Orbital velocities in the Schwartzschild geometry

[starthaus]

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

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 Second off, thanks to matheinste's notice, you already have withdrawn from your nonsense in post#251 by adding "$$r=const.$$" to your hack which "validates" it but makes it vacuouslty correct.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:

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


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

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.

 

[Altabeh]

kev,
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.

It is not your fault to be thinking of this nonsense that has been dragged into our consideration by starthaus. See the source introduced in my post 305 to get familiar with the Lagrangian procedure for geodesics. It's really helpful and pithy at the same time.

AB

 

[Altabeh]

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

Do you get the difference between "nonsense" and "correct"? All over this thread you have left behind a nonsense of yours and the last being the Lagrangian and its derivative. I have to add this to the list of your hacks I sent in a private message to you. LOL!

Besides, his "solution" is incorrect.

It behooves oneself to give reasons for claiming incorrectness of a solution otherwise the whole reasoning is nonsense as is yours right now!

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

Did you take those introductory courses in everything? If not, then stop playing the role of tutor here. With all those nonsense claims your knowledge won't be empty of leaks EVER.

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

Basic stuff that doesn't have anything to do with your old fallacy. You've been declared as finished in your post 251.

-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.

Another nonsense. The whole problem which was completely explained to you long time ago is that the initial conditions are all point-wise as is the proper velocity of the momentarily at rest particles. Go read the textbooks I gave you when discussing this. Please do not go for hacks/twists when you're stuck, or better say, blocked.

Basic stuff

Then why don't you understand it, huh?

AB

 

[Altabeh]

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.

Here I think mentors would come to check these mind-made theories of starthaus and don't let him publish wrong ideas/nonsense claims like the "old fallacy" $$dr/ds=0\Rightarrow d^2r/ds^2=0$$. We are doing our best we can to stop such fallacies to be made but as far as I know, individuals here are not allowed to disperse their own ideas specially when they are full of hacks.

AB

 

[yossell]

Hi Altabeh,

thanks for your comment in 316 - I had wondered at your post 305, but I admit it was too pithy for me to be sure I understood. I couldn't get hold of the reference - though the book's in Amazon, the preview that was available to me didn't have the reference.

Of your example, I don't see why the Lagrangian you write is, as you claim, either -1, 0 or 1. It looks to me like a function of position and velocity vectors, and will vary accordingly. What am I missing?

 

[espen180]

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?

Here's a nice counter-example.

Take the equation of a plane:

$$L=ax+by+cz+d=0$$ (letter L chosen for comparative purposes)

It is obvious that L is a constant (it is equal to zero), but the partial derivatives are

$$\frac{\text{d}L}{\text{d}x}=a$$

$$\frac{\text{d}L}{\text{d}y}=b$$

$$\frac{\text{d}L}{\text{d}z}=c$$All the equation above says is that

$$\text{d}L=\left(\frac{\text{d}L}{\text{d}x}\right)_{yz}\text{d}x+\left(\frac{\text{d}L}{\text{d}y}\right)_{xz}\text{d}y+\left(\frac{\text{d}L}{\text{d}z}\right)_{xy}\text{d}z=0$$

And equally, for the lagrangian, it is true that

$$\text{d}L=\sum_{i=0}^N \left(\frac{\text{d}L}{\text{d}q_i}\right)_{q_j\,,\,j\neq i}\text{d}q_i=0$$

where $$q_i$$ are the independent coordinates. The equation says nothing about the individual partial derivatives, it only restricts them as a group.

 

[yossell]

Hi espen180,

thanks for your example. Let me try and express my worries about over this example.

One can define a function of three variables, L(x, y, z) as:
L(x, y, z) = ax + by + cz + d

More explicitly: for every triple <x, y z>, the value of the function at this triple is ax + by + cz + d.

It is typical to avoid an explosion of notation hide the variables, keep them explicit, and just write L for the function - but we should remember they are there.

It makes perfect sense to take partial derivatives of this function wrt x, y and z, as you do in lines 5-7 and get the answers you say.

But it does not make sense to say: let this function be the zero function. You can't do it: you've already defined the function, as above, and it's just not the zero function.

When defining a plane, it *does* make sense to talk about the set of points <x y z> which are such that ax + by + cz + d = 0. This, of course, defines a plane in 3 dimensions. But I don't know what it means to start differentiating a plane, or differentiating anything like ax + by + cz + d = 0.

I may of course be very wrong - but I hope this worry makes sense.

 

[yuiop]

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.

Here is a simple demonstration that although the derivative of a constant function is zero, it does not necessarily follow that the partial derivative of a constant function is zero.

Consider the function $f(x,y) = 2y^2 - x^2 = 1$

Since f(x,y) = 1, the function f is a constant function.

