Easy Lorentz Transform Problem

[Kreizhn]

Hey all,

Simple question, so hopefully it gets answered quickly.

I hate SR, but am marking an assignment on it. Here's the set-up

Alice is standing still on a train traveling at 3/5c and Bob is on the platform at rest. A series of lights are set up along the track, yada yada. Bob sees the lights flash every $T_0$ seconds, so what is the time measured by Alice?

So clearly, Alice will see a slower time by a factor of $\gamma = \frac45$ and will measure $\frac45 T_0$. However, I have a student who is using Lorentz transformations. If Alice's frame is (t',x') and Bob's is (t,x) then
$$\Delta t' =\gamma(\Delta t-\frac v{c^2} \Delta x )$$
He claims that $\Delta x = 0$ since Bob is not moving, and calculates the time difference to be $\frac54 T_0$ instead. Clearly this must be wrong because Alice should experience a slower amount of time. Is the assumption that $\Delta x = 0$ wrong because in Alice's frame, Bob is in fact moving?

 

[harrylin]

Hey all,

Simple question, so hopefully it gets answered quickly.

I hate SR, but am marking an assignment on it. Here's the set-up

Alice is standing still on a train traveling at 3/5c and Bob is on the platform at rest. A series of lights are set up along the track, yada yada. Bob sees the lights flash every $T_0$ seconds, so what is the time measured by Alice?

So clearly, Alice will see a slower time by a factor of $\gamma = \frac45$ and will measure $\frac45 T_0$. However, I have a student who is using Lorentz transformations. If Alice's frame is (t',x') and Bob's is (t,x) then
$$\Delta t' =\gamma(\Delta t-\frac v{c^2} \Delta x )$$
He claims that $\Delta x = 0$ since Bob is not moving, and calculates the time difference to be $\frac54 T_0$ instead. Clearly this must be wrong because Alice should experience a slower amount of time. Is the assumption that $\Delta x = 0$ wrong because in Alice's frame, Bob is in fact moving?

Right. The statement "Bob is not moving" implies using the frame in which Bob is in rest, not Alice's rest frame. He may directly use the LT for Alice in rest, taking Alice's frame as (t,x) and Bob's as (t',x').

Note: watch out with "see", for what Alice will see is also Doppler shifted. Better to say "infer".

 

[Dale]

So clearly, Alice will see a slower time by a factor of $\gamma = \frac45$

$\gamma$ is always greater than 1.

and will measure $\frac45 T_0$.

No, in Alice's frame the flashes are going slow, so the time between successive flashes is $\frac{5}{4} T_0$.

However, I have a student who is using Lorentz transformations. If Alice's frame is (t',x') and Bob's is (t,x) then
$$\Delta t' =\gamma(\Delta t-\frac v{c^2} \Delta x )$$
He claims that $\Delta x = 0$ since Bob is not moving, and calculates the time difference to be $\frac54 T_0$ instead.

The student is correct. The Lorentz transforms are the best equations to use. I routinely caution against using the length contraction and time dilation formulas, they are too easy to misuse.

Clearly this must be wrong because Alice should experience a slower amount of time. Is the assumption that $\Delta x = 0$ wrong because in Alice's frame, Bob is in fact moving?

No, Bob's frame is the unprimed frame, so $\Delta x$ is the displacement in Bob's frame, which is indeed 0.

 

[GrayGhost]

Kreizhn,

The time transformation is ...

t' = gamma(t - vx/c²)​

you said ...

Is the assumption that delta x = 0 wrong because in Alice's frame, Bob is in fact moving?

