# Special Relativity, light down a rocket

• CAF123
In summary: S' are (x', t') as you have said. Would this mean that the length as measured in S of the rocket would be contracted? In summary, the light signal takes a time of ##L_o/c## to reach the end of the rocket in the rest frame of the rocket, and a time of ##L_o\sqrt{\frac{1-v/c}{1+v/c}}/c## in the rest frame of the observer. This is due to the effects of time dilation and length contraction.
CAF123
Gold Member

## Homework Statement

A rocket of length ##L_o## flies with constant velocity ##v## (in frame S' relative to a frame S in the x direction). At time t=t'=0, the capsule on the top of the rocket passes the point ##P_o## in S. At this moment, a light signal is sent from the top of the rocket to the bottom.
1)In the rest frame of the rocket, how long does it take the light signal to reach the end of the rocket?
2)In the rest frame of the observer, S, at which time does the signal reach the end of the rocket?

## Homework Equations

Lorentz transformations, proper length and proper time.

## The Attempt at a Solution

1)The wording of the question at the beginning is confusing, but I take it that S is the rest frame of the observer, S' is the rest frame of the rocket and S' moves wrt S at velocity v. I also took ##L_o## to be the proper length of the rocket (i.e the value measured in S', the rest frame of the rocket). Are these correct interpretations?

At ##t_o' = 0## the signal leaves the top of the rocket. At ##t_1'## it arrives at the bottom in frame S'. Speed of light same in all inertial frames (S' inertial since it moves with constant velocity v wrt another inertial frame) and ##L_o## the length of the rocket in S'. So ##t_1' = L_o/c## is the time taken in S'.

2)In S, as the rocket passes, the observer would see the rocket contract so expect time for signal to reach bottom in S to be smaller than that in S'. Why is ##t_1 = \gamma t_1'## not valid here?
Thanks.

yeah, for part 1) I agree. The wording is a bit unclear. But I think you have interpreted it the right way. and yes, that would mean ##L_0/c## is the time taken according to the rocket rest frame. For part 2), using the time dilation formula would not work (as you've guessed). To understand why, think about the spacetime coordinate difference in the two frames of reference. And in what (special) case does the time dilation formula apply? (it is not a general case). So, to answer the question, what is the equation you must use for the general case?

Hi BruceW,
BruceW said:
To understand why, think about the spacetime coordinate difference in the two frames of reference. And in what (special) case does the time dilation formula apply?
(it is not a general case)

Consider S' first. Then there are two events of interest - the light leaving the top of the rocket (denote by ##(x_o', t_o')## and the light reaching the bottom (denote by ##(x_1', t_1')##)
I think ##(x_o', t_o') = (P_{o,x}, 0)## and ##(x_1', t_1') = (P_{o,x} + L_o, L_o/c)##, where ##P_{o,x}## is the x component of ##P_o##. Is that right? I see your point and it is obvious that the rocket observer will notice that the light hitting the bottom and leaving the top are at two different locations within the rocket. I think it is correct to say that the time dilation formula is applicable when the two events of interest happen at the same spatial coordinate in the frame of reference.

In S, ##(x_o, t_o) = (P_{o,x}, 0)## as before and ##(x_1, t_1) = (P_{o,x} + l, t_1)## where ##l## is the length of the rocket in S. Can this be found? I think I can use the Lorentz transformation ##\Delta x \equiv l = \gamma(\Delta x' + v \Delta t') = \gamma(L_o + v L_o/c)##?

So, to answer the question, what is the equation you must use for the general case?
So $$t_1 - t_o = t_1 = \Delta t = \gamma ( \Delta t' + v\Delta x'/c^2) = \gamma (L_o/c + vL_o/c^2)$$

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The rear of the rocket in the S' frame is located at x' = -L0, and, as you said, the signal reaches this location at t' = L0/c. What does the inverse Lorentz Transform tell you about the coordinates of the event as reckoned from the S frame of reference?

