Orbital velocities in the Schwartzschild geometry

[starthaus]

That post was about the acceleration of a particle immediately after being dropped from rest. The word "rest" should justify the dr=0 in the geodesic equations.

Not if you are unable to derive the equations of motion.I don't know why you persist in hacking the metric. I thought that your whole thread is about deriving the equations of motion, no? I gave you the model of how this is done for circular orbits, the same exact model applies to radial motion. Once you get the equation of motion you can get the correct expressions for velocities, acceleration, whatever you may need.

 

[starthaus]

So kev is wrong that $$dr=0$$ for a stationary particle because espen180's post refers to radial motion? Still makes no logical sense.

I spent quite a few posts explaining to you why you don't hack the metric (nor the equation of motion) by hacking in $dr=0$ (or $\frac{dr}{d\tau}=0$) by hand. Do you think that you could spend your time in a more useful way helping espen180 get the correct equation of motion from the correct (unhacked) metric? The benefit would be obtaining the correct equation of motion, velocity and acceleration.

 

[starthaus]

I'm trying to use the tensor formulation of GR to calculate the velocity of a particle in a circular orbit around a black hole.

Here is the work I have done so far.

What concerns me is that I end up getting zero velocity when applying the metric to the differential equations I get from the geodesic equation. I wonder if I have made a miscalculation, but I am unable to find any, so maybe there is a misunderstanding on my part.

Any help is appreciated.

Post 53 answers your question. Now, that you see the general solution, how would you apply it to radial (instead of circular) motion?

 

[yuiop]

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.

Hi espen,

The equations are analytically solvable and all the information you need is already contained in the Schwarzschild metric for radial motion only:

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

Divide through by $(1-r_s/r) d\tau^2$ and rearrange:

$$c^2\left(\frac{dt}{d\tau}\right)^2 = \frac{c^2}{(1-r_s/r)} + \frac{1}{(1-r_s/r)^2} \left(\frac{dr}{d\tau} \right)^2$$

Substitute this expression for $(c dt/d\tau)^2$ into your second equation and you obtain:

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

Simplify the above:

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

This result means that the acceleration using these coordinates is independent of the falling velocity dr/ds because we have made no assumption of dr/ds=0. It also means that the acceleration is non-zero for any value dr/ds and so it is proved that the claim by Starthaus that the acceleration is zero when dr/ds=0, is false.

Now you may have heard that acceleration IS dependent on velocity and also that the time dilation factor is the product of time dilation due to gravitational potential and time dilation due to velocity. The above calculations seem to contradict this, but it all makes sense when you are careful about who makes the measurements. The following is an attempt to clarify various coordinate measurements and define a notation convention:

Schwarzschild coordinate measurement:

These are measurements made by a stationary observer at infinity and use the notation r, t, dr and dt.

Local measurements:

These are measurements made by a stationary observer at r of a free falling particle passing r and are indicated by the use of primed variables, e.g. r', t', dr' and dt'.

Co-moving measurements:

These are measurements made by an free-falling observer that is co-moving with respect to the free falling particle and local to the particle. In these coordinates the, free falling particle's velocity is always zero and the acceleration of the particle is always zero.
These measurements are indicated by the zero substript as $r_o, t_o, dr_o$ and $dt_o$.
In the Schwarzschild metric $dt_o$ means the same as the infinitesimal proper time interval $d\tau$ or $ds/c$.

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

Using these definitions and considering a free falling particle with radial motion only, the Schwarzschild coordinate acceleration according to an observer at infinity is:

$$\frac{d^2 r}{dt^2} = -\frac{GM}{r^2}\left((1-r_s/r)-\frac{3(dr/dt)^2}{(1-r_s/r)}\right)$$

(Note that this measurement of acceleration is velocity dependent.)

This equation can be obtained from the general equation I derived earlier in post #48:

... the general coordinate radial acceleration of a freefalling particle in the metric:

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

by setting the angular velocity constant $H_c$ to zero and substituting the full form of $K_c$ back in.

The acceleration of the free falling particle according to a local observer that is stationary at r is:

$$\frac{d^2r '}{dt' ^2} = - \frac{GM}{r^2}\left(\frac{1-(dr '/dt ')^2}{\sqrt{1-2GM/r}}\right)$$

The derivation of the above equation can found in post #https://www.physicsforums.com/showpost.php?p=2747788&postcount=345".

