Special Relativity, light down a rocket

[BruceW]

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.

 

[CAF123]

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:

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 for your help.

 

[Chestermiller]

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

 

[vela]

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.

 

[CAF123]

Hi Chestermiller,

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,

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!

 

[Chestermiller]

Hi Chestermiller,

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?

Yes. The reason that this is so important is that, in relativistic mechanics, the 4 acceleration is obtained by taking the derivative of the 4 velocity with respect to proper time. So observers in all inertial frames obtain the same acceleration and the same force.

 

[BruceW]

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 $d\tau = ds/c$.

yeah, exactly. The $d\tau$ is the proper time (along the path) according to any frame. And in the special frame where the particle is not moving, the coordinate time is simply equal to the proper time along the path $dt=d\tau$, since $dx$ is zero in that frame.

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?

yes, very true. Or you can think of it the other way around. The formula is valid in any frame, and in the special frame where the particle is not moving through space, the formula takes a nice, simple form. The formula is what I like to think is 'fundamental' to special relativity. You can derive the Lorentz transform, and pretty much any other formula in special relativity from the amazing formula $ds^2 = c^2 dt^2 - dr^2$.

 

[BruceW]

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?

Think about it for a little while. If we choose a frame where $\Delta t=0$ then the two spacetime events are separated by space only, and not time. Doesn't this sound like a pretty good description of a 'proper length' ? It is fairly analogous to 'proper time', except of course it is for events that are space-like separated, as you said.

 

[vela]

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?

No. As you noted, ds² is invariant, so it has the same value in all frames. ds²>0 doesn't imply dt is the proper time. (I assume when you refer to time, you're talking about dt.)

ds²>0 means the interval is time-like, as opposed to light-like (ds²=0) and space-like (ds²<0).

 

[CAF123]

No. As you noted, ds² is invariant, so it has the same value in all frames. ds²>0 doesn't imply dt is the proper time. (I assume when you refer to time, you're talking about dt.)

Ok, rather the proper time along a worldline between two space time events is the frame where the events happen simultaneous in space, so dx=0. And in any frame related to this frame via a Lorentz transformation, the sign of the invariant interval remains positive.

 

[CAF123]

Think about it for a little while. If we choose a frame where $\Delta t=0$ then the two spacetime events are separated by space only, and not time. Doesn't this sound like a pretty good description of a 'proper length' ? It is fairly analogous to 'proper time', except of course it is for events that are space-like separated, as you said.

If dt=0, then ds²=-dx². Is it analogous to 'proper time' in the sense that for two events in spacetime space like separated, all frames related to the frame that measured dt=0 will also agree on the proper length along the path? So, what I mean is all frames agree on ΔL_o = √|(c²dt² - (dx)²)|, where ΔL_o is the proper length along the path.

 

[BruceW]

you got it. The proper length along the path is the same in any reference frame. And in the special frame where both spacetime events happen at the same coordinate time, the spatial separation is equal to the proper length.

edit: well, the absolute value of the spatial coordinate separation is equal to the proper length. (according to that special frame of reference).

 

[Simon Bridge]

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.

Sounds like you worked out the time for the light to travel the contracted length of the spacecraft .

Lets say the rocket nose is a distance L, in frame S, from the tail.
If the tail is at x=0 at t=0 and the pulse starts traveling at t=0 too.

Working in the S frame:
In that frame, every part of the rocket is moving in the +x direction at speed v.

At t=0, the light has position $x_{light}=x_{tail}=0$ and the nose has position $x_{nose}=L$.

At time t>0, working in the S frame,
Write down the equation that tells you the new position of the light pulse.
Write down the equation that tells you the new position of the nose of the rocket.

Using those equations:
At what time (still in S) is the pulse position the same as the nose position?

 

[CAF123]

Hi Simon Bridge,

Sounds like you worked out the time for the light to travel the contracted length of the spacecraft .

Do you mean to say my answer before was incorrect?

Write down the equation that tells you the new position of the light pulse.
Write down the equation that tells you the new position of the nose of the rocket.

$x_{\text{light}} = ct$ and $x_{\text{nose}} = L + vt$.

Using those equations:
At what time (still in S) is the pulse position the same as the nose position?

Equating the above gives $t = L/(c-v)$. Since the rocket nose moves 'away' from the light pulse, the c-v factor makes sense. But I am not really sure what I calculated here - surely in S, the rocket is contracted along its length? Thanks.

 

[CAF123]

Hi BruceW,

you got it. The proper length along the path is the same in any reference frame. And in the special frame where both spacetime events happen at the same coordinate time, the spatial separation is equal to the proper length.

edit: well, the absolute value of the spatial coordinate separation is equal to the proper length. (according to that special frame of reference).

