr/dt=0 we are saying dr/dt is zero at that instant but we are saying nothing about what the value of dr/dt is at any other instant, so dr/dt=0 does not define whether we mean dr/dt is zero for all time or just at that instant (it could be either). If on the other hand we say dr/dt=0 AND $d^2r/dt^2=0$ then it is clear that we mean dr/dt=0 for all time.

 

[yuiop]

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

...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.

It is quite easy. My method directly obtains the simplified general solution:

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

For circular motion the radius is constant by definition and so it follows by simple logic that the radial acceleration $d^2r/ds^2$ must also be zero. Therefore for the special case of circular motion the following is true:

$$0 = -\frac{m}{r^2} +\frac{d\phi^2}{ds^2}(r-3m)$$

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

Simpler than how you obtained the same result in https://www.physicsforums.com/showpost.php?p=2784409&postcount=252"

 

[yuiop]

As to your post 26, you show no derivation whatsoever, you simply agree with espen180's result based on the geodesic formalism.

It's called teamwork

I simply simplified his result and transformed it into local coordinates that could be compared with know results and demonstrated that his document was correct (which was not sure about at the time). This thread could have ended right there but you insisted on telling Espen that his results are incorrect and making everyone spend a lot of effort for hundreds of posts proving to you that your objections are groundless. You "improved" on our solution for orbital motion in post #26 by introducing an alternative (but wrong) solution in post #53. However it has not all been in vain. We would not have had the amusing introduction and failed defence of the now famous Starthaus calculus fallacy. LOL

So, can you solve the five exerciises I gave you?

@Espen. Have you done the homework set for you by "Prof." Starthaus yet? ROFL.

 

[starthaus]

Perhaps this might help. When we say dr/dt=0 we are saying dr/dt is zero at that instant

No, "we" are not saying anything of this nonsense. The $$\frac{dr}{dt}$$ in a differential equation is a function, not the value of that function in a point.

but we are saying nothing about what the value of dr/dt is at any other instant, so dr/dt=0 does not define whether we mean dr/dt is zero for all time or just at that instant (it could be either). If on the other hand we say dr/dt=0 AND $d^2r/dt^2=0$ then it is clear that we mean dr/dt=0 for all time.

You need to take a few calculus classes.

 

[starthaus]

For the special case of a particle in purely radial free fall $d\phi/dt=0$ and:

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

Passed the third test.

Err, no. Not even close to the correct formula, even the dimensions are ridiculously wrong. This is caused by the fact that your "general" formula is incorrect as well.

 

[starthaus]

when he assumes $$dr(s)/ds=0$$ this only holds for a special set of $$s$$'s

Nowhere do I assume such nonsense. Why do you have such a difficulty passing basic calculus that teaches you

1. $$\frac{d^2f}{dx^2}=\frac{d}{dx}(\frac{df}{dx})$$

2. $$\frac{df}{dx}=0$$ substituted into 1. produces $$\frac{d^2f}{dx^2}=0$$

I tried by using different variables in the hope that you'll remember the introductory class to calculus. I made the variables appear the same exact way as in your basic book.

 

[yuiop]

Err, no. Not even close to the correct formula. This is caused by the fact that your "general" formula is incorrect as well.

My solution agrees with the solution provided by Espen. How do our solutions compare with yours? Oops, you don't have a solution.

My solution agrees with equations (4) and (5) of mathpages http://www.mathpages.com/rr/s6-07/6-07.htm which is written by a real professor (Kevin brown). Have you already forgotten the pounding given to you by Rolfe2 the last time you claimed the derivations in the mathpages website were flawed? I guess the wounds have healed eh?

What do you propose the solution should be? Let me guess. That is an exercise left for the students by "Prof" Startaus. When they provide you with the solution, you will claim you taught it to them. You are so transparent.

 

[starthaus]

Moving on to the general case for coordinate acceleration as promised, this is the derivation based on the one I started in post #48.

