
#1
Feb1912, 07:24 PM

P: 8

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
Feb1912, 07:52 PM

P: 260

[tex](c\Delta \tau)^2 =(c\Delta t)^2(\Delta x)^2=(c\Delta t)^2(v\Delta t)^2=(\Delta t)^2(c^2v^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
Feb2012, 12:31 AM

P: 8

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=spacetime interval? Because what I know is that instead of putting (C)(Delta proper time), we put delta S, or the spacetime interval. 



#4
Feb2012, 01:05 AM

P: 260

coordinate time and proper time. 



#5
Feb2012, 03:43 PM

Emeritus
Sci Advisor
P: 7,445

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 spacetimeinterval. Spacetimeintervals 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 spacetime interval in terms of coordinate changes is flat spacetime. Spacetime itnervals can be timelike, or spacelike. If a spacetime interval is timelike, it represents some wristwatch time. A spacetime interval is computed between endpoints along some particular curve joining t hem. The spacetime interval computed along the curve is equal to the wristwatch time, also known as proper time, elapsed from some observer following the spacetime 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 spacetime interval between them. In this case, it's easy  we know that the spacetime 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 spacetime interval, and the proper time, was NOT 1 second, but something larger, due to corrections from the metric. 



#6
Feb2012, 05:11 PM

P: 8

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 



#7
Feb2012, 07:03 PM

P: 260

The proper time separating the events would be (cΔτ)^{2}=(cΔt)^{2}(Δx)^{2}. 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. [itex]\Delta x'=\gamma (\Delta xv\Delta t)[/itex] [itex]\Delta t'=\gamma (\Delta tv\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. 



#8
Feb2012, 08:04 PM

P: 5,634

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
Feb2012, 09:06 PM

PF Gold
P: 363

to the proper time ? 



#10
Feb2012, 09:11 PM

P: 2,078





#11
Feb2112, 10:40 AM

P: 5,634

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: 



#12
Feb2112, 03:32 PM

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Sci Advisor
P: 7,445

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 spacetime, 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
Feb2112, 05:02 PM

P: 226

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 newtonianesque 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
Feb2112, 06:54 PM

PF Gold
P: 706

Imagine that you lived in a massfree universe, full of massless "clocks", and they were all moving apart moreorless 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 propertime. 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 moreorless arbitrary perspective of the clock in the center. 



#15
Feb2112, 07:40 PM

P: 226

The closest thing to this would be the proper time in the middle of a supervoid. 



#16
Feb2112, 09:24 PM

PF Gold
P: 706

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 speedoflight expanding sphere of clocks no matter where it was, unless it was at one edge of the explosion. 



#17
Feb2112, 09:38 PM

P: 226

In the first case I simply put forward that the central clock is actually central, nonmoving 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
Feb2112, 09:46 PM

PF Gold
P: 363

If the spacetime interval is spacelike then it is numerically equal to the proper distance ? 


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