
#73
Oct2210, 05:09 PM

Emeritus
Sci Advisor
P: 7,443

A more detailed calculation confirms to my satisfaction that the ferminormal distance will give the same result as the measurement of distance via a Born rigid ruler. (I felt I needed to confirm that the timelike worldlines of the accelerating observers at constant ferminormal distance were orthogonal to the spacelike geodesics along which the distance was measured. This will be true if the metric in these coordinates is diagonal away from the origin, which appears to be the case. Using the same series approach but including higher order terms, I convinced myself that metric was in fact diagonal.) However, I don't believe that passionflower's distance will be equivalent to the distance measured by dropping a Born rigid ruler. 



#74
Oct2210, 06:12 PM

P: 3,967

Just a quick note to acknowledge that Passionflower pointed out a major mistake in my calculations for the motion of a falling body with an initial velocity at infinity in post #69. I have now edited that post and hopefully the equations are now correct. Thanks for checking Passionflower. It is good to know that someone is checking the posted equations so that they might be a useful and accurate reference to someone in the future .




#75
Oct2210, 07:10 PM

P: 3,967





#76
Oct2210, 07:31 PM


#77
Oct2210, 08:14 PM

Sci Advisor
P: 8,470

Is it true that each hovering observer's local surface of simultaneity in their local inertial rest frame is parallel to a line of constant Schwarzschild time and varying radius? If so, maybe Passionflower's distance would be the proper distance along a single spacelike curve where the tangent at each point along the curve is parallel to the local surface of simultaneity of a freefalling observer at that point?




#78
Oct2210, 10:10 PM

P: 1,555

Hopefully we can work out the kinks because then we have a generic velocity formula for a free falling observer. Then it is simply a matter of finding the Lorentz factor based on the obtained velocity v(r) and divide (because it is length contraction! ) the integrand by the Lorentz factor to get the correct integrand to obtain a proper distance in all free falling radial scenarios (except for acceleration that is, but I think it is better to get the kinks out free falling motion first). By the way, the parametric approach has the advantage that we can extend things to a solution with a rotating black hole by applying some adjustments. As far as I am concerned that would then cover most areas where the notion of distance makes some sense, as the concept of proper (e.g. physical) distance in nonstationary spacetimes do not make much sense to me (except perhaps in things like FLRW models where distances are 'pumped up' in time). To me, but that is likely because I am simple minded, the generic distance discussion is rather academic for arbitrary nonstationary spacetimes as there is really not much to calculate. By the way once this is sorted out it would be really interesting to lay out (in a separate topic) all the Doppler shifts (including a decomposition into kinematical and gravitational components) for all radial motion wrt a stationary local observer, an observer from infinity and of course wrt other free falling observers. As far as I know, no such generic formula is available. 



#79
Oct2210, 10:44 PM

P: 3,967

Starting with the Schwarzschild falling coordinate velocity I gave in #69: [tex] \frac{dr}{dt} = (12m/r) \sqrt{\frac{2m}{r}(1V^2) +V^2} [/tex] and using dr'/dr = 1/sqrt(12m/r) and dt/dt' = 1/sqrt(12m/r) to relate local measurements of the stationary observer to the observer at infinity such that dr/dt = dr'/dt' (12m/r), the local coordinate velocity is: [tex] \frac{dr'}{dt'} = \sqrt{\frac{2m}{r}(1V^2) +V^2} [/tex] This can be written using A = 1/sqrt(1v^2) as: [tex] \frac{dr'}{dt'} = \sqrt{\frac{2m/r +V^2A^2}{A^2} } [/tex] [tex]\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{2m/r + A^2  A^2 +V^2A^2}{A^2} } [/tex] (because A^2A^2 =0) [tex]\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{2m/r + A^2  A^2(1V^2)}{A^2} } [/tex] [tex]\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{2m/r + A^2  1 }{A^2} [/tex] [tex]\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{ A^2  (1 2m/r) }{A^2} [/tex] [tex]\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{ A^2  g_{\it tt} }{A^2} [/tex] 



#80
Oct2210, 11:01 PM

P: 1,555

Since: [tex] {dt \over d\tau} = {A \over g_{tt} } [/tex] It should be staightforward to convert this coordinate velocity to proper velocity. By the way it would be helpful is someone could attach the relevant page from Hartle, unfortunately I do not have it: http://books.google.com/books?id=azZ...ed=0CC0Q6AEwAA 