Chet

Hi Chestermiller,
Chestermiller said:
The rear of the rocket in the S' frame is located at x' = -L0, and, as you said, the signal reaches this location at t' = L0/c.
Do you mean to say the two events of interest are located at (Po,x,0) and (Po,x-Lo, Lo/c)? I think I made a sign error in my previous post.
What does the inverse Lorentz Transform tell you about the coordinates of the event as reckoned from the S frame of reference?
See my previous post but instead, correcting the sign error, I have ##\Delta x = \gamma (\Delta x' + v \Delta t') = \gamma(-L_o + v L_o/c)##. So is this the length of the rocket in the frame S?

Similarly, the time interval in S for the two events is $$\Delta t = \gamma(\Delta t + v \Delta x'/c^2) = \gamma(L_o/c - v L_o/c^2).$$

CAF123 said:
Hi Chestermiller,

Do you mean to say the two events of interest are located at (Po,x,0) and (Po,x-Lo, Lo/c)? I think I made a sign error in my previous post.

See my previous post but instead, correcting the sign error, I have ##\Delta x = \gamma (\Delta x' + v \Delta t') = \gamma(-L_o + v L_o/c)##. So is this the length of the rocket in the frame S?

Similarly, the time interval in S for the two events is $$\Delta t = \gamma(\Delta t + v \Delta x'/c^2) = \gamma(L_o/c - v L_o/c^2).$$
Yes. But, please factor out the L0/c, and do some cancellation of terms with those γ's.

Chet

Yes, so $$\Delta t = \frac{L_o}{c} \sqrt{\frac{1-v/c}{1+v/c}}$$ which is always less than ##L_o/c## since v<c (I think this was to be expected, in my first post I noted that in S the rocket would be contracted and so in that frame the time taken for the light signal to reach the other end would be shorter, so I think the result makes sense.)

I also mentioned in my first post that we had a proper length in S'. But the space time coordinates in each of the two frames are:

S': light leaving top (Po,x, 0) ; light hitting bottom (Po,x-Lo, Lo/c)

S: light leaving top (Po,x, 0) ; light hitting bottom (Po,x-l, t1), where ##l = \gamma(-L_o + vL_o/c)## and ##t_1 = \Delta t## given above.

Neither events have a common spatial or temporal coordinate in common so all distances/ time measured in this process are said to not be proper. Is that correct? When it is said ##L_o## is the length of the rocket, is that meant to be it's rest length, assumed to be measured by an observer when the rocket is grounded?

CAF123 said:
Yes, so $$\Delta t = \frac{L_o}{c} \sqrt{\frac{1-v/c}{1+v/c}}$$ which is always less than ##L_o/c## since v<c (I think this was to be expected, in my first post I noted that in S the rocket would be contracted and so in that frame the time taken for the light signal to reach the other end would be shorter, so I think the result makes sense.)

I also mentioned in my first post that we had a proper length in S'. But the space time coordinates in each of the two frames are:

S': light leaving top (Po,x, 0) ; light hitting bottom (Po,x-Lo, Lo/c)

S: light leaving top (Po,x, 0) ; light hitting bottom (Po,x-l, t1), where ##l = \gamma(-L_o + vL_o/c)## and ##t_1 = \Delta t## given above.

Neither events have a common spatial or temporal coordinate in common so all distances/ time measured in this process are said to not be proper. Is that correct? When it is said ##L_o## is the length of the rocket, is that meant to be it's rest length, assumed to be measured by an observer when the rocket is grounded?
No. In an object's rest frame, the coordinates of its two ends don't change with time, so they don't have to be measured at the same time. L0 is the proper length of the rocket.

Chet

Chestermiller said:
No. In an object's rest frame, the coordinates of its two ends don't change with time, so they don't have to be measured at the same time. L0 is the proper length of the rocket.
Ok, so while the rocket is flying, in the rest frame put a light emitter in the middle of the rocket and send light pulses towards the top and bottom of the rocket. If there are clocks at the ends, they will both read a common time T, say and from that deduce the length.

