coordinate time and proper time.


by Moataz
Tags: coordinate, proper, time
Moataz
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#1
Feb19-12, 07:24 PM
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Hello,

So, I have just started studying relativity, and I am confused about some basic concepts in relativity. So, the book we use says that that time has three different kinds, proper or path time, coordinate time and spacetime intervals.

I understand that coordinate time is the same as spacetime interval if the space seperation between the two events is zero (from the metric equation) The book also says that the spacetime interval is the proper time measured by a clock moving between the two event at a constant speed. This is clear as well. However, I find it hard to find a connectin between proper time and coordinate time. I know there is one since spacetime intervals, which are special case of proper time intervals, are connected to coordinate as I mentioned earlier.

So can anyone clarify this for me? Thanks.
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elfmotat
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#2
Feb19-12, 07:52 PM
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Quote Quote by Moataz View Post
Hello,

So, I have just started studying relativity, and I am confused about some basic concepts in relativity. So, the book we use says that that time has three different kinds, proper or path time, coordinate time and spacetime intervals.

I understand that coordinate time is the same as spacetime interval if the space seperation between the two events is zero (from the metric equation) The book also says that the spacetime interval is the proper time measured by a clock moving between the two event at a constant speed. This is clear as well. However, I find it hard to find a connectin between proper time and coordinate time. I know there is one since spacetime intervals, which are special case of proper time intervals, are connected to coordinate as I mentioned earlier.

So can anyone clarify this for me? Thanks.
Well, think about the spacial separation someone would measure for a clock that passes between two events. (The clock obviously reads off its own proper time.) If the clock is moving at constant velocity, then the proper time of the clock is related to an observer's coordinate time by:

[tex](c\Delta \tau)^2 =(c\Delta t)^2-(\Delta x)^2=(c\Delta t)^2-(v\Delta t)^2=(\Delta t)^2(c^2-v^2)[/tex]

Simplifying this gives:

[tex]\Delta t=\gamma \Delta \tau [/tex]


If the clock is moving along some crazy path then you would have to use an integral:

[tex]\Delta \tau =\int \sqrt{1- \frac{1}{c^2} \left ( \left (\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left (\frac{dz}{dt}\right)^2\right )}dt[/tex]
Moataz
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Feb20-12, 12:31 AM
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So, I just wanna make sure I understand this correctly. In the first equation you provided, delta T (or the coordinate time) depends only on the delta X in that frame since C is always constant?

Also, are you saying delta proper time times C=space-time interval? Because what I know is that instead of putting (C)(Delta proper time), we put delta S, or the space-time interval.

elfmotat
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Feb20-12, 01:05 AM
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coordinate time and proper time.


Quote Quote by Moataz View Post
So, I just wanna make sure I understand this correctly. In the first equation you provided, delta T (or the coordinate time) depends only on the delta X in that frame since C is always constant?

Also, are you saying delta proper time times C=space-time interval? Because what I know is that instead of putting (C)(Delta proper time), we put delta S, or the space-time interval.
Yes, [itex](\Delta s)^2=(c\Delta \tau )^2=(c\Delta t)^2-(\Delta x)^2[/itex]. This comes from the fact that, for the clock, Δx=0.
pervect
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Feb20-12, 03:43 PM
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Coordinates are just labels on a map. I use "map" here to describe any mathematical representation of the world. Usually in relativity the map is specified by a metric.

A space coordinate might be, for example, a lattitude and a longitude. A time coordinate would be a similar label - for example atomic time, TAI time, which is a "high precesion coordinate time", see the wiki http://en.wikipedia.org/w/index.php?...ldid=476895504

The job of the metric is basically to convert changes in coordinates to distances - it represents a set of scale factors that turn coordinate changes on the map into a displacement.

But there is an important wrinkle - distances in relativity are observer dependent. So the "distance" a metric returns is not an actual distance, but a space-time-interval. Space-time-intervals can be thought of as the time readings, or distance readings, made by one particular observer. I think another poster has discussed the equation that gives you the space-time interval in terms of coordinate changes is flat space-time.