Now the partial derivative of f with respect to x is:

$$\frac{\delta}{\delta x} f(x,y) = \frac{\delta}{\delta x} (1) = \frac{\delta}{\delta x} (2y^2 - x^2) = 2x$$

The partial derivative of f with respect to y is:

$$\frac{\delta}{\delta y} f(x,y) = \frac{\delta}{\delta y} (1) = \frac{\delta}{\delta y} (2y^2 - x^2) = 4y$$

Clearly the partial differentiation of the constant function f(x,y)=1 with respect to x or y is not zero, except at the point where x=0 or y=0. It is easy to demonstrate that x or y can take other values.

Solve the original equation for x:

$$2y^2 - x^2 = 1$$

$$\Rightarrow x = \sqrt{2y^2-1}$$

When y=1, x = sqrt(2*1^2-1) = 1.

When y=5, x = sqrt(2*5^2-1) = sqrt(2*25-1) = sqrt(50-1) = sqrt(49) = 7.

Obviously x can take on many other values depending on the value of y.

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

 

[espen180]

When defining a plane, it *does* make sense to talk about the set of points <x y z> which are such that ax + by + cz + d = 0. This, of course, defines a plane in 3 dimensions. But I don't know what it means to start differentiating a plane, or differentiating anything like ax + by + cz + d = 0.

Think about it this way:

A line is given by

$$ax+by+c=0$$ (every line can be represented in 2 dimensions by a coordinate transformation)

By the equation, ax+by+c is always zero. Can be still talk about the slope of the line? Sure! The slop is equal to $$\frac{\text{d}x}{\textd}y}$$ .

All surfaces can be represented in the form $$f(x,y,z,...)=0$$. We often want to know the change of some coordinates when we change others. For that we have to differentiate f wrt. the coordinates. An example is the gradient of a manifold, which shows the direction of the steepest "slope".

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

Isn't r an independent coordinate? (This might just be a typo. I'm no Lagrange guru.).

 

[yuiop]

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

Isn't r an independent coordinate? (This might just be a typo. I'm no Lagrange guru.).

It would have perhaps been better for me to say something like: When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those variables that the metric is independent of. This is probably still not close to a formal statement. Perhaps I can explain it like this:

Consider the Schwarzschild metric:

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

Only the variables r and $\theta$ appear in their "raw" form so the metric is directly dependent of those variables, but the variables s, t, and $\phi$ only appear in the ds, dt, $d/phi$ forms respectively and it said that the metric is "independent" of s,t and $\phi$ in their "raw" from.

 

[yossell]

kev,
thanks. My worry (or mental block) with what you say is pretty much the same as my worry about espen180's post.

`consider the function $f(x,y) = 2y^2 - x^2 = 1$' is problematic.
You can consider the two-place function of x and y given by f(x y) = 2y^2 - x^2;
you can consider the two-place (constant) function of x and y given by f(x y) = 1
you can consider the set of points <x y> that satisfy the equation: 2y^2 - x^2 = 1
But you can't say that the whole function f(x y) is 1 for all x and y, and that it's 2y^2 - x^2 for all x and all y.

And are you really sure you want to say that the derivative of the constant function f(x y) = 1 is not zero (apart from at x = y = 0)? I thought we'd agreed that the derivatives of constant functions *were* zero everywhere. Since the function doesn't change, the limit of
f(x +dx, y) - f(x, y)/dx is the limit of k - k/dx (as the function is constant) = 0.

espen180, thanks for your response. It's true that a line is given by ax + by + c = 0. This is because it corresponds to a 1-place function f(x) = {c - ax}/b, and then *this* function is what is differentiated to find the slope of the line. Again, there is a notational issue - but it's another notational issue: the fact that in Leibniz notation, y is used sometimes as the name of a function, sometimes as a variable.

Anyway, thanks again for the posts and the spirit of your discussion with me. I see I'm in a minority here, so it could well be my own mental block, and I'll get some sleep now and study your posts again.

 

[espen180]

It seems you find the fact that a function of independent coordinates is set equal to a constant.

I resolution is that when such a restraint is set on the system, the coordinates are no longer fully independent. If you have s function of 4 variables, you may only choose 3 of them freely. The fourth is determined from the values of the other three.

 

[sf222]

My solution agrees with equations (4) and (5) of mathpages http://www.mathpages.com/rr/s6-07/6-07.htm...

That web page deals with purely radial motion, explaining how the acceleration is expressed in terms of Schwarzschild coordinates, but another page from the very same site gives a simple derivation of the equations of motion for general motion, which seems to be what you guys are interested in: http://www.mathpages.com/rr/s6-02/6-02.htm

 

[starthaus]

Here's a nice counter-example.

Take the equation of a plane:

$$L=ax+by+cz+d=0$$ (letter L chosen for comparative purposes)

It is obvious that L is a constant (it is equal to zero), but the partial derivatives are

$$\frac{\text{d}L}{\text{d}x}=a$$

$$\frac{\text{d}L}{\text{d}y}=b$$

$$\frac{\text{d}L}{\text{d}z}=c$$All the equation above says is that

$$\text{d}L=\left(\frac{\text{d}L}{\text{d}x}\right)_{yz}\text{d}x+\left(\frac{\text{d}L}{\text{d}y}\right)_{xz}\text{d}y+\left(\frac{\text{d}L}{\text{d}z}\right)_{xy}\text{d}z=0$$

And equally, for the lagrangian, it is true that

$$\text{d}L=\sum_{i=0}^N \left(\frac{\text{d}L}{\text{d}q_i}\right)_{q_j\,,\,j\neq i}\text{d}q_i=0$$

where $$q_i$$ are the independent coordinates. The equation says nothing about the individual partial derivatives, it only restricts them as a group.