Are the space time coordinates of the emission of light at the top and the receiving of light at the bottom in S and S' given in my previous post correct? Thanks.

CAF123 said:
Ok, so while the rocket is flying, in the rest frame put a light emitter in the middle of the rocket and send light pulses towards the top and bottom of the rocket. If there are clocks at the ends, they will both read a common time T, say and from that deduce the length.
I don't know why you would do it this way, instead of just using a tape measure. If you were measuring the width of your kitchen, would you use light signals and clocks? The rocket is at rest within its own rest frame.
Are the space time coordinates of the emission of light at the top and the receiving of light at the bottom in S and S' given in my previous post correct? Thanks.
Yes. They are perfect.

Chet

Chestermiller said:
I don't know why you would do it this way, instead of just using a tape measure. If you were measuring the width of your kitchen, would you use light signals and clocks? The rocket is at rest within its own rest frame.
That would make sense, but in what circumstance would you use light signals and clocks?

Is it also correct to say that ##\Delta t' = L_o/c## is a proper time since it was measured in the rest frame of the rocket? It seems intuitive to be the case, but ##\Delta x' \neq 0## when I look at the space time coordinates.

Thanks.

CAF123 said:
Is it also correct to say that ##\Delta t' = L_o/c## is a proper time since it was measured in the rest frame of the rocket? It seems intuitive to be the case, but ##\Delta x' \neq 0## when I look at the space time coordinates.

Thanks.
You've got to think about what spacetime events you are talking about. I'm guessing that here, you are talking about (light leaves front of spaceship, time t1) and (light reaches back of spaceship, time t2). Clearly, this is not a proper time or proper length, as you say, because the spacetime difference is non-zero for both the time and space coordinates. To get a proper length or time, it is not enough to say that something was measured in the rest frame. It must be measured in the rest frame, and have one of either space or time difference be equal to zero.

BruceW said:
You've got to think about what spacetime events you are talking about. I'm guessing that here, you are talking about (light leaves front of spaceship, time t1) and (light reaches back of spaceship, time t2). Clearly, this is not a proper time or proper length, as you say, because the spacetime difference is non-zero for both the time and space coordinates. To get a proper length or time, it is not enough to say that something was measured in the rest frame. It must be measured in the rest frame, and have one of either space or time difference be equal to zero.
I respectfully beg to differ. In the rest frame of an object, you don't need to measure the coordinates of both ends of the object at the same time. This is because, in the object's rest frame, the coordinates of its ends don't change with time. So the proper length of the rocket is L0, since it is always at rest within its own rest frame.

Chet

CAF123 said:
That would make sense, but in what circumstance would you use light signals and clocks?
If you were trying to calibrate the clocks
Is it also correct to say that ##\Delta t' = L_o/c## is a proper time since it was measured in the rest frame of the rocket? It seems intuitive to be the case, but ##\Delta x' \neq 0## when I look at the space time coordinates.

Thanks.
##\Delta t' = L_o/c## is the amount of proper time that elapses at each end of the rocket during the time interval that the signal travels from one end of the rocket to the other.

Chestermiller said:
I respectfully beg to differ. In the rest frame of an object, you don't need to measure the coordinates of both ends of the object at the same time. This is because, in the object's rest frame, the coordinates of its ends don't change with time. So the proper length of the rocket is L0, since it is always at rest within its own rest frame.

Chet
yeah, that's true. but if the spaceship is stretching or something, then we have the more general case where it's ends do change with time. Also, I feel it is more intuitive to think about proper length (between two spacetime events) as the spatial coordinate difference, in a frame where the time coordinate difference is zero.

Chestermiller said:
##\Delta t' = L_o/c## is the amount of proper time that elapses at each end of the rocket during the time interval that the signal travels from one end of the rocket to the other.

Intuitively, it seems correct to say that ##\Delta t' = L_0/c## is a proper time (and you have suggested above that this indeed is the case), but more generally, is the proper time not defined as the time in a frame in which two events happen at the same spatial location? In which case, neither in S or S' are the spatial components the same for the events of interest.
Thanks.

