Orbital velocities in the Schwartzschild geometry

[espen180]

You realize that this is mathematically and physically impossible?

A particle cannot have an acceleration if it has zero velocity?

 

[starthaus]

A particle cannot have an acceleration if it has zero velocity?

\frac{d^2r}{d\tau^2}=\frac{d}{d\tau}(\frac{dr}{d\tau})

What does this tell you? You and kev are starting to worry me.

 

[espen180]

@#58
Ah, thank you very much.

 

[espen180]

\frac{d^2r}{d\tau^2}=\frac{d}{d\tau}(\frac{dr}{d\tau})

What does this tell you? You and kev are starting to worry me.

The condition was that that \frac{dr}{d\tau} was momentarily zero, not constantly. We are talking about free fall here.

 

[starthaus]

The condition was that that \frac{dr}{d\tau} was momentarily zero, not constantly. We are talking about free fall here.

Either you (or kev) have written this nonsense before. Do you even understand differential equations? Do you understand the meaning of the symbols? I think you and kev complement each other in terms of mathematical "skills".
If you want the correct solution, I gave you a complete one in post 53.

 

[espen180]

Either you (or kev) have written this nonsense before. Do you even understand differential equations? Do you understand the meaning of the symbols? I think you and kev complement each other in terms of mathematical "skills".
If you want the correct solution, I gave you a complete one in post 53.

Post #53 treats circular orbital motion. Post #56 treats a particle released from rest at r, pure radial motion. Therefore dr/ds changes with time, so dr/ds=0 initially doesn't mean d²r/ds²=0.

 

[starthaus]

Post #53 treats circular orbital motion.

This is what you asked for in the OP. Have you forgotten what type of problem you were trying to solve?
This is also what you are trying to solve in your writeup. Have you changed your mind?

Post #56 treats a particle released from rest at r, pure radial motion.

First off, I don't think that you got the right equation (you simply copied the geodesic equation from orbital motion). Second off, you switched gears in the middle of the thread. Third off, you are mixing an initial condition (\frac{dr}{d\tau}|_{\tau=0}=0) with the general condition \frac{dr}{d\tau}=0. You are using the general condition (i.e.\frac{dr}{d\tau}=0 everywhere) in order to drop terms from your differential equation. You have done this error before meaning that you don't understand differential equations.
The radial motion problem was already solved in another thread. I already showed kev the solution for this problem.

 

[espen180]

This is what you asked for in the OP. Have you forgotten what type of problem you were trying to solve?
This is what you are trying to solve in your writeup.

Is it against the rules to discuss two topics in one thread?

First off, I don't think that you got the right equation (you simply copied the geodesic equation from orbital motion). Second off, you switched gears in the middle of the thread. Third off, you are mixing an initial condition (\frac{dr}{d\tau}|_{\tau=0}=0) with the general condition \frac{dr}{d\tau}=0. You are using the general condition (i.e.\frac{dr}{d\tau}=0 everywhere) in order to drop terms from your differential equation. You have done this before meaning that you don't understand differential equations.
The radial motion problem was already solved in another thread. I already showed kev the solution for this problem.

I actually went back to the general geodesic equations and started from there, imposing neccesary restrictions.

I am well aware that the solution only holds initially and not after that. In my opinion, it is, however, useful to consider the instantaneous case before tackling the general case.

 

[starthaus]

Is it against the rules to discuss two topics in one thread?

No, it isn't against the rules. Would be nice if you acknowledged that you received the solution to your OP and if, for clarity purposes, you opened a separate thread about radial-only motion.

I actually went back to the general geodesic equations and started from there, imposing neccesary restrictions.

If you want to do this correctly, start with the appropriate metric:

ds^2=\alpha dt^2-\frac{dr^2}{\alpha}
\alpha=1-2m/r

From the above, you can get the equation of motion.

I am well aware that the solution only holds initially and not after that. In my opinion, it is, however, useful to consider the instantaneous case before tackling the general case.

No, it isn't useful until you understand the difference between initial and general conditions.

 

[espen180]

If you want to do this correctly, start with the appropriate metric:

ds^2=\alpha dt^2-\frac{dr^2}{\alpha}
\alpha=1-2m/r

From the above, you can get the equation of motion.

Doing that produces the same result as putting the conditions into the general geodesic equation.

