Schwarzschild equation of motion: initial conditions

In summary,The Schwarzschild equation of motion, where coordinate length is differentiated by proper time is as far as I know.
  • #36
m4r35n357 said:
Just to be clear, I am looking for an expression which calculates ##v/c## (where c is the speed of a light pulse instantaneously in the same place and moving in the same direction) for a particle in curved spacetime. I think I already have it, so can you tell me what is wrong with it?
Please have a look at this article: https://www.physicsforums.com/threa...initial-conditions.873588/page-2#post-5487369
 
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  • #37
Stollaxel Stoll said:
Are you aware that that post is in this very thread, was aimed at me, and that I have already replied to it?

I don't think it answers my question, but if it does, then I would guess they are saying I need to translate (tetrad-style) the coordinate ##v/c## into a local frame (that I don't have). I have access to the proper time, three coordinates, and their derivatives.

I don't see why I should have to do this, are you saying the coordinate ##v/c## is not defined or valid?

I would need to know for sure up front before embarking on this because it would be a major exercise (I'll need to use a Boyer-Lindquist tetrad, and I don't fully understand tetrads!).

The code is currently 25kB of HTML and Javascript (including the integrators), and any changes I make for this are just for a velocity display.

Do you understand my doubts?
 
  • #38
m4r35n357 said:
the coordinate v/c
v/c is not a coordinate at all. This is the velocity in terms of the speed of light!

m4r35n357 said:
I would need to know for sure up front before embarking on this because it would be a major exercise (I'll need to use a Boyer-Lindquist tetrad, and I don't fully understand tetrads!).
You need to give some examples what velocities you get with different initial conditions, otherwise it's hard to tell if you are using the right or the wrong equations.
 
  • #39
Stollaxel Stoll said:
v/c is not a coordinate at all. This is the velocity in terms of the speed of light!
Exactly. That what I have been asking about all along. Please see third paragraph of post #35.

s/coordinate/global/ above
 
  • #40
m4r35n357 said:
A particle has a coordinate velocity in the global coordinates. A pulse of light at the same point moving in the same direction has a velocity in the same global coordinates. I just want to divide the former by the latter, no more than that. I would like to know what proportion of light speed my particle is moving at.
Let's make an easy example for transversal motion. Take an observer at r=10GM/c². Then the Newtonian orbital velocity would be vn = √(GM/r) = c/√10, so the einsteinian orbital velocity is ve = vn/√(1-rs/r) = c/3. If the shell observer throws a particle that orbits at r=10GM/c² he observes this particle to have a velocity of c/3. An observer at infinity would see the particle orbit with the Newtonian orbital velocity of c/√10 since the higher local velocity cancels with the gravitational time dilatation with respect to an observer at infinity. The particle itself would have a rapidity of ve/c/√(1-ve²/c²) = 1/√8. Which one you use depends only on the system you describe.
 
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  • #41
m4r35n357 said:
I'm just not using a local frame at all, so I don't have, want or need a local velocity.
What is the question then? If you don't want the local frame stay in your coordinate frame, what else? As long as you know what you are doing...
 
  • #42
m4r35n357 said:
Sorry George, I've left this stewing for a day in my mind, and I'm still not sure what you are telling me here. Am I supposed to be using the last equation (in your notation I was looking for ##V = something##)? BTW I don't have access to the ##v## that Yukterez uses so any expression containing that would involve rewriting my program. I reiterate that my ##V## is strictly an output of the program (I have access to ##\frac {d t} {d \tau}= \gamma##, ##\frac {d r} {d \tau}##, and ##\frac {d \phi} {d \tau}##, these and the metric components are my only ingredients), and not part of the simulation.

And I still don't understand what is wrong with ##V = \sqrt {1 - \frac {1 } { \gamma^2} }##. It seems to produce reasonable results under every situation that I can throw at it.

Just to be clear, I am looking for an expression which calculates ##v/c## (where c is the speed of a light pulse instantaneously in the same place and moving in the same direction) for a particle in curved spacetime. I think I already have it, so can you tell me what is wrong with it?

