Orbital velocities in the Schwartzschild geometry

[yuiop]

I don't understand why you insist in continuing to embarrass yourself by showing your ignorance in terms of calculus. Especially since you have been shown several derivations that do not employ the $$dr=0$$ hack.

What is even more embarassing is that even with my limited knowledge of calculus (which is improving all the time) I can out perform you in terms of arriving at correct solutions. I have done several derivations in this thread (and other threads) with and without the so called dr=0 "hack". Unlike you I have the flexibilty to work with either method and understand the domain of validity of each method.

While you "demonstartion" is correct, it simply illustrates your inability to tell the difference between a function and its value in a point. You could easily remedy this if you took a class in calculus. You can't really pretend that you're doing physics when you fail basic calculus. It is really simple, kev, calculus 101 teaches you that if $$f(x)=constant$$ then
$$\frac{df}{dx}=0$$ for all x. There is no way around it.

This is a nice attempt to create your own straw man argument here, but it is simply a distraction from the simple fact that your assertion that the acceleration of a particle at its apogee is zero, because its velocity is zero, is simply wrong.

I can tell the difference between a function and its value at a point and can apply either with equal ease. You on the other hand are unable to answer a simple question like what is the acceleration of a particle at its apogee, because as far as you are concerned there is no such thing as "when dr=0".

Although calculus is important for doing physics and I am working on improving my calculus abilities, I think a basic understanding of physics and algebra is an even more fundamental prerequisite and you seem to lack these.

 

[George Jones]

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[yuiop]

calculus 101 teaches you that if $$f(x)=constant$$ then
$$\frac{df}{dx}=0$$ for all x. There is no way around it. Can you generalize this to the case$$\frac{df}{dx}=constant$$ implies $$\frac{d^2f}{dx^2}=0$$ for all x?

Let us say we have a function of x such that f(x)= x^2.

When x=0 then f(x) = 0.
When x=1 then f(x) = 1.
When x=2 then f(x) = 4 and so on..

Now the first statement (When x=0 then f(x)=0) does not imply (f(x) = constant) and nor does it imply that (f(x) = 0 for all x). So when I say at the apogee x=0 and f(x)=0, I am not saying f(x)=0 for all x. You have to understand the context. Sometimes people use notation like (When x=0 then f(x=0) = 0) to make it clearer that (f(x)=0 for x=0) does not imply (f(x)=0 for all x) and maybe this is what is confusing you.

 

[starthaus]

This is a nice attempt to create your own straw man argument here, but it is simply a distraction from the simple fact that your assertion that the acceleration of a particle at its apogee is zero, because its velocity is zero, is simply wrong.

This is not what I've been telling you. I've been showing you that you don't understand the implication $$f(x)=const$$ => $$\frac{df}{dx}=0$$ in order to help you undestand why you can't just plug in $$dr=0$$ in the geodesic equation for radial motion.

I can tell the difference between a function and its value at a point and can apply either with equal ease.

The fact that you persist demonstrates that you still can't.

 

[starthaus]

Let us say we have a function of x such that f(x)= x^2.

That's a bad start. What you have is $$f(x)=\frac{dg}{dx}=0$$. What can you infer about $$\frac{df}{dx}$$?

 

[yuiop]

That's a bad start. What you have is $$f(x)=\frac{dg}{dx}=0$$. What can you infer about $$\frac{df}{dx}$$?

What I had was $$f(x)=x^2$$.

This infers $$\frac{df}{dx} = 2x$$

Which in turn infers $$\frac{d^2f}{dx^2} = 2$$

When x=0,

$$f(x)=0$$

$$\frac{df}{dx} = 0$$

$$\frac{d^2f}{dx^2} = 2$$

 

[starthaus]

What I had was $$f(x)=x^2$$.

What you have is $$f(x)=constant=0$$

As in the case of radial motion, you are not allowed to hack in $$\frac{dr}{ds}=0$$ in the equation of motion:

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

If you do, you'll get the wrong result.
You are allowed to set terms to 0 in the metric (indicating that there is no motion in the respective direction) but you are not allowed to set terms to 0 in the differential equation that describes the motion. Because if you hack in terms like $$\frac{dr}{ds}=0$$ in the equation of motion you end up with the wrong equation of motion.

 

[Altabeh]

I was certain all along that dr/dt=0 does not imply that d^2r/dt^2=0 but I did not know how to prove it until I came across this article http://mathforum.org/library/drmath/view/65095.html in Dr math. Seeing as how Starhaus has not acknowledged his mistake I will assume he still does not get it and will elaborate on it.