Thanks. It is my understanding though that for space-like separated events, no signal can possibly casually connect them since this would require a signal traveling greater than the speed of light. I can see two ways of getting this: using the invariant interval equation ds²>0 and noting that it implies such a Lorentz boost to another frame would require v>c. Or geometrically, space-like separated events are illustrated geometrically under the line ct=x on a Minkowski diagram, which means x > ct, again impossible.

So is it correct to say that if two events are spacelike separated as viewed by two frames, any Lorentz transformation to another frame from either of these two frames can have the events such that they are non-causually connected?

Similarly, if a frame views two events as being time-like separated, then in any Lorentz boost to another frame from this frame, the events too are time-like separated, which means they can be connected via a signal and thus are causally connected?

 

[BruceW]

Hi BruceW,

Thanks. It is my understanding though that for space-like separated events, no signal can possibly casually connect them since this would require a signal traveling greater than the speed of light. I can see two ways of getting this: using the invariant interval equation ds²>0 and noting that it implies such a Lorentz boost to another frame would require v>c. Or geometrically, space-like separated events are illustrated geometrically under the line ct=x on a Minkowski diagram, which means x > ct, again impossible.

yep. sounds good. And so because of this, we need to use a slightly indirect method to measure the proper length of paths through spacetime. i.e. for timelike paths, we can just send a clock along some path. But we cannot do this for spacelike paths. So for example, to measure the proper length of a (straight) spacelike path, we can hold start and finish flags at the spatial location of the start position and end position, and hold them fixed with time, according to some reference frame, and then travel between these two points, or send a light signal off a mirror, e.t.c. And from this, we can calculate the proper length. So it is a slightly indirect method. But it is still valid, as long as we have set everything up correctly.

So is it correct to say that if two events are spacelike separated as viewed by two frames, any Lorentz transformation to another frame from either of these two frames can have the events such that they are non-causually connected?

as viewed by two frames? It just needs to be spacelike separated according to one frame, and then it must be spacelike separated by all frames. And yes, those events will be non-causally connected, according to every frame, because they are spacelike separated according to every frame. (which is nice).

Similarly, if a frame views two events as being time-like separated, then in any Lorentz boost to another frame from this frame, the events too are time-like separated, which means they can be connected via a signal and thus are causally connected?

yep.

 

[Simon Bridge]

$x_{\text{light}} = ct$ and $x_{\text{nose}} = L + vt$.

Equating the above gives $t = L/(c-v)$. Since the rocket nose moves 'away' from the light pulse, the c-v factor makes sense.

That was easy wasn't it?
So long as you work in the same inertial frame, the regular laws of physics must apply. All the values in the equation you just worked out are from the S frame so you didn't need any special relativity stuff to work the maths.

But I am not really sure what I calculated here -

You have the equation for the time in the S frame for the light signal to travel from it's emitter in the tail to the receiver in the nose. i.e. t=Δt

But the light pulse, in your problem statement, goes the other way: from nose to tail. This calculation was for tail to nose. So you will need to rework the calculation by the same steps. Enjoy :)

surely in S, the rocket is contracted along its length?

Length contraction only happens to the other guy: the observer in S does not see any length contraction. But you are right in a way: you are not given L, you are given L₀. You should express the equation you got in terms of that.
What is the relationship between L and L₀?


You've seen the time-dilation derivation where the light pulse goes perpendicular to the direction of motion? Remember that the pulse had to travel further in one frame than in the other? This is the same thing with the pulse going along the direction of motion.

 

[CAF123]

Length contraction only happens to the other guy: the observer in S does not see any length contraction. But you are right in a way: you are not given L, you are given L₀. You should express the equation you got in terms of that.
What is the relationship between L and L₀?

Why is there no length contraction of the rocket in frame S? I believe the situation is the same as a stationary observer on a platform, observing a train flying past him. If the speed of the train is relativistic, then from the platform perspective, the train is length contracted. But, in the train frame, isn't also the platform contracted?

Using purely Lorentz transformations, I derived the time for the signal to traverse the rocket in S as $t = \frac{L_o}{c} \sqrt{\frac{1-v/c}{1+v/c}}$. I noticed that this was less than $L_o/c$ for all v<c, and my interpretation was that in S, the rocket length is contracted so as viewed from S, it takes less time for the signal to reach the other end.