Starting with Schwarzschild metric and assuming orbital motion in a plane about the equator such that $\theta = \pi/2$ and $d\theta = 0$

$$ds^2=\alpha dt^2-dr^2/\alpha-r^2d\phi^2, (\alpha=1-2M/r)$$

Divide both sides by $\alpha dt^2$ and rearrange so that the constant (1) is on the LHS:

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

The metric is independent of $\phi$ and t, so there is a constant associated with coordinate angular velocity $(H_c)$ which is obtained by finding the partial derivative of L with respect to $d\phi/dt$

$$\frac{\delta{L}}{\delta(d\phi/dt)} = \frac{r^2}{\alpha} \frac{d\phi}{dt} = H_c$$

The metric is independent of s and t, so there is a constant associated with time dilation $(K_c)$ which is obtained by finding the partial derivative of L with respect to $ds/dt$

What "partial derivative of L"? You resurrected your old hack of attempting to "calculate" partial derivatives of a... constant. You just declared $$L=1$$ a few lines above, when will you stop with the ugly hacks?

 

[starthaus]

My solution agrees with the solution provided by Espen. How do our solutions compare with yours? Oops, you don't have a solution.

Wrong guess, you saw it in https://www.physicsforums.com/blog.php?b=1957 about 5 weeks ago, including the correct derivation. Try reading it again, you might learn something.

 

[yuiop]

What "partial derivative of L"? You resurrected your old hack of attempting to "calculate" partial derivatives of a... constant. You just declared $$L=1$$ a few lines above, when will you stop with the ugly hacks?

Surely you mean my beautiful, compact and elegant methods that produce quick precise and correct results in just a few steps?

 

[yuiop]

Wrong guess, you saw it in my blog about 5 weeks ago, including the correct derivation. Try reading it again, you might learn something.

You do not have a general solution in terms of coordinate time for both radial and angular motion in your blog. I derived it first.

 

[starthaus]

You do not have a general solution in terms of coordinate time for both radial and angular motion in your blog. I derived it first.

LOL, you also need to have a valid derivation and valid results, remember? You have neither.

 

[starthaus]

$$\Rightarrow \frac{d^2r}{dt^2}= \alpha\left( r \frac{d\phi^2}{dt^2}-\frac{M}{r^2}+ \frac{3M}{\alpha^2 r^2} \frac{dr^2}{dt^2}\right)$$

This is the fully general form of acceleration in Schwarzschild coordinates in terms of coordinate time and gives Espen something to compare his results against in his latest document.

No, it is just a collection of errors brought about by yet another of your incorrect "derivations".

 

[espen180]

Err, no. Not even close to the correct formula, even the dimensions are ridiculously wrong. This is caused by the fact that your "general" formula is incorrect as well.

The dimensions are wrong? Dear me, please tell me you know that in units where c=1, velocity is dimensionless.

@Espen. Have you done the homework set for you by "Prof." Starthaus yet? ROFL.

Let's see him do it himself.
Many of those aren't analytically solvable anyway.

No, "we" are not saying anything of this nonsense. The $$\frac{dr}{dt}$$ in a differential equation is a function, not the value of that function in a point.



You need to take a few calculus classes.

Each function (r, t, etc) are functions of a parameter, here s.

The differential equations is a relationship f(r(s),t(s),...,r'(s),t'(s),...,r''(s),t''(s),...)=0 which relates the values of the functions for different values of s. That last part is important.

 

[Altabeh]

Nowhere do I assume such nonsense. Why do you have such a difficulty passing basic calculus that teaches you

1. $$\frac{d^2f}{dx^2}=\frac{d}{dx}(\frac{df}{dx})$$

2. $$\frac{df}{dx}=0$$ substituted into 1. produces $$\frac{d^2f}{dx^2}=0$$

I tried by using different variables in the hope that you'll remember the introductory class to calculus. I made the variables appear the same exact way as in your basic book.