CAF123 said:
Intuitively, it seems correct to say that ##\Delta t' = L_0/c## is a proper time (and you have suggested above that this indeed is the case), but more generally, is the proper time not defined as the time in a frame in which two events happen at the same spatial location? In which case, neither in S or S' are the spatial components the same for the events of interest.
Thanks.
I'm going to ask one of the relativity mavens in Homework Helpers to put their two cents in on this.

Chet

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CAF123 said:
Intuitively, it seems correct to say that ##\Delta t' = L_0/c## is a proper time (and you have suggested above that this indeed is the case), but more generally, is the proper time not defined as the time in a frame in which two events happen at the same spatial location? In which case, neither in S or S' are the spatial components the same for the events of interest.
Thanks.
##\Delta t' = L_0/c## is the proper time that has elapsed for a clock, which is attached to the rocket (i.e. stationary in the rocket rest frame). for example, we could organise things so that the clock at the rear of the rocket starts ticking when the light is emitted at the front of the rocket (according to the rocket frame). And, we know that the light takes time ##\Delta t' = L_0/c## to reach the clock (according to the rocket frame). So for the clock, ##\Delta t' = L_0/c## proper time has passed, since the clock is also in the same reference frame as the rocket.

edit: I think the most important thing to keep in mind is which worldline are you talking about. If we are talking about the worldline of the light beam, then the spacetime interval along that worldline is zero. But, if we are talking about a clock attached to the back of the spaceship, then it's spacetime interval will be nonzero. So it depends on which worldline you are talking about.

edit again: I think I've used slightly incorrect terminology. I think worldline is only for paths through spacetime that are timelike. So the path of the beam of light is not a worldline, so what I really meant is just the more general 'path through spacetime'.

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Chestermiller said:
I'm going to ask one of the relativity mavens in Homework Helpers to put their two cents in on this.
Thanks.
BruceW said:
##\Delta t' = L_0/c## is the proper time that has elapsed for a clock, which is attached to the rocket (i.e. stationary in the rocket rest frame). for example, we could organise things so that the clock at the rear of the rocket starts ticking when the light is emitted at the front of the rocket (according to the rocket frame). And, we know that the light takes time ##\Delta t' = L_0/c## to reach the clock (according to the rocket frame). So for the clock, ##\Delta t' = L_0/c## proper time has passed, since the clock is also in the same reference frame as the rocket.

This helps me see it intuitively but, in the reference frame of the rocket S', the light leaves the rocket at the top when x1' = Po,x and hits the bottom when x2' = Po,x-Lo with a corresponding time interval Lo/c. The proper time is the time in some frame between two events which occur at the same spatial coordinate. Since x1' is not equal to x2' , according to the definition, the time between the two events in this frame is not proper. So I get intuitively why it is a proper time, but I don't see how it agrees formally with the definition. Thanks.

Edit: I now see your edit, but this would suggest to me that now the proper time is in the frame of the light beam, not the rocket.

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between any two spacetime events you can calculate a spacetime interval. And if ##| \Delta t | < | \Delta x |## then that spacetime interval is a proper length. So clearly, you can't just choose any two spacetime events and say the length of the spaceship in its rest frame is equal to any proper length. A similar thing goes for proper time. The time which passes according to a clock is equal to the proper time along the worldline of the object. You can't just choose any worldline and associate it to the proper time of the clock.

The reason I thought the proper time was in the frame of the light beam was because in that frame the difference in spatial coordinates is zero. I understand that the proper time is the time measured by a clock moving along the worldline of the object.

Why can't we say, in our example, that the light beam is the object and attach a clock to it's frame of reference. Then in that frame the difference in spatial coordinates between the two events is zero, and from this infer that the proper time is in the frame of the light?

Sorry for labouring the point, I just wish to make it clear.