Per Alice, Bob is indeed moving and so too are the lamps. It doesn't matter that the scenario has multiple lamps stationed along the track, because any single lamp is all that is required to answer the question at hand. It's not that delta x = 0 because it's in Bob's frame. The first event is the flash of the lamp which occurs "at" the origin (at the lamp). The 2nd event is the next flash event, which occurs "at" the same moving lamp. Delta x signifies the "considered location" in space wrt the moving lamp as the reference. IOWs, its a spatial offset from the lamp at the 2nd event, and any location may be considered if desired. Since the 2nd event considered "is the flash event itself", which of course must occur AT the moving lamp, delta x = 0. The solution follows ...

t' = gamma(t - vx/c²)
T₀ = (5/4)(t - (0.6)(0)/(1)²)
T₀ = (5/4)t
(4/5)T₀ = t​

GrayGhost

 

[Dale]

t' = gamma(t - vx/c²)
T₀ = (5/4)(t - (0.6)(0)/(1)²)
T₀ = (5/4)t
(4/5)T₀ = t​

Gray Ghost, you misread the problem, you have $T_0$ and t and t' mixed up here. $T_0$ is the time between flashes in Bob's frame, and Bob's frame is the unprimed frame. See:

Bob is on the platform ... lights are set up along the track, ... Bob sees the lights flash every $T_0$ seconds ... Alice's frame is (t',x') and Bob's is (t,x)

So in your above calculations you need to substitute $T_0$ for t and solve for t', not substitute $T_0$ for t' and solve for t.

 

[harrylin]

Right. The statement "Bob is not moving" implies using the frame in which Bob is in rest, not Alice's rest frame. He may directly use the LT for Alice in rest, taking Alice's frame as (t,x) and Bob's as (t',x').

Note: watch out with "see", for what Alice will see is also Doppler shifted. Better to say "infer".

Oops, sorry Kreizhn, I answered the wrong way round - you managed to confuse me.

You wrote:

Alice will see a slower time by a factor of $\gamma = \frac45$ and will measure $\frac45 T_0$.

However, measuring a "slower" time implies that your own clock measures a longer time period than T₀.

The handiest approach is just that of your student, taking the frame in which the clocks that flash light are in rest so that the horizontal coordinate x=constant. Then you get:

$$\Delta t' =\gamma(\Delta t - 0 )$$

or $$\Delta t'$$ = 5/4 T₀

That is correct.

Cheers,
Harald

 

[Kreizhn]

Hey guys,

Thanks for the replies. Sorry I messed up the reciprocal of $\gamma$. Unfortunately, this has only proven to make me more confused.

After conferring with the professor, he indicates that Alice measures $\frac 45 T_0$. His research is in cosmology, and in fact there have been threads on pf discussing results of his papers. Maybe he is made a calculation error?

Furthermore, there is http://Galileo.phys.Virginia.EDU/classes/252/time_dil.html" site which also indicates that the Alice will measure $\frac45 T_0$. So there does not seem to be a consensus.

This is why I hate SR.

 

[Kreizhn]

The profs solution is as follows:

In the rest frame the time between the flashes is $T_0$, so let the time of the ith flash for Bob be given by $t_i^{(B)}$ so that $t_1^{(B)} = 0$ and $t_2^{(B)} = T_0$. Using the Lorentz transform, we have
$$t = \gamma \left( t' + \frac{v}{c^2} x' \right)$$
Now for Alice, both event occur at the same place, i.e. just outside her window so
$$\begin{align*}<br /> t_1^{(B)} &= \gamma\left( {t_1'}^{(A)} + \frac v{c^2} {x_1'}^{(A)} \right) \\<br /> &= \gamma{t_1'}^{(A)} \\<br /> t_2^{(B)} &= \gamma {t_2'}^{(A)}\end{align*}$$
So that $\Delta t^{(B)} = \gamma \Delta {t'}^{(A)}$.
The result then follows.

 

[Kreizhn]

Anyway, essentially every resource I have found, including other posts on PF indicate that the time measured by the train-goer will be LESS than that measured by a rest observer. Hence the solution is $\frac45 T_0$ and not $\frac54 T_0$.