Let's see if I understand the calculation for V, first, by an example. Suppose we have a particle that is instantaneously at rest in Schwarzschild coordinates. Then we have ##dr/d\tau = d\theta/d\tau = d\phi / d\tau = 0##. We can solve for ##dt / d\tau =\frac{1}{\sqrt{1-2M/r}} ##. Then would we say that for this particle from post 31 we'd calculate

$$\gamma = \frac{1}{\sqrt{1-2M/r}} \quad V = \sqrt{1 -1/\gamma^2} = \sqrt{1 - (1-2M/r)} = \sqrt{2M/r}$$

Is this calculatio of your number V correct?
 
  • #43
pervect said:
√2M/r
That's what I get (for the radial velocity in river-flow coordinates, where the motion of a particle at rest equals Newton's escape velocity at that coordinate).
 
  • #44
pervect said:
Is this calculatio of your number V correct?
I think this is very close to what I am asking, yes. Can I just ask before I confuse myself, is there any implied punctuation in your equations (is it one long equality or two equations side by side, I would hope the latter)?
 
  • #45
pervect said:
Is this calculatio of your number V correct?

OK, I'm going to rewind to my original attempt as it might be clearer what I am trying to do. Starting from a Kerr (Boyer-Lindquist) metric:

##
d\tau^2 = g_{tt}dt^2 + 2 g_{t\phi}dtd\phi + g_{rr}dr^2 + g_{\theta\theta}d\theta^2 + g_{\phi\phi}d\phi^2
##

4-velocity norm

##
1 = g_{tt}\frac {dt^2} {d\tau^2} + 2 g_{t\phi}\frac {dt} {d\tau} \frac {d\phi} {d\tau} + g_{rr}\frac {dr^2} {d\tau^2} + g_{\theta\theta}\frac {d\theta^2} {d\tau^2} + g_{\phi\phi}\frac {d\phi^2} {d\tau^2}
##

let

##
\gamma = \frac {dt} {d\tau}
##

##
1 = \gamma^2 g_{tt} + 2 \gamma^2 g_{t\phi} \frac {d\phi} {dt} + \gamma^2 g_{rr}\frac {dr^2} {dt^2} + \gamma^2 g_{\theta\theta}\frac {d\theta^2} {dt^2} + \gamma^2 g_{\phi\phi}\frac {d\phi^2} {dt^2}
##

##
\frac {1 - \gamma^2g_{tt}} {\gamma^2} = 2 g_{t\phi} \frac {d\phi} {dt} + g_{rr}\frac {dr^2} {dt^2} + g_{\theta\theta}\frac {d\theta^2} {dt^2} + g_{\phi\phi}\frac {d\phi^2} {dt^2} = v^2
##

OK, so now I've got two formulae for ##v## (is it ##v/c##?). The one using ##g_{tt}## is simpler. Trouble is, this expression is not bounded between 0 and 1 when I use it. Hence my arm-waving justification (cop-out) in post #15 for using the simpler (modulo known metric signature errors!) Minkowski-type expression:

##
v^2 = \frac {\gamma^2 - 1} {\gamma^2} = 1 - \gamma^2
##

Now let me try to paraphrase my cop-out. The velocity terms are all products of ("square rooted") metric components and coordinate velocities. At any point the metric/potential is identical for particles and light, so should (?) cancel when I divide ##v## by ##c##. This just leaves the ##\gamma## term as the only contributor to the speed. The simpler form is clearly bounded by 0 and 1.

Hope that's enough of the grubby details, please pick it apart for me!
 
  • #46
Stollaxel Stoll said:
That's what I get (for the radial velocity in river-flow coordinates, where the motion of a particle at rest equals Newton's escape velocity at that coordinate).

