# On the Physical Separation of Time

by Anamitra
Tags: physical, separation, time
P: 621
 Quote by JesseM But "physical distance" is just an arbitrary phrase you made up to describe that integral, which has no meaning apart from your definition--correct? If so you should be careful to use the full phrase "physical distance" and not just something more vague like "distance along the x-axis".
Just think of two points an ordinary sphere[three dimensional-spatial]

$${\int {{d}{\theta}}}$$ gives an angular distance, when integration is performed between the limits $${(}{\theta}_{1}{,}{\theta}_{2}{)}$$
While the integral,
$${\int {r}{d}{\theta}}{=}{\int{\sqrt{{g}_{\theta\theta}}{d}{\theta}}$$
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.
P: 8,470
 Quote by Anamitra Just think of two points an ordinary sphere[three dimensional-spatial] $${\int {{d}{\theta}}}$$ gives an angular distance, when integration is performed between the limits $${(}{\theta}_{1}{,}{\theta}_{2}{)}$$ While the integral, $${\int {r}{d}{\theta}}{=}{\int{\sqrt{{g}_{\theta\theta}}{d}{\theta}}$$ Conforms to the idea of distance[it has the dimension of length] along the curve r=constant and phi=constant
And what if you pick some different curve with varying phi and/or r? In that case this integral will not necessarily correspond to the geometric length of the curve, do you agree?
 Quote by Anamitra "Physical distance" is not an arbitrary phase. It has a well defined physical concept behind it.
Can you explain what that "well defined physical concept" actually is?
P: 621
 Quote by JesseM And what if you pick some different curve with varying phi and/or r? In that case this integral will not necessarily correspond to the geometric length of the curve, do you agree?
I am measuring the distance along the axis defined by r=const and phi=constant[only theta changes along this direction]
P: 8,470
 Quote by Anamitra I am measuring the distance along the axis defined by r=const and phi=constant[only theta changes along this direction]
This seems like an artificial restriction since there is no mathematical reason that the integral couldn't be evaluated along paths where r and/or phi are allowed to vary. Besides, when it comes to defining "physical distance" and "physical time" in spacetime, you didn't mention any requirement that all the other coordinates besides the one being integrated must be constant, so if the sphere is supposed to be an analogy to what you're talking about in spacetime, imposing such requirements in one case but not the other seems to make the analogy break down.

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 well-defined concept in mind that you used to derive the equation.
 P: 621 We can always calculate the physical distance along an axis.Let us consider the points$${(}{r}{,}{\theta}{,}{\phi}{)}$$ and $${(}{r}{,}{\theta}^{'}{,}{\phi}^{'}{)}$$ We may calculate the distance between them along r=const,phi=$${\phi}$$,[taken const] or r=constant ,phi=$${\phi}^{'}$$, [taken constant] between the aforesaid coordinates using the second integral in #55 We may also calculate the distance between them along r=const,theta=$${\theta}$$ [taken constant] or r=constant ,theta=$${\theta}^{'}$$, [taken constant] between the aforesaid points using the integral: $${\int{ \sqrt{g}_{\phi\phi}}{d}{\phi}$$ 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]
 P: 621 If one wants to calculate the physical distance along some path connecting $${(}{r}{,}{\theta}{,}{\phi}{)}$$ and $${(}{r}{,}{\theta}^{'}{,}{\phi}^{'}{)}$$ He/she can use the integral $${\int{dL}}{=}{\int{\sqrt{{g}_{rr}{dr}^{2}{+}{g}_{\theta\theta}{{d}{\the ta}}^{2}{+}{g}_{\phi\phi}{d}{{\phi}}^{2}}}$$ Along the specified path lying on the surface of the sphere
P: 8,470
 Quote by Anamitra If one wants to calculate the physical distance along some path connecting $${(}{r}{,}{\theta}{,}{\phi}{)}$$ and $${(}{r}{,}{\theta}^{'}{,}{\phi}^{'}{)}$$ He/she can use the integral $${\int{dL}}{=}{\int{\sqrt{{g}_{rr}{dr}^{2}{+}{g}_{\theta\theta}{{d}{\the ta}}^{2}{+}{g}_{\phi\phi}{d}{{\phi}}^{2}}}$$ Along the specified path lying on the surface of the sphere
Obviously you can get the length along a path using the whole metric, and similarly in spacetime you can get the coordinate-invariant proper time along a timelike path, or the coordinate-invariant proper distance along a spacelike path, by integrating the whole spacetime metric along that path, i.e. $$\int{\sqrt{{g}_{tt}{dt}^{2}{+}{g}_{xx}{{dx}^{2}{+}{g}_{yy}{dy}^{2}}{+}{ g}_{zz}{dz}^{2}}$$. And obviously if you pick a path of constant r and phi, so dr = dphi = 0 all along the path, then the total spherical integral in 3D space which you wrote above reduces to $$\int \sqrt{ g_{\theta\theta} \, d\theta^2 } = \int \sqrt{g_{\theta\theta}} \, d\theta$$, and similarly if we pick a spacelike path through spacetime with constant t, y, and z coordinate, then the integral in that case would reduce to $$\int \sqrt{ g_{xx} \, dx^2 } = \int \sqrt{g_{11}} \, dx$$ which is what you wrote down earlier. But when you wrote down this integral, you just said it was the "distance along the x-axis", you didn't specify that it applies only to paths where the x-coordinate varies while the t, y, and z coordinates are constant. Did you mean to imply that restriction? (please answer this question yes or no) I didn't think you were implying this, since you seemed to say earlier that your notion of "physical time" could be applied to arbitrary paths, not just paths of constant position coordinate (i.e. t varying while x, y, z stay constant). And if you did intend for your integral to define "distance along the x-axis" for arbitrary paths, not just paths where 3 coordinates are held constant, then the analogy to distance in 3D space doesn't work, because the integral $$\int \sqrt{g_{\theta\theta}} d\theta$$ doesn't define "distance" for arbitrary paths in space, it only defines distance along paths where r and phi are constant.
P: 621
 Quote by JesseM But when you wrote down this integral, you just said it was the "distance along the x-axis", you didn't specify that it applies only to paths where the x-coordinate varies while the t, y, and z coordinates are constant. Did you mean to imply that restriction? (please answer this question yes or no) I didn't think you were implying this, since you seemed to say earlier that your notion of "physical time" could be applied to arbitrary paths, not just paths of constant position coordinate (i.e. t varying while x, y, z stay constant). And if you did intend for your integral to define "distance along the x-axis" for arbitrary paths, not just paths where 3 coordinates are held constant, then the analogy to distance in 3D space doesn't work, because the integral $$\int \sqrt{g_{\theta\theta}} d\theta$$ doesn't define "distance" for arbitrary paths in space, it only defines distance along paths where r and phi are constant.
By x-axis we mean paths along which the other coordinates do not change.
Similarly for the time axis we have to consider a fixed spatial location. Interval is given by the integral: $${\int {\sqrt{g}_{00}}{dt}}$$