I get a similar answer by the method you outlined, but not quite the same. Here it is:
In frame S, $x_{\text{tail}} = -L + vt$ and $x_{\text{light}} = -ct$ so again equating gives $t=L/(c+v)\,\,\,\, (1)$. L is the length in S. To relate to S', use a Lorentz boost to that frame which in general is $\Delta x = \gamma(\Delta x' + v\Delta t')$ and in this case, we have $L = \gamma(-L_o + vL_o/c)$. Subbing this in for L into (1) gives $$t = \frac{L_o}{c+v} \sqrt{\frac{1-v/c}{1+v/c}},$$ a slight difference in the denominator there.

@BruceW: Let me try to apply what I learned here to the problem at hand. Consider the path through spacetime of the light signal. Between such a path we have two spacetime events and in frame S' (the rocket) I calculated that those events happen at (P_o,x, 0) (the emission as viewed in S') and (P_0,x-l, t₁), where l = γ(-L_o + vL_o/c) and t₁= formula in #7. So, the path of the light signal from top to bottom in S' is not a proper time since Δx is not zero.

 

[Simon Bridge]

Why is there no length contraction of the rocket in frame S?

Because frame S is moving at the same speed as frame S.

If you go faster and faster, you do not notice yourself getting thinner or your watch getting slower.

If the speed of the train is relativistic, then from the platform perspective, the train is length contracted. But, in the train frame, isn't also the platform contracted?

Loosely put - yes.
Technically, the observer on the platform and the observer on the train disagree about the lengths of the train and of the platform.

If the train observer says the train is length $L_T$ and the platform is length L_P, and the platform observer says they are $L_T^\prime$ and $L_P^\prime$ respectively - what is the relationship between the two sets of measurements?

L would be the length-contracted value for L_0 for the observer in S.

As for the difference you found betwen the two methods - did you properly take account of the motion of the rocket in S.
If you are happy with that, then by all means go for it and see.
It may be that I missed something of course.
Let me know how you got on.

 

[CAF123]

Because frame S is moving at the same speed as frame S.

If you go faster and faster, you do not notice yourself getting thinner or your watch getting slower.

Ah, sorry I was not careful enough with my words. I thought you meant as viewed from S, the length of the rocket is not contracted.

Loosely put - yes.
Technically, the observer on the platform and the observer on the train disagree about the lengths of the train and of the platform.

If the train observer says the train is length $L_T$ and the platform is length L_P, and the platform observer says they are $L_T^\prime$ and $L_P^\prime$ respectively - what is the relationship between the two sets of measurements?

In general, a spacetime coordinate in S, say (x,t) is related to a space time coordinate (x',t') by the transformation x' = γ(x - vt), where v is the speed of S' relative to S. By inverting, we get x = γ(x'+vt') since -v is the speed of S relative to S'.
In this case, L_T' = γ(L_T-vt) and similarly for L_P'.

As for the difference you found betwen the two methods - did you properly take account of the motion of the rocket in S.
If you are happy with that, then by all means go for it and see.
It may be that I missed something of course.
Let me know how you got on.

I can't really see the error. I thought it might have had to do with t and using the corresponding Lorentz transformation to get a value involving the t. But this would not make sense since t in those equations was derived by already working in S to begin with. L is the length of the rocket as viewed from S but, to relate to the parameters in the question, I should reexpress L in terms of L_o, which can be found by the Lorentz transformation (which I already did). Thanks.

 

[CAF123]

Even though I already have the correct answer using the Lorentz transformations from the start, can you see where I made the error in the other method outlined above?

@BruceW: Did you catch my last message? (#48 bottom)

Thanks.

 

[BruceW]

I haven't looked back at the original problem for a while now. But now I do, uh... yeah, it looks like maybe you have gotten the velocity the wrong way around or something like that. To clarify, the rocket is moving at velocity v to the right, according to the S frame, yes? and the light beam is sent in a leftward direction from the nose of the rocket to it's end?

I think that maybe you made a slight mistake in the Lorentz transform which you used to derive your formula in post 7. Remember that the standard Lorentz formula
\Delta t&#039; = \gamma (\Delta t - v\Delta x/c^2)
works on the assumption that $v$ is the velocity of frame S' with respect to frame S.

 

[CAF123]

Hi BruceW,
Just to clarify the formula in #7 is correct, (according to the answer key) so there is an error with what I did following Simon's method.

Yes, the inverse transform is $\Delta t = \gamma(\Delta t'+ v \Delta x '/c^2)$ where, as viewed in S', S moves with velocity -v. What do you mean for me to do?

Thanks.

