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Are proper travel distances contracted by SR?

  1. May 25, 2009 #1
    Imagine a hypothetical large region of empty, non-expanding space whose spacetime is flat. At the start of the scenario, a rocket sits stationary relative to our galaxy at a large proper distance "x", in the common inertial frame. Then at t=0 the rocket quickly accelerates to a constant .99c velocity, radially away from our galaxy, for a time interval t=100, at which point the rocket streaks by the (stationary) proper distance "y". Both of the (stationary) proper distances "x" and "y" have been previously measured by travellers from our galaxy, who left massless flags behind to mark the distances. Marker "y" emits a light burst when it observes the rocket passing by very close to the marker. Time and simultaneity are measured in our galaxy's inertial frame. I have two questions:

    1. Is the proper distance traveled by the rocket, from marker x at t=0 to marker y at t=100, Lorentz contracted as measured in our galaxy's frame?

    2. During the time period between t=0 and t=1, is the proper distance from our galaxy to marker x Lorentz contracted as measured in our galaxy's frame?

    I don't want to talk about the Lorentz contraction of the rocket itself. Nor about time dilation or simultaneity in the rocket's frame.

    I think the answer to both my questions above is no. I don't see how the passage of an object moving at relativistic speeds could cause Lorentz contraction of proper distances between markers which themselves are at rest in the observer's frame.
  2. jcsd
  3. May 25, 2009 #2


    Staff: Mentor

    Proper distance, like proper time, is frame invariant.
  4. May 25, 2009 #3


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    That isn't a distance; that is a duration. The (straight line) separation between those events is timelike, not spacelike -- it's measured in seconds, not meters.

    More generally,

    what you are describing is not a proper distance; it's essentially the coordinate distance according to the galaxy's "inertial frame". Proper distance is a property of a specific pair of (spacelike-separated) points in space-time, or more generally, of a path through spacetime.

    However, massless flags don't identify points -- they identify lines (called "worldlines"). They don't exist for a single instant of time, but instead they exist for all times.

    In order to compute a distance, you have to choose specific points to compare -- a typical convention is to take whatever coordinate chart you're using, and choose a specific coordinate time, and use the points on the flag's worldlines that occur at that specific coordinate time. Now that we've selected points, we can compute a distance; we call this the "coordinate distance".
  5. May 25, 2009 #4
    Since Einstein arrived a conclusion of cosmological expansion from his equations (albeit GR) which he then tried to compensate for with a cosmological constant, until Hubble released his results showing that the cosmological expansion could be observed, I would think that assuming non-expanding space is akin to assuming that the speed of light in vacuo is not a constant.

    Expanding space is a feature of relativity, I would think that if you remove it, you are no longer talking about relativity.



    PS Additionally, you can't explain the problem presented without taking into account the whole scenario, so not wanting to talk about time dilation makes it a tad difficult.
  6. May 26, 2009 #5
    OK, let me simplify the scenario. I don't want to swing at an SR flea with a Minkowski sledgehammer here, and I think the answer should be straightforward.

    Let's say Alice and Bob are both at rest in an inertial frame slightly more than 1 ly apart, and a stationary rigid 1 ly long tape measure stretches from Bob most of the way to Alice. Alice then begins accelerating rapidly directly toward Bob, such that by the time she passes close by the end of the tape measure (putting her exactly 1 ly away from Bob, as Bob reckons it), she achieves a constant velocity toward Bob which remains at .99c until she passes close by Bob. Alice emits a light signal when she passes the 1 ly end of the tape measure and again when she passes Bob. Alice subsequently decelerates and joins up with Bob, where they compare (a) the elapsed time on Bob's clock between the receipt of Alice's two light signals, and (b) the elapsed time on Alice's clock between her emission of the two light signals.

    After subtracting the light travel time, Bob calculates that Alice's trip along the length of the tape measure required just over 1 year on his clock, c / .99. So Bob concludes that, unsurprisingly, Alice's motion didn't cause the length of the tape measure to become Lorentz contracted in his frame.

    Alice measures a shorter elapsed time (between the emission of the two pulses) on her clock than Bob calculated, which she knows is because her clock was time dilated relative to Bob's clock.

    Alice also expects that the tape measure was Lorentz contracted in her frame. Does that mean that the elapsed time shown on her clock was shortened (relative to Bob's measurement) by a combined factor of (time dilation * tape measure Lorentz contraction)?
  7. May 26, 2009 #6
    No, she does not "know" that. There is no such thing as absolute rest. Bob is not at absolute rest. Alice's frame is exactly as good a rest frame as Bob's is. That is the entire point of Relativity. There is no absolute sense in which either clock is slower than the other.

    In Alice's reference frame, it is Bob's clock, NOT her own, that is time dilated. Alice agrees, however, when she arrives at Bob that the total time on her clock is less than Bob's because the length of the tape measure was much shorter than 1 light-year due to length contraction.
  8. May 26, 2009 #7
    I'm not sure about that, because Alice is the one who undergoes deceleration before rejoining Bob to compare clocks. So in that way it's like the 2nd leg of the twins paradox.

    Then your saying that the effect of the length contraction reduces the elapsed time on Alice's clock by more than the time dilation increases it? How much more?
  9. May 26, 2009 #8
    She need not decelerate at all. They can compare clocks as they pass each other. If she does decelerate, it makes no significant difference because she is right next to Bob when she does it.

