# Coordinate time and proper time.

1. Feb 19, 2012

### Moataz

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.

2. Feb 19, 2012

### elfmotat

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:

$$(c\Delta \tau)^2 =(c\Delta t)^2-(\Delta x)^2=(c\Delta t)^2-(v\Delta t)^2=(\Delta t)^2(c^2-v^2)$$

Simplifying this gives:

$$\Delta t=\gamma \Delta \tau$$

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

$$\Delta \tau =\int \sqrt{1- \frac{1}{c^2} \left ( \left (\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left (\frac{dz}{dt}\right)^2\right )}dt$$

3. Feb 20, 2012

### Moataz

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.

4. Feb 20, 2012

### elfmotat

Yes, $(\Delta s)^2=(c\Delta \tau )^2=(c\Delta t)^2-(\Delta x)^2$. This comes from the fact that, for the clock, Δx=0.

5. Feb 20, 2012

### pervect

Staff Emeritus
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. Feb 20, 2012

### Moataz

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 $$\Delta x=0$$ in some reference frame,

$$(\Delta s)^2=(c\Delta \tau )^2=(c\Delta t)^2$$

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 $$\Delta x=0$$

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 $$\Delta x\neq 0$$ 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 $$c\Delta t$$ is used, then a deep quantity such as spacetime will show up?

Thank you again

7. Feb 20, 2012

### elfmotat

Yes.

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.

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.

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.

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.

It comes right from the Lorentz transformation:

$\Delta x'=\gamma (\Delta x-v\Delta t)$

$\Delta t'=\gamma (\Delta t-v\Delta x/c^2)$

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.

Last edited: Feb 20, 2012
8. Feb 20, 2012

### Naty1

Also check at least the introductions to 'proper time' and 'coordinate time' in Wikipedia

9. Feb 20, 2012

### morrobay

So if c is taken to be 1 , then is the spacetime interval always numerically equal
to the proper time ?

10. Feb 20, 2012

### Matterwave

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. $ds=\sqrt{-d\tau^2}$).

11. Feb 21, 2012

### Naty1

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:

12. Feb 21, 2012

### pervect

Staff Emeritus
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. Feb 21, 2012

### salvestrom

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

14. Feb 21, 2012

### JDoolin

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.

Last edited by a moderator: Apr 15, 2017
15. Feb 21, 2012

### salvestrom

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. Feb 21, 2012

### JDoolin

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.

Last edited: Feb 21, 2012
17. Feb 21, 2012

### salvestrom

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. Feb 21, 2012

### morrobay

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. Feb 21, 2012

### JDoolin

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. Feb 22, 2012

### Naty1

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....

21. Feb 24, 2012

### JDoolin

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.

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:

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

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. Feb 24, 2012

### yuiop

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. Feb 25, 2012

### JDoolin

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?

Last edited: Feb 25, 2012
24. Feb 25, 2012

### ghwellsjr

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. Feb 26, 2012

### Moataz

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.