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

When x=0:

f(x=0) = 0
f'(x=0) = 0
f''(x=0) = 2 ... Ta da! ... Non-zero!

Quite a simple proof. You should be able to get it now surely?

Obviously mathematically correct and has nothing to do with nonsense claims like since it's the value at a point so it doesn't apply in general. Because one can find points at which the first derivative of a function iz zero while the second isn't, the claim $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ is not generally correct.

This one another time shows how limited the guy's knowledge of basic algebra and calculus is.

AB

 

[Altabeh]

What I had was $$f(x)=x^2$$.

This infers $$\frac{df}{dx} = 2x$$

Which in turn infers $$\frac{d^2f}{dx^2} = 2$$

When x=0,

$$f(x)=0$$

$$\frac{df}{dx} = 0$$

$$\frac{d^2f}{dx^2} = 2$$

The "Second Derivative" method we use to find extrema of a given function says that

Step A. Let f(x) be a differentiable function on a given interval and let f'' be continuous at stationary point. Find f'(x) and solve the equation f'(x) = 0 given. Let x = a, b, ... be solutions.

Step B. Case (i) : If f''(a) < 0 then f(a) is maximum. Case (ii): If f''a)> 0 then f(a) is minimum.

This basic material is way below the level of discussion, but since some people sound quite unfamiliar to it, was recalled here.

AB

 

[yuiop]

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.

If the solution using the Langragian method is supposed to be the same as the that obtained from the geodesic method, then yes, this is what you should be getting if you calculate the Langragian expression carefully. This is what Espen got from the geodesic mthod. It is not what you obtained from Langragian. What you obtained in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" was:

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

which evaluates to:

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

which is definitely not the same as the solution obtained by Espen. Is this your way of admitting that you made a mistake and did not evaluate the lagrangian method carefully?

Hint:

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

The term $$\frac{d}{ds}(\frac{2/\alpha}{dr/ds})$$

does not appear in your expression derived from the langragian.

Hint:

$$\frac{d}{ds}(\frac{2/\alpha}{dr/ds}) \ne \frac{d}{ds}(\frac{1}{\alpha}*\frac{2dr}{ds})$$

 

[starthaus]

If the solution using the Langragian method is supposed to be the same as the that obtained from the geodesic method,

It isn't "supposed", it IS.

then yes, this is what you should be getting if you calculate the Langragian expression carefully. This is what Espen got from the geodesic mthod. It is not what you obtained from Langragian.

...because you did not finish the calculations. Use the hint I gave you. at post 136. If you calculate the derivatives correctly you will find out that , contrary to your claims, the lagrangian method has produced the same exact equation of motion as the geodesic method, as it should.

 

[starthaus]

Obviously mathematically correct and has nothing to do with nonsense claims like since it's the value at a point so it doesn't apply in general. Because one can find points at which the first derivative of a function iz zero while the second isn't, the claim $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ is not generally correct.

Let's try again:
$$\frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds})$$.
Since you have been given that $$\frac{dr}{ds}=0$$ FOR ALL VALUES OF s, what can you infer about $$\frac{d^2r}{ds^2}$$?

This one another time shows how limited the guy's knowledge of basic algebra and calculus is.

AB

You miss the point completely. Read post 157 then try comparing the solutions to the two ODE's shown here:

1. $$\frac{d^2r}{ds^2}+A\frac{dr}{ds}+B=0$$

and

2. $$\frac{d^2r}{ds^2}+B=0$$

After you do that, explain to kev why is that you can't simply hack $$\frac{dr}{ds}=0$$ into the first equation. It might help to remember that $$\frac{dr}{ds}$$ represents a function, not the value of a function in a point as the two of you use it.

 

[espen180]

1. $$\frac{d^2r}{ds^2}+A\frac{dr}{ds}+B=0$$

and

2. $$\frac{d^2r}{ds^2}+B=0$$

After you do that, explain to kev why is that you can't simply hack $$\frac{dr}{ds}=0$$ into the first equation. It might help to remember that $$\frac{dr}{ds}$$ represents a function, not the value of a function in a point as the two of you use it.

You do realize what the equation means? It relates the values of the functions for different values of a geodesic parameter.

As a reminder, a geodesic is a curve $$g\left(t(p),r(p),\theta(p),\phi(p)\right)=g_0\left(s(p),0,0,0\right)$$, where $$p$$ is the parameter of the curve.

Assuming there is such an r (or value of the geodesic parameter) where dr/ds is zero (an example of such an r is the apogee radius, or the point of release from rest), at that r, equations 1 and 2 degenerate into each other. This is Algebra II tops.