Grayghost, I'm not sure if I understand what you mean by the lamps moving, or whether it affects the problem. The problem was stated such that there are a series of lamps which flash simultaneously in the rest frame. I guess we take some liberty in assuming that the lamps are perfectly spaced so that Alice can observe each flash. In any case, the lamps from the first and second flashes are then different. Does this affect your reasoning?

Again guys, I'm sorry if I seem ignorant about this stuff. There's a reason I went into QM instead of GR.

 

[Kreizhn]

Ah, I think I get why the reasoning of the student is incorrect. $\Delta x$ must represent the displacement of the events we're considering, which in this case is Alice measuring a flash. Since the train is moving, $\Delta x = \frac35 c T_0$, which represents the distance traveled by Alice between the first flash at the origin and the second flash. With this, we also get $\Delta t' = \frac45 T_0$.

 

[JesseM]

It seems this word-problem just isn't stated clearly enough, it's not clear if you're asking about the time between flashes of any individual light in Alice's frame, or the time between flashes of two neighboring lights which go off simultaneously in Bob's frame.

 

[Kreizhn]

Yeah, I guess I could have worded that better. I was in a rush and it seemed like an obvious question so I guess that's where the confusion stems from.

However, I think I did state

Bob sees the lights flash every $T_0$ seconds, so what is the time measured by Alice?

I should have mentioned that the light flash simultaneously and that we're looking for the time between any two flashes for Alice, so that was my mistake. However, I do not see how one might be confused about the time measured between flashes in Bob's frame.

 

[JesseM]

I should have mentioned that the light flash simultaneously and that we're looking for the time between any two flashes for Alice, so that was my mistake.

Still not clear--any two flashes of what? Two flashes of a single light, or two flashes of neighboring lights? Based on the prof's answer it seems to be neighboring lights, not a single light (and if it were meant to be a single light, it would seem totally unnecessary to the word-problem to mention that there were a bunch of lights at regular spacings along the tracks, that alone indicates they're probably asking about the time between flashes of neighboring lights).

 

[Kreizhn]

Hmm...two simultaneous flashings of a single light? Seems like the answer to your question could have been inferred from the nature of that statement.

Again, I apologize for the ambiguity of the original statement. However, the corrections of my previous post should have made it possible to infer the obvious meaning. As you mentioned, the fact that there are a series of lights and the vacuousness of the notion of a light flashing simultaneously with itself should have allowed you to infer the situation.

 

[Dale]

This is not very difficult and I am sorry that you are confused on the subject, but the correct answer is very clear.

If the lights are separated by a distance of $L_0$ in Bob's (unprimed) frame and if the time between flashes is $T_0$ in Bob's frame, then the nth flash of the mth light occurs at the event:
$$t=nT_0$$
$$x=mL_0$$

Lorentz transforming into Alice's (primed) frame we get:
$$t'=\gamma n T_0-\frac{\gamma L_0 m v}{c^2}$$
$$x'=\gamma \left(L_0 m-n T_0 v\right)$$

For m=0 and n=0,1 we find clearly that
$$\Delta t' = \gamma T_0$$

Note that the professor is wrong at this point:

Now for Alice, both event occur at the same place, i.e. just outside her window so
$$\begin{align*}<br /> t_1^{(B)} &= \gamma\left( {t_1'}^{(A)} + \frac v{c^2} {x_1'}^{(A)} \right) \\<br /> &= \gamma{t_1'}^{(A)} \\<br /> t_2^{(B)} &= \gamma {t_2'}^{(A)}\end{align*}$$

From the OP it is clear that the events occur at the same place in Bob's frame:

Bob is on the platform at rest. A series of lights are set up along the track

 

[JesseM]

Hmm...two simultaneous flashings of a single light?