Space-time itnervals can be timelike, or spacelike. If a space-time interval is timelike, it represents some wristwatch time. A space-time interval is computed between endpoints along some particular curve joining t hem. The space-time interval computed along the curve is equal to the wristwatch time, also known as proper time, elapsed from some observer following the space-time curve, or worldline, between the two endpoints.

It might be helpful to take an actual example.

Suppose point #1 is at noon TAI time at some day at sea level on the north pole, and point #2 is 1 second past noon TAI time on the same day at the same spot.

We've specified the coordinates (the position and the coordinate time) of two events, and now we want to find the space-time interval between them. In this case, it's easy - we know that the space-time interval is one second, that a clock running at sea level at the north pole will tick at the same right as TAI time does.

If we modify the problem so that the clock is located well above sea level, we would find that the space-time interval, and the proper time, was NOT 1 second, but something larger, due to corrections from the metric.
Moataz
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#6
Feb20-12, 05:11 PM
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Thank you guys. That is very helpful. I still have some questions though. I guess I am gonna need more time to absorb the concept here :)

@elfmotat:
So whenever [tex]\Delta x=0[/tex] in some reference frame,

[tex](\Delta s)^2=(c\Delta \tau )^2=(c\Delta t)^2[/tex]


Is this always true? at least in the context of special relativity?

Also, in a problem I found it says suppose there are two fire crackers. And we put a clock in between the two firecrackers, but not half way. The two firecrackers explode.

So, my thinking is: in the reference frame of the clock, where the two firecrackers are stationery to the clock, the time the clock will measure between the two events (explosion of the two fire crackers) is the coordinate time and also proper time since [tex]\Delta x=0[/tex]

However, if we supposed the fire crackers along with the clocks were moving with a constant velocity with respect to another ref frame, then in that frame the time the previous clock will measure is the coordinate time but not the proper time because [tex]\Delta x\neq 0[/tex] between the two events (clock receives light from one explosion then after some time receives the light from the other firecracker)


Is my understanding of the problem correct? or no? Also, can we safely say proper time is a special case of coordinate time? And space time interval is a special case of proper time?

@pervect: We have not actually studied the spacelike and timelike intervals yet. But may I ask you, if you know or you know a good link, who came up first with the idea that if [tex]c\Delta t[/tex] is used, then a deep quantity such as spacetime will show up?

Thank you again
elfmotat
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Feb20-12, 07:03 PM
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Quote Quote by Moataz View Post
@elfmotat:
So whenever [tex]\Delta x=0[/tex] in some reference frame,

[tex](\Delta s)^2=(c\Delta \tau )^2=(c\Delta t)^2[/tex]


Is this always true? at least in the context of special relativity?
Yes.

Quote Quote by Moataz View Post
Also, in a problem I found it says suppose there are two fire crackers. And we put a clock in between the two firecrackers, but not half way. The two firecrackers explode.

So, my thinking is: in the reference frame of the clock, where the two firecrackers are stationery to the clock, the time the clock will measure between the two events (explosion of the two fire crackers) is the coordinate time and also proper time since [tex]\Delta x=0[/tex]
But Δx isn't zero. Δx is the separation between the firecrackers. One of the firecrackers will explode at some position x1, and the other will explode some time Δt later at position x2. Δx=x2-x1 cannot possibly be zero, because the firecrackers are (as you said) separated by some distance and stationary in this frame.

The proper time separating the events would be (cΔτ)2=(cΔt)2-(Δx)2.


Quote Quote by Moataz View Post
Also, can we safely say proper time is a special case of coordinate time?
In a sense, yes. Proper time is the time between two events as measured by a clock that passes through both events, i.e. when the spacial separation between the events is zero.


Quote Quote by Moataz View Post
And space time interval is a special case of proper time?
Actually, I would say the reverse is true. When, for example, two events occur simultaneously in some frame (i.e. Δt=0) we call the spacial separation between the events the proper length, which is again equal to the interval. So I would say that proper time and proper length are special cases of the spacetime interval.


Quote Quote by Moataz View Post
@pervect: We have not actually studied the spacelike and timelike intervals yet.
A spacelike interval is one in which Δx>cΔt, i.e. the spacial separation between the events is larger than the distance that beam of light could have traveled in Δt. Two events that are separated by a spacelike interval cannot be causally connected (i.e. one cannot have caused the other) because information can't travel faster than c.

