Orbital velocities in the Schwartzschild geometry

[yuiop]

$$\frac{d}{dt}(K/\alpha)=\frac{d}{ds}(K/\alpha)\frac{ds}{dt}=-\frac{K}{\alpha^2}\frac{2m}{r^2}\frac{dr}{dt}$$

See if you can figure the intermediate steps.

I deleted the question because 1) I am perfectly capable of carrying out the calculation myself and 2) no one is disputing that K is a constant with respect to s. The fact that I obtain the same result by treating K as a constant with respect to r is (to me) ample proof that K is a constant with respect to r.

I don't claim to derive everything from base principles. I take results from various authors and documents and extrapolate or extend them a little and draw my own conclusions and cross check my conclusions against known results. Sometimes I get it wrong and hope to learn from those times. The advantage of a forum is that you get feedback before you go too far off track. I am not a physicist or a mathematician. This is just a hobby for me and I like to pick the brains of people like Dalespam, Altabeh, Espen, etc, etc that are more advanced than me and see what I can learn from them and cross relate their results to my work and other references and generally build up "the big picture". That is just my way of doing things. Sorry if you do not like it.

Finally a question. Do you withdraw your claim that K is a function of r?

Yes, no or not sure?

 

[yuiop]

I wasn't going to put in all the intermediate calculations.

Of course you weren't going to. That would be too helpful.

 

[yuiop]

I'm getting closer to a solution, but the expressions are ridicculously big, so doing the calculus/algebra/proofreading is a pain.

Here's how far I've gotten:
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" (Section 5, page 7)

I'm somewhat worried that I have an r-derivative raised to the 4th power. I can't see how it would disappear later on, either, so maybe I have an error somewhere.

Hi Espen,

From my earlier post:

Re-insert the full form of H back into the equation:

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

(which is the same as (49) in your document) it would seem that (57) in your document should be:

$$\left(\frac{dr}{dt}\right)^2 \cdot \frac{d^2r}{ds^2} = \left(\frac{dr}{dt}\right)^2 \left(-\frac{m}{r^2} + (r-3m)\frac{d\phi^2}{ds^2}\right)$$

using units of G=c=1.

You seem to have an extra (dt/ds)^2 expression on the right.

Equation (30) is fine in the context it is given in and as long as it not used to derive the general case (and you don't seem to have done so) it is not a problem.

 

[espen180]

Hi Espen,

From my earlier post:
(which is the same as (49) in your document) it would seem that (57) in your document should be:

$$\left(\frac{dr}{dt}\right)^2 \cdot \frac{d^2r}{ds^2} = \left(\frac{dr}{dt}\right)^2 \left(-\frac{m}{r^2} + (r-3m)\frac{d\phi^2}{ds^2}\right)$$

using units of G=c=1.

You seem to have an extra (dt/ds)^2 expression on the right.

I just have one question regarding this. Why are you multiplying d²r/ds² by (dr/dt)²?

I used (dt/ds)² to convert (dø/ds)² to (dø/dt)².

 

[yuiop]

I just have one question regarding this. Why are you multiplying d²r/ds² by (dr/dt)²?

I used (dt/ds)² to convert (dø/ds)² to (dø/dt)².

My mistake. I misread the tau as an r. (That is why prefer the s symbol). Taking that into account and your conversion factor then it would seem that the first line of (57) is indeed correct.

I will go and polish my glasses and dig deeper. Sorry bout that

 

[starthaus]

I deleted the question because 1) I am perfectly capable of carrying out the calculation myself and 2) no one is disputing that K is a constant with respect to s. The fact that I obtain the same result by treating K as a constant with respect to r is (to me) ample proof that K is a constant with respect to r.

It is ample proof that you are an incorigible hacker. You demanded to see how the equation was derived, I showed you the rigorous derivation , yet you persist in the hack.

BTW:

1. The equation I derived is not identical to your original one from post 277, it includes your post 277 as a particular case.

2. The equation in itself is pretty worthless since it is just a transformation from the simpler equation written in terms of proper time to uglier one written in terms of coordinate time.

3. The equation written in coordinate time can, through a lot of mathematical skill, lead to the equation of motion. I have challenged you , espen180 and Altabeh to find the expressions of $$r(s)$$, $$\frac{dr}{ds}$$, $$\frac{d^2r}{ds^2}$$, in other words to see if you can integrate the differential equation, if not for general orbits (way outside your skills) at least for radial motion. So far, nothing.

