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B The difference between the Lorentz transformation & time dilation/length contraction

  1. Mar 25, 2016 #1
    Hey guys,

    In what circumstance or scenario would you use Lorentz transformations as a opposed to time dilation or length contraction? The reason that I ask this is because in all of the problems that I have worked with, the observer is always stationary relative to the event. For example, if Anna is on earth and is observing Bobby in a rocket ship flying towards Planet B at a significant portion of the speed of light. Then the distance and time that Anna observes/measures is ##x'## (distance from Anna) and ##t'## (time difference Bobby travels between two intervals as related to Bobby) while ##x=0## (Anna is stationary). If you know the time that Anna observes and the speed of the rocket. Wouldn't the equations ##\Delta x' = \gamma (\Delta x - v \Delta t)## and ##\Delta t' = \gamma \left(t - \frac{v \Delta x}{c^2} \right)## resolve to the length contraction formula and time dilation formulas when ##x = 0## is plugged in for ##x##?

    This same consolidation would happen from Bobby's frame of reference because he is stationary relative to himself and Anna is moving away at the same but negative velocity. Thus, ##x=0## and same situation would occur. If you could, please include an example which would not give the same answer when using the Lorentz transformations as opposed to time dilation or length contraction.

    Thank you very much for your help,

    Kind regards,
    Jonathan


    P.S. If you have any helpful hints, tips, tricks, or anything that helps you to remember what variable relates to who and who's reference frame, I would greatly appreciate it. I have a unit test on Wednesday and an exam in 43 days.
     
  2. jcsd
  3. Mar 25, 2016 #2

    PeterDonis

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    You are confusing the coordinates ##x, t## of a single event with the difference in coordinates ##\Delta x, \Delta t## between two events. I suggest going back and re-writing your formulas, paying careful attention to the difference between these two concepts. The Lorentz transformation equations apply to the coordinates of a single event--they allow you to transform the coordinates ##x, t## of an event in one frame to the coordinates ##x', t'## of the same event in another frame. The time dilation and length contraction formulas involve the differences in coordinates ##\Delta x, \Delta t##, and those do not transform according to the Lorentz transformation. In fact, when you factor in relativity of simultaneity, you will find that, in order to determine the length of an object in a frame in which it is moving, you need to consider a different pair of events than the pair you would use to determine the object's length in its rest frame.
     
  4. Mar 25, 2016 #3
    I'm not exactly sure what you are asking. The Lorentz transformation contains all the necessary math to give time dilation and length contraction. If you're asking when can you just use the time dilation or length contraction formulas instead of the full Lorentz transformation, then I would say you can use the former in specific situations. Thus, if there is a stationary clock in one frame that shows a particular time interval Δt, the time dilation formula will directly give you the time interval Δt' in the other frame. Similarly, if there is an object of length L that is stationary in one frame, the length contraction formula will directly give you the length L' in the other frame.

    Is that what you are asking?
     
  5. Mar 25, 2016 #4
    If I had a specific ##t## value or ##x## value, couldn't I just make it a ##\Delta x## or ##\Delta t## by subtracting ##0## from it? My question really is, if making a calculation from the observer's reference frame, wouldn't his/her ##x## value just be ##0## because he is stationary relative to himself? So, what is the point in the Lorentz transform equations if half of it disappears because you have to substitute ##0## in anyway?
     
  6. Mar 25, 2016 #5

    PeterDonis

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    Mathematically, yes, but physically, the two still mean different things; ##x, t## are still coordinates of a single event, while ##\Delta x, \Delta t## are the differences in coordinates between two events. By subtracting ##0## to make ##x, t## into ##\Delta x, \Delta t##, you are simply defining the second event to be the event at the origin of coordinates.
     
  7. Mar 25, 2016 #6

    PeterDonis

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    For events that happen at the observer's spatial location, yes. But the observer's frame is not limited to describing only events that happen at the observer's spatial location.

    You only substitute ##0## for events that happen at the observer's spatial location. But what about events that do not happen at the observer's spatial location? They will have ##x## values that are not ##0##. The Lorentz transformation applies to them too.
     
  8. Mar 25, 2016 #7
    I'm very sorry, I don't quite understand what you mean by the observer's spatial location, can you please give me an example and how the lorentz transform equation would correspond to the values in the example?
     
  9. Mar 25, 2016 #8

    PeterDonis

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    In your example, you have Anna, who remains on Earth, and Bobby, who flies from Earth to Planet B in his rocket ship. Let's suppose that Planet B is at rest relative to Earth, and is 10 light-years away from Earth. And let's suppose that Bobby travels at 86.6% of the speed of light relative to Anna, Earth, and Planet B; we'll use units where the speed of light is 1, so Bobby's speed is 0.866 (you can think of this as light-years per year). And let's say that the time when Bobby leaves Earth is time zero according to both Anna's and Bobby's clocks. (We also assume that Planet B's clocks are synchronized with Earth's.)