First off, as I guessed, you couldn't proceed to understand my posts 274 and 281 and actually ignored them because they are a little bit beyond "basic caluclus" where you understanding still suffers being leaky. It seems like you need to take language courses too because when I said "startaus assumes $$dr(s)/ds=0$$" this only means either r is constant everywhere or r is point-wise constant and if the first part "which you clearly assume" sounds nonsense, then congratualtions; you're contradicting yourself! LOL! Second off, thanks to matheinste's notice, you already have withdrawn from your nonsense in post#251 by adding "$$r=const.$$" to your hack which "validates" it but makes it vacuouslty correct. So you have submitted to our corrections and actually you're announced as "finished" by now. Third off, we were really lucky to wrap this up after 251 posts but I strongly suggest you to take introductory courses in everything AGAIN! LOL!

AB

 

[Austin0]

Perhaps this might help. When we say dr/dt=0 we are saying dr/dt is zero at that instant but we are saying nothing about what the value of dr/dt is at any other instant, so dr/dt=0 does not define whether we mean dr/dt is zero for all time or just at that instant (it could be either). If on the other hand we say dr/dt=0 AND $d^2r/dt^2=0$ then it is clear that we mean dr/dt=0 for all time.

Hi kev I have followed this thread for some time and though most of the math is way over my head I have found it fascinating.
There is a point that has occurred to me and I hope I am not being extremely naive in expressing it.

This question regarding dr/dt=0 and $d^2r/dt^2=0$ has seemed to be somewhat integral to the controversy.
Using your example of the tossed ball in the garden:
1) Isnt it a correct view that as soon as the ball leaves the hand it is in freefall ; Is in inertial motion along a geodesic with inertial motion constant everywhere along the path?
In this view $d^2r/dt^2$ either does not apply or is everywhere zero

2) If you view it in the context of coordinate velocity and acceleration then at apogee dr/dt--->0 as dt---->0 but does dr/dt=0 actually exist? It seems to me to be the mathematical equivalent of the classical instantaneously motionless arrow.
If you do assume this durationless instant then wouldn't $d^2r/dt^2=0$ also apply everywhere along the path?

Forgive me if this is really dumb.
Thanks

 

[Altabeh]

What "partial derivative of L"? You resurrected your old hack of attempting to "calculate" partial derivatives of a... constant. You just declared $$L=1$$ a few lines above, when will you stop with the ugly hacks?

Yeah, now the set of finite nonsense claims of this rookie turns out to be "infinite"; clearly the Lagrangian "$$L=g_{ab}\dot{x}^a{x}^b$$" is either 0, or 1 or -1 but yet the Lagrangian is differentiated when introduced in the Euler-Lagrange equations. See for example:

Hobson M., Efstathiou G., Lasenby A. General relativity: An introduction for physicists (CUP, 2006) pages 78-80.

You're already finished; don't go for nonsense claims.

AB

 

[yossell]

kev,
(This thread is a little more heated than I would like but...here goes)

thanks, I think I agree with your points and distinctions in your post 282, though I do think the notation is confusing: df/dt = 0 and df/dt|x= 0 might be better.

But if it's agreed that the derivative of constant function is zero, then I guess I do feel worried about starthaus' objection to your derivation in post 277.

Is it the case that your L is a constant function? It appears to be - Am I mistaken in this? If it is a constant function, then taking its derivative does, it seems to me, result in the zero function.

Maybe this isn't the way to see it - but I'm not seeing, in this part of the derivation, where steps like 'set t = a and treat all differential equations as valid only at a' comes in. But I understand that there may be some context that I'm missing.

 

[starthaus]

First off, as I guessed, you couldn't proceed to understand my posts 274 and 281

I don't know why you have a mental block with such a basic thing, so I'll simplify it further for you:

-if $$y(x)=0$$ for all x then $$\frac{dy}{dx}=0$$ for all x

-if you have a differential equation in $$y$$ and its derivatives, then, you are not allowed to put in by hand $$y=0$$ without implying immediately that the higher derivatives of $$y$$ are also null.

