Coordinate time and proper time.

In summary: MoreIn summary, the book says that time has three different kinds: proper time, coordinate time, and spacetime intervals. Proper time is the same as spacetime interval if the space separation between the two events is zero. Coordinate time is the time measured by a clock moving between the two events at a constant speed. Spacetime intervals are special case of proper time intervals, and are connected to coordinate as I mentioned earlier.
  • #1
Moataz
8
0
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 separation 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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  • #2
Moataz said:
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 separation 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 spatial 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]
 
  • #3
So, I just want to 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.
 
  • #4
Moataz said:
So, I just want to 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.
 
  • #5
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?title=International_Atomic_Time&oldid=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.
 
  • #6
Thank you guys. That is very helpful. I still have some questions though. I guess I am going to 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
 
  • #7
Moataz said:
@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.

Moataz said:
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.
Moataz said:
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 spatial separation between the events is zero.
Moataz said:
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 spatial 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.


Moataz said:
@pervect: We have not actually studied the spacelike and timelike intervals yet.

A spacelike interval is one in which Δx>cΔt, i.e. the spatial 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.
Moataz said:
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.
 
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  • #8
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.
 
  • #9
elfmotat said:
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 ?
 
  • #10
morrobay said:
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]).
 
  • #11
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.
 
  • #12
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.
 
  • #13
Naty1 said:
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 no one else here sees it this way is what's caused the arguements. :P
 
  • #14
salvestrom said:
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 no one 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...

milneexplosion_equipartition_of_momentum.gif


...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.
 
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  • #15
JDoolin said:
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.
 
  • #16
salvestrom said:
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.
 
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  • #17
JDoolin said:
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.
 
  • #18
Matterwave said:
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 ?
 
  • #19
salvestrom said:
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.

Unfortunately, I don't feel free to discuss this issue on these forums, due to previous infractions. If you'd like to discuss it further, you can send me a private message, or comment on my blog.
 
  • #20
Apparently that Wikipedia piece IS rather obtuse ...


I'm not going to pursue interpretating those two paragraphs any further [not worth it] , but I wondered when I read them if their comments were trying to point out that clock times vary within a reference frame dependent on different gravitational potentials...

...a clock located at a 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...
 
  • #21
JDoolin said:
Unfortunately, I don't feel free to discuss this issue on these forums, due to previous infractions. If you'd like to discuss it further, you can send me a private message, or comment on my blog.

The main thing I don't want to do is "speculate" on this. However, now that the weekend has come, and I was able to put a good five hours of work into the problem, I can show you the idea without any hand-waving.

salvestrom said:
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.

Here, I have set up a distribution of 5000 "clocks" and given them each a rapidity in the x-direction between -3 and 3, and a rapidity in the y-direction between -3 and 3. I picked out several random clocks and gave them colors. Here is the explosion from the blue-clock's perspective:

ClockExplosionBlue.gif


And here is the same explosion from the Yellow clock's perspective:

ClockExplosionYellow.gif


But the point I wanted to make was, if I had, for instance, selected rapidities in the domain of (-100,100) instead of between (-3,3) then there would hardly be any noticeable difference in the distribution for the selected clocks.

So in fact, I guess I mis-spoke, because it's not the number of clocks that makes a difference, but the width of the rapidity space.

If you'd like to see all 11 perspectives, see here
 
  • #22
JDoolin said:
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.

It is my hunch that for a universe worth of clocks similar in size and density to ours, light emitted outwards from clocks near the edge would curve back inwards so that even an observer near the edge would still see an apparent expanding sphere of clocks and would not be aware that they were indeed near the edge of the sphere.
 
  • #23
yuiop said:
It is my hunch that for a universe worth of clocks similar in size and density to ours, light emitted outwards from clocks near the edge would curve back inwards so that even an observer near the edge would still see an apparent expanding sphere of clocks and would not be aware that they were indeed near the edge of the sphere.

That's why I keep saying the clocks are massless. I was hoping to avoid questions about light "curving," so we could discuss the OP's original question, regarding coordinate time, and proper time.

But, let's say, for instance these 5000 clocks have some negligible mass. There would be a problem at t=0 because they occupy a point, and the negligible masses at zero distance would have an infinite force between them.

But, can you somehow relate what you are saying to the animations; actually it doesn't make any sense to me in regards to the diagrams. How would you see an apparent expanding sphere of clocks if you were at the edge of a flat front? If you look in one direction and see a bunch of clocks moving away, and if you look the other way and see no clocks, how could you avoid being aware that you were near the "edge" of the clocks?
 
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  • #24
JDoolin said:
That's why I keep saying the clocks are massless.
There's no such thing as a massless clock. That's why time has no meaning for massless particles like photons. You can only build a clock with massive particles and they cannot travel at the speed of light.
 
  • #25
I have a simple question. From what I understand about relativity and proper time (I am beginner btw) is that when a clock moves at a huge speed, then it will measure different time than the coordinate time in some other reference frame. So, does that mean it really slows down? as in the difference between its ticks is observed to be slower compared to the other synchronized clocks? For example if I matched my wristwatch against a clock and then traveled at the speed of light then stopped and saw another clock that was already synchronized with the first clock, then I will see that my wristwatch does not match anymore. So it seems to me that what happened is just not a matter of mathematical manipulations but rather a physical reality (wristwatch moving slower) is that true?

Also on wikipedia it says second is 'the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom' but that seems absolute to me?

Also, last question when people say synchronizing two clocks. Does this mean just making sure they have the same reading at the time of checking? or does it mean they have the same pace of ticking? For example if two clocks read 12:00 but one ticks faster does this mean they are not synchronized?

Thank you.
 
  • #26
Also, my book says that this formula only works in inertial frames.

[tex]\Delta \tau = \int \sqrt[]{1-v^2}dt ; v=speed[/tex]

Bu that does not make sense because v is in terms of t and it is supposed to change over time which means there must be acceleration!?
 
  • #27
Moataz said:
Also, my book says that this formula only works in inertial frames.

[tex]\Delta \tau = \int \sqrt[]{1-v^2}dt ; v=speed[/tex]

Bu that does not make sense because v is in terms of t and it is supposed to change over time which means there must be acceleration!?
"Inertial frame" means the observer is not accelerating. It doesn't mean that the object being observed can't accelerate.
 

What is coordinate time?

Coordinate time is a measure of time used in the theory of relativity to describe the position of an event in space-time. It is based on the concept that time is relative and can be measured differently by observers in different reference frames.

What is proper time?

Proper time is the time experienced by an observer who is in the same reference frame as the event being measured. It is considered the most accurate measure of time because it takes into account the effects of relativity and is not affected by the observer's motion or position.

What is the difference between coordinate time and proper time?

The main difference between coordinate time and proper time is that coordinate time is measured by an observer in a different reference frame, while proper time is measured by an observer in the same frame as the event. Coordinate time is also affected by an observer's motion and position, while proper time is not.

Why is it important to distinguish between coordinate time and proper time?

It is important to distinguish between coordinate time and proper time because they provide different perspectives on the same event. Coordinate time is useful for describing the sequence of events from different reference frames, while proper time is useful for accurately measuring time intervals and understanding the effects of relativity.

How is coordinate time and proper time related to the theory of relativity?

The theory of relativity, specifically the theory of special relativity, explains the relationship between coordinate time and proper time. It states that time is relative and can be measured differently by observers in different reference frames, and that proper time is the most accurate measure of time as it is not affected by an observer's motion or position.

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