 

[starthaus]

or the point of release from rest), at that r, equations 1 and 2 degenerate into each other. .

...but they DO NOT degenerate into each other ANYWHERE else over the WHOLE domain of definition of the function $$r=r(s)$$. This is the whole point as to why you, Al68,kev should not be HACKING $$\frac{dr}{ds}=0$$ into the equation of motion. In the ODE, $$\frac{dr}{ds}$$ represents a FUNCTION, not a value, so you are not allowed to set it to 0 (or any other value) because you destroy the equation. Read post 157.

This is Algebra II tops

No, it is calculus 101. And, after all the explanations, you still get it wrong.

 

[espen180]

...but they DO NOT degenerate into each other ANYWHERE else over the WHOLE domain of definition of the function $$r=r(s)$$. This is the whole point as to why you, Al68,kev should not be HACKING $$\frac{dr}{ds}=0$$ into the equation of motion. In the ODE, $$\frac{dr}{ds}$$ represents a FUNCTION, not a value, so you are not allowed to set it to 0 (or any other value) because you destroy the equation. Read post 157.

You are wrong. Here is an example.

Look at these equations, in which I have only changed the derivative to the fuction itself:

1. $$f^{\prime\prime}(x)+af(x)+b=0$$

2. $$f^{\prime\prime}(x)+b=0$$

I argue that assuming there is such an x that f(x)=0, at that x the solutions degenerate and f''(x)=-b for both equations, not f''(x)=0 as you claim.

Solution of 1: $$f(x)=-\frac{b}{a}+c_1\sin\left(\sqrt{a}x\right)+c_2\cos\left(\sqrt{a}x\right)=-\frac{b}{a}+c_3\sin\left(\sqrt{a}x+c_4\right)$$

Solution of 2: $$-\frac{b}{2}x^2+c_1x+c_2$$

Double derivative of 1: $$f^{\prime\prime}=-c_3a\sin\left(\sqrt{a}x+c_4\right)$$

Double derivative of 2: $$f^{\prime\prime}=-b$$

When the solution to 1 is zero: $$x_0=\frac{\arcsin\left(\frac{b}{c_3a}\right)-c_4}{\sqrt{a}}$$

Plugging x₀ into double derivative of 1: $$-c_3a\sin\left(\sqrt{a}x_0+c_4\right)=-c_3a\sin\left(\arcsin\left(\frac{b}{c_3a}\right)\right)=-b$$

as expected.

 

[starthaus]

I argue that assuming there is such an x that f(x)=0, at that x the solutions degenerate and f''(x)=-b for both equations,

at that x...but NOWHERE ELSE in the domain of definition of f(x). This is the part that you, kev, Al68, Altabeh seem unable to grasp. You are being asked to find out the general equation of motion, your hack produces ONLY the acceleration at the point of release.

Exercise: using your hack try finding

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

 

[espen180]

at that x...but NOWHERE ELSE in the domain of definition of f(x). This is the part that you, kev, Al68, Altabeh seem unable to grasp. You are being asked to find out the general equation of motion, your hack produces ONLY the acceleration at the point of release.

The problem statement clearly spesified where the solution needed to be valid; at the apogee radius or drop point.

In a more general solution, such an assumption is not made, as can be seen in many posts here where work to reach such a solution has been/is being done.

 

[espen180]

Exercise: using your hack try finding

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

That's like asking what a photon's rest frame is like. You are trying to implement a model outside it's area of validity.

 

[starthaus]

The problem statement clearly spesified where the solution needed to be valid; at the apogee radius or drop point.

In what post did you specify this?
Even if your claim is true, what precluded you to understand what I have been telling you for tens of posts?

 

[starthaus]

That's like asking what a photon's rest frame is like.

It is good that you are starting to see the hack for what it is.

You are trying to implement a model outside it's area of validity.

The exercise I gave you is fully solvable with the tools I have given you, you simply need to stop trying to hack the ODE.

 

[Altabeh]

$$\frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds})$$.
Since you have been given that $$\frac{dr}{ds}=0$$ FOR ALL VALUES OF s, what can you infer about $$\frac{d^2r}{ds^2}$$?

Nothing can be inferred about $$\frac{d^2r}{ds^2}$$ because kev gave you a very keen counter-example to the hack $$dr/ds=0\Rightarrow d^2r/ds^2=0$$. We don't know what form $$r$$ would have as a function of some (affine) parameter so that we are not allowed to generally start talking about all values of $$s$$ and that whether your hack works or not.