Basic stuff, you should give it some thought.

 

[starthaus]

kev,
(This thread is a little more heated than I would like but...here goes)

thanks, I think I agree with your points and distinctions in your post 282, though I do think the notation is confusing: df/dt = 0 and df/dt|x= 0 might be better.

Correct, I told kev this about 200 posts ago. He (and Altabeh) still don't get the difference.

 

[starthaus]

The dimensions are wrong? Dear me, please tell me you know that in units where c=1, velocity is dimensionless.

Read (and understand) post 293. In the RHS kev has combined dimensionless $$\alpha$$ with dimensionfull $$\frac{dr^2}{dt^2}$$. Besides, his "solution" is incorrect.

Let's see him do it himself.
Many of those aren't analytically solvable anyway.

Wrong, they all have symbolic solutions. But you need to know how to do it. So, there is a challenge for you, you can stop rolling your eyes and you can roll your sleeves and start working on solving the exercise.

 

[espen180]

Read (and understand) post 293. In the RHS kev has combined dimensionless $$\alpha$$ with dimensionfull $$\frac{dr^2}{dt^2}$$. Besides, his "solution" is incorrect.

If you knew what you were talking about, you would have known that the expression had c² in the denominator of the (dr/dt)², but since c=1, it cancels, so it is dimensionless.

 

[starthaus]

If you knew what you were talking about, you would have known that the expression had c² in the denominator of the (dr/dt)², but since c=1, it cancels, so it is dimensionless.

If you weren't so impertinent, you'd have known how to get the analytic formulas for $$\frac{dr}{dt}$$ and $$\frac{d^2r}{dt^2}$$. Armed with that, you could even derive $$r(t)$$.

 

[espen180]

I have derived them in my paper. If you disagree with them, you should state the correct ones, or you are making groundless arguments and breaking PF rules.

 

[starthaus]

I have derived them in my paper. If you disagree with them, you should state the correct ones, or you are making groundless arguments and breaking PF rules.

By massaging the geodesic you have only managed to get a relationship between $$\frac{d^2r}{dt^2}$$ and $$\frac{dr}{dt}$$. Not very useful since you are unable to find any of the analytic formulas for the challenges:$$\frac{d^2r}{dt^2}$$ , $$\frac{dr}{dt}$$. You could read them in https://www.physicsforums.com/blog.php?b=1957 from May 28

 

[yuiop]

Hi kev I have followed this thread for some time and though most of the math is way over my head I have found it fascinating.
There is a point that has occurred to me and I hope I am not being extremely naive in expressing it.

It is not naive. You are touching on an aspect of calculus most people have not thought about or do not like to think about.

This question regarding dr/dt=0 and $d^2r/dt^2=0$ has seemed to be somewhat integral to the controversy.
Using your example of the tossed ball in the garden:
1) Isn't it a correct view that as soon as the ball leaves the hand it is in freefall ; Is in inertial motion along a geodesic with inertial motion constant everywhere along the path?

Yes, the ball is in inertial motion. This question regarding dr/dt=0 and $d^2r/dt^2=0$ is integral to the "Starhaus fallacy" controversy but it not integral to my derivations because I have not used $dr/dt=0 \Rightarrow d^2r/dt^2=0$ in any of my derivations. If you go back a couple of pages in this thread you will see a very straightforward proof by a George (a senior member and moderator) that $dr/dt=0 \Rightarrow d^2r/dt^2=0$ is false.

In this view $d^2r/dt^2$ either does not apply or is everywhere zero

It does apply. It is acceleration measured using clocks and rulers. To someone co-moving with the ball the acceleration is indeed zero because to the co-moving observer the ball remains stationary, but if you are still standing on the lawn of your garden, the ball has velocity and acceleration relative to you. This acceleration is constantly downwards everywhere along its path, slowing its upward velocity on the way up and after the apogee increasing its downward velocity on the way down.