I don't claim to derive everything from base principles. I take results from various authors and documents and extrapolate or extend them a little and draw my own conclusions and cross check my conclusions against known results. Sometimes I get it wrong and hope to learn from those times. The advantage of a forum is that you get feedback before you go too far off track. I am not a physicist or a mathematician. This is just a hobby for me and I like to pick the brains of people like Dalespam, Altabeh, Espen, etc, etc that are more advanced than me and see what I can learn from them and cross relate their results to my work and other references and generally build up "the big picture". That is just my way of doing things. Sorry if you do not like it.

This is fine. What I do not understand is your reluctance to drop the hacks and to pick up rigorous approaches.

Finally a question. Do you withdraw your claim that K is a function of r?

Yes, no or not sure?

Why do you have to persist in horrible hacks? Can you ever learn how to derive things cleanly? I just showed you how to do it in the previous post.

 

[starthaus]

My mistake. I misread the tau as an r. (That is why prefer the s symbol). Taking that into account and your conversion factor then it would seem that (57) is indeed correct. I think I might have been led on a "bum steer" by you know who. Should have known better. :P

Err, no. (57) is definitely wrong. It is amusing that you fail to see the error especially after a few minutes after I showed you the solution. So, I think that the "bum steer" is coming from you.

 

[yuiop]

1. The equation I derived is not identical to your original one from post 277, it includes your post 277 as a particular case.

Dead wrong. my general equation, (derived long before your equation) contains all the information contained in your equation.

2. The equation in itself is pretty worthless since it is just a transformation from the simpler equation written in terms of proper time to uglier one written in terms of coordinate time.

Equations in terms of proper time can get pretty ugly. The orbit of a photon for example, contains division by zero and infinities. It is much easier to measure the coordinate velocity of a particle and use the measurement directly in calculations, rather than have to also calculate the proper time time of the particle as well.

3. The equation written in coordinate time can, through a lot of mathematical skill, lead to the equation of motion. I have challenged you , espen180 and Altabeh to find the expressions of $$r(s)$$, $$\frac{dr}{ds}$$, $$\frac{d^2r}{ds^2}$$, in other words to see if you can integrate the differential equation, if not for general orbits (way outside your skills) at least for radial motion. So far, nothing.

I think you are losing the plot here. Espen (the OP) has requested assistance with his calculation on orbital velocities. This is not the homework forum. It is not about setting a series of challenges with a sprinkling of dead-end diversions and red herrings. It is not just about your ego. We like to be collaborate and be cooperative in this forum. Your style would probably be better appreciated in the homework forum. Have you considered moving there?

 

[starthaus]

Dead wrong. my general equation, (derived long before your equation) contains all the information contained in your equation.

Err, no. The general solution contains $$K$$ as a parameter, yours doesn't. Only if you set $$K$$ to particular values one recovers your equation. So, your solution is a subset of mine, like it or not.

Equations in terms of proper time can get pretty ugly.

Not the one that I am talking about. Compare it to the much uglier one expressed in coordinate time.

I think you are losing the plot here. Espen (the OP) has requested assistance with his calculation on orbital velocities.

Of which, despite yoour help he hasn't calculated any. It would be nice if you concentrated on helping espen180 in getting eqs (30) and (57) right.

 

[espen180]

It would be nice if you concentrated on helping espen180 in getting eqs (30) and (57) right.

Time and time again! You keep saying this! Yet, when asked to substansiate your claims, you back away, meaning you are either not in possesion of an answer (in other words, a groundless assertion) or purposefully withholding it. In both cases, such a claim has no merit.

Claim <-> Proof

These go hand in hand. You have yet to provide the latter.

 

[yuiop]

Err, no. The general solution contains $$K$$ as a parameter, yours doesn't. Only if you set $$K$$ to particular values one recovers your equation. So, your solution is a subset of mine, like it or not.

So you substituted $(\alpha/K)$ for $(ds/dt)$. Big deal!

I gave both versions, with and without constants:

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 $ds/dt$

$$\frac{\delta{L}}{\delta(ds/dt)} = \frac{1}{\alpha} \frac{ds}{dt} = K_c$$
.
.
.
$$\frac{d^2r}{dt^2}= \frac{\alpha^2 H_c^2}{r^4}(r-5M) +\frac{M}{r^2}(2\alpha-3K_c^2\alpha^2)$$

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

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

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:

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

which after a bit of algebra simplifies to:

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

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.

