- #36
Dale
Mentor
- 35,319
- 13,540
And the part of The Pot will be played by starthaus.starthaus said:You need to learn how to stop making false claims.
And the part of The Pot will be played by starthaus.starthaus said:You need to learn how to stop making false claims.
DaleSpam said:And the part of The Pot will be played by starthaus.
My results for proper acceleration agree with yours.kev said:Does that seem about right?
DaleSpam said:My results for proper acceleration agree with yours.
So, using the convention that timelike intervals squared are positive, in Schwarzschild coordinates given by:
[tex](t,r,\theta,\phi)[/tex]
The line element is:
[tex]ds^2=\left(1-\frac{R}{r}\right) c^2 dt^2-\left(1-\frac{R}{r}\right)^{-1}dr^2-r^2 d\theta^2- r^2 \sin ^2(\theta ) d\phi^2[/tex]
where
[tex]R=\frac{2GM}{c^2}[/tex]
And the metric tensor is:
[tex]\mathbf g = \left(
\begin{array}{cccc}
c^2 \left(1-\frac{R}{r}\right) & 0 & 0 & 0 \\
0 & -\left(1-\frac{R}{r}\right)^{-1} & 0 & 0 \\
0 & 0 & -r^2 & 0 \\
0 & 0 & 0 & -r^2 \sin ^2(\theta )
\end{array}
\right)[/tex]
In this Schwarzschild metric, the Christoffel symbols
[tex]\left(
\begin{array}{cccc}
\left\{\Gamma _{t t}^t,\Gamma _{t r}^t,\Gamma _{t \theta }^t,\Gamma _{t\phi
}^t\right\} & \left\{\Gamma _{r t}^t,\Gamma _{r r}^t,\Gamma _{r \theta }^t,\Gamma
_{r \phi }^t\right\} & \left\{\Gamma _{\theta t}^t,\Gamma _{\theta r}^t,\Gamma
_{\theta \theta }^t,\Gamma _{\theta \phi }^t\right\} & \left\{\Gamma _{\phi
t}^t,\Gamma _{\phi r}^t,\Gamma _{\phi \theta }^t,\Gamma _{\phi \phi
}^t\right\} \\
\left\{\Gamma _{t t}^r,\Gamma _{t r}^r,\Gamma _{t \theta }^r,\Gamma _{t \phi
}^r\right\} & \left\{\Gamma _{r t}^r,\Gamma _{r r}^r,\Gamma _{r \theta }^r,\Gamma
_{r \phi }^r\right\} & \left\{\Gamma _{\theta t}^r,\Gamma _{\theta r}^r,\Gamma
_{\theta \theta }^r,\Gamma _{\theta \phi }^r\right\} & \left\{\Gamma _{\phi
t}^r,\Gamma _{\phi r}^r,\Gamma _{\phi \theta }^r,\Gamma _{\phi \phi
}^r\right\} \\
\left\{\Gamma _{t t}^{\theta },\Gamma _{t r}^{\theta },\Gamma _{t \theta }^{\theta
},\Gamma _{t \phi }^{\theta }\right\} & \left\{\Gamma _{r t}^{\theta },\Gamma
_{r r}^{\theta },\Gamma _{r \theta }^{\theta },\Gamma _{r \phi }^{\theta }\right\}
& \left\{\Gamma _{\theta t}^{\theta },\Gamma _{\theta r}^{\theta },\Gamma
_{\theta \theta }^{\theta },\Gamma _{\theta \phi }^{\theta }\right\} &
\left\{\Gamma _{\phi t}^{\theta },\Gamma _{\phi r}^{\theta },\Gamma _{\phi
\theta }^{\theta },\Gamma _{\phi \phi }^{\theta }\right\} \\
\left\{\Gamma _{t t}^{\phi },\Gamma _{t r}^{\phi },\Gamma _{t \theta }^{\phi
},\Gamma _{t \phi }^{\phi }\right\} & \left\{\Gamma _{r t}^{\phi },\Gamma
_{r r}^{\phi },\Gamma _{r \theta }^{\phi },\Gamma _{r \phi }^{\phi }\right\} &
\left\{\Gamma _{\theta t}^{\phi },\Gamma _{\theta r}^{\phi },\Gamma _{\theta
\theta }^{\phi },\Gamma _{\theta \phi }^{\phi }\right\} & \left\{\Gamma _{\phi
t}^{\phi },\Gamma _{\phi r}^{\phi },\Gamma _{\phi \theta }^{\phi },\Gamma
_{\phi \phi }^{\phi }\right\}
\end{array}
\right)[/tex]
are given by:
[tex]\left(
\begin{array}{cccc}
\left\{0,\frac{R}{2 r^2-2 r R},0,0\right\} & \left\{\frac{R}{2 r^2-2 r
R},0,0,0\right\} & \{0,0,0,0\} & \{0,0,0,0\} \\
\left\{\frac{c^2 (r-R) R}{2 r^3},0,0,0\right\} & \left\{0,-\frac{R}{2 r^2-2 r
R},0,0\right\} & \{0,0,R-r,0\} & \left\{0,0,0,(R-r) \sin ^2(\theta )\right\} \\
\{0,0,0,0\} & \left\{0,0,\frac{1}{r},0\right\} & \left\{0,\frac{1}{r},0,0\right\} &