No, I meant two successive flashings of a single light. That seemed to be what the student in your original post was probably thinking--if an individual light flashes every $$T_0$$ seconds in Bob's frame, then it will flash every $\frac54 T_0$ seconds in Alice's frame (and the student also said $$\Delta x = 0$$ in Bob's frame, which would also suggest a single light). If the word-problem didn't clearly specify it was talking about the time between flashes of neighboring lights, I think the student deserves at least partial credit, if not full credit for a reasonable interpretation of a badly-stated problem.

Meanwhile if we are indeed talking about the time between flashes of neighboring lights, you can't use the time dilation equation at all, unless the speeds are just right so that each light is at the same position in Alice's frame at the moment it flashes (the time dilation equation can only be used when the two events happen at the same position in one of the frames you're using). Did the word-problem specify anything about the distance between the lights in Bob's frame? If not, just assume the separation in Bob's frame is $$\Delta x$$, and the time between flashes of neighboring lights in Bob's frame is 0, so in that case the time between flashes in Alice's frame would be:

$$\Delta t' = \gamma * (0 - v\Delta x /c^2) = (3/4)*\Delta x/c^2$$

But in this case $$T_0$$ doesn't enter into it...I suppose it is also possible the word-problem is asking about the time in Alice's frame between two successive flashes of neighboring lights in Bob's frame (though in this case the answer would depend on whether the first flash or the second flash is the one further in the +x direction in Bob's frame).

Perhaps you could post the exact wording of the problem here?

However, the corrections of my previous post should have made it possible to infer the obvious meaning.

I can't really see any interpretation under which your answer of $\frac45 T_0$ would make sense, unless you left out some specific information about the separation between the lights in Bob's frame (I also notice that you say in post #8 that "Now for Alice, both event occur at the same place", which is very puzzling--was it part of the statement of the problem that successive flashes of neighboring lights occur at the same position in Alice's frame, implying that in Bob's frame the separation between lights must be $\frac35 c T_0$? If so this is a really crucial aspect of the problem, you can't leave it out in your summary!)

 

[JesseM]

This is not very difficult and I am sorry that you are confused on the subject, but the correct answer is very clear.

If the lights are separated by a distance of $L_0$ in Bob's (unprimed) frame and if the time between flashes is $T_0$ in Bob's frame, then the nth flash of the mth light occurs at the event:
$$t=nT_0$$
$$x=mL_0$$

Lorentz transforming into Alice's (primed) frame we get:
$$t'=\gamma n T_0-\frac{\gamma L_0 m v}{c^2}$$
$$x'=\gamma \left(L_0 m-n T_0 v\right)$$

For m=0 and n=0,1 we find clearly that
$$\Delta t' = \gamma T_0$$

Since you choose m=0 for both events, apparently you are talking about two successive flashes of the same light, but in his last post to me Kreizhn seemed to indicate (though he didn't clearly state, which is frustrating given that I asked for a specific answer) that he was talking about time between flashes of neighboring lights, though even here it's not clear whether the two events are supposed to be flashes that are simultaneous in Bob's frame (m=0,1 and n=0) or successive in Bob's frame (m=0,n=0 and m=1,n=1, or perhaps even m=0,n=1 and m=1,n=0).

 

[Dale]

It is certainly possible that he is giving different problems to different people, and that he is getting different correct answers to each problem.

 

[Kreizhn]

Okay okay, I'm sorry. My haste has clearly led to chaos, so may we perhaps start from scratch? I will post the problem verbatim:

Note that the role of Bob has switched.

Alice and Bob are riding on a forward moving train that is moving on a long straight track at speed v=3/5 c. Alice is in the back of the train and Bob is in the front. The rest-length of the train is L0. Eve is at rest on the tracks and has setup os a long series of lights along the side of the track. Alice and Bob are looking out the side-windows of the train and can only see one light at a time (the light just outside their respective windows at any given time). Eve has set up the lights such that they all flash simultaneously in her reference frame.
[Note: As I mentioned earlier, I think we have taken the liberty of assuming that are appropriately spaced such that the windows correspond to flashes, otherwise the problem becomes essentially unsolvable].
Two consecutive flashes occur $T_0$ apart as measured by Eve using her watch.

