Orbital velocities in the Schwartzschild geometry

[starthaus]

It's true that there is a differentiation error, but that doesn't invalidate the proof since the result was unchanged.

$$\frac{H_c^2}{4r^4}(r-3M)=0$$

still gives

$$r=3M$$

No, it is not a "differentiating error", it is numerology.Apparently you share the same misconception with kev, that constants can be differentiated resulting into non-zero algebraic expressions.

 

[yuiop]

It's true that there is a differentiation error, but that doesn't invalidate the proof since the result was unchanged.

$$\frac{H_c^2}{4r^4}(r-3M)=0$$

still gives

$$r=3M$$

I was aware that

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

should be

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

but since the above equation is effectively

$$0 = 2 r^2 \frac{d\phi}{dt}$$

it is perfectly valid to divide both sides by 2. As long as the substitution is done correctly (as Espen has shown) it does not matter what multiple you use. The Lagrangian does not tell you the numerical value of the constants, it just tells you that they are in fact constants. I am not a mathematician so I can not explain it rigorously. I am just observing how others do it.

 

[starthaus]

I was aware that

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

should be

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

but since the above equation is effectively

$$0 = 2 r^2 \frac{d\phi}{dt}$$

it is perfectly valid to divide both sides by 2. As long as the substitution is done correctly (as Espen has shown) it does not matter what multiple you use.

that is a minor error.
Why don't you address the major error?

 

[yuiop]

that is a minor error.
Why don't you address the major error?

It's not.

 

[starthaus]

It's not.

So, according to you differentiating $$0$$ produces a polynomial ? You really need to learn calculus.

 

[yuiop]

So, according to you differentiating $$0$$ produces a polynomial ? You really need to learn calculus.

Are you saying that you can not differentiate the left hand side of ax^2 + bx + c = 0 because the right hand side is zero?

 

[starthaus]

Are you saying that you can not differentiate the left hand side of ax^2 + bx + c = 0 because the right hand side is zero?

If you do that , you get something as meaningless as you have been getting in your "derivation', i.e. 2ax+b=0.

You declared $$L=0$$. Otherwise, you can't execute all the manipulations you have been executing. So, you can't change your mind and differentiate it and obtain a non-zero result. This is calculus 101.

 

[espen180]

If you do that , you get something as meaningless as you have been getting in your "derivation', i.e. 2ax+b=0.

You declared $$L=0$$. Otherwise, you can't execute all the manipulations you have been executing. So, you can't change your mind and differentiate it and obtain a non-zero result. This is calculus 101.

This makes sense. But still, even though you say the derivations are erronous, they are producing correct results, so there must be something compensating for the errors. For example, you say that post #8 that I am wrong when obtaining three differential equations, but how do you then explain that they give the correct orbits?

 

[Al68]

This makes sense. But still, even though you say the derivations are erronous, they are producing correct results...

Imagine that. Someone can technically violate someone else's "rules" of calculus and get the correct result? Seriously, that's done routinely. And as has been pointed out in other threads, there are often ways to prevent the "broken rule" from rendering the result incorrect.

 

[starthaus]

$$({dr}/{dt})^2 = (1-2M/r)(1-2M/r -H_c^2/r^2)$$

Differentiate the above with respect to r and divide by 2 (this is the same as differentiating dr/dt with respect to t, but is much quicker and simpler) to obtain the radial acceleration of a photon in the metric:

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

.

Nope, your "differentiation" skills are still bad, in addition to the errors I flagged down earlier you are missing a whole bunch of terms in the above differentiation.Show your steps and you'll find out what terms are missing.
You simply put in the desired answer by hand. According to Al68 and espen180, the glaring errors don't matter, it is important that you got the "expected" result.

 

[Al68]

According to Al68 and espen180, the glaring errors don't matter, it is important that you got the "expected" result.

I didn't say errors don't matter. I said textbook calculus rules were broken routinely. I meant on purpose, not by "error".

Sorry if that wasn't clear.

 

[yuiop]

Nope, your "differentiation" skills are still bad, in addition to the errors I flagged down earlier you are missing a whole bunch of terms in the above differentiation.Show your steps and you'll find out what terms are missing.
You simply put in the desired answer by hand. According to Al68 and espen180, the glaring errors don't matter, it is important that you got the "expected" result.

