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#37
Nov211, 09:09 PM

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Peter I mean by demonstrating using a test situation and formulas to prove it. It is so because it is so, is just a tautology.
I just added some formulas to my prior posting to show you what I am getting at. 


#38
Nov211, 09:15 PM

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You are correct that freely falling observers are different; in fact, since they *are* moving inward at escape velocity, you could argue that your time dilation formula should apply to them, *not* to static observers. Unfortunately, you left out an important piece of information: a freely falling observer is changing his r coordinate as well as his t coordinate, so to calculate his rate of time flow you have to integrate dtau along his worldline using both the dt^2 and the dr^2 terms in the line element; you can't just look at g_00 like you can for static observers. This is easier to do in Painleve coordinates, particularly if you want to carry the computation inside the horizon. You've posted such calculations in these forums before, so I know you know how to do them; such a calculation is also given on the Wikipedia page on Painleve coordinates: http://en.wikipedia.org/wiki/Gullstr...A9_coordinates 


#39
Nov211, 09:21 PM

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#40
Nov211, 09:27 PM

P: 1,555

How about this: Let's take three observers: rs=1 O1 = Stationary at R1 O2 = Free falling at escape velocity at R1 O2 = Free falling at 0.5* escape velocity at R1 If we consider their resp tau differentials at that location can we prove their differences are gravitational instead of velocity based? Can you show it with math? 


#41
Nov211, 09:41 PM

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(1) Observers at radius r on a disk rotating with angular velocity [itex]\omega[/itex], in flat spacetime, have a linear velocity v and proper acceleration a of: [tex]v = \omega r[/tex] [tex]a = \frac{\omega^{2} r}{1  \omega^{2} r^{2}} = \frac{\omega v}{1  \omega^{2} r^{2}} = \omega \frac{v}{1  v^{2}}[/tex] It should be obvious from the above that I can first pick whatever v I like (subject to the constraint 0 <= v < 1), and then adjust [itex]\omega[/itex] appropriately so as to make a assume any value > 0 that I want, by adjusting r in concert with [itex]\omega[/itex] so as to keep their product constant and equal to v. So I can choose v and a independently and achieve any combination of the two within the constraints, as I said. (2) Observers "hovering" at a constant radial coordinate r above a gravitating body experience time dilation and proper acceleration of: [tex]\frac{d\tau}{dt} = \gamma = \sqrt{1  \frac{2 M}{r}} = \sqrt{1  2 U}[/tex] [tex]a = \frac{M}{r^{2} \sqrt{1  \frac{2 M}{r}}} = \frac{U}{r \gamma} = \frac{1}{r} \frac{U}{\sqrt{1  2 U}}[/tex] So again, it should be obvious that I can pick any value for [itex]\gamma[/itex] that I like (subject to the constraint that [itex]\gamma > 1[/itex]), which fixes the ratio M/r = U, and then use the second formula to adjust r appropriately, adjusting M in concert with r to keep U constant, to make a assume any value > 0 I want. So again I can choose [itex]\gamma[/itex] and a independently and achieve any combination of the two within the constraints, as I said. 


#42
Nov211, 09:43 PM

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#43
Nov211, 09:46 PM

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#44
Nov211, 10:03 PM

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#45
Nov211, 10:06 PM

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I am fully aware of your point I merely wanted to demonstrate that from another perspective it is velocity based instead of gravitational based. 


#46
Nov211, 10:09 PM

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#47
Nov211, 10:16 PM

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That was all I wanted to show. :) 


#48
Nov211, 10:16 PM

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http://en.wikipedia.org/wiki/Gullstr...A9_coordinates As that line element is written, it mixes two different radial coordinates, so I'm not sure what it's supposed to mean physically. I don't believe such a frame is valid globally. I see the Wiki page references TaylorWheeler, I'll have to break open my copy and reread their discussion of it. Edit: Oops, read "TaylorWheeler" and immediately thought "Spacetime Physics" without reading the actual book title on the Wiki page. I don't have Exploring Black Holes but it's on my list to get. 


#49
Nov211, 10:52 PM

P: 1,555

You might also want to search for Lemaître observers.
Sulu: Captain! The stars. they are gone! We can't navigate! Spock: Captain, I am locating an unidentified large object rapidly approaching us, with increasing speed. (Lemaitre) Captain: Scottie, full speed reverse. Scottie: Aye captain! Captain, we keep the engines at full speed and we are now just staying ahead of this 'thing'. Soon we will deplete our dilithium crystals. (Static) Spock: Captain, I have a theory, it might work. Scottie: We are just about to burn all of them out captain! Spock: We should stop the engines and give some small lateral impulse power. Captain: Sulu do it now! Sulu: Captain it works, but now this thing is following us at a constant distance. (Hagihara) Captain: Following us? Is it intelligent Spock? It all depends on the point of view! 


#50
Nov711, 02:36 AM

P: 153

This velocity of escape is calculated with a classical kinetic energy : U = 1/2mv^2 However shouldn't we use a relativistic expression : U = mc2*(1/Sqrt(1v2/c2)  1) ? Then we find the velocity throuh equating with the potential energy : GMm/r2 


#51
Nov711, 01:54 PM

P: 1,555

The energy of a test observer in a Schwarzschild solution is constant as long as it travels on a geodesic.
You can use the following equality: [tex]\Large E=\sqrt {1{\frac {{\it r_s}}{r}}}{\sqrt {1{v}^{2}}} ^{1}[/tex] So if you know the energy at one location it is very easy to calculate either the velocity or position at another location or velocity. For instance it is easy to see that the energy of an observer free falling at escape velocity is exactly 1, just plugin v_{escape} in the equation above. From the energy we can get the apogee of a test radial observer, because when the velocity is zero we get: [tex]\Large E=\sqrt {1{\frac {r_{{s}}}{r_{{{\it ap}}}}}} [/tex] Or alternatively we can get the apogee from the velocity at a given r value: [tex]\Large {\frac {r_{{s}}}{r_{{{\it ap}}}}}= \left( {\frac {r_{{s} }}{r}}{v}^{2} \right) \left( {v}^{2}1 \right) ^{1} [/tex] 


#52
Nov811, 01:08 AM

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#53
Nov811, 01:47 AM

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The exact form of g_00 is (12M/r) in Schwarzschild coordinates, and geometric units. Of course it's worth remembering that often other coordinates are used. For instance isotropic coordinates, which make the speed of light the same in all directions for small M/r and are used in the PPN approximations would make
g_00 = [(1M/2r) / (1+M/2r) ]^2 This still approaches 12M/r if you series expand it in 1/r, however ... (12M/r + 2M^2/r^2  (3/2) M^3 / r^3 + ...) (12U) is symbolically identical to the first expression in Schwarzschild coordinates, but the r in that expression stands for the Newtonian radius, not the Schwarzschild r coordinate, so the resemblance is formal and not an identity. 


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