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The ratio between temporal lengths is frame-invariant

  1. May 7, 2013 #1
    Hi,
    I am a researcher in (analytical) philosophy. I am writing a paper on the conventionalism vs. objectvism about temporal metric, which is focused on a problem that arises before any relativistic consideration. However, in a note, I just wrote the following "the ratio between temporal lengths is frame-invariant". I tried to demonstrate it from the assumption that the space-time interval (square of dS^2 - dT^2) is frame invariant, but I got stuck :-)

    Could you confirm (or reject) my claim?
    Btw - testimony by authority would suffice

    Thanks a lot!
     
  2. jcsd
  3. May 7, 2013 #2

    ghwellsjr

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    A "ratio between temporal lengths" must include two temporal lengths but a space-time interval is just one length between two events, so I'm not sure what you have in mind. If you're just trying to make a comment about the length of a temporal space-time interval between two events and how it's frame invariant, the easiest way to express it is to say that it can be measured with a real inertial (non-accelerating) clock that is present at both events. This statement doesn't even invoke any frames so it is frame invariant. It doesn't even invoke any requirement for synchronization of clocks, since there is just one.

    If you want a testimony by authority, this is the theme of the first chapter of the second edition of Taylor and Wheeler's Spacetime Physics.
     
  4. May 7, 2013 #3
    Thanks a lot ghwellsjr. I try to reformulate my question, because I am afraid I did not express myself clearly. Suppose you measure with a clock the lengths of two non instantaneous events and calculate the ratio between the results that you have obtained. As far as I understand, if you transform the results of your two measurement from the frame you are in to some other inertial frame, in STR you will obtain different temporal lengths (although if the relative velocity between the frames is small, the differences is negligible). My question is: will the ratio between the two temporal lengths stay the same? Again, as far as I understand the temporal interval between events, namely the root of the ration between the square of the spatial distance between them minus the square of the temporal distance between them, stays the same when you shift from reference frame to reference frame. Will also the ratio between the temporal lengths of the two non-instantaneous events? I hope what I am saying makes some sense :) Thanks anyway!
     
  5. May 7, 2013 #4

    Nugatory

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    Events do not have lengths; an event is a single point in spacetime identified by four numbers, three to position it in space and one to position it in time. For example: At 45 degrees latitude, 20 degrees longitude, 5000 feel above sea level, at high noon GMT, I sneezed.

    So I expect that when you speak of "non-instantaneous" events and their "length" you're thinking of two events and the interval between them: for example, "the light leaves the source" and "the light arrives at my eye".

    If that's the case... The spacetime interval between any two events is the same for all observers, so the ratio between two such events will also be the same for all observers. It's what we call a "frame-independent" quantity. However, different observers moving at different velocities relative to one another or in interestingly different gravitational fields will in general not agree about the amount of time between the two events or the distance between the points in space at which they happened.
     
  6. May 7, 2013 #5

    ghwellsjr

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    If you are talking about a single clock at rest in an inertial frame making multiple measurements of different time intervals and then transforming those time intervals into another inertial frame moving with respect to the first one, then yes, those time intervals will maintain the same ratios in the second frame as they had in the first frame. In fact, they will all be multiplied by the gamma factor determined by the relative speed between the two frames.

    But keep in mind that this second set of time intervals cannot be measured by a single clock in the new frame so you can't just assume that you can apply the same process to get back to the measurements in the original frame. That would end up with a gamma squared result instead of getting back the same intervals that you started with.
     
  7. May 7, 2013 #6
    Thanks Nugatory,
    it is apparent that in the philosophical jargon and in the physics one, the term "event" has a different (although related) meaning. Anyway, yes —*with (temporal) length of an event(philosophical term) E, I mean the temporal distance between the event(physical term) e1, which can be seen as the beginning of E and the event(physical term) e2, which can be seen as the ending of E. Thus the temporal length of the event(philosophical term) of my "giving a long sip to my cup of tea" is the distance between two events(physical terms) its beginning and its ending. Now. Firstly, I made a mistake when I wrote "Again, as far as I understand the temporal interval between events", I wanted to say, as in my previous post "the space-time interval". Secondly, I am afraid I still haven't had a answer. I know that different observers moving at different velocities relative to one another will in general not agree about the amount of time between two events. What I wanted to know is the following. Suppose you have two such observers, O1 and O2, who measure the amount of time between two pairs of events, <e1, e2> and <e3, e4>. Let us call O1(e1, e2) the amount of time between e1 and e2 as measured by O1; O2(e1, e2) the amount of time between e1 and e2 as measured by O2, and so on. I know that in general O1(e1, e2) will be different from O2(e1, e2). My question is: what about the two ratio R1 = O1(e1, e2)/O1(e3, e4) and R2 = O2(e1, e2)/O2(e3, e4); will R1 and R2 be the same? (And of course in general for any observer / frame of reference)? Thanks a lot for your patience!
     
  8. May 7, 2013 #7

    PAllen

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    It seems that what you mean by, e.g. O1(e1, e2) , is what physicists call a coordinate time difference between the events. In that case, you supposition is simply false. You can have O1(e1,e2)=O1(e3,e4), while O2(e1,e2) ≠ O2(e3,e4). It would just be an accident for these ratios to be the same.

    To make sure I'm understanding your question, I'll give a trivial example. I am watching two rockets moving away from each other, and from me, at the same speed in opposite directions. I happen to see (and and also model using an inertial frame) that they both turn on a signal beacon for one second. However, each of the rockets will measure that their own beacon was turned on for less than a second, and for less time than the other rocket's beacon. What everyone agrees on is that each rocket's clock measures the same time for the beacon interval; but each rocket observer will measure the other the other rocket's beacon as being on longer than their own.
     
  9. May 7, 2013 #8

    DEvens

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    The jargon used in physics has been carefully chosen to be unambiguous. You could do worse than adopting it.

    Anyway, to answer your question: No, not in general.

    The only case it would be true would be if the e1 and e2 were points on the history of one particle, and e3 and e4 were points on the history of another particle, and the two particles were at rest relative to each other. Indeed, it's just another way of stating that observers will agree if two particles are moving relative to each other.

    Think of two spaceships that start together and move apart at a relative velocity of 0.994987 c. And let the intervals be from the point the ships were together, until each ship has registered 1 year of ship time. Ship A will say "we've had 1 year of ship time, but you have a gamma of 10, so can only have had 1/10 of a year." And ship B will say the same thing. So Ship A will say A/B is 10, and ship B will say A/B is 1/10. And if the two spaceships leave from the Earth, then an Earth observer thinks they each have the same time, so he says the ratio of A/B is 1.

    It's good that you are seeking invariants. Invariants are important aspects of relativity in particular and science in general. They are related to symmetry, which is one of the more powerful tools in scientific discussion. But invariants are usually derived from careful examination of the math. So the first thing you ought to study in this context is the Lorentz transform and the Lorentz group and the invariants that arise from it.
    Dan
     
  10. May 7, 2013 #9

    Dale

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    No, they will not be the same in general.
     
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