The acceleration of the free falling particle, using a mixture of distance measured by the Schwarzschild observer at infinity and time as measured by a co-free-falling observer is:

$$\frac{d^2r}{dt_o^2} = -\frac{GM}{r^2}$$

This is the expression I derived at the start of this post from your acceleration equation which is in turn obtained from the derivation in your document http://sites.google.com/site/espen180files/Schwartzschild.pdf?attredirects=0&d=1 (using equation (9))

The acceleration of the free falling particle, using distance and time measured by a co-free-falling observer is:

$$\frac{d^2r_o}{dt_o^2} = 0$$

This I suspect is something like the acceleration derived by Starhaus, but he is using co-moving in an inconsistent way (the observer is not co-rotating in his orbital calculations) and he using the symbol dr to mean distance measured by a free falling observer when everyone else is using dr to mean distance measured by the Schewarzschild observer. This is why Starthaus said the velocity dr/dt of a stationary particle at r is not zero, because by his definition of dr/dt being measured by a free falling observer, the stationary particle does not appear to be stationary. Starthaus never defines his variables in physical terms, never checks the physical implications or conclusions of his equations and never checks if the symbols he is using have a different physical meaning to the symbols being used by everyone else. He just blindly applies calculus to symbols (defined differently to everyone else) and when his results do not look the same as everyone elses (not surprising really) he declares everyone else to be wrong. (For someone who is supposed to be good at calculus, his basic algebra is surprisingly shaky too.)

 

[yuiop]

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.

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

In a different thread https://www.physicsforums.com/showpost.php?p=2737116&postcount=215 you said:

$$\frac{d^2r}{dt^2}=-\frac{m}{r^2}(1-2m/r)(3\frac{1-2m/r}{1-2m/R}-2})$$

Now R is the apogee height or release point and when R=r the velocity dr/dt is zero. Setting R=r gives the acceleration at apogee as:

$$\frac{d^2r}{dt^2}= -\frac{m}{r^2}(1-2m/r)$$

This is non-zero, so you are contradicting yourself when you claim that dr/dt=0 means that d^2r/dt^2 must also be zero.

 

[yuiop]

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.

The time dilation equation is not correct. Start with the Schwarzschild metric for radial motion only:

$$c^2 dt_o^2 = (1-r_s/r) c^2 dt^2 - \frac{1}{(1-r_s/r)} dr^2$$

$$\left(\frac{dt_o}{dt}\right)^2 = (1-r_s/r) - \frac{(dr/dt)^2}{c^2(1-r_s/r)}$$

$$\left(\frac{dt_o}{dt}\right)^2 = (1-r_s/r) \left(1 - \frac{(dr/dt)^2}{c^2(1-r_s/r)^2} \right)$$

$$\left(\frac{dt_o}{dt}\right)^2 = (1-r_s/r) \left(1 - \frac{(dr '/dt ')^2}{c^2} \right)$$

$$dt_o = dt \sqrt{\left(1-\frac{r_s}{r}\right) \left(1 - \frac{(dr '/dt ')^2}{c^2}\right) }$$

Of course you now have a mixture of dt, $dt_o$ and dt' but this is not a problem. I demonstrated how to get around this in the earlier post.

 

[starthaus]

Hi espen,

The equations are analytically solvable and all the information you need is already contained in the Schwarzschild metric for radial motion only:

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

Divide through by $(1-r_s/r) d\tau^2$ and rearrange:

$$c^2\left(\frac{dt}{d\tau}\right)^2 = \frac{c^2}{(1-r_s/r)} + \frac{1}{(1-r_s/r)^2} \left(\frac{dr}{d\tau} \right)^2$$

Substitute this expression for $(c dt/d\tau)^2$ into your second equation and you obtain:

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

Simplify the above:

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

Very good, you finally managed to reproduce the equation of motion from my blog. If you knew how to use the Euler-Lagrange method you could have done it in two lines weeks ago.

This result means that the acceleration using these coordinates is independent of the falling velocity dr/ds because we have made no assumption of dr/ds=0.

Good, so you finally understood that the hack $$\frac{dr}{ds}=0$$ is not correct and that you could get the equation of motion without resorting to this hack. This is what I've been telling you repeatedly.

