
#55
Dec3010, 12:16 AM

P: 621

[tex]{\int {{d}{\theta}}}[/tex] gives an angular distance, when integration is performed between the limits [tex]{(}{\theta}_{1}{,}{\theta}_{2}{)}[/tex] While the integral, [tex]{\int {r}{d}{\theta}}{=}{\int{\sqrt{{g}_{\theta\theta}}{d}{\theta}}[/tex] Conforms to the idea of distance[it has the dimension of length] along the curve r=constant and phi=constant "Physical distance" is not an arbitrary phase. It has a well defined physical concept behind it. I would request the mentors to comment on the issue. 



#56
Dec3010, 12:24 AM

Sci Advisor
P: 8,470





#57
Dec3010, 12:27 AM

P: 621





#58
Dec3010, 01:18 AM

Sci Advisor
P: 8,470

Finally, if you can't actually explain what the "well defined physical concept" behind your notion of "physical distance" is supposed to be, then it follows that you don't actually have any welldefined concept in mind that you used to derive the equation. 



#59
Dec3010, 02:00 AM

P: 621

We can always calculate the physical distance along an axis.Let us consider the points[tex]{(}{r}{,}{\theta}{,}{\phi}{)}[/tex] and [tex]{(}{r}{,}{\theta}^{'}{,}{\phi}^{'}{)}[/tex]
We may calculate the distance between them along r=const,phi=[tex]{\phi}[/tex],[taken const] or r=constant ,phi=[tex]{\phi}^{'}[/tex], [taken constant] between the aforesaid coordinates using the second integral in #55 We may also calculate the distance between them along r=const,theta=[tex]{\theta}[/tex] [taken constant] or r=constant ,theta=[tex]{\theta}^{'}[/tex], [taken constant] between the aforesaid points using the integral: [tex]{\int{ \sqrt{g}_{\phi\phi}}{d}{\phi}[/tex] For the two parallel lines[of latitude lying between the meridians concerned] we get different values. This is quite natural [I am referring to a common,3D spherical surface,obviously a spatial one] 



#60
Dec3010, 02:20 AM

P: 621

If one wants to calculate the physical distance along some path connecting [tex]{(}{r}{,}{\theta}{,}{\phi}{)}[/tex] and [tex]{(}{r}{,}{\theta}^{'}{,}{\phi}^{'}{)}[/tex]
He/she can use the integral [tex]{\int{dL}}{=}{\int{\sqrt{{g}_{rr}{dr}^{2}{+}{g}_{\theta\theta}{{d}{\the ta}}^{2}{+}{g}_{\phi\phi}{d}{{\phi}}^{2}}}[/tex] Along the specified path lying on the surface of the sphere 



#61
Dec3010, 12:29 PM

Sci Advisor
P: 8,470





#62
Dec3010, 03:19 PM

P: 621

Similarly for the time axis we have to consider a fixed spatial location. Interval is given by the integral: [tex]{\int {\sqrt{g}_{00}}{dt}}[/tex] But what am I to understand by the time interval between events occurring at distant points A and B if I am working in a laboratory at A.?Proper time would not help me since I am not ready to run between the events with a clock in my hand. Coordinate time difference may not have the unit of time itself. The time interval in such a situation may be conveniently defined by the integral:[tex]{\int {\sqrt{g}_{00}}{dt}}[/tex] and this may be a path dependent quantity.I have explained this in #49 In fact time interval between a pair of events is meaningful only if signals can be passed between them.You may consider [as an example]a pair of events A and B which have a spacelike separation. If you are stationed at A the event B[at a spacelike separation] exits hypotheticallyits there in your imagination or you can have it in the papers of mathematics![since you can never have any information from it]If you are in a closed isolated room you can only speculate what is happening in the outside world you can never observe them.Quite meaningless to say when did such events occur? If signals can be passed between a pair of events[considering a finite separation] you may define the time difference by the line integral I have given.The value may be path dependentthat should not usher in any type of contradiction. 