    Then we have the following events, with their coordinates in Anna's frame:

    Event O: Bobby leaves Earth. Coordinates ##(x, t) = (0, 0)##.

    Event B: Bobby arrives at Planet B. Coordinates ##(x, t) = (10, 11.55)##.

    So clocks on Planet B will read 11.55 years when Bobby arrives there. This just means, of course, that relative to Anna, Earth, and Planet B, Bobby travels 10 light-years in ##10 / 0.866 = 11.55## years.

    Now we transform these two events to Bobby's frame. I picked Bobby's speed so that ##\gamma = 2##; thus, we have:

    Event O: ##(x', t') = \gamma \left( x - v t \right), \gamma \left( t - v x \right) = 2 \left( 0 - 0 \right), 2 \left( 0 - 0 \right) = (0, 0)##.

    Event B: ##(x', t') = \gamma \left( x - v t \right), \gamma \left( t - v x \right) = 2 \left( 10 - 0.866 * 11.55 \right), 2 \left( 11.55 - 0.866 * 10 \right) = (0, 5.77)##.

    Notice that the ##x'## coordinate of Event B is zero; this is what we expect, since Bobby's ##x'## coordinate should always be zero and Bobby is present at Event B. Another way of saying this is that Event B happens at Bobby's spatial location; this is what I meant by "spatial location" in my previous post.

    Notice also that the ##t'## coordinate of Event B is half of the ##t## coordinate of Event B. This is also what we expect, since Bobby's clock is time dilated by a factor of ##\gamma = 2## compared to Anna's, Earth's, and Planet B's clock.

    However, there is a key difference between Bobby and Anna (and Earth and Planet B) with respect to these two events: only Bobby is present at both events, i.e., Bobby is the only one in the scenario for whom both events happen at his spatial location. And that means that, when we interpreted the ##t'## coordinate of Event B as telling us that Bobby's clock is time dilated by a factor of 2 relative to Anna's clock (or Earth's clock or Planet B's clock), we had to make an additional assumption: we had to assume that the event at which Planet B's clock reads zero is simultaneous with the event at which Anna's (and Earth's) clock reads zero.

    In other words, we have implicitly identified a third event, which we'll call event P:

    Event P: the event at which Planet B's clock reads zero. Coordinates ##(x, t) = (10, 0)##.

    We then say that this event happened "at the same time" as Event O, the event where Bobby left Earth; and since 11.55 years elapse on Planet B's clock between this event and Event B, where Bobby arrives at Planet B, we say that Bobby's trip took 11.55 years according to Planet B's clock. But Bobby's clock only reads half that, 5.77 years, when he arrives, so Bobby's clock must be time dilated by a factor of 2 relative to Planet B's clock.

    But there's more, because now we have identified a third event, and we need to look at its coordinates in Bobby's frame. They are:

    Event P: ##(x', t') = \gamma \left( x - v t \right), \gamma \left( t - v x \right) = 2 \left( 10 - 0 \right), 2 \left( 0 - 0.866 * 10 \right) = (20, - 17.32)##.

    Put aside the ##x'## coordinate for now and just look at the ##t'## coordinate. It is negative! What does that mean? It means that, in Bobby's frame, Planet B's clock is out of sync with Anna's and Earth's clock. In fact, Planet B's clock is 17.32 years ahead of Anna's and Earth's clock--it must be, because it reads zero at Event P, when, according to Bobby's frame, clocks synchronized with Anna's and Earth's clocks ought to read minus 17.32 years. So from Bobby's point of view, the reason Planet B's clock is ahead of his when he arrives at Planet B (it reads plus 11.55 years, when only 5.77 years elapse between Event O and Event B on Bobby's clock) is that Planet B's clock started out ahead. (This is a manifestation of the relativity of simultaneity.)

    Now go back and look at the ##x'## coordinate of Event P. It is 20 light-years. And in Bobby's frame, he stays motionless at ##x' = 0## while Planet B moves towards him at 0.866 light-years per year. So Planet B has to cover 20 light-years, and it takes 17.32 + 5.77 = 23.1 = 20 / 0.866 years to do it. Everything works out as it should. But we have only used the Lorentz transformation so far; in a follow-up post I'll look at how length contraction and time dilation are manifested in this scenario.
     