It also means that the acceleration is non-zero for any value dr/ds and so it is proved that the claim by Starthaus that the acceleration is zero when dr/ds=0, is false.

What I've been telling you is something totally different. If you insist on hacking the metric by putting $$\frac{dr}{ds}=0$$ by hand as you've been doing, then, by virtue of elementary calculus, you'd get $$\frac{d^2r}{ds^2}=0$$.

This I suspect is something like the acceleration derived by Starhaus, but he is using co-moving in an inconsistent way (the observer is not co-rotating in his orbital calculations)

You mean that you don't understand the general solution at post 56? Why don'y you say so? I can explain it to you.

and he using the symbol dr to mean distance measured by a free falling observer when everyone else is using dr to mean distance measured by the Schewarzschild observer.

Umm, no. The definition is given by the Schwarzschild metric itself.

This is why Starthaus said the velocity dr/dt of a stationary particle at r is not zero,

No, what I said is that you can't hack the metric by putting $$\frac{dr}{ds}=0$$ by hand as you have been doing. That's all.

 

[yuiop]

Your post #53 is very suspect. Let's go through it.

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 work out the equations of motion for a photon (which by definition always has ds=0) then when you divide both sides by ds^2 you end up with:

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

$$\frac{0}{0} =\alpha \frac{dt^2}{0}-\frac{1}{\alpha}\frac{dr^2}{0}-r^2d\frac{\phi^2}{0}$$

which is undefined on the left and a bunch of infinite terms on the right. This is a very shaky foundation for a rigorous derivation.

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 defined:

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

where $K_c$ is the constant with respect to cordinate time. It is not incorrect just because it is not the same as the constant with respect to proper time. I never intended $K = K_c$. They are two different things. In fact $K = 1/K_c$.

$H$ and $H_c$ are two different things too.

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

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

Let's break down your steps:

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

You then introduce R without defining it, but in our previous discussions we have used R to mean the radial coordinate when the particle is at its apogee or perigee and dr/ds=0. You then claim that when dr/ds=0 that the terms containing dr/ds cancel out (this is not always true so it is a false assumption) so that you obtain:

$$0-(2m/r^2(dt/ds)^2-0-2r(d\phi/ds)^2)=0$$

which solves to:

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

and not as you claim:

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

You can add this to the list of basic algebra blunders you have already made in this thread.

 

[starthaus]

In a different thread https://www.physicsforums.com/showpost.php?p=2737116&postcount=215 you said:

$$\frac{d^2r}{dt^2}=-\frac{m}{r^2}(1-2m/r)(3\frac{1-2m/r}{1-2m/R}-2})$$

Now R is the apogee height or release point and when R=r the velocity dr/dt is zero. Setting R=r gives the acceleration at apogee as:

$$\frac{d^2r}{dt^2}= -\frac{m}{r^2}(1-2m/r)$$

Yes, I solved this in my blog weeks ago. Did you just get around to reading it?

This is non-zero, so you are contradicting yourself when you claim that dr/dt=0 means that d^2r/dt^2 must also be zero.

Well, what I've been telling you is something entirely different :
-that you can't hack the metric by putting $$\frac{dr}{dt}=0$$ by hand as you've been doing. Once you do this hack, it has the consequence that $$\frac{d^2r}{dt^2}=0$$ curtesy of basic calculus. This is not about the relationship between instantaneous speed and instantaneous acceleration but between the function $$\frac{dr}{dt}$$ and the function $$\frac{d^2r}{dt^2}$$. Do you understand the difference?

 

[starthaus]

Your post #53 is very suspect. Let's go through it.



If you work out the equations of motion for a photon (which by definition always has ds=0) then when you divide both sides by ds^2 you end up with:

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

$$\frac{0}{0} =\alpha \frac{dt^2}{0}-\frac{1}{\alpha}\frac{dr^2}{0}-r^2d\frac{\phi^2}{0}$$

which is undefined on the left and a bunch of infinite terms on the right. This is a very shaky foundation for a rigorous derivation.

LOL. You really need to take a class in calculus. If you want the lightlike metric, you only need to set $$ds=0$$. If you manage to do this correctly, you will be rewarded by getting:

$$\alpha dt^2-dr^2/\alpha=0$$

I snipped the rest of your nonsense.

