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

#235
May2810, 01:14 PM

P: 1,568





#236
May2810, 01:26 PM

P: 801

What's relevant here is that in post 1, kev clearly defined [tex]a_0[/tex] as the proper acceleration measured by an accelerometer by a local stationary observer. 



#237
May2810, 01:36 PM

P: 3,967

The best I can do is demonstrate that using k as a constant that is indepent of r produces reasonable results and using k as a function of r produces incorrect results. Let us initially assume the value of k is [tex]\sqrt{12M/R}[/tex] where M and R are constants of a given trajectory and R is the apogee. This gives: [tex]\frac{dt}{ds} = \frac{\sqrt{12M/R}}{(12M/r)} [/tex] where r is a variable of the trajectory. Now if the location of the particle (r) coincides with the apogee (R) this gives [tex]\frac{dt}{ds} = \frac{1}{\sqrt{12M/r}} [/tex] in this specific instance and is what we would expect for the difference between coordinate time and proper time for a stationary particle at R. If the apogee of the particle is at infinity the equation reduces to: [tex]\frac{dt}{ds} = \frac{1}{(12M/r)} [/tex] which is also what we would expect the time dilation ratio to be for a particle falling from infinity to r and takes both the gravitational and velocity time dilation components into account. This result has been obtained in many other threads. Now let us assume k is a function of (r) such that [tex]k=\sqrt{12M/r}[/tex] where r is a variable. This gives: [tex]\frac{dt}{ds} = \frac{\sqrt{12M/r}}{(12M/r)} = \frac{1}{\sqrt{12M/r}} [/tex] This is an incorrect result because it says the time dilation ratio is independent of the height of the apogee and of the instantaneous velocity of the falling particle and so is not generally true. It is easy to see that time dilation ratio of the falling particle does depend on the velocity of the falling particle by taking the radial Schwarzschild solution: [tex]ds^2 = \alpha dt^2  \alpha^{1} dr^2 [/tex] and solving for ds/dt: [tex]\frac{ds}{dt} = \sqrt{\alpha  \frac{1}{\alpha} \frac{dr^2}{dt^2}} [/tex] From the above it obvious that ds/dt is only independent from dr/dt when [itex]dr^2/(\alpha dt ^2)= 0[/itex] which is not generally true. That should be enough to convince any reasonable person that k is not a function of the variable r. Treating k as constant is consistent with what we know about the Schwarzschild metric. If the constant k is inserted into the Schwarzschild metric it is easy to obtain the falling velocity in terms of proper time as: [tex]\frac{dr}{ds} = \sqrt{k^2 \alpha} = \sqrt{(12M/R) (12M/r)} = \sqrt{2M/r  2M/R}[/tex] This allows us to calculate the terminal velocity of a falling particle at r when released from a height R. Setting the apogee R to infinite gives [tex]dr/ds = \sqrt{2M/r}[/tex] which is the Newtonian escape velocity at r and is related to the conversion of potential energy to kinetic energy. I guess a formal proof of the constancy of k would be based on energy considerations. I leave that to a better person. 



#238
May2810, 01:38 PM

P: 1,568

1. [tex]\frac{m}{r_0^2}[/tex] (my derivation based on lagrangian mechanics and K.Brown's) 2. [tex]\frac{m}{r_0^2\sqrt{12m/r_0}}[/tex]? 



#239
May2810, 01:54 PM

P: 3,967

[tex]\frac{d^2r}{dt^2}=\frac{m}{r^2}(12m/r)[/tex] or: [tex]\frac{d^2r}{dt^2}=\frac{m}{r_0^2}(12m/r_0)[/tex] (the two equations are synonymous at that instant). This agrees with the equation given for coordinate acceleration in post #1 for the specific instance of a particle at its apogee and with the equation beautifully derived by Dalespam for the same specific instance. 



#240
May2810, 02:03 PM

P: 3,967





#241
May2810, 02:15 PM

P: 1,568





#242
May2810, 02:26 PM

P: 801





#243
May2810, 02:34 PM

P: 18

By the way, it isn't a good idea to talk about "the coordinate" variables, because there are infinitely many coordinate systems, and even the socalled "proper" variables are equal to coordinate variables for some suitable choice of coordinates. To avoid confusion, it's best to just say exactly what you mean (assuming you know what you mean). 



#244
May2810, 04:32 PM

P: 1,568

[tex]U=(dx/d\tau,dy,d\tau,dz/d\tau,d(ct)/d\tau)[/tex] (see also Rindler, p.99, Moller p.288)) IF what you were saying were true, the coordinate acceleration a would not show up in the definition of the fourvector [tex]A[/tex], nor would we be able to calculate proper acceleration [tex]a_0[/tex] from the conditon [tex]A=(a_0,0)[/tex] for [tex]u=0[/tex] 



#245
May2810, 04:34 PM

P: 1,568





#246
May2810, 04:53 PM

P: 801

Otherwise, I will choose a third option allowed by GR: 3. [tex]a_0=2\pi r[/tex], where [tex]a_0[/tex] is defined as the circumference of the moon. 



#247
May2810, 05:23 PM

P: 18





#248
May2810, 05:33 PM

P: 3,967

The acceleration of the particle as it passes the surface observers is measured by the Schwarzschild observer at infinity using Schwarschild radial coordinates and Schwarzschild coordinate time. Let's call this the coordinate acceleration of the particle [tex]a = (d^2r/dt^2)[/tex]. The observers on the shell measure the acceleration using their local measuring devices. Let's call this the local acceleration [tex]a' = (d^2r'/dt' ^2)[/tex] The ratio between [tex](d^2r/dt^2)[/tex] and [tex](d^2r'/dt' ^2)[/tex] will always be: [tex]a' = a\gamma^3[/tex] where [itex]\gamma = 1/\sqrt{(12M/r)} [/itex] and r is the radius of the massive body, to first order and for a sufficiently local measurement, for a particle dropped from any height greater than r. (as far I can tell) 



#249
May2810, 05:45 PM

P: 1,568





#250
May2810, 05:56 PM

P: 1,568

[tex]a=\frac{d^2r}{dt^2}=\frac{m}{r^2}(12m/r)(3\frac{12m/r}{12m/r_0}2})[/tex] [tex]a_0=\frac{d^2r}{d\tau^2}=\frac{m}{r^2}[/tex] (Even if we accepted that the above might be of the form: [tex]a_0=\frac{d^2r}{d\tau^2}=\frac{m}{r^2 \sqrt{12m/r}}[/tex] it still isn't true) Why does this nonsense about [tex]a_0=a\gamma^3[/tex] has so much fascination for you? 



#251
May2810, 06:13 PM

P: 3,967

Do you agree that a' as measured in the MCIF is equivalent in magnitude to the proper acceleration of the particle? i.e [tex]a' = a_0[/tex] Do you agree that [tex]a_0 = d^2x'/dt'^2[/tex] and not [tex]a_0 = d^2x/dt'^2\;\; [/tex]? (i.e. the local or proper acceleration is measured using local rulers and clocks and not coordinate distance) 



#252
May2810, 06:28 PM

P: 18




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