I think the problem here is that you think there is one proper time (distance), but proper time (distance) generally depends on the space-time path connecting two events. So asking if ##L_0/c## is the proper time doesn't really make sense because you haven't specified what path you're considering. If you're talking about the path of the photon, the proper time would be 0 because the photon follows a light-like path through space-time.

There is, however, the notion of proper length, which is different than the notion of proper distance. The proper length of the rocker is ##L_0##, the length of the rocket measured in its rest frame.

1 person
Hi vela, do you mean to say there is a notion of the proper length but not the proper time and that there exists a proper time in every frame?

Please also criticize/comment on the below:

The worldline of the rocket passes through a series of spacetime events, two of which are ##(P_{o,x}, 0)## and ##(P_{o,x}-L_o, L_o/c)## which represent the emission of light from the top and the receiving of light at the bottom, respectively, in the frame of the rocket. So the clock co-moving with this frame measures that the time for the signal to travel the length of the rocket (##L_0## in the rocket frame) is ##L_o/c##.

The proper time between the two events above is a time in a frame where the spatial coordinates of the two events are common. In the rocket frame, this is not the case so this is not a proper time in the rocket frame.

Thanks.

BruceW said:
I think the most important thing to keep in mind is which worldline are you talking about.
I would have said to be careful about which observer you are talking about.
The original problem statement does not appear to be all that carefully worded so it needs to be interpreted in the context of the other lessons OP has received.

We have a transmitter T and a detector D in a lab on a rocket.

The fact that the rocket is relativistic wrt some other observer is irrelevant to the time between transmission and detection in the frame of the rocket. This is what chestermiller is saying in:
I respectfully beg to differ. In the rest frame of an object, you don't need to measure the coordinates of both ends of the object at the same time. This is because, in the object's rest frame, the coordinates of its ends don't change with time. So the proper length of the rocket is L0, since it is always at rest within its own rest frame.
... for this particular experiment, ##L_0## is the separation of D and T in the reference frame of the lab.

But BruceW has a "but":
...but if the spaceship is stretching or something, then we have the more general case where it's ends do change with time.
... so, according to someone stationary in the lab, ##L_0## changes with time (as measured on his watch - with the usual allowances for simultaniety: which seems to be the heart of Bruce's objection).

Notice though, in the actual problem statement (see post #1) - whichever frame the length of the rocket is measured in, the length of the rocket does not change. Thus, whatever it's merits, BruceW's "but" is moot.

Shouldn't we be asking ourselves: "How does the author of the question expect the student to handle it?"

CAF123 said:
1)The wording of the question at the beginning is confusing,
It's a bit lib - yeah.

... but I take it that S is the rest frame of the observer, S' is the rest frame of the rocket and S' moves wrt S at velocity v. I also took ##L_o## to be the proper length of the rocket (i.e the value measured in S', the rest frame of the rocket). Are these correct interpretations?

At ##t_o' = 0## the signal leaves the top of the rocket. At ##t_1'## it arrives at the bottom in frame S'. Speed of light same in all inertial frames (S' inertial since it moves with constant velocity v wrt another inertial frame) and ##L_o## the length of the rocket in S'. So ##t_1' = L_o/c## is the time taken in S'.
Which is fine.

2)In S, as the rocket passes, the observer would see the rocket contract so expect time for signal to reach bottom in S to be smaller than that in S'. Why is ##t_1 = \gamma t_1'## not valid here?
Not always ... in the time it takes the light pulse to go from the transmitter (tail) to the detector (nose) in the S frame, the rocket has moved ... so the detector at the nose is farther along on the journey. The time between events would be (b-a)/c where a is the position (in S) of the tail when the pulse was sent, and b is the position of the nose (also in S) when the pulse was received.

That help?
... I think this is the core of the problem.

