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Dispute over basic relativity question

  1. Feb 12, 2010 #1
    The following problem is part of a homework from last year. I never did understand the solution, and it's been nagging at me ever since. I've discussed the problem with my tutor and with my friends, and they all come to the same answer. They have tried (patiently) to explain this answer to me over and over, but I fail to understand their reasoning.

    Below, I have described as clearly as I can my attempt at the problem. Could somebody explain to me in very simple terms where I go wrong?

    Ok, so part A: 7.5 miles. That's fine. You can ignore the part about speeding tickets.

    Part B is the problem. The official solution is, "The student has longer than 30 minutes to spend in the restaurant, according to his watch". I can't get my head around that. My answer is that the student has exactly 30 minutes in the restaurant. My reasoning is below.

    • Let the classroom-restaurant frame be [itex]S[/itex].
    • While the student is travelling between the classroom and restaurant at 30 mph, let his frame be [itex]S'[/itex].
    • Assume that the classroom clock is synchronised with the restaurant clock. Call the time on these clocks [itex]t[/itex].
    • Call the time on the student's wristwatch [itex]t'[/itex].
    • Let [itex]{t_0} = {t'_0} = [/itex] 12:00.
    1. Student departs the classroom at [itex]{t_0} = {t'_0} = [/itex] 12:00.

    2. Student travels 7.5 mi from the classroom to the restaurant at 30 mph (w.r.t. the [itex]S[/itex] frame). [itex]\gamma=[/itex] 1.51.

    2. Student arrives at the restaurant at [itex]t=[/itex] 12:15, [itex]t'=[/itex] 12:10 (approx.). Upon reaching the restaurant, the student stops his car, thus exiting the [itex]S'[/itex] frame and entering the [itex]S[/itex] frame.

    3. The student's watch is now currently ticking with [itex]S[/itex]-frame minutes. Suppose the student stays in the restaurant for 30 of these minutes.

    4. Student departs the restaurant at [itex]t=[/itex] 12:45, [itex]t'=[/itex] 12:40 (approx.).

    5. Student travels 7.5 mi from the restaurant back to the classroom at 30 mph.

    6. Student arrives at the classroom at [itex]t=[/itex] 13:00, [itex]t'=[/itex] 12:50 (approx.).

    It seems to me that if the student spends longer than 30 minutes in the restaurant then he arrives after [itex]t=[/itex] 13:00 and is therefore late for class. However, my friends and tutor disagree (adamantly!). What is the real solution?
  2. jcsd
  3. Feb 12, 2010 #2

    Vanadium 50

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    The key is "according to the student's watch". The assumption is that the student will not correct for relativity.
  4. Feb 12, 2010 #3


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    But the first part of the calculation was explicitly based on the idea that the student would spend a half an hour in the restaurant and 15 minutes of travel each way. Are you saying that in the second part the student would forget this, and since it only took 15*sqrt(1 - 0.75^2) = 9.92 minutes according to his watch to get to the restaurant, he'll assume he can spend 60 - 2*9.92 = 40.16 minutes in the restaurant, which will actually end up making him 10.16 minutes late for class? This seems like a pretty convoluted solution, you'd naturally think that whatever assumptions were made in the first part of the problem would carry over to the second part.
  5. Feb 12, 2010 #4

    Ben Niehoff

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    I think the wording is ambiguous. Once the student is in the rest frame of the restaurant, he has exactly 30 ticks on his watch in which to eat, or else he will be late for class. The problem is that his watch reads 12:10 when he arrives, so he will think he has 40 minutes, if he does not take relativity into account. But if he takes 40 minutes and then travels back, he will think he is precisely on time when in fact he is 10 minutes late.

    It is unclear to me, a priori, whether the phrase "according to his watch" means "according to the rate at which his watch ticks", or "according to the time his watch reads when he arrives". Usually one only talks about time differences, and the first interpretation is the most common in relativity problems. A "clock" is generally taken to mean a device that ticks at a uniform rate in its own rest frame; whereas this problem is calling for a "clock" to be a device that reads out a specific time of day.

    If the question wanted the answer given, it would have been clearer to ask, "How much time will the student assume he has to eat, given the reading on his watch, if he does not take into account relativity?"

