So there is this very simple situation in one of my exercices:

In the earth's frame of reference a tree is at the origin and a pole is at $x=20$km. Lightning strikes at both the tree and the pole at $t=10$ microseconds. The lightning strikes are observed by a rocket traveling in the positive x-direction at $0.5c$.

1) At what time does the lighting strikes take place in the rocket's reference frame?.

I understand the concepts of time dilation, length contraction and etc, but the questions bring me confusion sometimes because they are not very well formulated in my sense. In this exercise I have difficulty understanding what do they really actually mean by 'the time in the rocket's reference frame.'

First it could mean that in the earth's reference frame what is the dilated time that an observer in A (earth) would measure for B (spaceship). An analogy could be that an observer in A measures that it takes his twin 16 years (dilated time) to age by the proper time of 8 years. So the proper question would be what is the dilated time $(t')$ that observer A measures, if it follows correctly the analogy. So we stay in Earth's reference frame and we are only measuring $t'$ as measured by an observer A (and not the time that take place in the rockets frame of reference which is different is my sense as explained below.)

Now a second meaning could be what is the proper time that someone traveling in the spaceship in his OWN frame of reference measures. Following the analogy, the time that it takes someone to go back to earth in the spaceship is 8 years because he measures his own proper time (which is different from the dilated time measured by an observer A on earth).

So when we use the equation $t'= \gamma(t-vx/c^2)$ or the one for position what do we really mean by $t'$? What I think is it is $t'$(dilated time) as measured in frame A because that is what we do in time dilation for example: When the twin measures proper time 8 and gamma factor 2 so $t'=16$ but here we are still measuring dilated time of B IN Earth's frame of reference A and not proper time in the spaceship reference frame B.

So here is my confusion. Does in the question they just 1) what they really mean is at what time does the lightining take place in spaceship B as measured by frame A.

So how do I get over this confusion?


Your confusion comes from overthinking the issue in terms of time dilation and length contraction rather than by just thinking in terms of what each observer would measure. In this problem, we have 2 frames of reference, the Earth's frame, $E$, and the spaceship's frame, $S$. Attached to $E$ is a coordinate system $(x,y,z,t)$ and attached to $S$ is a coordinate system $(x',y',z',t')$. An observer in $E$ uses the $(x,y,z,t)$ coordinate system to make measurements and observations and similarly an observer in $S$ uses $(x',y',z',t')$ to make measurements and observations. In this sense, $t$ is the time elapsed since $t=0$ in frame $E$ and $t'$ is the time elapsed since $t'=0$ in frame $S$.

Any given event, $P$, in space time can be described by a set of 4 coordinates. In the $E$, event $P$ has coordinates $(x_P,t_P)$ where I've neglected the $(y,z)$ coordinates for simplicity and since this problem is a two dimensional problem. In $S$, event P has the coordinates $(x'_P,t'_P)$. So we say event $P$ happened at displacement $x_P$ and at time $t_P$ in $E$, while it happened at displacement $x'_P$ and time $t'_P$ in $S$. In this language, the question the book is asking then is: "Given two events $P_1$ and $P_2$ (lightning strikes) which happen in frame $E$ at displacements and times $(x_{P_1},t_{P_1})=(0~\text{km},10~\mu\text{s})$ and $(x_{P_2},t_{P_2})=(20~\text{km},10~\mu\text{s})$ respectively, at what time(s) $t'_{P_1},t'_{P_2}$ do they occur in $S$?"

All that is required, then, is to make a relationship between $(x,t)$ and $(x',t')$ for any given pairs of $(x,t)$ and $(x',t')$. Generally $(x,t)$ and $(x',t')$ will be related by a Poincare transformation which would include translations, rotations, and Lorentz boosts. For this one dimensional problem, we can get rid of the rotations, and for simplicity we can get rid of the translations by setting $(x,t)=0$ and $(x',t')=0$ to be the same space time point (this is simply saying that we set the origin of the two frames to coincide). Given these simplifications, we are left with only a single dimensional Lorentz transformation: $$x'=\gamma(x-vt)$$$$t'=\gamma\left(t-\frac{vx}{c^2}\right)$$

You are given the two pairs of $x$ and $t$, it is sufficient here to simply plug and chug to get the pair of $t'$.

| cite | improve this answer | |
  • $\begingroup$ OK I think I understand better, it is just that it is confusing because t' is actually t in S frame of reference because in S own frame of reference he is stationnary so it brings a lot of confusing. Also regarding time dilation when t'=yt, t' is dilated time (for S in E's frame) but it is not the time that S actually mesure for himself. For example A is in earth B travels in space, gamma is 2, proper time is 8 so t'=2x8=16 but this t' is what A mesure but in reality in B's frame he mesure 8 years. So how do I reconcile these two counter-intuitive t'... $\endgroup$ – ValenciaG. Jul 20 '18 at 22:07
  • 1
    $\begingroup$ Don't get into the business of renaming t' to t if you are in S. That will lead you to a ton of confusion. S measures t' period. E measures t period. The relation between the two is a Lorentz boost. One of the things you have to give up when going to special relativity is the notion that simultaneity is universal. It seems you are still stuck on this fact (everybody gets stuck here for a while as they are learning relativity). Ponder this fact for a while, and things may become clearer. $\endgroup$ – enumaris Jul 20 '18 at 22:19
  • $\begingroup$ OK i seem to understand. The use of t' is not the same in relativity and simultaneity? In time dilation would you tell me S mesure t' period for example 16 years. But that t' actually correspond to dilated time but inside the ship his clock measure 8 years. Sorry but mixing Time dilation with Simulteaneity is a lot of confusion and still time dilation is a derivation of Lorentz transformations. What I think is that if we find t=10s and t'=15s, I would think that t' actually corresponds to dilated time because that is what the time dilation tell us but inside the ship the clock measure 10s... $\endgroup$ – ValenciaG. Jul 20 '18 at 22:26
  • 1
    $\begingroup$ Your confusion appears to be too broad for me to answer in the comments. The only advice I can really give is to put less emphasis on "dilated time" and just think in terms of "what time does E measure?" and "what time does S measure"? If you have a specific question, you should probably post a new question. Otherwise, it might be best to go back to some lectures/books to clear up your misunderstandings. $\endgroup$ – enumaris Jul 20 '18 at 22:34
  • $\begingroup$ Will do ! thank you for your explanations helped me to clear up some things! $\endgroup$ – ValenciaG. Jul 20 '18 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.