Alice then sees the two consecutive flashes while looking out her side windows. What is the time interval between the two flashes as measured by Alice using her own watch?

While it might be inexcusable, I hope now that you can see why I chose to only briefly recapitulate the question: it is a novel.

 

[JesseM]

It's odd that the problem even mentions "Bob" given that he seems to have nothing to do with the problem! Anyway, start from Eve's frame, let's say that Alice is moving in the +x direction of Eve's frame, and that at t=0 she is at position x=0 and next to a light which flashes there. Then at 3/5c it will take her a time of L0 / (3/5c) to reach the position of the next light in the row at position x=L0, so she'll be at the position of the next light at t=(5/3)*L0/c, and as per your parenthetical note let's assume this next light flashes at the moment Alice passes it. If we know the x and t coordinate of some event in Eve's frame, the t' coordinate of the same event in Alice's frame is given by this part of the Lorentz transform:

t' = gamma*(t - vx/c^2)

So if the first flash is at (x=0, t=0) in Eve's frame, it's also at t'=0 in Alice's frame. Meanwhile if the second flash is at (x=L0, t=(5/3)*L0/c) in Eve's frame, in Alice's frame the time is:

t' = (5/4) * ((5/3)*L0/c - (3/5)c*L0/c^2) = (5/4) * ((5/3)*L0/c - (3/5)*L0/c)
= (5/4) * ((25/15)*L0/c - (9/15)*L0/c) = (5/4) * (16/15)*L0/c = (4/3) * L0/c

So that's one answer, the time in Alice's frame is (4/3)*L0/c. But note that this answer doesn't involve T0 at all! If we again assume that T0 in Eve's frame is timed just right so that each light flashes when Alice is next to it, then since the time in Eve's frame between Alice passing successive lights was t=(5/3)*L0/c, it must be true that T0 = (5/3)*L0/c, or L0/c = (3/5)*T0, and plugging that into the above gives t' = (4/3)*(3/5)*T0 = (4/5)*T0, just as your professor said.

And in fact, your professor's mode of solving the problem is probably the simplest one--if each flash is outside Alice's window as it happens, then each flash occurs at the same position in Alice's frame, so if the time in Alice's frame is $$\Delta t'$$ the it must be related to the time in Eve's frame $$\Delta t$$ by $$\Delta t = \gamma * \Delta t'$$ according to the time dilation equation (where the time interval that's multiplied by gamma is always the time interval in the frame where the events happened at the same position), which can also be derived from the Lorentz transformation equation $$\Delta t = \gamma * (\Delta t' + v \Delta x' /c^2)$$ with $$\Delta x' = 0$$. And we know $$\Delta t$$ in Eve's frame is T0, so that means $$T_0 = \gamma * \Delta t'$$ or $$\Delta t' = T_0 / \gamma$$ which again gives (4/5)*T0.

 

[Kreizhn]

JesseM, thanks for the reply, and thanks everyone for your patience!

I could see the reasoning of the question clearly but my confuddled description of it clearly led to much confusion and frustration for all. Hopefully this resolves it for good.

The reason it mentions Bob though is because there are many parts to the question. We have only analyzed the first part. In the spirit of ensuring there were no more possible ambiguities, I stated the question verbatim regardless of each component's relevance to the task at hand.

 

[harrylin]

Hey guys,

Thanks for the replies. Sorry I messed up the reciprocal of $\gamma$. Unfortunately, this has only proven to make me more confused.

After conferring with the professor, he indicates that Alice measures $\frac 45 T_0$. His research is in cosmology, and in fact there have been threads on pf discussing results of his papers. Maybe he is made a calculation error?

Furthermore, there is http://Galileo.phys.Virginia.EDU/classes/252/time_dil.html" site which also indicates that the Alice will measure $\frac45 T_0$. So there does not seem to be a consensus.