  10. Mar 25, 2016 #9
    So, if I was to use length contraction and time dilation as I normally would, in Anna's reference frame, she perceives herself to be stationary and measures the distance between Earth and planet B to be 10 light-years. If I was to use the length contraction formula ##L=\frac{L_0}{\gamma}## if would get:
    $$L = \frac{10}{2}$$ $$L = 5$$ Which means that Bobby would measure the length between the two planets as 5 light years as opposed to 10 light years. Using the time dilation formula:
    $$\Delta t' = \gamma \Delta t$$ $$11.55 = 2\Delta t$$ $$\Delta t = 5.77$$ I'm not sure if I used these formulas in the right way. I was inclined to multiplying the Lorentz factor by the ##\Delta t##, which is 11.55, to get the time experienced by bobby which would have been 23.1 light years which wouldn't have made sense. I doubt very much this is the way to do it. Where did I go wrong, and how would you do it?
     
  11. Mar 25, 2016 #10

    Dale

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    In all circumstances. I always use the Lorentz transform, never time dilation and very rarely length contraction. The time dilation and length contraction formulas automatically fall out of the Lorentz transform whenever appropriate, and using the Lorentz transform helps you avoid misapplying the simplified formulas to scenarios where they don't work. Also it resolves relativity of simultaneity issues which are usually neglected in most paradoxes.

    I always recommend the Lorentz transform over length contraction and time dilation formulas.
     
  12. Mar 25, 2016 #11

    PeterDonis

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    Yes, but you have to be careful to take relativity of simultaneity into account. Remember that, to Bobby, the "distance" between the two planets has to be measured at the same time. For example, we know that when Bobby's clock reads zero, he is spatially co-located with Earth (he is just about to leave). So if we knew where Planet B was when Bobby's clock reads zero, i.e., relative to Bobby's frame when ##t' = 0##, then we would know the distance between the planets in Bobby's frame.

    How can we figure out where Planet B is when Bobby's clock reads zero? Well, we know that at Event P, which I defined in my last post, Planet B is at ##x' = 20##, and that event happens at ##t' = - 17.32## in Bobby's frame. So to figure out Planet B's ##x'## coordinate when ##t' = 0##, we just need to subtract ##17.32 * 0.866 = 15## from ##20##, to get--##5##! So it works out; the distance between the planets, in Bobby's frame, *is* 5 light-years, which is length contracted by a factor of 2 from the 10 light-years distance in Anna's/Earth's/Planet B's frame.

    But note what we had to do to figure this out: we had to find an event, call it Event Q, which is at ##t' = 0## in Bobby's frame and is at Planet B's spatial location at that time in Bobby's frame. And we figured out that Event Q has coordinates ##(x', t') = (5, 0)##. But now suppose we transform these coordinates back to the unprimed (Anna/Earth/Planet B) frame. This transformation is the inverse of the one we used before, i.e., ##x = \gamma \left( x' + v t' \right)##, ##t = \gamma \left( t' + v x' \right)##. So we have:

    Event Q: ##(x, t) = \gamma \left( x' + v t' \right), \gamma \left( t' + v x' \right) = 2 \left( 5 + 0 \right), 2 \left( 0 + 0.866 * 5 \right) = (10, 8.66)##.

    So we see that ##x = 10## for Event Q, which is what we expect since Planet B is present at this event. But we also see that ##t = 8.66##, which means that Planet B's clock reads 8.66 years at this event. Again, this is a manifestation of relativity of simultaneity; the two events that Bobby uses to determine the distance from Earth to Planet B, in his frame, are not simultaneous according to Earth/Planet B clocks, since one happens when those clocks read zero and the other happens when they read 8.66 years. This is why you can't just use length contraction by itself; you have to take relativity of simultaneity into account as well, or you will end up making mistakes.

    The time experienced by Bobby between what two events? Remember that the 5.77 years is the time experienced by Bobby between Event O, where he leaves Earth, and Event B, where he arrives at Planet B. We obtained the 5.77 years by Lorentz transforming the coordinates of those two events from the Earth frame to Bobby's frame, and then taking the coordinate differences in Bobby's frame. This was made easy by the fact that Event O is the origin in both frames, so its coordinates are ##(0, 0)## in both frames. But conceptually we still performed two separate operations: we transformed the coordinates of each event separately, and then took differences. You can't do it the other way around, or you will make mistakes.

    Suppose, however, that we wanted the answer to a different question: how much time elapses, on Bobby's clock, between the event where Planet B's clock reads 10 years earlier than Event B, and Event B? In other words, suppose we define Event R as the event which has coordinates ##(10, 1.55)## in the Earth/Planet B frame; note that ##1.55## is 10 years earlier than the ##11.55## years that Planet B's clock reads when Bobby arrives there at Event B.