You miss the point completely. Read post 157 then try comparing the solutions to the two ODE's shown here:

1. $$\frac{d^2r}{ds^2}+A\frac{dr}{ds}+B=0$$

and

2. $$\frac{d^2r}{ds^2}+B=0$$

After you do that, explain to kev why is that you can't simply hack $$\frac{dr}{ds}=0$$ into the first equation. It might help to remember that $$\frac{dr}{ds}$$ represents a function, not the value of a function in a point as the two of you use it.

You're the one who misses the point and doesn't listen to my notes. By now it's crystal clear that you're not even familiar with ODE's and boundary conditions. Actually a particle following a geodesic near any gravitating body would be considered momentarity at rest at any point* which means that along the geodesic $$dt/ds=0$$ for the particle. This never does mean that the particle is no longer going to have an acceleration but the object now has a uniform rest acceleration. If you have read a little about this, then you wouldn't dare to start giving us nonsense claims/equations. For example, you can read about this in this book "Relativity" By J. Rice. Purchase the book and don't waste our time by your hacks in calculus and algebra.

*For another usage in GR see for example "A first course in GR" by B. Schutz p. 254.

 

[starthaus]

Nothing can be inferred about $$\frac{d^2r}{ds^2}$$ because kev gave you a very keen counter-example to the hack $$dr/ds=0\Rightarrow d^2r/ds^2=0$$.

Let's try again, since you have been given that the function $$\frac{dr}{ds}=0$$ FOR ALL VALUES OF s, what can you infer about $$\frac{d^2r}{ds^2}$$?
Keep in mind that $$\frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds})$$.

We don't know what form $$r$$ would have as a function of some (affine) parameter so that we are not allowed to generally start talking about all values of $$s$$ and that whether your hack works or not.

Irrelevant.

You're the one who misses the point and doesn't listen to my notes. By now it's crystal clear that you're not even familiar with ODE's and boundary conditions.

This is not about boundary conditions. Since when do you plug boundary conditions straight into the ODE? LOL

Actually a particle following a geodesic near any gravitating body would be considered momentarity at rest at any point* which means that along the geodesic $$dt/ds=0$$ for the particle.

Irrelevant in finding the trajectory $$r=r(t)$$ or the velocity $$v=v(t)$$ or the acceleration $$a=a(t)$$.

If you think otherwise, using your hack, find these:

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

The hack is only good only for finding the acceleration at the release point. Worthless for anything else.

 

[espen180]

The hack is only good only for finding the acceleration at the release point. Worthless for anything else.

Again, that is everything it was meant to do.

 

[starthaus]

Again, that is everything it was meant to do.

It is good that you are starting to realize the reality.
In what post did you set the problem to only finding the acceleration at the release point?
I asked you this question, why don't you answer it?
Can you solve any of these:

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

You have been given all the tools.

 

[espen180]

It is good that you are starting to realize the reality.
In what post did you set the problem to only finding the acceleration at the release point?
I asked you this question, why don't you answer it?

Posts #50, #64, #66, #68 and #77 all specify that dr/ds=0 was a momentary, inital condition. The fact that the acceleration obtained only holds at that instant is so basic that it shouldn't need mentioning. Aren't you reading the thread?

 

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

...because you are trying to hack your way to the solution.

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.

This is post #50. You are still struggling with the circular orbits. There is no mention of any radial motion yet. Nor is there any problem statement. I told you to split the threads but you insisted in co-mingling them.

 

[starthaus]

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

This is post #64. Still no mention of the problem statement, just an attempt to deflect my criticism that you are hacking $$\frac{dr}{d\tau}=0$$ into the general equation of arbitrary orbits. I don't think it is worth continuing.

 

[espen180]

This is post #50. You are still struggling with the circular orbits. There is no mention of any radial motion yet. Nor is there any problem statement. I told you to split the threads but you insisted in co-mingling them.

I'm sorry, "acceleration of a particle dropped from rest at r" must have been a little too vague for you.

 

[espen180]

This is post #64. Still no mention of the problem statement, just an attempt to deflect my criticism that you are hacking $$\frac{dr}{d\tau}=0$$ into the general equation of arbitrary orbits. I don't think it is worth continuing.

Are you saying that a paticular family of geodesics are not a subset of the set of all geodesics?

 

[starthaus]

I'm sorry, "acceleration of a particle dropped from rest at r" must have been a little too vague for you.

Yes, I missed that, you have just embarked on your journey of hacking the general equation. You have just switched to a simpler problem and , since you are hacking, you got the wrong answer.You could have saved a lot of wasted time, we already solved this eons ago, see here. Both radial and circular motion.

So, can you solve the five exerciises I gave you?