 

[yuiop]

Very good, you finally managed to reproduce the equation of motion from my blog. If you knew how to use the Euler-Lagrange method you could have done it in two lines weeks ago.

I have known all along that the equations given by Espen in his document are correct. You have been telling him he has been doing it wrong and he has lost confidence in his own calculations and abilities and that is why I have gone to the trouble to explain again that there is nothing wrong with his latest document.

Good, so you finally understood that the hack $$\frac{dr}{ds}=0$$ is not correct and that you could get the equation of motion without resorting to this hack. This is what I've been telling you repeatedly.

It is not incorrect. It is just a special case at apogee and the conclusions drawn from it are only valid at apogee.

You mean that you don't understand the general solution at post 56? Why don'y you say so? I can explain it to you.

If I could be bothered I could show you that your equation leads to the acceleration being zero for all r. This contradicts the claim you are now making that the acceleration is GM/r^2. It is obvious from this result that the acceleration is non-zero even when the dr/dt=0 because the acceleration in this form is independent of dr/dt.

 

[Mentz114]

Hi Espen,

using software like Maxima you can reduce errors ( or at least check your results ). I've listed a Maxima script that does something similar to your calculation. It might be of interest.

/*  
Schwarzschild : orbital motion

save as Scwarz-orbit.mac

Load from the file menu - > File|Batch file
*/
kill(all);
load(ctensor);
/* set some flags */
cframe_flag: false;
ratchristof: true;
ratriemann : true;
ratfac : true;
ctrgsimp: true;
/* define the dimension */
dim: 4;
/* list the coordinates */
ct_coords: [t,r,theta,phi];
/* set up the metric */
/* assign to lg a matrix of zeros  ':' means 'assign' */
lg:zeromatrix(4,4);
/* now add the Schwarzschild coefficients */
lg[1,1]:-c^2*(1-2*m/r);
lg[2,2]:1/(1-2*m/r);
lg[3,3]:r^2;
lg[4,4]:r^2*sin(theta)^2;
/* make the inverse matrix */
ug:invert(lg);
/* get Christoffels */
christof(mcs);
cgeodesic(true);

/** for orbital motion set r,phi derivs to 0 **/
geod[1]:subst(Tdotdot, 'diff(t,s,2), geod[1])$
geod[1]:subst(0, 'diff(r,s,1), geod[1])$

geod[2]:subst(0, 'diff(r,s,2), geod[2])$
geod[2]:subst(0, 'diff(r,s,1), geod[2])$
geod[2]:subst(Thetadot, 'diff(theta,s,1), geod[2])$
geod[2]:subst(0, 'diff(phi,s,1), geod[2])$
geod[2]:subst(tdot, 'diff(t,s,1), geod[2])$

geod[3]:subst(0, 'diff(r,s,2), geod[3])$
geod[3]:subst(0, 'diff(r,s,1), geod[3])$
geod[3]:subst(Thetadot, 'diff(theta,s,1), geod[3])$
geod[3]:subst(0, 'diff(phi,s,1), geod[3])$
geod[3]:subst(tdot, 'diff(t,s,1), geod[3])$

geod[4]:subst(0, 'diff(r,s,2), geod[4])$
geod[4]:subst(0, 'diff(r,s,1), geod[4])$
geod[4]:subst(Thetadot, 'diff(theta,s,1), geod[4])$
geod[4]:subst(0, 'diff(phi,s,1), geod[4])$
geod[4]:subst(tdot, 'diff(t,s,1), geod[4])$

/* these are all = 0 giving 4 diff. equations */
geod[1];
geod[2];
geod[3];
geod[4];

solve([geod[2] ], [Thetadot]);

 

[starthaus]

I have known all along that the equations given by Espen in his document are correct.

..up to the point where you introduce the hack $$dr=0$$

If I could be bothered I could show you that your equation leeds to the acceleration being zero for all r.

You really need to take the time and to understand how the results were derived. It will require that you learn calculus, the theory of differential equations and the lagrangian approach.

 

[espen180]

#109-#111

This is great! I'll take some time to study these posts closely.

Hi Espen,

using software like Maxima you can reduce errors ( or at least check your results ). I've listed a Maxima script that does something similar to your calculation. It might be of interest.