    I think you should just put the question out of your mind; the course is over anyway. You obviously understand the physics involved, and the rest is simply arguing the semantics of the question. Even if the course were still in session and the points mattered, I've noticed many professors seem to be surprisingly unsympathetic to the notion that their wording of questions might have multiple logical interpretations.
  6. Feb 12, 2010 #5
    I should point out that, while I can't remember word-for-word everything that my friends and tutor have explained to me in our discussions, if any of their answers had been as short and simple as "The assumption is that the student will not correct for relativity" then I feel certain that the discussion would have ended then and there.

    What happened was that, at the time the homework was set, I submitted the above solution to the question (but obviously with different wording etc.) and it was marked wrong. Today I was thinking about the question and I showed it to my friends, and we got to talking about it. I wrote down my same solution again, and they looked at it, and they both concluded that my chain of reasoning is wrong. So either they're right, or I'm right, or there has been an incredible series of repeat-misunderstandings.

    I can see the point you're getting at, Vanadium 50. The question "According to the student’s watch, how much time does he actually have for being in the restaurant?" is ambiguous, right? It could mean:

    "What is the longest time he can spend in the restaurant and still make it back to the classroom before the classroom clock strikes 13:00?"

    Or it could mean:

    "What is the longest time he can spend in the restaurant and still make it back to the classroom before his wristwatch strikes 13:00?"

    But the disagreement we were having was not simply a matter of semantics. It goes deeper than that. I wish I could remember how they tried to explain things to me! But you know how it is, it's so difficult to remember things that you don't understand. I will encourage my friends to post here; that way you can hear their explanations first-hand.
  7. Feb 12, 2010 #6

    Ben Niehoff

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    To further argue semantics (:D), I think the idea that people living in a world with c = 40 mph would not take relativity into account is totally preposterous. If relativity were part of our everyday experience for our entire lives, then we would have developed an intuition for it. We would probably also have a more sophisticated time-keeping system than merely saying "class begins at 13:00 sharp!"
  8. Feb 12, 2010 #7
    I pressed submit before I noticed your answer, but thank you, Ben! Excellent answer. I was worried that if I didn't understand this, the most basic of relativity questions, then I didn't understand anything about relativity at all. It was probably all a big misunderstanding, after all, then.

    (All things considered, if we're going to start assuming things about the question, then it would seem reasonable that a person who lives in a universe where c = 40 mph would be completely accustomed to the effects of relativistic speed, and therefore would know to account for time dilation when planning his journeys. But, yes, you're right, this question in the past and I thank you for freeing me of the burden of worrying about it!)
  9. Feb 12, 2010 #8

    Ben Niehoff

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    I think JesseM has a valid point, also. The student obviously knows how to take relativity into account in part A. Otherwise, he should travel until his watch reads 12:15, which would let him travel over 11 miles!
  10. Feb 12, 2010 #9
    I'll definitely refresh the page more often from now on...
  11. Feb 12, 2010 #10
    Yes, that's right! Well spotted. So the fact that "30 minutes" is not the answer makes even less sense to me now than it did before. But Ben Niehoff assures me that I managed to at least grasp the basic physics. I'm happy with that.
  12. Feb 12, 2010 #11


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    According to his watch the travel time is 10 min. So if he had a 1 hour lunch break according to his watch then he would have 40 minutes to eat at that restaurant. I.e. if the duration of the lunch break were determined by proper time of the student instead of proper time of the school then he would have 40 minutes.

    I think it is a really sloppy wording and that it is no surprise it tripped you up. It would have tripped me up too, my instinct was to give the answer 30 minutes with relativistic effects being completely irrelevant. I wouldn't lose any sleep over a poorly worded question, it won't be the last and it seems like you understand the physics fine.
  13. Feb 13, 2010 #12
    There is away that the student can depart at 1200 hrs (according to the school clock), go to a restauraunt 7.5 miles away from the school and spend over 40 minutes (according to the student's watch) having a relaxing lunch at the restauraunt and still get back to the school in time, without risking getting a speeding ticket. See if you can figure that out :devil:
  14. Feb 13, 2010 #13
    Is the student allowed to go to a rotating restaurant?
  15. Feb 13, 2010 #14
    Well the question never specified that the restaraunt was at rest in the reference frame of the school, or if the school and restaurant are in flat spacetime, so there are many ways that relativity could be used so that the student could have a longer lunch break and still get back to school on time :P
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