0
$\begingroup$

I am new to the field of special theory of relativity and while understanding its concepts the following question popped into my mind...

Suppose there is an object O moving with a very high velocity, let us assume c/2 (where c is the velocity of light), with respect to an inertial frame of reference S. Two light beams L1,L2 are produced in S former along the direction of motion of O and the latter in the opposite direction. Assume two arbitrary points P1,P2 in the light beams L1 and L2 respectively. How much time does it take for points P1 and P2 to cross the whole length of the object with respect to an observer at rest in S?

According to my calculation the time taken by P1 and P2 should be same. Here's my explanation ...

Assume I am a still observer in the frame of reference of O. Let the absolute length of the object be L. I observe that P1 crossing the length of O with a speed c in time t=L/c .I observe the same in case of P2 taking the same time t with respect to me because the speed and the length are same. Now if t seconds have passed for me in observing P1 cross L then some T seconds must have passed for an observer who is still in S where T>t since the time passes fast in S. Now same amount of time(T) passes in case of P2 since again t duration corresponds to the same T in S. So the time taken by P1 and P2 is same for crossing the length of the object O with respect to a still observer in S.

But our general intuition tells that P1 should have taken more time than P2 since P1 is moving along the direction of motion of O and the other opposite to the direction of motion O. Is my explanation correct? If I am wrong where have I gone wrong.

$\endgroup$

closed as off-topic by John Rennie, ACuriousMind, Kyle Kanos, Ryan Unger, Neuneck Aug 5 '15 at 9:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, ACuriousMind, Kyle Kanos, Ryan Unger, Neuneck
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is your progress so far researching and solving this question? $\endgroup$ – Asher Aug 3 '15 at 8:52
  • 1
    $\begingroup$ Welcome to PSE! Please provide some work you have done and your current reasoning in this problem. This site is not for solving problems for you, but rather to help on the process. Please check our policy for this cases. $\endgroup$ – rmhleo Aug 3 '15 at 9:04
  • $\begingroup$ Sorry for the inconvenience, now I have edited my question. $\endgroup$ – Yash Aug 3 '15 at 9:43
  • $\begingroup$ Note that questions of the form Is this correct are not good formats for this Q&A site because the answer (Yes|No) is too short to be a valid answer. It would be best if you wrote the question to be about clarifying your understanding than asking if it is correct. $\endgroup$ – Kyle Kanos Aug 3 '15 at 12:42
  • $\begingroup$ @Yash, you need to remember one thing. The claim that light always travels at c is based on a 2-way measurement. And there is no other way of measuring the speed of light. Taking time to understand the consequences of this fact brings true understanding of the issue of the constancy of lightspeed and removes many paradoxes. Obviously, many people prefer to stay in the Neverland. It's so impressive and fires the imagination ... $\endgroup$ – bright magus Aug 4 '15 at 5:26
0
$\begingroup$

You are right, in the frame where $O$ is at rest, the time $P1$ and $P2$ take to travel across $O$ will be the same, call these times $t_1=t_2$ But it is also true that in $S$, $P1$ will take an longer time to cross $O$ than $P2$.

How is this possible?

There are various ways to begin understanding how. Here is one.

Go back to the rest frame of $O$, and assume $P1$ and $P2$ enter $O$ at the same time. They travel at the same speed, so they will both exit at the same time. The two pair of events ("$P1$ enters $O$","$P2$ enters $O$") and ("$P1$ exits $O$","$P2$ exits $O$") are each simultaneous in this frame: events that happen at the same time, but at different places.

I said "in this frame" because special relativity, when one transforms between coordinate frames, time and space coordinates mix. This results in relativity of simultaneity: if two events appear to happen at the same time but at different places on the axis of a particular frame, , they will appear to happen at different times in all frames traveling along that axis. Simultaneity is a property of (it is relative to) the frame.

The way the coordinates change according to Lorentz Transformations is so that in $S$,the two pair of events ("$P1$ enters $O$","$P2$ enters $O$") and ("$P1$ exits $O$","$P2$ exits $O$") are not simultaneous.
Additionally, the intervals of time between ("$P1$ enters $O$","$P1$ exits $O$") as measured in $S$ will be larger than that between ("$P2$ enters $O$","$P2$ exits $O$").

At this time, it is normal that this is still confusing. Try looking at the "Ladder Paradox" article on Wikipedia. This is very similar to your question. Another advice: I find spacetime diagrams are very useful guides for intuition.

Good luck.

$\endgroup$
  • $\begingroup$ theory of simultaneity it is.. thank you @Andrea Di Biagio $\endgroup$ – Yash Aug 3 '15 at 17:20
0
$\begingroup$

Yash, Imagine S is the Sun sending two photons, P1 and P2, and the object O is represented by two asteroids, O1 and O2, equidistant from each other all the time and from S at $t_0$ - they are moving in the same direction at the same velocity (c/2). So for some time one will be moving towards the Sun and one away from it.

So you are right that the speed of light is the same in all frames of reference. However, the distance travelled by the two photons, P1 and P2, before they reach their destination asteroids, O1 and O2 respectively, will not be the same. Until photon P1 reaches asteroid O1, it will have moved, say, away from the Sun, while until the photon P2 reaches asteroid O2 it will have moved toward the Sun.

Therefore, lightspeed being the same for two photons P1 and P2, they will not reach the ends of the object O at the same time, since each of them will travel a different distance.

$\endgroup$
  • $\begingroup$ Here you consider two photons traveling in the same direction with an object that's changing size, but in OP's question the photons travel in opposite directions through an object that is of constant size. $\endgroup$ – Andrea Aug 3 '15 at 11:16
  • 2
    $\begingroup$ @AndreaDiBiagio: Nope, the two photons in my explanation are travelling in opposite directions, and size of the object O is not changing ("the object O is represented by two asteroids, O1 and O2 - equidistant from S and moving in the same direction at the same velocity (c/2), and one moving towards the Sun and one away from it" - the two asteroids O1 and O2 represent two ends of an imaginary, in my case, object O). $\endgroup$ – bright magus Aug 3 '15 at 11:19

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