#63
Dec3010, 04:42 PM

Sci Advisor
P: 8,470





#64
Dec3010, 05:04 PM

Sci Advisor
P: 8,470

In any case, it seems to me that if you do choose to use a coordinate system where the tcoordinate is spacelike rather than timelike, then in that case the integral [tex]\int \sqrt{g_{tt}} \, dt[/tex] will not actually have units of time! If we define the metric in such a way that ds^2 is positive for timelike intervals and negative for spacelike intervals, then this means that if you integrate the metric along a path (i.e. evaluate the integral [tex]\int \sqrt{ds^2}[/tex] along the path) and get a real number then the path's length in spacetime has "units" of time, while if you get an imaginary number then the path's length in spacetime has "units" of distance. So, let's consider a simple example of a coordinate system where the tcoordinate is spacelike: let's suppose in Minkowski spacetime we use a coordinate system where the "taxis" was identical to the "xaxis" of an ordinary SR inertial frame, and likewise the "xaxis" of this coordinate system was identical to the "taxis" of the SR inertial frame, and both coordinate systems had identical y and z axes. In this case the metric for this new coordinate system would have [tex]g_{tt} = g_{yy} = g_{zz} =1/c^2[/tex] and [tex]g_{xx} = 1[/tex]. If we picked a path of constant x,y,z coordinates and varying tcoordinate and evaluated [tex]\int \sqrt{g_{tt}} \, dt[/tex], then the result will be an imaginary number, which means the answer has units of distance, not time (since this coordinate system is just an SR inertial frame with the x and t axes switched, then we should get the same answer as if we evaluated the integral [tex]\int \sqrt{g_{xx}} \, dx[/tex] using the Minkowski metric, where [tex]g_{xx} = 1/c^2[/tex]). 



#65
Dec3010, 11:42 PM

P: 621

One can consider a very simple case: An event occurs at point[spatial] B and is reported at A[spatial point].
Events:1)Transmission of signal from B 2)Reception of the same signal at A This signal can reach A by several paths. So we are considering a number of events at AA1,A2,A3,.........They have the same spatial coordinates but different temporal coordinates. For each pair [Event at B which includes the time part, Ai] ,the time interval is=Time taken by the light ray along the path concerned=[tex]{\int {\sqrt{{g}_{00}}dt}}[/tex] [please see #49 for the calculations] Each set [event at B, Ai] may have a different value for the time interval [depending on the path of travel of the light ray]but this interval is being correctly measured/depicted by the integral[ which is nothing but the time of travel of the light ray between the events] Each event is ,of course, observed at a different instant of time as I have said in #49. The real problem lies in the fact that if a pair of events[spacetime points] are specified, it may not always be possible to connect them by a light raythey may not lie lie on a null geodesic.In such a situation we may use mirrors to guide the ray from one point to the other.This may be possible in a huge number of cases. 



#66
Dec3010, 11:57 PM

P: 621

A pair of events may not lie on a null geodesic.[One may consider a spacelike separation by way of example]But we can always change the temporal coordinate at the reception point to enable a connection by a null geodesic.[ I believe ,this has been suggested by Jesse herselfplease do correct me if I am mistaken!]This may help us.
But for a certain period of time we remain ignorant of the information at the reception center 



#67
Dec3110, 12:24 AM

Sci Advisor
P: 8,470





#68
Jan311, 04:47 AM

P: 621

This is in relation to #67
A pair of events (t,x,y,z) and (t',x',y',z') with spatial separation are considered.We cannot connect them by a signal.But at the spatial location (x,y,z) we have the event (t1,x,y,z) at a later point of time. Now the events (t1,x,y,z) and (t',x',y',z') may be connected by a null geodesic to get information from the second point. But one has to wait for the coordinate interval (t1t) or the physical interval g(00)(t1t) to get the informationwhich may be a hundred years or more. [Change in temporal coordinate:t1t] The coordinate system is not being changed. 



#69
Mar2011, 07:38 AM

P: 205





#70
Mar2011, 09:09 AM

P: 621

The integral in the last posting is not path dependentit simply represents the time of travel of a light ray between a pair of points along some specified null geodesic. This matter has been explained in #49,#65
The issue may be recast in the following manner: Suppose you are standing in your laboratory and a pair of events occur in some distant galaxies in curved spacetime.What do you understand by the time difference between the two events? [Your laboratory is also in a region of sufficient spactime curvature] Since you in your laboratory, you cannot travel between the events along a spacetime path.Proper time will not help you. Coordinate time should not be used since the speed of light in vacuum changes if coordinate time is considered.You may take the example of the Schwarschild Geometry. Physical time could have been a useful concept if it was independent of path.But it is not. In most cases the spacetime points do not correspond to conjugate pointsthey cannot be connected by several null geodesics. Physical time gives us a unique picture in many cases. 



#71
Mar2011, 09:26 AM

P: 10

Anamitra, how do you interpret this?
http://www.physicsforums.com/showthread.php?t=172288 Here we have an example of something is both moving and stationary, depending on how you interpret it. So time means something different with distant objects. Cooperstock was at pains to explain this in his book "General Relativistic Dynamics". The correct understanding is essential for the correct interpretation of distant processes. edit: Here is Cooperstock's paper on "General Relativistic Velocity" http://arxiv.org/abs/0712.0019 and here is the application http://arxiv.org/abs/1101.3224 drl 


Register to reply 
Related Discussions  
Why is physical timeasymmetry vs dynamical law timesymmetry a problem?  General Physics  8  
Separation of variables and the separation constant  Calculus & Beyond Homework  4  
Separation of time and space  Differential Equations  6  
what is the physical meaning for time?  General Physics  6  
calculate separation of galaxies as a function of time  General Astronomy  12 