The third equation down contains only the single variable r and the constant parameters of the trajectory, $K_c$ and $H_c$. In fact it is only if I set $H_c$ to particular values, that I recover your solution, so your solution is a subset of mine by your definition.

 

[starthaus]

Time and time again! You keep saying this! Yet, when asked to substansiate your claims, you back away, meaning you are either not in possesion of an answer (in other words, a groundless assertion) or purposefully withholding it. In both cases, such a claim has no merit.

Claim <-> Proof

These go hand in hand. You have yet to provide the latter.

You need to look at your equations and compare with the ones already derived. Maybe kev can help you find the error in (57), it is quite obvious.

 

[starthaus]

So you substituted $(\alpha/K)$ for $(ds/dt)$. Big deal!

Err, no. Look again.

 

[espen180]

Maybe kev can help you find the error in (57), it is quite obvious.

It's exactly this I'm talking about. Learn some pedagogy!

 

[starthaus]

It's exactly this I'm talking about. Learn some pedagogy!

There is value in working towards discovery, even if it is in order to discover your own mistakes. Demanding (in a very impolite manner) that others do your own work is not the right approach.

 

[espen180]

There is value in working towards discovery, even if it is in order to discover your own mistakes.

I can certainly appreciate such a philosophy, but saying "It's wrong" without further clarification is, and I hope you agree, insufficient when correcting someone. It simply makes it frustrating for the receiver when an answer is withheld, and I am certain that you'd feel the same had you been exposed to it.

 

[yuiop]

You need to look at your equations and compare with the ones already derived. Maybe kev can help you find the error in (57), it is quite obvious.

Hi Espen, I have double checked and triple checked all 3 lines of (57) now and can assure you it is perfect. Ignore Starthaus. It is another red herring.

 

[yuiop]

Err, no. The general solution contains $$K$$ as a parameter, yours doesn't. Only if you set $$K$$ to particular values one recovers your equation. So, your solution is a subset of mine, like it or not.

You have shot yourself in the foot there by declaring that a general solution is incomplete, if it does not contain the constant parameters of the trajectory. By that definition, your solution is incomplete. I will complete your solution for you:

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

Combine the above with the first Euler-Lagrange equation

$$\alpha\frac{dt}{ds}=K$$

and, without any hacky assumptions about $$K$$ you will obtain that :

$$\frac{d^2r}{dt^2}=(r-3m)(\frac{d\phi}{dt})^2-(\frac{\alpha}{K})^2\frac{m}{r^2}+\frac{2m}{\alpha*r^2}(\frac{dr}{dt})^2$$

The following can be directly obtained from the Schwarzschild metric:

$$\frac{dr^2}{dt^2} = \alpha \left(\alpha - \frac{ds^2}{dt^2} - r^2\frac{d\phi^2}{dt^2} \right)$$

The second Euler-Langrange equation is

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

Substitute the above expressions (along with the first Euler-Lagrange constant) into your incomplete solution and you obtain, after simplification;

$$\frac{d^2r}{dt^2}=\frac{\alpha^2}{r^4}\frac{H^2}{K^2} (r-5m) +\frac{m}{r^2}(2\alpha - 3\alpha^2 /K^2 )$$

There you go. You now have a fully complete solution just like the one I derived ages ago. It is not as nice, because you have a mixture of K and H in the first term while in my solution the constant parameters are cleanly isolated into separate terms.

The Lagrange constants in your solution, are related to my constants by the relations:

$$K_c = \frac{1}{K} = \frac{1}{\alpha} \frac{ds}{dt}$$

and

$$H_c = \frac{H}{K} = \frac{r^2}{\alpha} \frac{d\phi}{dt}$$

 

[starthaus]

You have shot yourself in the foot there by declaring that a general solution is incomplete, if it does not contain the constant parameters of the trajectory. By that definition, your solution is incomplete. I will complete your solution for you:

LOL, give it a rest.

The following can be directly obtained from the Schwarzschild metric:

$$\frac{dr^2}{dt^2} = \alpha \left(\alpha - \frac{ds^2}{dt^2} - r^2\frac{d\phi^2}{dt^2} \right)$$

The second Euler-Langrange equation is

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

You are learning. Notice that $$\frac{dH}{dr}=2r\frac{d\phi}{ds}$$. You need to give pause and think about this.