This is why I hate SR.

Again, Dalespam gave the right answer (and my correction of myself was right too).

I now quickly looked at the reference you gave: it uses the reciprocal of the usual factor - perhaps that is what confused you. And your professor got confused to, it seems (if you indeed accurately reproduced the problem - see below!).

And I always thought that I'm the only one who easily mixes up + and -, and x and 1/x.

To elaborate more on the setting of the problem you sketched: apparently the idea was that the train passes next to each flash light when they flash. That makes the problem easy to picture. [*Edit: however, see next!]
The flash lights are fixed in the frame in which Bob is in rest, so that the clocks in that frame do not move - those flashlights are fixed clocks in that frame which means, x=constant.

The outcome as we already gave in detail, is that the time period of the moving flash lights (moving according to Alice) is increased, according to Alice; thus each moving clock appears to tick slower, just as you thought.

Edit: just when it struck me that the answer is the inverse for the clock readings on the subsequent clocks that Alice passes, I found your exact reproduction of the puzzle, here once more:

"Alice and Bob are riding on a forward moving train that is moving on a long straight track at speed v=3/5 c. Alice is in the back of the train and Bob is in the front. The rest-length of the train is L0. Eve is at rest on the tracks and has setup os a long series of lights along the side of the track. Alice and Bob are looking out the side-windows of the train and can only see one light at a time (the light just outside their respective windows at any given time). Eve has set up the lights such that they all flash simultaneously in her reference frame.
[Note: As I mentioned earlier, I think we have taken the liberty of assuming that are appropriately spaced such that the windows correspond to flashes, otherwise the problem becomes essentially unsolvable].
Two consecutive flashes occur LaTeX Code: T_0 apart as measured by Eve using her watch.
Alice then sees the two consecutive flashes while looking out her side windows. What is the time interval between the two flashes as measured by Alice using her own watch?"

Yes exactly - that turns the problem on its head!

Here we have the case of a single moving clock (in the train) that is compared with two clocks on the tracks. Then in fact my very first answer applies, it's the easiest to use the inverse Lorentz transformation, for x'=constant. For Alice the light clocks are very much more out-of-sync than that they run slow, so that even though each clock flashes slower according to Alice, still the time until the next flash of every next flashlight is shorter.

Does that help?

 

[JesseM]

I now quickly looked at the reference you gave: it uses the reciprocal of the usual factor - perhaps that is what confused you. And your professor got confused to, it seems (if you indeed accurately reproduced the problem).

Did you read post 19 where Kreizhn gave the exact wording of the problem? The professor's answer was correct for the problem as written, although Kreizhn's earlier descriptions were confusing.

 

[harrylin]

Did you read post 19 where Kreizhn gave the exact wording of the problem? The professor's answer was correct for the problem as written, although Kreizhn's earlier descriptions were confusing.

I did not read that one yet, there is a bandwidth problem ("The connection has timed out") and at the moment I'm lucky if I can read and reply to a single post of physicsforums.com - I'll try my luck now!

 

[Kreizhn]

Yeah, something seems to be going on in the back-end since PF is taking brutally long to load.

 

[JesseM]

I did not read that one yet, thewre is a bandwidth problem and at the moment I'm lucky if I can read and reply to a single post of physicsforums.com - I'll try now luck now!

Yes, I've been having that problem too, I checked the feedback forum and apparently there's an ongoing DOS attack on physicsforums right now.

 

[Dale]

While it might be inexcusable, I hope now that you can see why I chose to only briefly recapitulate the question: it is a novel.

Yes, but you changed the question significantly. Using my above notation, the question in the OP is looking for the time of flash n=1, m=0 in the primed frame, but this new question is looking for the time of flash n=1, m=1 in the primed frame (with the additional constraint that L0=vT0). Different questions, different answers. The prof is right, as you can see by substituting in m=1, n=1, and L0=vT0 and then simplifying.