\{0,0,0,-\cos (\theta ) \sin (\theta )\} \\
\{0,0,0,0\} & \left\{0,0,0,\frac{1}{r}\right\} & \{0,0,0,\cot (\theta )\} &
\left\{0,\frac{1}{r},\cot (\theta ),0\right\}
\end{array}
\right)[/tex]
So, without loss of generality the worldline of a particle at rest in these coordinates is given by:
[tex]\mathbf X = (t,r_0,0,0)[/tex]
Then we can derive the four-velocity as follows:
[tex]\mathbf U = \frac{d \mathbf X}{d \tau} = c \frac{d \mathbf X}{ds} = c \frac{d \mathbf X}{dt} \frac{dt}{ds} = c \; (1,0,0,0) \; \frac{1}{\sqrt{c^2 \left(1-\frac{R}{r}\right)}} = \left(\left(1-\frac{R}{r}\right)^{-1/2},0,0,0\right)[/tex]
And we can verify that the norm of the four-velocity is equal to:
[tex]||\mathbf U||^2=U_{\mu} U^{\mu}= g_{\mu\nu} U^{\nu} U^{\mu} = c^2[/tex]
Now we can derive the four-acceleration as follows:
[tex]A^{\mu}=\frac{DU^{\mu}}{d\tau}=\frac{dU^{\mu}}{d\tau}+\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}[/tex]
[tex]\frac{d \mathbf U}{d\tau}= c\frac{d \mathbf U}{ds}= c\frac{d \mathbf U}{dt}\frac{dt}{ds}= c \; (0,0,0,0) \; \frac{dt}{ds}=(0,0,0,0)[/tex]
There is only one non-zero component of:
[tex]\Gamma^{\mu}_{\nu\lambda}U^{\nu}U^{\lambda}=\Gamma^{r}_{tt}U^{t}U^{t}=\frac{c^2 (r-R) R}{2 r^3}[/tex]
So, substituting back in we obtain the four-acceleration in the Schwarzschild coordinates:
[tex]\mathbf A = \left(0,\frac{c^2 R}{2 r^2},0,0\right) = \left(0,\frac{G M}{r^2},0,0\right)[/tex]
Finally, the norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
[tex]-||\mathbf A||^2= -A_{\mu} A^{\mu}= -g_{\mu\nu} A^{\nu} A^{\mu} = \frac{c^4 R^2}{4 r^4 (1-R/r)} = \left( \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} \right)^2[/tex]
So this also agrees with your previous result. As expected, the proper acceleration for a particle at rest in the Schwarzschild coordinates goes to infinity as r goes to R and is imaginary for r<R.
DaleSpam said:What are you talking about? There is no centripetal acceleration, this is a particle which is stationary in the Schwarzschild coordinates, not stationary in a rotating reference frame.
.
If you want me to derive the proper acceleration for a particle which is moving in uniform circular motion in the Schwarzschild coordinates then I can certainly do that
DaleSpam said:I did read it, I just don't see how it relates to kev's question which is concerning a stationary particle in the Schwarzschild spacetime, not a rotating particle.
DaleSpam said:The correct centripetal acceleration for a stationary particle in the Schwarzschild metric is 0.
DaleSpam said:That isn't my link, it is an automatic link to the PF library. It is not really relevant since it is for a flat spacetime. In any case even in the limit R->0 (flat spacetime) for a stationary particle in the Schwarzschild metric [itex]\omega = 0[/itex] implies centripetal acceleration equals 0.
I think you are talking about the other thread where we were dealing with rotational motion in flat spacetime. None of your last 5 posts have any relevance whatsoever to this thread where we are dealing with stationary particles in a curved Schwarzschild spacetime. Certainly my derivation here never used the variables [itex]\omega[/itex] or [itex]\gamma[/itex].starthaus said:OBVIOUSLY...but NOT [tex]\gamma^2r\omega^2[/tex]. This is the point of what I have been telling you for days. There is no [tex]\gamma[/tex] in the correct derivation.