    We can easily Lorentz transform to find Event R's coordinates in Bobby's frame:

    Event R: ##(x', t') = \gamma \left( x - v t \right), \gamma \left( t - v x \right) = 2 \left( 10 - 0.866 * 1.55 \right), 2 \left( 1.55 - 0.866 * 10 \right) = (17.32, -14.23)##.

    Now we just take the difference in ##t'## between Event B and Event R: this is ##5.77 - (-14.23) = 20##. So 20 years elapses in Bobby's frame between these two events, which is 10 years (the elapsed time in the Earth/Planet B frame) times the time dilation factor of 2. But Event R does not happen at Bobby's spatial location, so this 20 years of elapsed time assumes the simultaneity convention of Bobby's frame--i.e., it assumes that Event R took place "at the same time" as Bobby's clock read ##-14.23## years--in other words, according to Bobby, Event R happened 14.23 years before he passed by Earth. (Here we assume that Bobby's speed is constant, i.e., that he was not at rest on Earth before he left it, but just passed by it in the course of traveling forever at a constant speed.) So once again, when applying the time dilation factor, you have to also take into account relativity of simultaneity, or you will make mistakes.
     
  13. Mar 26, 2016 #12
    I'm getting a little lost in all of the different events. So on a test, they would generally ask one question, such as, what is the time Booby takes to get to planet B that Anna observes? Or, what is the time and distance that Booby observes to get to planet B? etc. I'd imagine you would not need to use events O, B, P, Q, R for the one question. What are some possible questions and the events that you would need to identify to answer those questions?

    What I do understand is, the Lorentz transformations are used in all events and that time dilation and length contraction are just special cases of the Lorentz transform. I also understand that it is safer to use the Lorentz transformations rather than time dilation or length contraction. In addition, I also understand that the primed variables are assigned to the entity that is present at two or more different events and the non-primed variables are assigned to the entity at only one event. Am I right in my understanding?
     
  14. Mar 26, 2016 #13

    PeterDonis

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    For the first question you give, what time does Bobby take in Anna's frame to go from Earth to Planet B, you only need events O and B, and you don't need to do any Lorentz transformation because you just need the coordinates of the events in Anna's frame, which are already given in the problem. So the first question is trivial.

    For the second question--what time elapses on Bobby's clock between Earth and Planet B and what distance does Planet B travel in Bobby's frame (note that in Bobby's frame it is Planet B that moves, not Bobby)--you need events O and B, but you also need event Q, because you need to know where Planet B is, in Bobby's frame, when Bobby leaves Earth (or more properly when Earth leaves Bobby).

    That's the safest way to proceed, yes.

    Not really; they are derived quantities that you can calculate once you have done the Lorentz transforms, but you can't calculate them by just plugging in deltas to the Lorentz transform formulas. You have to Lorentz transform the coordinates of events, and then calculate the deltas (the differences between coordinates of events in the new frame).

    No. First, unprimed and primed variables simply refer to the coordinates in different frames; which observers, if any, are at rest in which frame depends on the problem. Second, there may be multiple entities (objects, observers) that are present at more than one event of interest. (For example, in the second question above, Bobby, who is at rest in the primed frame, is present at events O and B, and Planet B, which is at rest in the unprimed frame, is present at events Q and B.)
     
  15. Mar 26, 2016 #14
    What question would you pose to use events P, or R?
     
  16. Mar 26, 2016 #15
    I thought that planet B's clock reads 11.55 years when Bobby lands on it. What is 8.66 years in relation to the question?
     
  17. Mar 27, 2016 #16
    So, if you didn't want to use Length contraction or time dilation, you would need to use event Q and R, am I right? What other events would you not need to answer the question with purely the Lorentz transformations?
     
  18. Mar 27, 2016 #17

    PeterDonis

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    You can always answer any SR question by picking out appropriate events and using Lorentz transformations. You never need to use length contraction or time dilation--unless the question itself is asking you to compute length contraction and time dilation, but even then the safest way is to pick out appropriate events and use Lorentz transformations.
     
  19. Mar 27, 2016 #18

    PeterDonis

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    8.66 years is what the clock on Planet B reads when Bobby leaves Earth (or Earth leaves Bobby), according to Bobby's frame.

    This can be used to show that Planet B's clock is time dilated, relative to Bobby; just subtract 8.66 from 11.55, the reading on Planet B's clock when Planet B arrives at Bobby (remember that in Bobby's frame, Bobby is at rest and Planet B moves). You will get 2.89, which is half of 5.77 (approximately--a more accurate calculation using exact numbers will show that it is exactly half); in other words, in Bobby's frame, the time elapsed on Planet B's clock during the trip is half of the time elapsed on Bobby's clock.
     
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