 

[starthaus]

Doing that produces the same result as putting the conditions into the general geodesic equation.

Provided you derive the correct equations of motion. So far, you have gotten the wrong ones.

 

[starthaus]

After reading post #27, the similarity between the Newtonian expression and the one I arrived at makes it very difficult for me to believe it is incorrect. It also predicts that the photon sphere should be at r=3GM. Post #48 seems to confirm my belief here.


I tried to use the same approach to derive the coordinate acceleration of a particle dropped from rest at r relative to a stationary observer also at r. The result was
\frac{\text{d}^2r}{\text{d}t^2}=-\left(\frac{GM}{r^2}-\frac{2G^2M^2}{r^3c^2}\right)
which goes to 0 as r approaches the Schwartzschild radius, I'm unsure what to make of that. It also changes sign when r<2GM, so I doubt its validity in that region. Still, it approximates the Newtonian expression at large r, differing only by about 1.4 ppb at Earth's surface.

Does it look correct? If neccesary, .

No, it is not correct.

 

[starthaus]

I'll post my derivation:

Starting with the general case

\frac{\text{d}^2r}{\text{d}\tau^2}+c^2\left(\frac{r_s}{2r^2}-\frac{r_s^2}{2r^3}\right)\left(\frac{\text{d}t}{\text{d}\tau}\right)^2-\left(\frac{r_s}{2\left(1-\frac{r_s}{r}\right)r^2}\right)\left(\frac{\text{d}r}{\text{d}\tau}\right)^2-\left(r-r_s\right)\left(\frac{\text{d}\theta}{\text{d}\tau}\right)^2-\left(r-r_s\right)\sin^2\theta\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0

impose \frac{dr}{d\tau}=\frac{d\phi}{d\tau}=\frac{d\theta}{d\tau}=0 to obtain

\frac{\text{d}^2r}{\text{d}\tau^2}+c^2\left(\frac{r_s}{2r^2}-\frac{r_s^2}{2r^3}\right)\left(\frac{\text{d}t}{\text{d}\tau}\right)^2=0

Dropping the term in \frac{dr}{d\tau} is what introduces your error.

Now I argue that since v=0 initially, and the particle and observer are per assumption at the same location in space-time, d\tau=dt, giving the equation in #50.

This is wrong as well. Try using the metric I posted for you.

 

[espen180]

Yeah, I got

\frac{\text{d}^2r}{\text{d}\tau^2}=-\frac{GM}{r^2}=\frac{\frac{\text{d}r}{\text{d}t}}{1-\frac{2GM}{rc^2}}

 

[yuiop]

If \frac{dr}{d\tau}=0, what does this say about your differential equation? Do you still have a non-null \frac{\text{d}^2r}{\text{d}\tau^2}?

Yes.

You realize that this is mathematically and physically impossible?

A particle cannot have an acceleration if it has zero velocity?

\frac{d^2r}{d\tau^2}=\frac{d}{d\tau}(\frac{dr}{d\tau})

What does this tell you? You and kev are starting to worry me.

The condition was that that \frac{dr}{d\tau} was momentarily zero, not constantly. We are talking about free fall here.

... Third off, you are mixing an initial condition (\frac{dr}{d\tau}|_{\tau=0}=0) with the general condition \frac{dr}{d\tau}=0. You are using the general condition (i.e.\frac{dr}{d\tau}=0 everywhere) in order to drop terms from your differential equation. You have done this error before meaning that you don't understand differential equations.
The radial motion problem was already solved in another thread. I already showed kev the solution for this problem.

Dear starthaus,

Since you seem to have a very poor grasp of the most elementary physics, let us go back back to Newtonian physics and review the basics.

Consider a ball thrown vertically upwards. To a good aproximation, if the ball is not thrown too high, the acceleration is 9.8m/s. At the apogee (the maximum height of the ball's trajectory or even more basically when the ball stops going upwards and starts falling back down again) the average velocity is zero or in the infinitesimal limit the velocity is momentarily exactly zero. At this point dr/dt = 0 and the acceleration is non-zero (d^2r/dt^2 = 9.8m/s. If the acceleration was zero at the apogee the ball would remain at its maximum height and not fall down again. You can experimentally prove that this is not the case in your own back garden.