A timelike interval is one in which cΔt>Δx. Events connected by a timelike interval can be causally connected.

A null interval is one in which cΔt=Δx. This only happens for things that travel at c, i.e. photons.


Quote Quote by Moataz View Post
But may I ask you, if you know or you know a good link, who came up first with the idea that if [tex]c\Delta t[/tex] is used, then a deep quantity such as spacetime will show up?
It comes right from the Lorentz transformation:

[itex]\Delta x'=\gamma (\Delta x-v\Delta t)[/itex]

[itex]\Delta t'=\gamma (\Delta t-v\Delta x/c^2)[/itex]


If you calculate (cΔt')2-(Δx')2, you find that:

(cΔt')2-(Δx')2=(cΔt)2-(Δx)2

This quantity has the same value in both frames, meaning it doesn't matter which frame you calculate it from. You just give this quantity the symbol 's' and call it the spacetime interval.
Naty1
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#8
Feb20-12, 08:04 PM
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Also check at least the introductions to 'proper time' and 'coordinate time' in Wikipedia
for some good insights...read further if interested for additional ones.
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Feb20-12, 09:06 PM
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Quote Quote by elfmotat View Post
Yes, [itex](\Delta s)^2=(c\Delta \tau )^2=(c\Delta t)^2-(\Delta x)^2[/itex]. This comes from the fact that, for the clock, Δx=0.
So if c is taken to be 1 , then is the spacetime interval always numerically equal
to the proper time ?
Matterwave
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Feb20-12, 09:11 PM
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Quote Quote by morrobay View Post
So if c is taken to be 1 , then is the spacetime interval always numerically equal
to the proper time ?
Depending on your metric, it's either equal to the proper time, or equal to the squareroot of the negative of the square of the proper time (i.e. [itex]ds=\sqrt{-d\tau^2}[/itex]).
Naty1
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Feb21-12, 10:40 AM
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Actually, I followed my own advice and read further here:

http://en.wikipedia.org/wiki/Coordinate_time

under the second heading: Coordinate time, proper time, and clock synchronization

Can someone paraphrase or explain these two paragraphs:


But the coordinate time is not a time that could be measured by a clock located at the place that nominally defines the reference frame, e.g. a clock located at the solar system barycenter would not measure the coordinate time of the barycentric reference frame, and a clock located at the geocenter would not measure the coordinate time of a geocentric reference frame.[4] The coordinate times cannot be measured, but only computed from the (proper-time) readings of real clocks with the aid of the time dilation relationship shown in equation (2) (or some alternative or refined form of it).

Only for explanatory purposes it is possible to conceive a hypothetical observer and trajectory on which the proper time of the clock would coincide with coordinate time: such an observer and clock have to be conceived at rest with respect to the chosen reference frame (v = 0 in equation (2) above) but also (in an unattainably hypothetical situation) infinitely far away from its gravitational masses (also U = 0 in equation (2) above).[5] Even such an illustration is of limited use because the coordinate time is defined everywhere in the reference frame, while the hypothetical observer and clock chosen to illustrate it has only a limited choice of trajectory.
pervect
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Feb21-12, 03:32 PM
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I'd rewrite it pretty much totally, something along these lines:

It's conventional to define coordinates in such a way that the coordinate time advances at the same rate as the time kept by an actual clock, i.e. proper time, at the origin, but there's no reason you have to do this other than convention. Generalized coordinates are truly general. Generalized coordinates are just labels that you use to identify events in space-time, and you have complete freedom to use any set of labels that you like.

A corollary to this freedom is that truly general coordinates have no physical significance whatsoever, being just labels.
salvestrom
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Feb21-12, 05:02 PM
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Quote Quote by Naty1 View Post
Actually, I followed my own advice and read further here:

http://en.wikipedia.org/wiki/Coordinate_time

under the second heading: Coordinate time, proper time, and clock synchronization

Can someone paraphrase or explain these two paragraphs:
Those two paragraphs have got me into arguements more than once on these forums.