Thank you very much! I will check it out.

..up to the point where you introduce the hack $$dr=0$$

You really need to take the time and to understand how the results were derived. It will require that you learn calculus, the theory of differential equations and the lagrangian approach.

I'm not using the Lagrangian approach. In post #1 I said I am using the geodesic equation. I don't understand where your expression comes from, so I don't like using it.

 

[starthaus]

I'm not using the Lagrangian approach. In post #1 I said I am using the geodesic equation. I don't understand where your expression comes from, so I don't like using it.

The two approaches produce the same exact equations. Whether you use geodesic equations or the Euler-Lagrange equations, they are the same exact equations. Same number of equations, same exact form. The Lagrangian method produces the equations quicker and with less chances of making mistakes.

 

[yuiop]

You mean that you don't understand the general solution at post 56? Why don'y you say so? I can explain it to you.

I understand the solution given by Espen in https://www.physicsforums.com/showpost.php?p=2769019&postcount=56"

Starting with the general case

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

I am dubious about your version in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53"

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

 

[starthaus]

I am dubious about your version in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53"

You need to learn euler-lagrange method and differential equations. The solution is identical with the one produced by the geodesic equations.

 

[yuiop]

You need to learn euler-lagrange method and differential equations. The solution is identical with the one produced by the geodesic equations.

Let's see if the solutions are the same. Start with the solution given by Espen:

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

Using notation r_s= 2m, c=1, $d\tau = ds$, $\alpha = (1-2m/r)$ and conditions $\theta =0$ and $d\theta =0$ to put Espen's solution in the same format as your equation then the following is obtained:

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

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

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

Contrast to your equation:

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

As a calculus expert you should know

$$-2/\alpha*d/ds(dr/ds) \ne -d/ds(1/\alpha*2dr/ds)$$

so your solution is in error.

 

[starthaus]

Let's see if the solutions are the same.

They are the same, this is a known theorem . You need to study.

 

[Dale]

Even when you have a theorem that tells you that two solutions should be the same it is often wise to compare them to see if you made a mistake in one or the other.

 

[yuiop]

If you insist on hacking the metric by putting $$\frac{dr}{ds}=0$$ by hand as you've been doing, then, by virtue of elementary calculus, you'd get $$\frac{d^2r}{ds^2}=0$$.

This is the acceleration equation given by you (after I fixed a major problem with it):

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

With some basic algebra and substitutions (but no assumption of dr/ds=0) this simplifies to:

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

When $d\phi^2/ds^2 = 0$ and dr=0 the equation becomes:

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

so your claim that d/ds (dr/ds) must equal zero when dr/ds=0 is false.

What you are missing is that although the first derivative of zero is zero, the second derivative is not necessarily zero.

This is explained by Dr math like this:

Just like you can have a function whose y-value is 0, but whose slope
is not, you can have a function whose slope is 0 but whose slope is
constantly changing and therefore never has a double-derivative of 0.

Take this example:

f(x) = x^2
f'(x) = 2x
f''(x) = 2

You'll notice here that the derivative is 0 at x = 0, but the double-
derivative is always 2, which means that the slope is always increasing.

See http://mathforum.org/library/drmath/view/65095.html for a full explanation.

No need to thank me for the free calculus lesson

 

[starthaus]

Even when you have a theorem that tells you that two solutions should be the same it is often wise to compare them to see if you made a mistake in one or the other.

kev's counterproof is false

 

[starthaus]

This is the acceleration equation given by you (after I fixed a major problem with it):

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

With some basic algebra and substitutions (but no assumption of dr/ds=0) this simplifies to:

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

Umm, wrong.

When $d\phi^2/ds^2 = 0$ and dr=0 the equation becomes:

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

Also wrong.Try again. If yyou do it right, you should get:

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}$$

If you still can't get it right, feel free to look at the derivation I put in my blog about a month ago.

 

[yuiop]

Also wrong.Try again. If yyou do it right, you should get:

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}$$

This is only true for $d\phi/ds =0$. I thought you were the guy who thinks special cases are wrong and the more generalised solution (which I gave) is the more correct solution? Must have been thinking of someone else...