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:$${\int {\sqrt{g}_{00}}{dt}}$$
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 hypothetically--its 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 dependent--that should not usher in any type of contradiction.
P: 8,470
 Quote by Anamitra By x-axis we mean paths along which the other coordinates do not change. Similarly for the time axis we have to consider a fixed spatial location. Interval is given by the integral: $${\int {\sqrt{g}_{00}}{dt}}$$ 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:$${\int {\sqrt{g}_{00}}{dt}}$$ and this may be a path dependent quantity.
So you are trying to evaluate this integral along paths where the x,y,z coordinates are not constant, correct? If so, I think you can see why the analogy with "distance" in 3D space doesn't work--evaluating this integral along paths of non-constant x,y,z is analogous in 3D to evaluating the integral $$\int \sqrt{g_{\theta\theta}} \, d\theta$$ along paths where r and phi are not constant, and this integral does not give what we ordinarily define to be the "distance" along such a path in 3D space. So, your argument in post #55 doesn't work as a justification for why you think it makes sense to label the integral $${\int {\sqrt{g}_{xx}}{dx}}$$ as a "distance" even when evaluating along a path where the other coordinates aren't held constant. So, you still haven't provided any meaningful justification for your statement in post #55:
 "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.
If you just wanted to say that "physical distance" was an arbitrary label you came up with for integrals like $${\int {\sqrt{g}_{xx}}{dx}}$$ that would be fine with me, but there is no reason it is "natural" to label it this way given our prior notions of "distance".
P: 8,470
 Quote by Anamitra 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.
Ordinarily a physicist will choose a coordinate system where paths of constant position coordinate are actually timelike and paths of constant time coordinate are actually spacelike, and in this case the coordinate time difference will have units of time. It's true that in the case of Schwarzschild coordinates the t-coordinate becomes spacelike inside the horizon, but I would think (and someone more experienced in how these coordinate systems are normally used can correct me if I'm wrong) that Schwarzschild coordinates would mostly be used to analyze events outside the horizon, whereas if a physicist wanted to analyze events inside the horizon it would be more common to use a coordinate system where the t-coordinate is still timelike inside the horizon, like "free-fall coordinates" or Kruskal-Szekeres coordinates illustrated near the bottom of this page.

In any case, it seems to me that if you do choose to use a coordinate system where the t-coordinate is spacelike rather than timelike, then in that case the integral $$\int \sqrt{g_{tt}} \, dt$$ 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 $$\int \sqrt{ds^2}$$ 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 t-coordinate is spacelike: let's suppose in Minkowski spacetime we use a coordinate system where the "t-axis" was identical to the "x-axis" of an ordinary SR inertial frame, and likewise the "x-axis" of this coordinate system was identical to the "t-axis" 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 $$g_{tt} = g_{yy} = g_{zz} =-1/c^2$$ and $$g_{xx} = 1$$. If we picked a path of constant x,y,z coordinates and varying t-coordinate and evaluated $$\int \sqrt{g_{tt}} \, dt$$, 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 $$\int \sqrt{g_{xx}} \, dx$$ using the Minkowski metric, where $$g_{xx} = -1/c^2$$).
 Quote by Anamitra 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 hypothetically--its 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?
But even if the events are spacelike-separated so that no signal can pass between the events themselves, signals about each event might later reach an observer in the overlap of each event's future light cone, so this observer can learn the coordinates of each event and calculate things such as the coordinate time between them.
 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 A--A1,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=$${\int {\sqrt{{g}_{00}}dt}}$$ [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[space-time points] are specified, it may not always be possible to connect them by a light ray--they 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.
 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 herself--please 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
 Quote by Anamitra We are considering a stationary curved spacetime fabric. Temporal separation[Physical]is given by: $${T}_{2}{-}{T}_{1}{=}{\int \sqrt {g}_{00}{dt}$$ [Limits of integration extending from t1 to t2which are of course the coordinate times] The above integral is path dependent,in the general case[depending on the nature of g(00)].So the physical separation of time in general is not unique for a pair of events. To reconcile the matter ,g(00) should not depend on more than one coordinate[leaving aside t]or else[rather in a generalized way] the above integral should be independent of path.