Gravitational force and acceleration in General Relativity.by yuiop Tags: acceleration, force, gravitational, relativity 

#55
May1110, 08:03 PM

P: 3,967

It can be shown that the local velocity of a particle falling from infinity dr'/dt' or dr/ds is [itex]\sqrt{(2M/r)}[/itex] (where ds is the proper time of the particle and primed quantities are the measurements according to a stationary observer at r) and this falling velocity contributes a factor of [itex]1/\sqrt{(12M/r)}[/itex] to the total time dilation of the particle. For a stationary particle this velocity contribution to the time dilation has to be factored out and the time dilation of the stationary particle is simply [itex]1/\sqrt{(12M/r)}[/itex] and not [itex]1/(12M/r)[/itex]. Next you have to show how to obtain the relationship between dr' and dr and it not obvious how you are going to obtain that from the equations of motion that you have given. Anyway, lets complete your "derivation" and see where that gets us. [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] Now your value for dt/ds is [itex]1/\alpha[/itex] so this value is substituted in to obtain: [tex]\frac{dr}{ds} \; = \; \sqrt{ 1  \alpha} \; = \; \sqrt{1\left(1\frac{2M}{r}\right)} \; = \; \sqrt{\frac{2M}{r}}[/tex] The second derivative of dr/ds is: [tex] \frac{d^2r}{ds^2} \; = \; \frac{d(dr/ds)}{dr} * \frac{dr}{ds} \; = \; \frac{m}{r^2\sqrt{2M/r}} * \sqrt{\frac{2M}{r}} \; = \; \frac{m}{r^2} [/tex] Your "proper" result for the acceleration of a stationary particle is not correct and the reason your result differs from my result is that you are considering the motion of a particle with significant velocity while we are consdering a stationary or nearly stationary particle and also because you have not taken into the account the transformation between dr' and dr. 



#56
May1110, 08:13 PM

P: 3,967

It follows that your result is also wrong, because you have to yet to show how you have derived your result from the equations of motion and as Luke said, it is doubtful that you are able to do it. 



#57
May1110, 08:28 PM

P: 3,967





#58
May1110, 08:41 PM

P: 1,568





#59
May1110, 08:42 PM

P: 1,568





#60
May1110, 08:48 PM

P: 3,967

[tex]E = mc^2 = \frac{m_0 c^2}{\sqrt{1v^2/c^2}} = \sqrt{(m_0c^2)^2 +(\vec{p}c)^2} [/tex] By your "logic" [tex]E^2  (\vec{p}c)^2=(mc^2)^2[/tex] is "wrong" because you have not derived it. 



#61
May1110, 08:54 PM

P: 3,967

Several times you have stated that your final results differ from the equations I gave in #1. Now you have begrudgingly admitted that the equations in #1 are correct that means the results you have obtained are wrong. You also stated the results obtained by Moller differ from the results obtained by Dalespam. Now that you have admitted Dalespam's results are correct, you are in fact saying the results given by Moller are wrong. Are you going to write to Moller and explain to him where he is going wrong? 



#62
May1110, 09:02 PM

P: 3,967





#63
May1110, 09:04 PM

P: 1,568

Physics is not a collection of hacks. 



#64
May1110, 09:12 PM

P: 1,568





#65
May1110, 11:19 PM

P: 3,967

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 realise by now that your derivation was wrong. 



#66
May1110, 11:38 PM

P: 1,568





#67
May1110, 11:41 PM

P: 1,568





#68
May1210, 12:18 AM

P: 3,967

Do you think that reference is wrong? What do you think the equation should be? 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 = (12M/r)dt^2  (12m/r)^{1} dr^2  r^2 d\phi^2 [/tex] Divide both sides by ds^2. [tex]\frac{ds^2}{ds^2} = (12M/r) \frac{dt^2}{ds^2}  (12m/r)^{1} \frac{dr^2}{ds^2}  r^2 \frac{d\phi^2}{ds^2} [/tex] [tex]1 = (12M/r) \frac{dt^2}{ds^2}  (12m/r)^{1} \frac{dr^2}{ds^2}  r^2 \frac{d\phi^2}{ds^2} [/tex] 



#69
May1210, 12:29 AM

P: 1,568

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. It can be derived either through covariant derivatives or , more directly, through using the metric. 



#70
May1210, 12:47 AM

P: 3,967

For your convenience, here is the start of your derivation; 



#71
May1210, 12:57 AM

P: 1,568

Let me give you the correct start that solves the problem in three lines: [tex]\vec{F}=grad\Phi[/tex] where [tex]\Phi[/tex] is the potential that you can read straight off the metric. You have two more lines to write, see if you can do it all by yourself, I have already shown you how to do that for centripetal acceleration, you only need to copy the method. 



#72
May1210, 01:01 AM

P: 3,967




Register to reply 
Related Discussions  
Gravitational force vs. relativity  Special & General Relativity  11  
General Question about Gravitational Potential & General Relativity  Special & General Relativity  4  
Special & General Relativity, time dilation, acceleration  Advanced Physics Homework  2  
Calculate acceleration after getting the Gravitational force between two bodies  General Physics  1  
General Relativity: gravitational waves  Special & General Relativity  2 