 

[yuiop]

Umm, no. If you do your algebra correctly, you should get

$$d^2r/ds^2 + m\alpha/ r^3 (dt/ds)^2 - ...$$

You need to redo all your "calculations". If you do them right, you'll find out that the geodesic method produces the same answers as the lagrangian one. In the process , you might want to check espen180's equation as a starting point, I can't vouch for it being correct. I know that my equation is correct.

I have checked and $d^2r/ds^2 + m\alpha/ r^3 (dt/ds)^2 - ...$ is wrong.

You really need to work on your basic algebra skills.

 

[yuiop]

You might want to check your signs , you got them pretty screwed up.

They are correct. You need to pay close attention to the brackets I inserted.

 

[Altabeh]

If you don't know how differential equations describe the equations of motion, that's ok.

I was talking about something else and you unfortunately AGAIN couldn’t get the point as you never can when it comes to creating "Physical mold" of basic ODEs. Your problem is that you don't understand basic calculus and this got exposed itself to me after I asked you a simple question in https://www.physicsforums.com/showpost.php?p=2742039&postcount=306") about 300 posts earlier was in a void attempt at responding to me ignored by you twice in that thread.

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.

Unfortunately you have nothing to teach me!

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.

You're not missing because you can't probably see clearly. Yeah I know and of course I'm going to show which "brother" has been trolling all along since the beginning of this thread.

I assume that the observer who measures the orbital speed of a particle near a gravitating body is hovering so that his 4-velocity is given by

$$u^a=(\gamma_g,0,0,0),$$

where $$\gamma_g=\frac{1}{\sqrt{1-2m/r}}.$$ The orbiting particle itself has the following 4-velocity:

$$u_a=(E,0,0,L),$$

with E being the conserved energy of particle per unit mass and L is the orbital angular momentum of the particle per unit mass, again. But what are the precise expressions corresponding to each of these quantities? From the Newtonian theory of gravitation, we remember that the angular momentum per unit mass of a particle in an orbit at $$r$$ is the simple equation

$$L=(mr)^{1/2},$$

where $$m$$ stands for the gravitating body's mass. Now for the Schwarzschild spacetime, one from the Euler-Lagrange equations would get:

$$L=r^2\dot{\phi},$$

for a particle in a circular orbit at $$r$$ and again with a unit mass. Here the over-dot represents differentiation wrt the parameter of geodesics, e.g. $$s$$. For a circular motion, $$dr/ds=0,$$ and considering motion taking place in the plane for which $$\theta=\pi/2$$ so that the metric

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

gives

$$1=(1-2m/r)\dot{t}^2-r^2d\dot{\phi}^2.$$

On the other hand, from the first Euler-Lagrange equation and $$dr/ds=0,$$ it is obvious that

$$\ddot{t}=0;$$

thus yielding

$$\dot{t}=k=const.$$*

The second of Euler-Lagrange equations (the equation for the radial component) with the same assumptions would lead to the following expression for $$\dot{\phi}$$ (which is very incorrectly given a name like "proper angular velocity" or stuff like that by non-experts):

$$\dot{\phi}^2=\frac{m}{r^3}k^2.$$**

Introducing * and ** into the relationship derived from the Schwarzschild metric gives

$$1=(1-\frac{2m}{r})k^2-\frac{m}{r}k^2\rightarrow$$
$$k^2=\frac{1}{1-3m/r}.$$

Now the expression for $$L$$ reads

$$L=\sqrt{\frac{mr}{1-3m/r}}.$$

The first Euler-Lagrange equation,

$$\frac{d}{ds}[(1-2m/r)\dot{t}]=0$$

if integrated would have teh following simple solution:

$$E=\frac{1-2m/r}{\sqrt{1-3m/r}}=\frac{1}{\gamma_g^2\sqrt{1-3m/r}}=const.$$

To wit, the energy of the particle is a conserved quantity. Now what about the energy per unit mass of particle with respect to the hovering observer? Let such energy be denoted by$$\gamma$$ (with $$c=1$$), then projecting the 4-momentum of the particle onto the 4-velocity of the observer gives

$$\gamma=u^au_a={E}\gamma_g.$$

Recall that the theory of special relativity portrays $$\gamma=1/\sqrt{1-v^2}$$ to hold, when $$c=1$$, between any two inertial frames. So using the equation for $$E$$ above and solving this for the orbital velocity $$v$$ yields

$$v=(m/r)^{1/2}\gamma_g.$$

And we are done. Clearly, putting $$r=3m$$ leads to $$v=1=c$$ which stands for the orbital speed of photons.