Substitute the above expressions (along with the first Euler-Lagrange constant) into your incomplete solution and you obtain, after simplification;

$$\frac{d^2r}{dt^2}=\frac{\alpha^2}{r^4}\frac{H^2}{K^2} (r-5m) +\frac{m}{r^2}(2\alpha - 3\alpha^2 /K^2 )$$

This is totally uunecessary, you did not understand an iota from the method of getting the general solution, you must NOT use all three of the Euler-Lagrange equations because only two are independent. I am quite sure I tried to teach you this long ago.
Do yourself a favor, learn the formalism, ok?

 

[starthaus]

Hi Espen, I have double checked and triple checked all 3 lines of (57) now and can assure you it is perfect. Ignore Starthaus. It is another red herring.

Err, wrong. I'll give you a hint, compare the coefficients for $$(\frac{d\phi}{dt})^2$$

 

[Altabeh]

LOL, give it a rest.




You are learning. Notice that $$\frac{dH}{dr}=2r\frac{d\phi}{ds}$$




This is totally uunecessary, you did not understand an iota from the method of getting the general solution, you must NOT use all three of the Euler-Lagrange equations because only two are independent. I am quite sure I tried to teach you this long ago.
Do yourself a favor, learn the formalism, ok?

After a day being off, I feel now refreshed because the fallacies you've been feeding all of us were just a few hours looked overshadowed by the beauties of nature.

Well, now back to our own discussion, give me the proof of proposition I gave you in post #389. If you couldn't make out proving it, then here is a hint: Use the geodesic equation $$\nabla_b u^a u_b=0$$.

AB

 

[yuiop]

The following can be directly obtained from the Schwarzschild metric:

$$\frac{dr^2}{dt^2} = \alpha \left(\alpha - \frac{ds^2}{dt^2} - r^2\frac{d\phi^2}{dt^2} \right)$$

The second Euler-Langrange equation is

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

You are learning. Notice that $$\frac{dH}{dr}=2r\frac{d\phi}{ds}$$

LOL. That is an elementary calculus blunder. Shame on you!

Using the power rule:

$$\frac{dH}{dr}=2r\frac{d\phi}{ds} + r^2 \frac{d}{dr}\left(\frac{d\phi}{ds}\right)$$

You need to prove

$$r^2 \frac{d}{dr}\left(\frac{d\phi}{ds} \right)= 0$$

That is only true if $(d\phi/ds)=0$ or r=0 which it generally not the case for orbital motion, or if $(d\phi/ds)=0$ is not a function of (r). We can easily show that $(d\phi/ds)$ IS a function of r in the Newtonian limit when $ds \approx t$ because it is well known that angular momentum is conserved in Newtonian physics and $d\phi/dt$ changes with radius for a non circular orbit. You do realize GR has to agree with the Newtonian limit right? Try again.

Hi Espen, I have double checked and triple checked all 3 lines of (57) now and can assure you it is perfect. Ignore Starthaus. It is another red herring.

Err, wrong. I'll give you a hint, look at the coefficient for $$\frac{d\phi}{dt}$$. Not your day today.

The equations in section (57) of Espen's document using G=c=1 are:

------------------------------------

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

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

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

------------------------------------

What you are missing is that from the Schwarzschild metric:

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


Using this information I can expand on Espen's work to make it clear that his work in section (57) contains no errors:

------------------------------------

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

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

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

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

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

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

------------------------------------

Get it now? Will you now stop telling Espen that (57) is wrong?

So in one day you have posted a elementary calculus blunder, an elementary physics blunder and an elementary algebra blunder and you were trying to tell me I was having a bad day? LOL.

 

[starthaus]

LOL. That is an elementary calculus blunder. Shame on you!

Using the power rule:

$$\frac{dH}{dr}=2r\frac{d\phi}{ds} + r^2 \frac{d}{dr}\left(\frac{d\phi}{ds}\right)$$

Err, $$r$$ and $$\phi$$ are independent variables, so your $$\frac{d}{dr}\left(\frac{d\phi}{ds}\right)$$ is pure nonsense. You really need to stop winging the calculus rules. Especially when multivariate functions are not covered by the sw package you are using in attempting your differentiations.

 

[espen180]

Err, $$r$$ and $$\phi$$ are independent variables...