2) If you view it in the context of coordinate velocity and acceleration then at apogee dr/dt--->0 as dt---->0 but does dr/dt=0 actually exist? It seems to me to be the mathematical equivalent of the classical instantaneously motionless arrow.

First of all, dt---->0 applies everywhere along the path, not just at the apogee. I prefer to think of dr/dt as the *average* velocity over a very small time interval. It is possible to have an *average* velocity of *exactly* zero at the apogee. Consider a time interval of 2 seconds extending from 1 second before arriving at the apogee to 1 second after the apogee. Let us say the ball is 10m above the ground at t=1. At t=2 the ball is at apogee at 19.8m. At time t=3 the ball is back to 10m above the ground. The total distance displacement (dr) over the 2 second interval is (19.8-10)+(10-19.8) = 0. In other words the ball started at r=10 and finished up at r=10 over the 2 second interval. The average velocity (dr/dt) is then *exactly* zero over the 2 second interval. You can extend this argument to as small a time interval as you like.

What is confusing is that we often talk about the velocity is at a given "point" in time. Velocity is a distance interval divided be a time interval and the smaller the time interval the greater the accuracy of the result. This is often described as taking the limit as the time interval goes to zero. The trouble is that if we take the time interval to *exactly* zero we get an indeterminate result, because when dt=0 it folows that dr=0 and dr/dt = 0/0. We can however take the limit to as arbitrarily close to zero as we desire.

If you do assume this durationless instant then wouldn't $d^2r/dt^2=0$ also apply everywhere along the path?

If we consider the path of to be the sum of a series of durationless instants then the sum of the path would be zero and the total time for the ball to rise to its apogee and fall back down would be zero, which is obviously not the case. It is perhaps better to call the "durationless instant" an infinitesimally small time interval. Infinitesimally small is not necessarily exactly the same as a time interval of *exactly* zero length.

Forgive me if this is really dumb.
Thanks

It is not dumb. I am not even sure I have the correct answers for you. I have tried my best to be helpful but I am no expert on calculus and I usually use mathematical software to do the hard work for me without thinking too deeply about how calculus actually works. Your question probably goes to the heart of the fundamental theory of calculus and deserves its own thread in the calculus forum where it would get a more formal and more accurate treatment from real calculus experts (unlike me).

 

[yuiop]

For the special case of a particle in purely radial free fall $d\phi/dt=0$ and:

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


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

Passed the third test.

See equations (4) and (5) of mathpages http://www.mathpages.com/rr/s6-07/6-07.htm for an alternative proof of this last equation.

Err, no. Not even close to the correct formula, even the dimensions are ridiculously wrong. This is caused by the fact that your "general" formula is incorrect as well.

Read (and understand) post 293. In the RHS kev has combined dimensionless $$\alpha$$ with dimensionfull $$\frac{dr^2}{dt^2}$$. Besides, his "solution" is incorrect.

If my solution is incorrect, then so is your solution in your blog because it is easy to prove they are equaivalent using only simple algebra. Since you seem to have difficulties with algebra I will do it for you.

The solution given in section (11) of your blog is:

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

In section (1) of your blog you state $\alpha = (1-2M/r)$ and in section (9) of your blog you state $K = \sqrt{(1-2M/R)}$ so by simple substitution we obtain:

$$\Rightarrow \frac{d^2r}{dt^2} = -\frac{M}{r^2} \alpha \left(3\frac{\alpha}{K^2} -2 \right)$$

In section (6) of your blog you state $dr/dt = \sqrt{(\alpha^2 - \alpha^3/K^2)}$ and this can be rearranged to $K^2 = \alpha^3/(\alpha^2 - dr^2/dt^2)$ and after substition of this term into your solution we obtain:

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

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

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

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

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

I have now proved that your solution is exactly equivalent to my solution (and Espen's). If my solution is wrong then so is yours. If the dimensions of my solution are "ridiculously wrong" then so are yours. I suspect you failed to realize that the solutions are equivalent because you have difficulties with elementary algebra and so you accidently ended up rubbishing your own solution. LOL.