It's an invalid "counter-example". $$r(s)$$ is a function of a single variable. Basic calculus tells you that $$r(s)=0$$ implies $$r'(s)=0$$.

 

[starthaus]

Here is a simple demonstration that although the derivative of a constant function is zero, it does not necessarily follow that the partial derivative of a constant function is zero.

Consider the function $f(x,y) = 2y^2 - x^2 = 1$

Since f(x,y) = 1, the function f is a constant function.

Now the partial derivative of f with respect to x is:

$$\frac{\delta}{\delta x} f(x,y) = \frac{\delta}{\delta x} (1) = \frac{\delta}{\delta x} (2y^2 - x^2) = 2x$$

The partial derivative of f with respect to y is:

$$\frac{\delta}{\delta y} f(x,y) = \frac{\delta}{\delta y} (1) = \frac{\delta}{\delta y} (2y^2 - x^2) = 4y$$

Clearly the partial differentiation of the constant function f(x,y)=1 with respect to x or y is not zero, except at the point where x=0 or y=0. It is easy to demonstrate that x or y can take other values.

Solve the original equation for x:

$$2y^2 - x^2 = 1$$

$$\Rightarrow x = \sqrt{2y^2-1}$$

When y=1, x = sqrt(2*1^2-1) = 1.

When y=5, x = sqrt(2*5^2-1) = sqrt(2*25-1) = sqrt(50-1) = sqrt(49) = 7.

Obviously x can take on many other values depending on the value of y.

When calculating the constants of motion, we note that the metric is independent of the variable s, t and $\phi$ and we only take partial differentials of L with respect to those independent variables.

It's an invalid "counter-example". $$r(s)$$ is a function of a single variable. This is the first error.
The second error is that you did not define a function, you defined an equation. It will become clearer to you when you correct your "counter example" to look like this $$f(x)=2x^2=1$$
You, espen180 and Altabeh share the same basic misconception. It is clear that your misconception cannot be fixed.

 

[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 $d\phi/dt$

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

$$H_c=\frac{r^2}{\alpha} \frac{d\phi}{dt}=\frac{r^2}{1-2m/r}*\sqrt{\frac{m}{r^3}}$$

From the above, it is clear that, contrary to your repeated fallacious claims, $$H_c$$ is a function of $$r$$. Since $$H_c$$ is a function of $$r$$, your whole differentiation of $$L$$ as if $$H_c$$ did not depend on $$r$$ fails

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 $ds/dt$

$$\frac{\delta{L}}{\delta(ds/dt)} = \frac{1}{\alpha} \frac{ds}{dt} = K_c$$

You may be unpleasantly surprised by the fact that $$K_c$$ is also a function of $$r$$, so your "derivation" is based on a double fallacy.

Substitute these constants into the equation for L

$$1 =\alpha K_c^2 +\frac{1}{\alpha^2}\frac{dr^2}{dt^2}+\frac{\alpha}{r^2}H_c^2}$$

and solve for (dr/dt)^2:

$$\frac{dr^2}{dt^2} = (1-2M/r)^2 -(1-2M/r)^3 (K_c^2 + H_c^2/r^2)$$

Differentiate the above with respect to r and divide by 2 to obtain the general coordinate radial acceleration of a freefalling particle in the metric:

Err, no. $$H_c$$ is a function of $$r$$ and you are incorrectly treating it as a constant. So, your differentiation attempt is a failure. You have persisted in this error over several threads now.

$$\frac{d^2r}{dt^2}= \frac{\alpha^2 H_c^2}{r^4}(r-5M) +\frac{M}{r^2}(2\alpha-3K_c^2\alpha^2)$$

Re-inserting the full forms of $H_c$ and $K_c$ back in gives:

$$\frac{d^2r}{dt^2}= <br /> \frac{d\phi^2}{dt^2}<br /> (r-5M) +\frac{M}{r^2}(2\alpha-3<br /> \frac{ds^2}{dt^2} <br /> )$$

Now the (ds/dt)^2 term is a little inconvenient, but we can find an alternative form by solving the Schwarzschild metric to directly obtain:

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

and substituting this form into the equation above it to obtain:

$$\frac{d^2r}{dt^2}= <br /> \frac{d\phi^2}{dt^2}<br /> (r-5M) +\frac{M}{r^2}(2\alpha-3<br /> (\alpha - dr^2/(\alpha dt^2) - r^2d\phi^2/dt^2)<br /> )$$

which after a bit of algebra simplifies to:

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

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.


.

As pointed out repeatedly to you in several posts in this thread, your derivation fails since it is based on a gross fallacy.