#81
Oct2310, 12:15 AM

P: 3,967

[tex] \frac{dr}{dtau} = \sqrt{\frac{2m}{r} + \frac{ V^2}{(1V^2)}} [/tex] where tau is the proper time of the falling clock. This can be converted to the proper velocity as measured by a local observer at coordinate radius r by using dr'/dr = sqrt(12m/r) so that: [tex] \frac{dr'}{dtau} = \frac{1}{\sqrt{12m/r}} \sqrt{\frac{2m}{r} + \frac{ V^2}{(1V^2)}} [/tex] and this can be simplified using A = 1/sqrt(1V^2) and [itex]g_{tt} = (12m/r)[/itex] to: [tex] \frac{dr'}{dtau} = \sqrt{\frac{A^2  g_{tt}}{g_{tt}}} [/tex] The equation for the local velocity of the falling object given by Hartle is obviously not the proper velocity. [tex] \frac{dr'}{dt'} =\sqrt {{\frac {{A}^{2}g_{{{\it tt}}}}{{A}^{2}}}} [/tex] the local proper velocity is: [tex]\frac{dr'}{dtau} = \frac{dr'}{dt'}*\frac{dt'}{dt}*\frac{dt}{dtau} = \frac{A}{\sqrt{g_{tt}}} \sqrt {\frac {{A}^{2}g_{tt}}{A^2}} = \sqrt{\frac{A^2  g_{tt}}{g_{tt}}} [/tex] 



#82
Oct2310, 12:27 AM


#83
Oct2310, 12:54 PM

P: 3,967





#84
Oct2310, 01:13 PM

P: 1,555





#85
Oct2310, 02:16 PM

P: 3,967

[tex] v_{(r)} = \sqrt {2m \left( {r}^{1}{{\it r_0}}^{1}\right) } [/tex] which agrees with the mathpages equation I gave in #69 for the proper velocity dr/dtau: [tex] \frac{dr}{dt} = \sqrt{\frac{2m}{r}(1v_0^2) +2mv_0^2} [/tex] which does not agree with either the mathpages coordinate velocity or proper velocity: You did not specify if [itex]v_{(r)}[/itex] is intended to be a proper or coordinate velocity or if the observer that makes the measurement is local or at infinity. [EDIT] Actually looking back I am guilty of the same thing in #69 and did not make it clear that all measurements in that post are according to the Schwarzschild observer at infinity. 



#86
Oct2310, 03:38 PM

P: 1,555

Perhaps that is the source of some of the confusion we had communicating this. At all times I talked about the velocity as measured by a local stationary observer. So far I do not see anything wrong with the formula, and the plots look right as well. Perhaps you could present an alternative and then we can plot those out as well. I would not have been wrong the first time. :) 



#87
Oct2310, 04:03 PM

Emeritus
Sci Advisor
P: 7,443

The motion of the observers can't be ignored and is important to specify, because different observers have different notions of space and time. As far as using small, bornrigid rulers goes  if you are in an area of spacetime small enough that the effects of curvature are small, radars and rulers agree. This is why the SI folks were able to switch from a physical, rigid rod, to a radarlike method of counting the wavelengths of a particular frequency of a cesium standard without a major upheaval. When you consider how atoms maintain their distance, it's not surprising, the forces between nearby atoms are based on the same electromagnetic fundamentals that the radar is based on. One of the advantages of the radar concept for measuring distance is that it makes the implied concept of simultaneity more manifest than using rigid bars. Let one wolrdline hold the radar. The other worldline reflects the radar signal. Then the midpoint between the time of signal emission and its reception on the radar worldline is simultaneous with the reflection of the radar signal on the reflection worldline. This defines a short segment of a curve on the radar worldline, that propagates in a spatial direction. It's an alternate way of defining "orthogonal" without the math, though the 4vector approach remains more convenient in my opinion. And if you imagine repeating the process, by putting a bunch of radars on a bunch of worldlines, and adjusting the timing so that the short linesegments all connect, you'll generate the entire curve, the curve of simultaneity, the one that PF's distance is being measured along. 



#88
Oct2310, 04:33 PM

P: 6

I want to know that is it possible that we will get all our lost things when the big CRUNCH occur?




#89
Oct2310, 04:49 PM

P: 3,967

The local velocity of a falling particle with initial velocity v0 at infinity, according to an observer at r, was derived in #79 and is: [tex] \frac{dr'}{dt'} = \sqrt{\frac{2m}{r}(1v0^2) +v0^2} [/tex] The local velocity of a falling particle with apogee r0 according to an observer at r is: [tex] \frac{dr'}{dt'} = \sqrt{\frac{2m(r^{1}  r0^{1})}{(12m/r0)}} [/tex] [tex]\frac{dr'}{dt'} = \sqrt{\frac{2m(r^{1}(1v0^2)  r0^{1})}{(12m/r0)}+v0^2} [/tex] Just to be clear, I am not talking about proper velocity and I assume you are not either. 



#90
Oct2310, 04:52 PM

P: 3,967




Register to reply 
Related Discussions  
Discrete Math "Proper Subsets"  General Math  23  
Calculating the "proper mean lifetime" of pions  Advanced Physics Homework  0  
"Proper" description of uniform circular motion  Introductory Physics Homework  3  
"Proper" accelerations and rigid bodies.  Special & General Relativity  33  
The proper article to use in the sentence "A/An NaI detector"  General Discussion  39 