Orbital velocities in the Schwartzschild geometry

[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.

...because kev put in the result by hand. The calculus errors in his derivation were shown later in this thread. The result is correct, the derivation, as usual, isn't.

 

[espen180]

...because kev put in the result by hand. The calculus errors in his derivation were shown later in this thread. The result is correct, the derivation, as usual, isn't.

Yeah, regarding the derivation.

You told me to use the metric ds²=c²adt²-a^-1dr²-r²dφ² where a=1-r_s/r and use the christoffel symbols I got from that.

That gave me two equations which said that d²t/ds²=0 and d²φ/ds²=0. What troubles me is that I cannot see any way to get numerical values out of these equations. With the other method, I had a way to relate the orbital velocity to the radius, here I don't.

 

[starthaus]

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

\alpha=1-\frac{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

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

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

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

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

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

The other Euler-Lagrange equation is:

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

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

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

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

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

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

i.e.

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

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

Substitute d\phi/ds and dr=0 into the metric equation and you get:

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

The above makes sense only for R>3m

 

[yuiop]

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, I can post my derivation.

Yes it is corrrect.

\frac{\text{d}^2r}{\text{d}t^2}=-\left(\frac{GM}{r^2}-\frac{2G^2M^2}{r^3c^2}\right) = -\frac{GM}{r^2}\left(1 -\frac{GM}{r c^2} \right)

This is exactly the same as the result I obtained for "the initial coordinate acceleration of a test mass released at r" in another thread here https://www.physicsforums.com/showpost.php?p=2710548&postcount=1

It is also the same as the equation given by the mathspages website (See http://www.mathpages.com/rr/s6-07/6-07.htm ) for the coordinate acceleration "At the apogee of the trajectory, when r = R" if you use units of G=c=1.

 

[yuiop]

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.

...because kev put in the result by hand. The calculus errors in his derivation were shown later in this thread. The result is correct, the derivation, as usual, isn't.

There is not one single error (or any calculus for matter) in post #27.

Here is post #27 again. Point to an error if you can.

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

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

which gives the correct result that the velocity with respect to proper time (ds) of a particle orbiting at r=3GM/c^2 is infinite.

If you convert the above equation to local velocity as measured by a stationary observer at r by using ds = dt&#039;\sqrt{1-(rd\phi)^2/(cdt&#039;)^2} you get:

\frac{rd\phi}{dt&#039;} = c \sqrt{\frac{GM}{rc^2 - 2GM}

which gives the result that the local velocity of a particle orbiting at r=3GM/c^2 is c.

If you convert the equation to coordinate velocity using dt = dt&#039;\sqrt{1-2M/(rc^2)}[/itex] you get:<br /> <br /> \frac{rd\phi}{dt} = c \sqrt{\frac{GM}{r}<br /> <br /> which is the same as the Newtonian result (if you use units of c=1).<br /> <br /> All the above are well known solutions, so it seems all is in order with your revised document (from (9) onwards anyway - I have not checked the preceding calculations).

 

[espen180]

Yes it is corrrect.

\frac{\text{d}^2r}{\text{d}t^2}=-\left(\frac{GM}{r^2}-\frac{2G^2M^2}{r^3c^2}\right) = -\frac{GM}{r^2}\left(1 -\frac{GM}{r c^2} \right)

This is exactly the same as the result I obtained for "the initial coordinate acceleration of a test mass released at r" in another thread here https://www.physicsforums.com/showpost.php?p=2710548&postcount=1

I re-read your referenced post. I seems that the equation you arrived at is the acceleration measured by an observer at infinity, while the one I arrived at is for the acceleration measured by a local stationary observer.

Therefore, unless I misread your post, our equations differ by a factor of \sqrt{1-\frac{2GM}{rc^2}}.

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

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.

 

[starthaus]

impose \frac{dr}{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

.

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}?

 

[yuiop]

I re-read your referenced post. I seems that the equation you arrived at is the acceleration measured by an observer at infinity, while the one I arrived at is for the acceleration measured by a local stationary observer.

Therefore, unless I misread your post, our equations differ by a factor of \sqrt{1-\frac{2GM}{rc^2}}.

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 {dr}=d\phi=d\theta=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

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.

You are making the assumption that d\tau=dt when the particle is stationary in the metric but it easy to show that is not correct.