Two equations side by side, one equation for ##\gamma##, the other equation for V
 
  • #47
pervect said:
Two equations side by side, one equation for ##\gamma##, the other equation for V
I assume this reply was aimed at me ;)

That is comforting news as it appears that the equation in post #31 is OK. I still wonder what I was "calculating" in post #45 though with the ##g_{tt}## term . . . and yes I did notice that I messed up an equation, it should of course be

##
v^2 = 1 - \frac {1} {\gamma^2}
##
 
  • #48
m4r35n357 said:
OK, I'm going to rewind to my original attempt as it might be clearer what I am trying to do. Starting from a Kerr (Boyer-Lindquist) metric:

##
d\tau^2 = g_{tt}dt^2 + 2 g_{t\phi}dtd\phi + g_{rr}dr^2 + g_{\theta\theta}d\theta^2 + g_{\phi\phi}d\phi^2
##

4-velocity norm

##
1 = g_{tt}\frac {dt^2} {d\tau^2} + 2 g_{t\phi}\frac {dt} {d\tau} \frac {d\phi} {d\tau} + g_{rr}\frac {dr^2} {d\tau^2} + g_{\theta\theta}\frac {d\theta^2} {d\tau^2} + g_{\phi\phi}\frac {d\phi^2} {d\tau^2}
##

let

##
\gamma = \frac {dt} {d\tau}
##

OK so far
##
1 = \gamma^2 g_{tt} + 2 \gamma^2 g_{t\phi} \frac {d\phi} {dt} + \gamma^2 g_{rr}\frac {dr^2} {dt^2} + \gamma^2 g_{\theta\theta}\frac {d\theta^2} {dt^2} + \gamma^2 g_{\phi\phi}\frac {d\phi^2} {dt^2}
##

I don't follows this at all, I get:
##1 = \gamma^2 g_{tt} + 2 \gamma g_{t\phi} \frac {d\phi} {dt} + g_{rr}\frac {dr^2} {dt^2} + g_{\theta\theta}\frac {d\theta^2} {dt^2} + g_{\phi\phi}\frac {d\phi^2} {dt^2}##

...

OK, so now I've got two formulae for ##v## (is it ##v/c##?). The one using ##g_{tt}## is simpler. Trouble is, this expression is not bounded between 0 and 1 when I use it. Hence my arm-waving justification (cop-out) in post #15 for using the simpler (modulo known metric signature errors!) Minkowski-type expression:

##
v^2 = \frac {\gamma^2 - 1} {\gamma^2} = 1 - \gamma^2
##

I'm really not following this, I see a "v" has appeared at the right of your expresssion, somehow, but I'm not sure how it got there.

The logical consequence of your cop-out the way I see it is that an observer with constant r, theta, phi does not have a "velocity" of zero. I would agree you've calculated a number, but I don't understand why you think this number represents a velocity - for instance, it doesn't seem to have any physical interpretation as the velocity between a pair of worldlines. And the "velocity" of an observer with constant spatial coordinates isn't zero.

However, if we keep the factor of ##g_{tt}## in your equation, I see something that happens to work out to be equivalent for the expression for "little v":

$$\Gamma = \frac{ \frac{dt}{d\tau} }{\sqrt{g_{tt} } }$$
$$v = \sqrt{1-\frac{1}{\Gamma^2}}$$

"It happens to work out" is a bit vague, so I'll give a more formal but brief sketch of why it works out. If we have two four velocities ##p## and ##q##, ##p \cdot q = g_{ij} \, p^i \, q^j## is a measure of the angle between p and q. In flat Minkowskii space-time, ##p \cdot q = \gamma##, modulo sign issues. So ##p \cdot q## it's a function of relative velocity, expressed in a coordinate independent manner.

Since the 4-velocity v of a shell observer has the components ##\left[1/\sqrt{|g_{tt}|},0,0,0\right]## the dot product of ##\left[1/\sqrt{|g_{tt}|},0,0,0\right]## with u is ## u^0 / \sqrt{g_{tt}} = \left(dt / d\tau \right) / \sqrt{g_{tt}} ##

"Little v" does has a physical interpretation as the velocity between shell observer with constant r, theta, phi coordinates relative to the velocity of an observer with a 4-velocity u as measured by either the shell observer , or the observer with 4-velocity u.

You point out that "little v", in your application (which isn't Schwarzschild) sometimes has a magnitude that's greater than 1. I believe this happens because in your application, an observer with constant r, theta, phi coordinates does not have a timelike worldline.
 
  • #49
pervect said:
I don't follows this at all, I get:
##1 = \gamma^2 g_{tt} + 2 \gamma g_{t\phi} \frac {d\phi} {dt} + g_{rr}\frac {dr^2} {dt^2} + g_{\theta\theta}\frac {d\theta^2} {dt^2} + g_{\phi\phi}\frac {d\phi^2} {dt^2}##
I was just attempting to go from "velocities" in terms of ##\tau## to coordinate velocities, that's where I got all the ##\gamma##s from. Not to worry, as I'm not using that expression anyway, and my speeds look fine.