CAF123 said:
The reason I thought the proper time was in the frame of the light beam was because in that frame the difference in spatial coordinates is zero. I understand that the proper time is the time measured by a clock moving along the worldline of the object.
proper time does not depend on the frame of reference. What you mean is "I thought the proper time is the proper time along the path of the light beam". Well, the spacetime interval along the path of the light beam is zero. So we should call it a spacetime interval, not proper time. And yes, a clock measures the proper time along the worldline of the clock.

CAF123 said:
Why can't we say, in our example, that the light beam is the object and attach a clock to it's frame of reference. Then in that frame the difference in spatial coordinates between the two events is zero, and from this infer that the proper time is in the frame of the light?

Sorry for labouring the point, I just wish to make it clear.
we can't attach a clock to the light beam, because a clock is an object, it moves slower than the speed of light. Also, we can't talk about the rest frame of the beam of light, because there is no frame of reference in which the light beam is not moving. But, we can say that the spacetime interval along the path of the beam of light is zero.

CAF123 said:
Hi vela, do you mean to say there is a notion of the proper length but not the proper time and that there exists a proper time in every frame?
there is a proper time or proper length (or spacetime interval) between any two spacetime events. It is very useful to learn about this kind of stuff, because the invariance of the spacetime interval is pretty much all you need to know, to understand special relativity.

CAF123 said:
Please also criticize/comment on the below:

The worldline of the rocket passes through a series of spacetime events, two of which are ##(P_{o,x}, 0)## and ##(P_{o,x}-L_o, L_o/c)## which represent the emission of light from the top and the receiving of light at the bottom, respectively, in the frame of the rocket. So the clock co-moving with this frame measures that the time for the signal to travel the length of the rocket (##L_0## in the rocket frame) is ##L_o/c##.
This is not quite right... a worldline represents the path of a point-like object through spacetime. So you can talk about the worldline of the front of the rocket, or the worldline of the back of the rocket, but you can't talk about the worldline of the front and back of the rocket together. That would be something like a worldsheet, but we don't need to talk about it for this problem, because I don't think it makes anything simpler.

Or perhaps, you are talking about the path of the beam of light through spacetime. Again, this is technically not a worldline, because it is not a time-like path. But it is a valid path through spacetime. So it does have a spacetime interval. And this spacetime interval is clearly zero (because a beam of light will always travel along a path of zero spacetime interval). Also, we can't get a clock to move with the beam of light.

CAF123 said:
The proper time between the two events above is a time in a frame where the spatial coordinates of the two events are common. In the rocket frame, this is not the case so this is not a proper time in the rocket frame.

Thanks.
yeah, the proper time along a worldline is equal to the time difference in a frame of reference where the spatial coordinates of the two events are the same. (this is for non-accelerating objects). The two events you mentioned have zero spacetime interval between them, so in fact, they can never be associated with a coordinate time in any frame of reference. (because there is no frame of reference in which the beam of light is not moving).

BruceW said:
So for the clock, ##\Delta t' = L_0/c## proper time has passed, since the clock is also in the same reference frame as the rocket.
So, for a clock attached to the back of the rocket, it is stationary in the rocket frame. But an observer next to the clock would not say that he saw the light be emitted from the top and received at the bottom at the same coordinate, would he?

Simon Bridge said:
Notice though, in the actual problem statement (see post #1) - whichever frame the length of the rocket is measured in, the length of the rocket does not change. Thus, whatever it's merits, BruceW's "but" is moot.

I thought that in S, the length of the rocket is contracted along it's length which means in S the time taken for the light to travel down the rocket should be less. At least, that is what I thought initially and the result I got for Δt in S (see #7) seems to agree.

Simon Bridge said:
I think this is the core of the problem.

The core of my problem is the following: in the rocket frame, the difference in spatial coordinates of the two events is not zero so the time between these two events in the rocket frame is not proper. (By, I believe, the definition of proper time). See also top quote above for why I am still thinking there is a contradiction.

yeah, the proper time along a worldline is equal to the time difference in a frame of reference where the spatial coordinates of the two events are the same. (this is for non-accelerating objects). The two events you mentioned have zero spacetime interval between them, so in fact, they can never be associated with a coordinate time in any frame of reference. (because there is no frame of reference in which the beam of light is not moving).
Could you explain this part a bit more? Do you mean to say that given that the spacetime interval vanishes always for any two events along the path of a light beam then there cannot be any frame of reference where the events are simultaneous (in time) or simulataneous in position. So there is no proper time for the path of the light beam between the top and bottom.