No, not really. Sure, you can set the initial conditions freely, but after that, the geodesic equation, in which every coordinate, as well as its first and second derivatives are related to the other coodinates, dictates their behavior, so no coordinates are free from the influence of other coordinates after you set the initial conditions.

 

[qbert]

You really need to stop winging the calculus rules.

yowsa. pot-to-kettle.

There's only ONE independent variable in this problem. The worldine
coordinate s. So what we have are a bunch of functions of 1-variable.
r = r(s), t = t(s)... Now listen up if H were a function it would be a
function of s! (and only s).

When we write
$$r^2\frac{d\phi}{ds} = H$$
What we mean is
$$r(s)^2 \frac{d\phi}{ds}(s) = H(s).$$

When we're solving a trajectory problem in mechanics we
are solving a system of ODE(<- these are NOT PDE's).
For this problem, the system of ODE's that we need to solve is
$$\frac{d^2r}{ds^2} = \ldots, \frac{d^2 t}{ds^2} = \ldots, \frac{d}{ds} \left( r^2 \frac{d\phi}{ds} \right)= 0.$$

This means H is independent of r, and $\phi$. it's a CONSTANT.

 

[starthaus]

even though it's late in the game --

i think a point has come up which hasn't been answered very well. So I'll
take a stab at it. You need to be careful when using the calculus of
variations to keep in mind which variables are dependent and which are
independent. In the current case, the only independent variable is the
worldline coordinate (t or s). Everything else is a function of (say t) t.
That means H can be at most a function of t. period. since we know
dH/dt = 0 it is the constant function. meaning a real, honest to goodness,
constant.

If I've expressed myself clearly -- this is enough to prove the result.
The rest is for the formula lovers out there.
S'pose $L = L(q_i(t), \dot{q}_i(t))$ then by the E-L eqns we know
[tex] \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right)
= \frac{\partial L}{\partial q_j}[/itex]

Now suppose that L is cyclic in one variable say $q_k$. That is
$$\frac{\partial L}{\partial q_k} = 0 = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right)$$

So that $\frac{\partial L}{\partial \dot{q}_k} = C$
Now suppose to the contrary that C were really a function of the "variables"
ie C = f(q_i).

Then we get
$$\frac{\partial f}{\partial q_j} = \frac{\partial^2 L}{\partial q_j \partial \dot{q}_k} = \frac{\partial^2 L}{\partial \dot{q}_k \partial q_j}$$

$$= \frac{\partial }{\partial \dot{q}_k} \left( \frac{\partial L}{\partial q_j} \right) = \frac{\partial}{\partial \dot{q}_k}\left( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} \right) = \frac{\partial}{\partial \dot{q}_j}\left( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_k} \right) = \frac{\partial}{\partial \dot{q}_k}\left( 0 \right) = 0.$$

This is one of the bestr answers I have seen in this thread. Unfortunately, you haven't taken the time to understand the dispute between me and kev. The derivation that you show above is based on the Euler-lagrange equations, so, your results are valid only only along the arc defined by:

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

kev is trying to hack his derivation by differentiating as if $$H$$ and $$K$$ are constant everywhere. This is obviously not true. To wit, in your example $$f(r)=r^2\frac{d\phi}{ds}$$ so, $$\frac{df}{dr}$$ isn't zero.

 

[starthaus]

yowsa. pot-to-kettle.

There's only ONE independent variable in this problem. The worldine
coordinate s. So what we have are a bunch of functions of 1-variable.
r = r(s), t = t(s)... Now listen up if H were a function it would be a
function of s! (and only s).

When we write
$$r^2\frac{d\phi}{ds} = H$$
What we mean is
$$r(s)^2 \frac{d\phi}{ds}(s) = H(s).$$

Good, so you just confirmed that kev's

$$\frac{dH}{dr}=2r\frac{d\phi}{ds} + r^2 \frac{d}{dr}\left(\frac{d\phi}{ds}\right)$$

is nonsense. Thank you

 

[starthaus]

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

------------------------------------

Get it now? Will you now stop telling Espen that (57) is wrong?

.

Err, wrong, post 430 gives you the correct (unhacked) solution. As you can see, the correct coefficient is $$r-3M$$ not $$r-2M$$.

 

[yuiop]

LOL. That is an elementary calculus blunder. Shame on you!