Now in the Schwarzschild metric dr/dt=0 means the vertical velocity of the test particle is momentarily zero at a given instant. (Recall that by definition dt means and infinitesimal interval of time - Please review an introductory textbook on calculus if you have forgotten this basic fact.) Setting dr/dt=0 says nothing about the motion of the particle in the next instant or in the previous instant. Therefore setting dr/dt=0 in the metric does not by itself, imply anything about (as you put it) whether dr/dt=0 is an "initial condition" or a "general condition". The only way you can determine if it is an initial condition (I prefer a momentary condition) or a general condition (I prefer to say a condition that does not vary over time) is by determining whether or not the acceleration of the particle is non-zero or zero. Setting dr/dt equal to zero, does not by itself imply d^2r/dt^2 must also be zero. If you think this is mathematically impossible, then you need to refresh your basic math skills as well as your basic physics knowledge.

 

[starthaus]

Dear starthaus,

Since you seem to have a very poor grasp of the most elementary physics, let us go back back to Newtonian physics and review the basics.

Personal attacks can't cover for your errors. If you don't know how to form and solve the equations of motion, there is always time to take a calculus class.

Now in the Schwarzschild metric dr/dt=0 means the vertical velocity of the test particle is momentarily zero at a given instant.

This is not what espen180 was doing. I think he finally understood his error. You need to be working a little harder to understand it.

(Recall that by definition dt means and infinitesimal interval of time - Please review an introductory textbook on calculus

Coming from you, this is rich :LOL:

Setting dr/dt equal to zero, does not by itself imply d^2r/dt^2 must also be zero.

What in \frac{d^2r}{dt^2}=\frac{d}{dt}(\frac{dr}{dt}) do you still struggle with?

 

[espen180]

This is not what espen180 was doing. I think he finally understood his error. You need to be working a little harder to understand it.

In my defense, I never intended dr/ds=0 to be true for s>0. I just wanted to derive that case before trying for the general case of nonzero dr/ds.

 

[starthaus]

In my defense, I never intended dr/ds=0 to be true for s>0. I just wanted to derive that case before trying for the general case of nonzero dr/ds.

True. I think that now you can explain to kev the error he's stuck in.

 

[yuiop]

What in \frac{d^2r}{dt^2}=\frac{d}{dt}(\frac{dr}{dt}) do you still struggle with?

Please re-read my last post slowly where I have tried to explain as simply as possible, why {dr}/{dt}=0 does not by itself imply d^2r/dt^2= 0.

 

[starthaus]

Please re-read my last post slowly where I have tried to explain as simply as possible, why {dr}/{dt}=0 does not by itself imply d^2r/dt^2= 0.

I don't need to, espen180 knows now how to explain to you your errors.I am certain he's now able to do a very good job explaining the many mistakes you're making.
Now, you wasted a lot of his time leading him astray, if you'd be so kind to stop posting nonsense, we can finish his second exercise, like we have finished the one set in the OP. If you listen carefully for a change, you might learn how to solve the problem.

 

[yuiop]

This is not what espen180 was doing. I think he finally understood his error. You need to be working a little harder to understand it.

What espen was doing was analysing the initial acceleration with the initial condition that dr/dt=0. The initial acceleration in the limit that the velocity is as close to zero as you desire, is accurately given in the equation given by espen. His final equation is not in error. He made a minor slip in the interpretation of the equation due to confusion over coordinate time versus proper time, but he quickly acknowledged that, indicating that it was a minor oversight rather than an ingrained misconception on his part.

It has been explained to you by numerous people that a special case is not an error, but just a limited case of a more general solution. You have wasted a lot of time and caused a lot of confusion in many threads by claiming that any none fully generalised solution is an error.

As I explained to you before, by your definition, SR (the special case of flat spacetime) is an error and the Schwarzschild solution (the special case of a non rotating non charged gravitational body) is an error. Please stop this irritating habit.

 

[starthaus]

It has been explained to you by numerous people that a special case is not an error, but just a limited case of a more general solution.

If you don't understand elementary calculus and how it applies to the equations of motion, it is ok. Why don't you listen for a little while and you'll learn how to solve the problems instead of hacking them.
You can start by understanding the solution for circular orbits, it contains the solution for radial motion as a particular case.