 

[BruceW]

hmm. strange. your formula in post #7 agrees with the answer key? I got a different answer. hmm, I agree with
\Delta t = \gamma(\Delta t&#039;+ v \Delta x &#039;/c^2)
where $v$ is positive, since S' is moving right (which we're calling positive) with respect to S. And $\Delta x' = L_0$ and $\Delta t'=L_0/c$ Is that the same as you used? From here, I get a different answer than your formula in post #7. I think that maybe the problem is meant to be interpreted differently, which is why the answer key does not agree with this method.

edit: I am an idiot. $\Delta x' = -L_0$

 

[BruceW]

Wait, wait wait. I'm so sorry. Of course, $\Delta x' = -L_0$. OK, I'm with you now. Yeah, I agree with your answer in post #7. haha, it's been a long day.

 

[CAF123]

Ok, thanks for the check. But I would still be interested in finding the error that led me to a denominator of c+v via Simon's method. Do you have any ideas?

 

[BruceW]

Using purely Lorentz transformations, I derived the time for the signal to traverse the rocket in S as $t = \frac{L_o}{c} \sqrt{\frac{1-v/c}{1+v/c}}$. I noticed that this was less than $L_o/c$ for all v<c, and my interpretation was that in S, the rocket length is contracted so as viewed from S, it takes less time for the signal to reach the other end.

There are two things which contribute to the time being less according to S than it is according to S'. Firstly, the rocket length is contracted, as compared to the length according to S'. Secondly, in S, the rocket tail is moving toward the light beam. And in S', the rocket tail is not moving toward the light beam.

I get a similar answer by the method you outlined, but not quite the same. Here it is:
In frame S, $x_{\text{tail}} = -L + vt$ and $x_{\text{light}} = -ct$ so again equating gives $t=L/(c+v)\,\,\,\, (1)$. L is the length in S. To relate to S', use a Lorentz boost to that frame which in general is $\Delta x = \gamma(\Delta x' + v\Delta t')$ and in this case, we have $L = \gamma(-L_o + vL_o/c)$. Subbing this in for L into (1) gives $$t = \frac{L_o}{c+v} \sqrt{\frac{1-v/c}{1+v/c}},$$ a slight difference in the denominator there.

You need to take more care with your Lorentz transform. Remember to keep in mind the two spacetime events which you are taking the difference of. Here, you have chosen the light being emitted at the head of the spaceship and the light being received at the tail of the spaceship in the S' frame. But these two events, according to the S frame are not equal to the length of the spaceship according to the S frame. In fact, the spacetime path is light-like, so it cannot correspond to any kind of length as viewed by any frame. You need to decide which two spacetime events according to S frame would correspond to the length of the spaceship according to the S frame, and then do a Lorentz transform to get this Length in terms of things you know.

 

[CAF123]

You need to take more care with your Lorentz transform. Remember to keep in mind the two spacetime events which you are taking the difference of. Here, you have chosen the light being emitted at the head of the spaceship and the light being received at the tail of the spaceship in the S' frame. But these two events, according to the S frame are not equal to the length of the spaceship according to the S frame.

In my other method, I think I just did a Lorentz transformation to get the length of the rocket in the S frame. Why was it valid there?

In fact, the spacetime path is light-like, so it cannot correspond to any kind of length as viewed by any frame. You need to decide which two spacetime events according to S frame would correspond to the length of the spaceship according to the S frame, and then do a Lorentz transform to get this Length in terms of things you know.

I am not sure how I would go about finding such events; the only two events that seem to be of interest are the light emission and receiving at the ends of the rocket. I wrote down the coordinates of those spacetime events in post #7 as viewed in S.

 

[CAF123]

@Simon Bridge: Since my equations for $x_{\text{light}}$ and $x_{\text{nose}}$ were derived in frame S, the resulting equation should not imply a faster than light movement. Yet the result I find for t (=L/(c+v)) implies this. Do you agree?

But I guess given that I then transform between frames in the next step (i.e substitute for L in terms of L_o), I should then be okay. But the result does not match up. I spoke to my professor and he said the problem was the fact that my equation for t implied a faster than light movement as written above. But I am not yet convinced given that I then make a transformation.

What do you think?
Thanks.

 

[BruceW]

In my other method, I think I just did a Lorentz transformation to get the length of the rocket in the S frame. Why was it valid there?

In your old method, you calculated
\Delta x = \gamma (\Delta x&#039; + v \Delta t&#039;) = \gamma(-L_o + v L_o/c)
which is the spatial length that the beam of light must travel to go from the front of the spaceship to the back of the spaceship, according to the S frame. But since the rocket is moving according to the S frame, this does not correspond to the length of the rocket according to the S frame.

I am not sure how I would go about finding such events; the only two events that seem to be of interest are the light emission and receiving at the ends of the rocket. I wrote down the coordinates of those spacetime events in post #7 as viewed in S.

Well, essentially, I'm just telling you to derive the standard formula for length contraction. But If you don't have so much time, you can just use the equation for length contraction.