Using the power rule:

$$\frac{dH}{dr}=2r\frac{d\phi}{ds} + r^2 \frac{d}{dr}\left(\frac{d\phi}{ds}\right)$$

You need to prove

$$r^2 \frac{d}{dr}\left(\frac{d\phi}{ds} \right)= 0$$

Err, $$r$$ and $$\phi$$ are independent variables, so your $$\frac{d}{dr}\left(\frac{d\phi}{ds}\right)$$ is pure nonsense. You really need to stop winging the calculus rules. Especially when multivariate functions are not covered by the sw package you are using in attempting your differentiations.

Note: I have completely re-edited this post to reflect the reminder by Espen that the anti-derivative of zero is a constant.

If as you claim, r and $\phi$ are independent variables, then it follows that

$$\frac{d}{dr}\left(\frac{d\phi}{ds}\right) =0 \qquad \qquad (1)$$

Taking the anti-derivative of both sides with respect to s gives:

$$\frac{d\phi}{ds} = C_{(r)} \qquad \qquad (2)$$

where $C_{(r)}$ is a constant with respect to r.

We agree that H is constant with respect to s so we obtain:

$$\frac{dH}{ds}= \frac{d}{ds}\left(r^2\frac{d\phi}{ds}\right) =0$$

$$\Rightarrow \frac{d\phi}{ds}\frac{d}{ds}(r^2) +r^2 \frac{d}{ds}\left(\frac{d\phi}{ds}\right) =0$$

$$\Rightarrow \frac{d\phi}{ds}\frac{d}{ds}(r^2) +r^2 \frac{dr}{ds}\frac{d}{dr}\left(\frac{d\phi}{ds}\right) =0 \qquad \qquad (3)$$

Substituting (2) into (3) gives:

$$\frac{d\phi}{ds}\frac{d}{ds}(r^2) +r^2 \frac{dr}{ds}\frac{d}{dr}\left(C_{(r)}\right) =0$$

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

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

Taking the anti-derivative of both sides with respect to s gives:

$$\Rightarrow r^2 = C_{(s)}$$

where $C_{(s)}$ is a constant with respect to proper time s.

This shows that Starthaus's claims are only valid for circular orbits where r is constant.

 

[espen180]

The antiderivative is ...=constant, but that shouldn't invalidate your proof. In fact, you can use it to prove that H is a constant.

 

[yuiop]

Err, wrong, post 430 gives you the correct (unhacked) solution. As you can see, the correct coefficient is $$r-3M$$ not $$r-2M$$.

In post #430 you give:

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

Espen gives:

$$\frac{ds^2}{dt^2} \cdot \frac{d^2r}{ds^2} = -\frac{M}{r^2} \alpha + \frac{M}{r^2\alpha}\frac{dr^2}{dt^2} + (r-2M)\frac{d\phi^2}{dt^2}$$

Do you not see that the left hand sides of the equations are different and so by elementary algebra we expect the right hand sides to be different?

You are talking about two different quantities. Wrong again.

 

[starthaus]

In post #430 you give:



Espen gives:



Do you not see that the left hand sides of the equations are different and so by elementary algebra we expect the right hand sides to be different?

You are talking about two different quantities. Wrong again.

Err, no. Bottom of post 430 says:

$$\frac{d^2r}{dt^2}=(r-3m)(\frac{d\phi}{dt})^2-(\frac{\alpha}{K})^2\frac{m}{r^2}+\frac{2m}{\alpha*r^2}(\frac{dr}{dt})^2$$

 

[espen180]

Err, no. Bottom of post 430 says:

$$\frac{d^2r}{dt^2}=(r-3m)(\frac{d\phi}{dt})^2-(\frac{\alpha}{K})^2\frac{m}{r^2}+\frac{2m}{\alpha*r^2}(\frac{dr}{dt})^2$$

Writing out the $$\frac{\alpha}{K}$$ term, what do you get?

 

[starthaus]

and I have proven:

$$\frac{d}{dr}\left(\frac{d\phi}{ds}\right) \ne 0$$

kev,

I understand that you did not understand calculus very well and you are desperately trying to cover your mathematical blunder. So, let's try an "intuitive" explanation, what you "proved" above is that the angular speed depends on radius. Do you grasp the enormity of your error?

 

[starthaus]

Writing out the $$\frac{\alpha}{K}$$ term, what do you get?

What is the puropse of this nonsensical request? The derivation is based on the fact that (see post 430)

$$\frac{ds}{dt}=\frac{\alpha}{K}$$