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.

How about you now to think about the idea that says "if you don't know what is meant by something, then stop being nonsense when answering?” You're not supposed to give us your nonsense prolongated and boring hints that are much of a pain to a student with a really basic knowledge of calculus and algebra. In the meanwhile, try to learn something about the difference between "proper" and "coordinate" quantities in GR! You can use my post over PF on this topic.

AB

 

[starthaus]

I was talking about something else and you unfortunately AGAIN couldn’t get the point as you never can when it comes to creating "Physical mold" of basic ODEs. Your problem is that you don't understand basic calculus and this got exposed itself to me after I asked you a simple question in https://www.physicsforums.com/showpost.php?p=2742039&postcount=306") about 300 posts earlier was in a void attempt at responding to me ignored by you twice in that thread.
Unfortunately you have nothing to teach me!
You're not missing because you can't probably see clearly. Yeah I know and of course I'm going to show which "brother" has been trolling all along since the beginning of this thread.

I assume that the observer who measures the orbital speed of a particle near a gravitating body is hovering so that his 4-velocity is given by

$$u^a=(\gamma_g,0,0,0),$$

where $$\gamma_g=\frac{1}{\sqrt{1-2m/r}}.$$ The orbiting particle itself has the following 4-velocity:

$$u_a=(E,0,0,L),$$

with E being the conserved energy of particle per unit mass and L is the orbital angular momentum of the particle per unit mass, again. But what are the precise expressions corresponding to each of these quantities? From the Newtonian theory of gravitation, we remember that the angular momentum per unit mass of a particle in an orbit at $$r$$ is the simple equation

$$L=(mr)^{1/2},$$

where $$m$$ stands for the gravitating body's mass. Now for the Schwarzschild spacetime, one from the Euler-Lagrange equations would get:

$$L=r^2\dot{\phi},$$

for a particle in a circular orbit at $$r$$ and again with a unit mass. Here the over-dot represents differentiation wrt the parameter of geodesics, e.g. $$s$$. For a circular motion, $$dr/ds=0,$$ and considering motion taking place in the plane for which $$\theta=\pi/2$$ so that the metric

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

gives

$$1=(1-2m/r)\dot{t}^2-r^2d\dot{\phi}^2.$$

...which is post 53. More correctly, it produces the Lagrangian
$$L=(1-2m/r)\dot{t}^2-r^2d\dot{\phi}^2.$$
Actually, the solution given in post 53 includes the solution for circular orbits as a particular case. Besides , it is much shorter.You are a few days late.

 

[Altabeh]

...which is post 53. Actually, the solution given in post 53 includes the solution for circular orbits as a particular case.You are a few days late.

This is one other nonsense you're making. Actually the point is that you don't read the whole of posts and ignore parts that sound unfamiliar to your probably "limited" skills in GR and this gives rise to many tensions in all threads you're involved in. Now I take your last post as "Yes, I now know I'm not the one who can do basic calculations in GR!" Haha!

AB

 

[starthaus]

This is one other nonsense you're making. Actually the point is that you don't read the whole of posts and ignore parts that sound unfamiliar to your probably "limited" skills in GR and this gives rise to many tensions in all threads you're involved in. Now I take your last post as "Yes, I now know I'm not the one who can do basic calculations in GR!" Haha!

AB

You mean that you don't recognize the Euler-Lagrange equations for arbitrary orbits? You should be familiar with the formalism, it is pretty simple.
Once you understand it, you can follow how I derived the solution for circular orbits in much fewer equations.
Same situation for radial motion. I hope you understood the solution from my blog.

 

[starthaus]

Let's see if the solutions are the same. Start with the solution given by Espen:
Using notation r_s= 2m, c=1, $d\tau = ds$, $\alpha = (1-2m/r)$ and conditions $\theta =0$ and $d\theta =0$ to put Espen's solution in the same format as your equation then the following is obtained:

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

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

If you calculate the lagrangian expression carefully, you should be getting exactly $$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

i.e. the same thing as the geodesic expression. Hint:

$$\frac{d}{ds}(\frac{2/\alpha}{dr/ds})=\frac{2}{\alpha^2}(\alpha\frac{d^2r}{ds^2}-(dr/ds)^2\frac{d\alpha}{dr})$$A nice consequence of the above general equation is that you can derive the equations of motion for

-circular orbits (see post 53)

-radial motion by making $$\frac{d\phi}{ds}=0$$$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 = 0$$

 

[Altabeh]

You mean that you don't recognize the Euler-Lagrange equations for arbitrary orbits?