Start with the full Schwarzschild metric:

c^2 d\tau^2 = (1-r_s/r)c^2 dt^2 - \frac{1}{(1-r_s/r)} dr^2 - r^2 d\theta^2 -r^2 \sin^2 (\theta) d\phi^2

Impose dr=d\phi=d\theta=0 to obtain:

c^2 d\tau^2 = (1-r_s/r)c^2 dt^2

d\tau = dt \sqrt{1-r_s/r}

(Not d\tau=dt)

In the Schwarzschild metric dt is always the time measured by the observer at infinity.

The time (dt') measured by a stationary local observer at r is:

dt &#039; = dt \sqrt{1-r_s/r}

and if the particle is stationary at r, then d\tau = dt&#039;

Now let's take your final equation:

\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

and use the correct value of (dt/d\tau)^2 so that:

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

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

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

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

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

This is the acceleration in terms of the proper time of the particle versus coordinate distance. (See the first equation of this mathpages page http://www.mathpages.com/rr/s6-07/6-07.htm).

The equation you gave in the earlier post:

\frac{\text{d}^2r}{\text{d}t^2}=-\left(\frac{GM}{r^2}-\frac{2G^2M^2}{r^3c^2}\right) = -\frac{GM}{r^2}\left(1 -\frac{GM}{r c^2} \right)

is the acceleration in terms of coordinate time versus coordinate distance (both measured by the observer at infinity).

Also see this very technical derivation by Dalespam (post #155) https://www.physicsforums.com/showthread.php?t=402135&page=10

 

[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.

 

[starthaus]

Yes.

You realize that this is mathematically and physically impossible?

 

[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.

 

[starthaus]

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.

You are on the right path (pay attention to the last term, the derivative wrt d/dr, you may have made a mistake , if you do the calculation right there will be some terms that will cancel out, at the end you will be rewarded by a very interesting (and correct) equation of motion.

An even easier approach, as I mentioned before, start with:

L=\alpha(dt/ds)^2-(dr/ds)^2/\alpha

and form the Euler-Lagrange equation wrt dr/ds and r (it is the same as one of the geodesic equations). We can continue tomorrow.

 

[Al68]

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

What is the value of \frac{dr}{dt} for a particle in freefall at the moment of apogee?

 

[starthaus]

What is the value of \frac{dr}{dt} for a particle in freefall at the moment of apogee?

If you don't understand ordinary differential equations, I can recommend a few good courses. In the meanwhile, try trolling other threads.

 

[Al68]

What is the value of \frac{dr}{dt} for a particle in freefall at the moment of apogee?

If you don't understand ordinary differential equations, I can recommend a few good courses. In the meanwhile, try trolling other threads.

LOL. You're such a sweetheart. How can you be so helpful and so pleasant at the same time? Thank you so much for your brilliant, congenial, and non-condescending answer.

 

[starthaus]

LOL. You're such a sweetheart. How can you be so helpful and so pleasant at the same time? Thank you so much for your brilliant, congenial, and non-condescending answer.

If you spent less time trolling and more time studying you would have known that, given the ODE:

\frac{d^2r}{dt^2}+A\frac{dr}{dt}+B=0 for any a&lt;t&lt;b

if you make \frac{dr}{dt}=0 for any a&lt;t&lt;b

this means

\frac{d^2r}{dt^2}=0 any a&lt;t&lt;b

meaning that:

B=0

So, in general, \frac{dr}{dt}=0 is not a solution for the ODE. This is why we try solving the ODE with known methods rather than doing silly things like setting \frac{dr}{dt}=0.

Even worse is the naive attempt by certain members of this forum at calculating \frac{d^2r}{dt^2} by inserting \frac{dr}{dt}=0 in the above ODE and declaring that \frac{d^2r}{dt^2}=-B.

Have you slept through your calculus classes or you never took any?
Now, if you could go troll other threads and leave me to help espen180 find the answer to his second question in this thread, that would be nice. We are almost done with the solution, if you have something to contribute, you are welcome but if you only plan to create trolling noise , than troll elsewhere.

 

[Al68]

If you spent less time trolling and more time studying you would have known that, given the ODE:

\frac{d^2r}{dt^2}+A\frac{dr}{dt}+B=0 for any a&lt;t&lt;b

if you make \frac{dr}{dt}=0 for any a&lt;t&lt;b

this means

\frac{d^2r}{dt^2}=0 any a&lt;t&lt;b

meaning that:

B=0

So, in general, \frac{dr}{dt}=0 is not a solution for the ODE. This is why we try solving the ODE with known methods rather than doing silly things like setting \frac{dr}{dt}=0.