DaleSpam said:I think you are talking about the other thread where we were dealing with rotational motion in flat spacetime. None of your last 5 posts have any relevance whatsoever to this thread where we are dealing with stationary particles in a curved Schwarzschild spacetime. Certainly my derivation here never used the variables [itex]\omega[/itex] or [itex]\gamma[/itex].
DaleSpam said:So, the pertinent question for this thread is: Do you find any error in my derivation of the proper acceleration for a stationary particle in the curved Schwarzschild spacetime?
DaleSpam said:Finally, the norm of the four-acceleration, which is equal to the magnitude of the proper acceleration, is given by:
[tex]-||\mathbf A||^2= -A_{\mu} A^{\mu}= -g_{\mu\nu} A^{\nu} A^{\mu} = \frac{c^4 R^2}{4 r^4 (1-R/r)} = \left( \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} \right)^2[/tex]
So this also agrees with your previous result. As expected, the proper acceleration for a particle at rest in the Schwarzschild coordinates goes to infinity as r goes to R and is imaginary for r<R.
starthaus said:Your derivation is correct, I never contested that.
So is my derivation for proper acceleration in a rotating frame.
kev said:Dalespam's result agrees with my statement in #1 and yet you say his result is correct and mine is wrong
Can we just clarify something? Do you agree that the results I posted in #1 are correct but you simply don't like the fact that I have not shown how they are derived?
starthaus said:...You need to start with the metric:
[tex]ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2[/tex]
[tex]\alpha=1-\frac{2m}{r}[/tex]
From this you construct the Lagrangian:
[tex]L=\alpha (\frac{dt}{ds})^2-\frac{1}{\alpha}(\frac{dr}{ds})^2-r^2(\frac{d\phi}{ds})^2[/tex]
...
starthaus said:... we can get immediately:
[tex] \frac{dt}{ds}=\frac{k}{1-\frac{2m}{r}}[/tex]
From the above, we can get the relationship between proper and coordinate speed:
[tex]\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}[/tex]
Differentiate one more time and you will obtain the correct relationship between proper and coordinate acceleration. ...
starthaus said:...
[tex]\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}[/tex]
Differentiate one more time and you will obtain the correct relationship between proper and coordinate acceleration. ...
starthaus said:DaleS derived his result, you didn't.
kev said:Can we just clarify something? Do you agree that the results I posted in #1 are correct but you simply don't like the fact that I have not shown how they are derived?
starthaus said:DaleS derived his result, you didn't.
kev said:So if I say E = mc^2, then that is "wrong" if I do not show how I derived the result ?
kev said:Ok I take that to mean that you agree the results posted in #1 are correct and you retract your statement in #2 that they are wrong.
starthaus said:You mean [tex]E^2 - (\vec{p}c)^2=(mc^2)^2[/tex] right?
starthaus said:Physics is not pasting together stuff you glean off the internet.
starthaus said:Physics is not pasting together stuff you glean off the internet.
kev said:In order to differentiate dr/ds we need to know its value. This can be obtained from the Lagrangian (with [itex]d\phi[/itex] set to zero).
[tex]L = \frac{1}{2} = \frac{1}{2} \left( \alpha \left(\frac{dt}{ds}\right)^2 - \frac{1}{\alpha}\left(\frac{dr}{ds}\right)^2 \right)[/tex]
(See section 13.1 of http://people.maths.ox.ac.uk/~nwoodh/gr/gr03.pdf )
Solve for dr/ds:
[tex]\frac{dr}{ds} \; = \; \sqrt{ \alpha^2 \left(\frac{dt}{ds}\right)^2 - \alpha}[/tex]
kev said:Now convert to ds to dt to get the coordinate result:
[tex] \frac{d^2r}{dt^2} = \frac{d^2r}{ds^2}\left(\frac{ds}{dt}\right)^2 \; [/tex]
.
starthaus said:Why on Earth would anyone do such an elementary calculus mistake? ...
kev said:Yep, I slipped at at the last step. This is how it should done.