Thanks for the rest of your post BTW which I'll mull over; as ever I get more to chew on than I expected!
 
  • #50
m4r35n357 said:
I was just attempting to go from "velocities" in terms of ##\tau## to coordinate velocities, that's where I got all the ##\gamma##s from. Not to worry, as I'm not using that expression anyway, and my speeds look fine.

Thanks for the rest of your post BTW which I'll mull over; as ever I get more to chew on than I expected!

One thing that didn't occur to me. If we have a 4-velocity ##\left[u^0, u^1, u^2, u^3 \right]## and take the dot product with a 4-velocity ##\left[\Gamma,0,0,0 \right]## in a general metric, we get:

##g_{00} u^0 \Gamma + g_{10} u^1 \Gamma + g_{20} u^2 \Gamma + g_{30} u^3 \Gamma##

So there are some additional terms to the dot product in your metric, due to the off-diagonal terms in your metric.
 
  • #51
pervect said:
So there are some additional terms to the dot product in your metric, due to the off-diagonal terms in your metric.
OK, thanks for pointing that out, something else to look at!
 
  • #52
When I have done calculations similar to this I have taken all vectors to be unity length [-1,0,0,0] (or m*[-1,0,0,0] and then used the metric tensor in sine/sinh form to rotate it to different viewpoints and relative velocities. As a picture/basis I refer almost all situations back to Flamm's paraboloid; which is the space-like slice in a frame where the center is stationary. It provides a basic reference frame that all other items can refer to; to avoid confusion in my simple mind. I did do a little research about a decade ago and came up with a "hyperbolic" path that looped around the black hole once before flying off to infinity :) Getting interesting paths did require a great deal of fine tuning since most paths just shot out or sank into the event horizon. I have tried to extend Flamm's paraboloid to pictorially show different geodesics, time evolution, but have never found a good picture/extrapolation into the third dimension (in this case).
 
  • #53
I am not sure if anybody has come up with this already, nor if this is the right place to ask, but I'll try anyway.
My aim is initialising a photon in Schwarzschild with a certain initial angle \phi between the radial and polar (or azimuthal) components of the velocity. How would I retrieve v^r, v^{\phi} in this case?
 
  • #54
fabsilfab said:
I am not sure if anybody has come up with this already, nor if this is the right place to ask, but I'll try anyway.
My aim is initialising a photon in Schwarzschild with a certain initial angle \phi between the radial and polar (or azimuthal) components of the velocity. How would I retrieve v^r, v^{\phi} in this case?
For the Schwarzschild case, I would just
1)look at Flamm's Paraboloid and then
2) pick/find your starting position,
3) Decide what coordinate system to use (I would use the local Schwarzschild coordinates but there are a lot more)
3) select the spatial direction for r , theta, phi in normal spherical coordinates with spatial length 1
4) Recast it into the selected relativistic coordinates (I even have some code (somewhere) )
5) Compute the time component and make sure the 4-space "length" is zero
6) Promulgate that vector forward by computing the null geodesic with those starting conditions.
Incidentally, Sagemanifolds has the tools to do this faster than you can read the above; although you have to learn their formalism/language, it's quite good. As with any computer implementation/tool you should have a firm idea about the answer before you trust any result. I have been done in so many ways you wouldn't believe it.
Anybody: please clarify or correct any of the above; constructively!
 
  • #56
Ibix said:
https://www.physicsforums.com/threads/null-geodesics-in-schwarzschild-spacetime.895174/

Do scroll all the way down - my initial solution had a couple of bugs.

Thanks, this was a very useful suggestion! I read through it, and the conclusion is that I can shoot a photon at a certain angle ##\psi## wrt the ##\phi=0## axis by setting $$ r u^\phi=\sin\psi,$$ $$\frac{u^r}{\sqrt{1-r_S/r}}=\cos\psi.$$
This indeed works. I can get proper Einstein rings knowing the initial shooting angle. Thanks a lot!
 

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