Have you slept through your calculus classes or you never took any?
Now, if you could go troll other threads and leave me to help espen180 find the answer to his second question in this thread, that would be nice.

Now you're just being too much of a sweetheart. How can you be so non-condescending and provide such a direct and specific answer to my question? I'm sure espen180 will appreciate such non-condescending and straightforward answers as much as I did.

 

[starthaus]

Now you're just being too much of a sweetheart. How can you be so non-condescending and provide such a direct and specific answer to my question? I'm sure espen180 will appreciate such non-condescending and straightforward answers as much as I did.

He does. Especiially since I answer his questions.

 

[Altabeh]

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.

Completely seconded. In particular we must add to this the fact that in https://www.physicsforums.com/blog_attachment.php?attachmentid=167&d=1275053682" the author bizarrly argues that d^2r/ds^2 is the proper acceleration of a radially freely falling particle and this has been explained to him several times by many people over other threads but instead of accepting "the error" he continued backing his wrong idea with no consequences.

Physics is not just "mathematical skills"; it requires you to have a skill of painting a physial picture first (the thing that you really sound to be unfamiliar with) and then coloring it with the mathematical hues.

If you spent less time trolling and more time studying you would have known that, given the ODE:

\frac{d^2r}{dt^2}+A\frac{dr}{dt}+B=0 for any a&lt;t&lt;b

if you make \frac{dr}{dt}=0 for any a&lt;t&lt;b

this means

\frac{d^2r}{dt^2}=0 any a&lt;t&lt;b

meaning that:

B=0

So, in general, \frac{dr}{dt}=0 is not a solution for the ODE. This is why we try solving the ODE with known methods rather than doing silly things like setting \frac{dr}{dt}=0.

Uh, this is worse than I thought it would be. Of course \frac{dr}{dt}=0 is not a solution for the ODE and this is 100% true. But you're missing the fact that there are other equations (basically conditions) involved so that differential equation is not alone to be given a condition (not a solution as you may wrongly have reckoned it to be), i.e. \frac{dr}{dt}=0, and this is by considering the motion to be momentarily or instantaneously at rest along the geodesic that particle follows. This has nothing wrong with it and therefore \frac{d^2r}{dt^2}=-B is very correct with the condition given.

Even worse is the naive attempt by certain members of this forum at calculating \frac{d^2r}{dt^2} by inserting \frac{dr}{dt}=0 in the above ODE and declaring that \frac{d^2r}{dt^2}=-B.

So if you don't have any idea as to what probably the whole "physical picture" is, please do not throw stones across the threads.

AB

 

[starthaus]

Physics is not just "mathematical skills"; it requires you to have a skill of painting a physial picture first (the thing that you really sound to be unfamiliar with) and then coloring it with the mathematical hues.

If you don't know how differential equations describe the equations of motion, that's ok.
If you want to learn, then it is not necessary to add your name to the list of trollers, wait a little for espen180 to use the tools I gave him and you'll learn.

Uh, this is worse than I thought it would be. Of course \frac{dr}{dt}=0 is not a solution for the ODE and this is 100% true. But you're missing the fact

No, I'm not missing anything, I am just pointing out that several of you are blissfully basking in the same elementary mistake. Instead of trolling, can you try deriving the equation of motion? It is really simple, you know.

i.e. \frac{dr}{dt}=0, and this is by considering the motion to be momentarily or instantaneously at rest along the geodesic that particle follows. This has nothing wrong with it and therefore \frac{d^2r}{dt^2}=-B is very correct with the condition given.

It is not necessary to resort to your hacks about "momentary" and "instantaneous" motion. If you knew how, you could have derived the general equation of motion, applicable for any t. How about you tried that instead on spending so much energy in ranting? Feel free to use the hints that I gave out in this thread.

 

[Al68]

If you don't know how differential equations describe the equations of motion, that's ok.
If you want to learn, then it is not necessary to add your name to the list of trollers, wait a little for espen180 to use the tools I gave him and you'll learn.




No, I'm not missing anything, I am just pointing out that several of you are blissfully basking in the same elementary mistake. Instead of trolling, can you try deriving the equation of motion? It is really simple, you know.



It is not necessary to resort to your hacks about "momentary" and "instantaneous" motion. If you knew how, you could have derived the general equation of motion, applicable for any t. How about you tried that instead on spending so much energy in ranting? Feel free to use the hints that I gave out in this thread.

Forum rules can be found here: https://www.physicsforums.com/showthread.php?t=5374.

Your repeated violations are too obvious and numerous to point out, and continue in this post.