Actually the problem is that I can't see your "virtual" (maybe imaginary) derivation of the formula of velocity of particles orbiting around a gravitating body from the perspective of a hovering observer other than just a "play-with-Euler-Lagrange-Equations" game at a very basic and incomplete level.

You should be familiar with the formalism, it is pretty simple.

What formalism? The more basic one I showed to you?

Once you understand it, you can follow how to get the solution for circular orbits in much fewer equations.
Same situation for radial motion.

You better start with basic and then go to hard. That is better for you and other students at your level!

AB

 

[starthaus]

Actually the problem is that I can't see your "virtual" (maybe imaginary) derivation of the formula of velocity of particles orbiting around a gravitating body from the perspective of a hovering observer other than just a "play-with-Euler-Lagrange-Equations" game at a very basic and incomplete level.

It is not my fault that you can't follow basic equations. Give it some time, it will come to you.

What formalism? The more basic one I showed to you?

Yes, it is "more basic", it is a bastardization of the one showed in post 53.

You better start with basic and then go to hard.

No need to, I started with the most general. I left the basics for you to rediscover. Unfortunately, you are a few days late.

 

[starthaus]

for a particle in a circular orbit at $$r$$ and again with a unit mass. Here the over-dot represents differentiation wrt the parameter of geodesics, e.g. $$s$$. For a circular motion, $$dr/ds=0,$$ and considering motion taking place in the plane for which $$\theta=\pi/2$$ so that the metric

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

gives

$$1=(1-2m/r)\dot{t}^2-r^2d\dot{\phi}^2.$$

You mean $$L=(1-2m/r)\dot{t}^2-r^2\dot{\phi}^2$$?
Congratulations, you have rediscovered post 2.

On the other hand, from the first Euler-Lagrange equation and $$dr/ds=0,$$ it is obvious that

$$\ddot{t}=0;$$

thus yielding

$$\dot{t}=k=const.$$*

You rediscovered post 6. You are on the right track.

The second of Euler-Lagrange equations (the equation for the radial component) with the same assumptions would lead to the following expression for $$\dot{\phi}$$ (which is very incorrectly given a name like "proper angular velocity" or stuff like that by non-experts):

$$\dot{\phi}^2=\frac{m}{r^3}k^2.$$**

Not very useful . A much more useful one is:

$$\omega=\frac{d\phi}{dt}=\sqrt{\frac{m}{r^3}}$$

So, the coordinate speed is:

$$v_c=r\frac{d\phi}{dt}=\sqrt{\frac{m}{r}}$$

and the proper speed is:

$$v_p=r\frac{d\phi}{d\tau}=r\frac{d\phi}{dt}\frac{dt}{d\tau}=\sqrt{\frac{m}{r}}*1/\sqrt{1-2m/r}$$

$$k^2=\frac{1}{1-3m/r}.$$

...and you rediscovered the formula already derived at the end of post 53.

 

[Altabeh]

It is not my fault that you can't follow basic equations. Give it some time, it will come to you.

And it's not my fault that you can't recognize basic special relativity and difference between proper and coordinate and finally how playing with equations would lead to a completely algebraically proven "formula"; not arrangement of terms and leaving $$dr/ds$$ unidentified so as to call the equation "general"! Haha!

Yes, it is "more basic", it is a bastardization of the one showed in post 53.

Which one? hehe!

No need to, I started with the most general. I left the basics for you to rediscover. Unfortunately, you are a few days late.

Actually you even didn't discover the "basics" so leave alone the general because your derivation is nothing but playing with a bunch of unidentified terms and stuff that can never be identified if not reduced to my formula! Besides, I suppose the students like you unable to deal with difficult material so I start giving primary lessens by making you get involved with basics. How can I teach a student that doesn't know the difference between proper and coordinate more complicated issues?

AB