CAF123 said:
So, for a clock attached to the back of the rocket, it is stationary in the rocket frame. But an observer next to the clock would not say that he saw the light be emitted from the top and received at the bottom at the same coordinate, would he?
true. the light is not emitted and received at the same coordinate.

CAF123 said:
I thought that in S, the length of the rocket is contracted along it's length which means in S the time taken for the light to travel down the rocket should be less. At least, that is what I thought initially and the result I got for Δt in S (see #7) seems to agree.
I think SimonBridge just means that the rocket length is constant with time according to any frame of reference. Not that the rocket length is the same in every frame of reference.

CAF123 said:
Could you explain this part a bit more? Do you mean to say that given that the spacetime interval vanishes always for any two events along the path of a light beam then there cannot be any frame of reference where the events are simultaneous (in time) or simulataneous in position. So there is no proper time for the path of the light beam between the top and bottom.
yeah, exactly. you seem to understand it pretty well. We can say the square of the spacetime interval is ##c^2 ( \Delta t)^2 - ( \Delta x)^2## And if this is positive, we take square root and call it a 'proper time'. If it is negative, we take the absolute value, then square root and call it a 'proper length'. And if it is zero, we know it describes the path of a beam of light. Well, there is also the special case when ##\Delta t## and ##\Delta x## are both zero, so the two events simply happen in the same place and time.

BruceW said:
true. the light is not emitted and received at the same coordinate.
But the time is still warranted 'proper' because ##c^2 (\Delta t)^2 - (\Delta x)^2## is positive?

yeah, exactly. you seem to understand it pretty well. We can say the square of the spacetime interval is ##c^2 ( \Delta t)^2 - ( \Delta x)^2## And if this is positive, we take square root and call it a 'proper time'.
If this is positive, then the two events are time-like separated. So, if I understand correctly, whenever this quantity is positive ##\Delta t## represents a proper time. When I said before that a proper time can occur in a frame where ##\Delta x = 0##, this is true and is the case where the quantity ##c^2 (\Delta t)^2 - (\Delta x)^2## is trivially greater than zero. But this is restrictive and the more general case is that a proper time exists in a frame provided the space time interval squared is positive?

If it is negative, we take the absolute value, then square root and call it a 'proper length'.
I did not really get this part - if the quantity is negative, then the events are space-like separated. The quantity can be made negative in a frame where ##\Delta t=0##. How does this relate to proper length?

Thanks!

CAF123 said:
If this is positive, then the two events are time-like separated. So, if I understand correctly, whenever this quantity is positive ##\Delta t## represents a proper time. When I said before that a proper time can occur in a frame where ##\Delta x = 0##, this is true and is the case where the quantity ##c^2 (\Delta t)^2 - (\Delta x)^2## is trivially greater than zero. But this is restrictive and the more general case is that a proper time exists in a frame provided the space time interval squared is positive?
This paragraph is not quite right. The square root of ##c^2 (\Delta t)^2 - (\Delta x)^2## Is the proper time. For a straight, timelike path between two spacetime events in special relativity, the proper time is defined to be
$$\sqrt{c^2 (\Delta t)^2 - (\Delta x)^2}$$
This is the nice mathematical definition which is good to learn. I guess you have not fully learned about this yet? I think they should teach this bit sooner rather than later, since it is so crucial to special relativity.

edit: whoops, I messed up, the definition of the proper time is:
$$(1/c) \sqrt{c^2 (\Delta t)^2 - (\Delta x)^2}$$
So that we have the correct units.