 

[espen180]

You can start by understanding the solution for circular orbits, it contains the solution for radial motion as a particular case.

It does? Doesn't circular motion assume dr/ds=0 for all s? Such that letting dø/ds->0 makes r go to infitiny?

 

[starthaus]

It does? Doesn't circular motion assume dr/ds=0 for all s? Such that letting dø/ds->0 makes r go to infitiny?

No, you only need to make \frac{d\phi}{ds}=0 in the third Euler-Lagrange equation I gave you in post 53. I solved for you both problems.

 

[espen180]

Okay, if I want to find an expression for the acceleration of a particle without assuming dr/ds=0 initially, I have to solve

\frac{\text{d}^2t}{\text{d}\tau^2}+\left(1-\frac{r_s}{r}\right)^{-1}\frac{r_s}{r^2}\frac{\text{d}t}{\text{d}\tau}\frac{\text{d}r}{\text{d}\tau}=0

\frac{\text{d}^2r}{\text{d}\tau^2}+c^2\left(\frac{r_s}{2r^2}-\frac{r_s^2}{2r^3}\right)\left(\frac{\text{d}t}{\text{d}\tau}\right)^2-\left(\frac{r_s}{2\left(1-\frac{r_s}{r}\right)r^2}\right)\left(\frac{\text{d}r}{\text{d}\tau}\right)^2=0

So I figure the first step is to substitute the dt/dτ. The problem is that

\text{d}t=\text{d}\tau \sqrt{\left(1-\frac{r_s}{r}\right)\left(1-\left(\frac{\frac{\text{d}r}{\text{d}t}}{c^2}\right)^2\right)}

which, if correct, would mean that I get a mix of derivatives of r wrt. t and τ. So I have no idea where to start, or if the equations are analytically solvable.

 

[starthaus]

Okay, if I want to find an expression for the acceleration of a particle without assuming dr/ds=0 initially, I have to solve

\frac{\text{d}^2t}{\text{d}\tau^2}+\left(1-\frac{r_s}{r}\right)^{-1}\frac{r_s}{r^2}\frac{\text{d}t}{\text{d}\tau}\frac{\text{d}r}{\text{d}\tau}=0

\frac{\text{d}^2r}{\text{d}\tau^2}+c^2\left(\frac{r_s}{2r^2}-\frac{r_s^2}{2r^3}\right)\left(\frac{\text{d}t}{\text{d}\tau}\right)^2-\left(\frac{r_s}{2\left(1-\frac{r_s}{r}\right)r^2}\right)\left(\frac{\text{d}r}{\text{d}\tau}\right)^2=0

So I figure the first step is to substitute the dt/dτ. The problem is that

\text{d}t=\text{d}\tau \sqrt{\left(1-\frac{r_s}{r}\right)\left(1-\left(\frac{\frac{\text{d}r}{\text{d}t}}{c^2}\right)^2\right)}

which, if correct, would mean that I get a mix of derivatives of r wrt. t and τ. So I have no idea where to start, or if the equations are analytically solvable.

You aren't listening...and you keep making new mistakes. Combine posts 84 and 53 and you'll get the solution you are after.

 

[Elkanah]

Espen180 wat were u xpectin 2 deriv at d end of ur calculatn?

 

[starthaus]

Espen180 wat were u xpectin 2 deriv at d end of ur calculatn?

He wanted the equation of motion for circular orbits. I solved it for him in post 53.
Now he wants the equation of motion for purely radial motion. He can use the solution at post 53 in order to find it.

 

[Elkanah]

Espen180, starthaus has said it all. Jst b careful enough not 2 continue making mistakes.

 

[espen180]

You aren't listening...and you keep making new mistakes. Combine posts 84 and 53 and you'll get the solution you are after.

I was writing #85 when you posted #84, so pardon me for not noticing it.

The equation in #53, after imposing dø/ds=0:

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

\frac{\text{d}}{\text{d}s}\frac{2}{\alpha}\frac{\text{d}r}{\text{d}s}=2\frac{\text{d}r}{\text{d}s}\frac{\text{d}}{\text{d}s}\frac{1}{\alpha}+\frac{2}{\alpha}\frac{\text{d}^2r}{\text{d}s^2}

It may be because I'm quite tired, but I think I need a push on solving this.