Assuming initial conditions of a stationary particle at infinity that radially freefalls:
[tex]1 = \alpha \left(\frac{dt}{ds}\right)^2 - \frac{1}{\alpha}\left(\frac{dr}{ds}\right)^2[/tex]
Solve for dr/dt:
[tex]\frac{dr}{dt} \; = \; \alpha \sqrt{\frac{2M}{r}} = \left(1-\frac{2M}{r}\right) \sqrt{\frac{2M}{r}} [/tex]
Differentiate dr/dt with respect to t:
[tex] \frac{d^2r}{dt^2} \; = \; \frac{d(dr/dt)}{dr} * \frac{dr}{dt} \;=\; \left(\frac{2m}{r^2}\sqrt{\frac{2M}{r}} -\frac{m}{r^2} \frac{\alpha}{\sqrt{2M/r}}\right) \alpha \sqrt{\frac{2M}{r}} \; = \; \frac{-m}{r^2} \left(1-\frac{2M}{r}\right) \left(1-\frac{6M}{r}\right) [/tex]
This is the correct equation for the Schwarzschild coordinate acceleration of a radially free falling particle, initially at rest at infinity.
Now it your turn to stop being a coward and show your derivation from the equations of motion for the proper/ coordinate acceleration of a stationary particle in Schwarzschild coordinates. My guess is that we will never see it, because you realize by now that your derivation was wrong.
kev said:[tex]L = \frac{1}{2} = \frac{1}{2} \left( \alpha \left(\frac{dt}{ds}\right)^2 - \frac{1}{\alpha}\left(\frac{dr}{ds}\right)^2 \right)[/tex]
kev said:Differentiate dr/dt with respect to t:
[tex] \frac{d^2r}{dt^2} \; = \; \frac{d(dr/dt)}{dr} * \frac{dr}{dt} \;=\; \left(\frac{2m}{r^2}\sqrt{\frac{2M}{r}} -\frac{m}{r^2} \frac{\alpha}{\sqrt{2M/r}}\right) \alpha \sqrt{\frac{2M}{r}} \; = \; \frac{-m}{r^2} \left(1-\frac{2M}{r}\right) \left(1-\frac{6M}{r}\right) [/tex]
This is the correct equation for the Schwarzschild coordinate acceleration of a radially free falling particle, initially at rest at infinity.
starthaus said:No, it isn't.
starthaus said:The equations of motion are all the way back in post #2. For someone who makes so many mistakes, you are quite abusive.
starthaus said:What gives you the bright idea that you can set the Lagrangian to a number?
L = constant. ... Since s is proper time, [itex]g_{ab}\dot{x}^{a} \dot{x}^{b} = 1[/itex], and therefore the third conservation law is L = 1/2 .
kev said:That exact equation is given here http://www.mathpages.com/rr/s6-07/6-07.htm
Do you think that reference is wrong? What do you think the equation should be?
You just asked for the equations of motion, I told you that I derived them at post #2. You got exactly what you asked for.You gave the equations of motion which is effectively giving nothing
because they are standard and quated in hundreds of online references. However you claim to be able to teach me how to derive the proper and coordinate acceleration of a stationary particle from those equations of motion, which would be impressive if you could actually do it but so far you have come up with nothing.
I got the idea from here: http://people.maths.ox.ac.uk/~nwoodh/gr/gr03.pdf section 13.1
That suggests to me that the Lagrangian is a constant.
Anyway, I can do it a different way and obtain the same result.
Using units of G=c=1 the Schwarzschild metric is;
[tex]ds^2 = (1-2M/r)dt^2 - (1-2m/r)^{-1} dr^2 - r^2 d\phi^2 [/tex]
[tex]\frac{ds^2}{ds^2} = (1-2M/r) \frac{dt^2}{ds^2} - (1-2m/r)^{-1} \frac{dr^2}{ds^2} - r^2 \frac{d\phi^2}{ds^2} [/tex]
[tex]1 = (1-2M/r) \frac{dt^2}{ds^2} - (1-2m/r)^{-1} \frac{dr^2}{ds^2} - r^2 \frac{d\phi^2}{ds^2} [/tex]
starthaus said:You just asked for the equations of motion, I told you that I derived them at post #2. You got exactly what you asked for.
starthaus said:I can do it in three lines.
starthaus said:You need to start with the metric:
[tex]ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2[/tex]
[tex]\alpha=1-\frac{2m}{r}[/tex]
From this you construct the Lagrangian:
[tex]L=\alpha (\frac{dt}{ds})^2-\frac{1}{\alpha}(\frac{dr}{ds})^2-r^2(\frac{d\phi}{ds})^2[/tex]
From the above Lagrangian, you get immediately the equations of motion:
[tex]\alpha \frac{dt}{ds}=k[/tex]
[tex]r^2 \frac{d\phi}{ds}=h[/tex]
whre h,k are constants.
From the first equation, we can get immediately:
[tex] \frac{dt}{ds}=\frac{k}{1-\frac{2m}{r}}[/tex]
From the above, we can get the relationship between proper and coordinate speed:
[tex]\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}[/tex]
Differentiate one more time and you will obtain the correct relationship between proper and coordinate acceleration.