I read it as this: every observer, stationary or in motion, has their own proper time. Coordinate time is some sort of hypothetical "proper time" of the universe in some newtonian-esque fashion: the time on a hypothetical clock at rest, far from us at infinity. In practice coordinate time is taken as the time on a clock at rest with respect to the observer in motion whose time dilation is to be determined. In "reality" both observers are experiencing some amount of time and gravitational dilation.

So more accurately, not realising that noone else here sees it this way is what's caused the arguements. :P
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Feb21-12, 06:54 PM
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Quote Quote by salvestrom View Post
Those two paragraphs have got me into arguements more than once on these forums.

I read it as this: every observer, stationary or in motion, has their own proper time. Coordinate time is some sort of hypothetical "proper time" of the universe in some newtonian-esque fashion: the time on a hypothetical clock at rest, far from us at infinity. In practice coordinate time is taken as the time on a clock at rest with respect to the observer in motion whose time dilation is to be determined. In "reality" both observers are experiencing some amount of time and gravitational dilation.

So more accurately, not realising that noone else here sees it this way is what's caused the arguements. :P
I wonder if we can get around the arguing by simply saying "imagine a massless universe."

Imagine that you lived in a mass-free universe, full of massless "clocks", and they were all moving apart more-or-less randomly like this...



...with the outermost shell moving at the speed of light.

Now find the dot in the center that is not moving. That clock represents a clock that is moving at the full speed of time. The clocks further away from the center that ARE moving in space are ticking at slower speeds in time. That's their proper time. Each clock has its own proper-time. The clocks at the furthest edges are not moving forward in time at all.

However, the coordinate time is based on the clock in the center.

But none of the other clocks care about the clock in the center. So the coordinate time isn't based on any clocks. It's just based on the more-or-less arbitrary perspective of the clock in the center.
salvestrom
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Feb21-12, 07:40 PM
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Quote Quote by JDoolin View Post
I wonder if we can get around the arguing by simply saying "imagine a massless universe."
If the central clock is stationary and the others are moving due to a ballistic explosion, then the central clock is far from arbitrary. If the central clock is stationary and the other clocks are moving due to spatial expansion then they are also stationary and will be recording the same time as the central clock, making the central clock arbitrary, but making all the clocks representative of a universal coordinate time, a value against which any moving clock might be calculated.

The closest thing to this would be the proper time in the middle of a supervoid.
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Feb21-12, 09:24 PM
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Quote Quote by salvestrom View Post
If the central clock is stationary and the others are moving due to a ballistic explosion, then the central clock is far from arbitrary.
If there are a finite number of clocks then the central clock is not arbitrary. If you have an infinite number of clocks, then the central clock is arbitrary.

But let's say there ARE a finite but (very very very) large number of clocks, and we can pick out one particular clock that is the center of the explosion. That middle clock is still not going to seem terribly significant to the others.

Each clock would still see roughly the same speed-of-light expanding sphere of clocks no matter where it was, unless it was at one edge of the explosion.
salvestrom
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Feb21-12, 09:38 PM
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Quote Quote by JDoolin View Post
If there are a finite number of clocks then the central clock is not arbitrary. If you have an infinite number of clocks, then the central clock is arbitrary.
I don't see how the number of clocks makes a difference. In the ballistic explosion the central non-moving clock is at the heart of the event, experiencing no velocity-based time dilation giving it a unique quality that none of the others have. In the spatial expansion version, I acknowledge that no clock is unique and infact suggest that all the clocks are synchronised since none of them have an actual velocity.

In the first case I simply put forward that the central clock is actually central, non-moving and not in a gravitational field making it an example of an otherwise hypothetical concept, that cannot be found so easily in our actual universe. But then, I put the same thing forward for the second example, but this time say that all the clocks are representative of the hypothetical coordinate clock.
morrobay
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Feb21-12, 09:46 PM
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Quote Quote by Matterwave View Post
Depending on your metric, it's either equal to the proper time, or equal to the squareroot of the negative of the square of the proper time (i.e. [itex]ds=\sqrt{-d\tau^2}[/itex]).
Then if the spacetime interval is timelike it is numerically equal to the proper time.
If the spacetime interval is spacelike then it is numerically equal to the proper distance ?


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