I have to admit you are right and I made a mistake in the calculations. I do not claim to have any calculus skills and simply use mathematical software (e.g. http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=calculus&s2=differentiate&s3=advanced. ) Unfortunately I plugged in the wrong expression:

$$({dr}/{dt})^2 = (1-2M)(1-2M/r -H_c^2/r^2)$$

instead of:

$$({dr}/{dt})^2 = (1-2M/r)(1-2M/r -H_c^2/r^2)$$

and got a seemingly correct solution by a bizarre self cancelling double error (GIGO). I will try and fix that. You claim to be good at calculus and have an understanding of the Euler-Lagrange formalism for the equations of motion, so maybe you could demonstrate the correct solution to us?

 

[yuiop]

You claim to be good at calculus and have an understanding of the Euler-Lagrange formalism for the equations of motion, so maybe you could demonstrate the correct solution to us?

Seeing as how Starthaus has failed to rise to the challenge (presumably because the solution is not in one of his textbooks) I will make this second attempt to find an alternative direct derivation of the coordinate orbital velocity and orbital radius of a photon.

Starting with Schwarzschild metric and assuming orbital motion in a plane about the equator such that $\theta = \pi/2$ and $d\theta = 0$

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

$$\alpha=1-\frac{2m}{r}$$

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 $d\phi/dt$

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

The metric is independent of s and t, so there is a constant associated with time dilation $(K_c)$ which is obtained by finding the partial derivative of L with respect to $ds/dt$

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

Substitute these constants into the equation for L and solve for (dr/dt)^2:

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

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

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

Special case of an orbiting photon:

For a photon, ds=0 and therefore $K_c=0$ and for a circular orbit dr/dt =0. Substituting these values into the equation for $(dr/dt)^2$ and solving for $H_c^2$ gives:

$$H_c^2 = r^2/\alpha$$

For a circular orbit we also require the acceleration to be zero. Substituting $H_c^2 = r^2/\alpha$ and $K_c=0$ into the equation for the general coordinate radial acceleration (a) gives:

$$0 = \frac{\alpha}{r^2}(r-5M) +\frac{2M\alpha}{r^2}$$

and solving for r gives r=3M.

Still using a = 0 and $K_c=0$ and substituting the expanded version of [itex]H_c[/tex] back into the equation for (a) gives:

$$0 = \frac{d\phi^2}{dt^2}(r-5M) +\frac{2M}{r^2}(1-2M/r)$$

Solve the above for the coordinate tangential velocity of an orbiting photon:

$$\frac{r d\phi}{dt} = \sqrt{\frac{-2M}{r}\frac{(r-2M)}{(r-5M)}}$$

At r=3M this gives:

$$\frac{r d\phi}{dt} = \frac{1}{\sqrt{3}}$$

which is in agreement with the equation I obtained in #27 based on Espen's derivation:

If you convert the equation to coordinate velocity using [tex]dt = dt'\sqrt{1-2M/(rc^2)}[/itex] you get:

$$\frac{rd\phi}{dt} = c \sqrt{\frac{GM}{r}$$

The local velocity according to a stationary observer at r=3M is:

$$\frac{r d\phi}{dt} \frac{dt}{dt'} = \frac{r d\phi}{dt'} = \frac{1}{\sqrt{3}}\frac{1}{\sqrt{1-2M/r}} = 1$$

 

[starthaus]

Seeing as how Starthaus has failed to rise to the challenge (presumably because the solution is not in one of his textbooks)

No, it is because I am tired of correcting your errors and hacks.

 

[espen180]

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.

 

[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 $$ds = dt'\sqrt{1-(rd\phi)^2/(cdt')^2}$$ you get:

$$\frac{rd\phi}{dt'} = c \sqrt{\frac{GM}{rc^2 - 2GM}$$

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 [tex]dt = dt'\sqrt{1-2M/(rc^2)}[/itex] you get:

$$\frac{rd\phi}{dt} = c \sqrt{\frac{GM}{r}$$

which is the same as the Newtonian result (if you use units of c=1).

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:

$$dt ' = dt \sqrt{1-r_s/r}$$

and if the particle is stationary at r, then $d\tau = dt'$

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.