Yes, sorry, so ##ds^2 = c^2 dt^2 - dr^2 = c^2 dt^2 - v^2 dt^2##. In frame of particle, we have a proper time, since there ##v=0## so ##ds^2 = c^2 d\tau^2 ##(change symbol to note that it is a proper time). Then ##d\tau = ds/c##. Quick question though: This formula for the proper time assumes we are in a frame where the particle is at rest (so v=0). But we can have a proper time in a frame where this is not the case. So why then is the formula valid? Is it because between any two spacetime events, there is only one proper time, but there may exist many frames where this time is observed. Since there is only one proper time, which is therefore invariant between said frames, the formula can be derived by considering the case v=0 and then it automatically holds in the other subset of frames where this time is observed?

Did you have any comments on this:
CAF123 said:
I did not really get this part - if the quantity is negative, then the events are space-like separated. The quantity can be made negative in a frame where ##\Delta t=0##. How does this relate to proper length?

The proper time between two events is the elapsed time that would be measured on the clock of an observer at rest in a unique inertial frame of reference, such that he is physically present at both events. The beauty of the equation ds2=c2(dt)2-(dx)2=c2(dτ)2 is that can be used by observers in any other frame of reference to reckon the proper time between the two events (and obtain the same value as the guy in the frame of reference that is moving at such a velocity that he can be physically present at both events).

Chet

CAF123 said:
Yes, sorry, so ##ds^2 = c^2 dt^2 - dr^2 = c^2 dt^2 - v^2 dt^2##. In frame of particle, we have a proper time, since there ##v=0## so ##ds^2 = c^2 d\tau^2 ##(change symbol to note that it is a proper time). Then ##d\tau = ds/c##. Quick question though: This formula for the proper time assumes we are in a frame where the particle is at rest (so v=0). But we can have a proper time in a frame where this is not the case. So why then is the formula valid? Is it because between any two spacetime events, there is only one proper time, but there may exist many frames where this time is observed. Since there is only one proper time, which is therefore invariant between said frames, the formula can be derived by considering the case v=0 and then it automatically holds in the other subset of frames where this time is observed?
No, there is not only one proper time between two spacetime events. The proper time depends on the world line. Look at the twin paradox for example. The proper time for the twin who stays on the planet is different than the proper time for the twin in the rocket, and that's why when they meet again years later, one is older than the other.

Given a world line connecting two spacetime events, every observer will agree on the proper time regardless of their state of motion. That's because of the invariance of the interval.

Hi Chestermiller,
Chestermiller said:
The proper time between two events is the elapsed time that would be measured on the clock of an observer at rest in a unique inertial frame of reference, such that he is physically present at both events. The beauty of the equation ds2=c2(dt)2-(dx)2=c2(dτ)2 is that can be used by observers in any other frame of reference to reckon the proper time between the two events (and obtain the same value as the guy in the frame of reference that is moving at such a velocity that he can be physically present at both events).
Suppose there exists two events in spacetime connected by a worldline such that in one frame, the events happen simultaneous in time. Then since ds2>0, this time is a proper time. Do you mean to say that all frames will agree on the proper time between these events τ = (√c2dt2 - (dx)2 )/c? If so, then in any frame, the coordinate events are such that the above combination of dt and dx is always the same, yes?

Hi vela,
vela said:
No, there is not only one proper time between two spacetime events. The proper time depends on the world line. Look at the twin paradox for example. The proper time for the twin who stays on the planet is different than the proper time for the twin in the rocket, and that's why when they meet again years later, one is older than the other.

Given a world line connecting two spacetime events, every observer will agree on the proper time regardless of their state of motion. That's because of the invariance of the interval.
I think there is something still bothering me: If dx and dt of two spacetime events satisfy the condition that ds2>0 in some frame, then this is a frame in which the time between the events is said to be proper, yes? If that is the case, then given that ds2 is invariant, any Lorentz transformation to a frame also has a dx and dt satisfying the sign of the invariant interval. So does this not imply all times are proper in every frame? Obviously I made a mistake here.

Thanks!

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