    In her frame, her own clock is NOT time dilated. Her clock reads a small amount because she traveled hardly any distance. Bob's clock reads a large amount because he doesn't set it correctly due to simultaneity failure. In other words, in Alice's frame, his clock is not properly synchronized with the clock on the end of the tape measure where the experiment began. He missets his clock, according to Alice, because when he receives Alice's light pulse that she sent off in the beginning, he incorrectly sets his clock to read 1 year, when the distance that the light pulse traveled was nowhere near that large.
  10. May 26, 2009 #9


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    Here's the trick: due to the Relativity of Simultaneity, according to Alice, Bob's clock already reads a time of almost a year the instant she passes the end of tape.

    Assume there is a clock at the end of the tape that is synchronized to Bob's clock in Bob's frame.

    In Bob's frame, Alice passes the end of the tape when all the clocks read zero, and takes ~1 year to travel the length of the tape, while only aging ~1/7 of a year(~52 days) due to time dilation. Thus her clock reads ~52 days upon reaching Bob and Bob's reads ~ 1 year.

    In Alice's frame, she is even with the end of the tape when that clock and her clock reads zero, but at that instant, Bob's clock reads ~ 7.5 days short of a year due the Relativity of Simultaneity.
    The length of the tape has been length contracted to 1/7 of a light year in her frame, so it only takes ~52 days by Alice's clock from passing the end of the tape to reaching Bob. During which time, ~7.5 days pass on Bob's clock due to time dilation. Adding this elapsed time to the time that was already on Bob's clock when she passed the end of the tape, gives a reading of ~1 year on Bob's clock when they meet compared to the ~52 days on hers. This is the same end result bob gets in his frame.
  11. May 26, 2009 #10
    So if Bob follows your advice and subtracts only the Lorentz contracted light travel distance from the elapsed time on his clock between signals, Bob and Alice will exactly agree on the elapsed time, and both will agree that in Bob's frame, Alice traveled 1 LY in a tiny fraction of 1 Bob-year. We have achieved superluminal travel simply by causing the separation between the start and finish line to conveniently contract temporarily while we step over it (!)
  12. May 26, 2009 #11
    I didn't give Bob any advice. He could subtract the contracted distance if he likes (he can also doodle, write a poem, or whatever else, and it would be exactly as helpful), but that would miss the point of synchronizing his two clocks: the one at the end of the measuring tape and his own. If he subtracts anything other than the measuring tape distance, then his two clocks aren't synchronized in his own frame. His calculation will be completely meaningless, since the calculation is distance/("time when Alice arrives" - "time when race began"). In order to make that calculation, the clock that measures the race beginning must be synched with the clock that measures her arrival. That is the entire point of sending your light flash, right?

    When Bob receives the synch signal, he MUST reset his clock to 1 year, because in his reference frame, and in the reference frame of the measuring tape, that is the distance that the light ray traveled. (Let's also say that it's an easy enough matter to add a notation to his logbook about when he did that, so that we can talk about times before he received the synch signal, as in, "Alice crossed the starting line when Bob's clock read zero.")

    There is NO WAY that Bob can synchronize his two clocks so that they are also synchronized in Alice's frame. It is a fundamental property of spacetime, just like length contraction and time dilation. Clocks that are synched in one reference frame can NEVER be synched in anyone else's. In Bob's reference frame, Alice started the race when both his clocks read zero. Alice, however, disagrees. In Alice's frame, the clock at the starting line read zero when she got there (after all, it set itself to zero when she arrived), but at that same moment, Bob's clock already read just under 1 year (in particular, it reads 1 year * v/c). So during the trip, Bob's clock is running slow, but it was set way, way too far ahead to begin with. When she arrives at Bob, of course Bob's clock reads more than a year. He set it wrong. But there was no other choice. Bob CANNOT set his clock to satisfy Alice. It is not that Bob was stupid. It is that two clocks, in synch according to Bob, read differently according to Alice.

    Alice and Bob COMPLETELY DISAGREE on which spacetime events constitute "now." To Bob, his clock reading zero and the starting line clock reading zero happen at the same moment. To Alice, those events occur at completely different moments.
    Last edited: May 26, 2009
  13. May 26, 2009 #12


    Staff: Mentor

    Here a and b are both proper times, not proper distances. As I mentioned above, all frames agree on proper times.
    a = sqrt(0.01²-0²) = 0.01 years
    b = sqrt(1.01²-1²) = 0.14 years
  14. May 26, 2009 #13
    Is it perhaps worthwhile to make sure everyone knows what proper time, proper length, coordinate time and coordinate length are?

    http://en.wikipedia.org/wiki/Proper_time" [Broken] - In relativity, proper time is time measured by a single clock between events that occur at the same place as the clock.

    http://en.wikipedia.org/wiki/Proper_distance" [Broken] - In relativistic physics, proper length is an invariant quantity which is the rod distance between spacelike-separated events in a frame of reference in which the events are simultaneous.

    http://en.wikipedia.org/wiki/Coordinate_time" [Broken] - In the theory of relativity, it is convenient to express results in terms of a spacetime coordinate system relative to an implied observer. An event is specified by one time coordinate and three spatial coordinates. The time measured by the time coordinate is referred to as coordinate time to distinguish it from proper time.

    Coordinate distance is not described on wikipedia but we can extrapolate thus - In the theory of relativity, it is convenient to express results in terms of a spacetime coordinate system relative to an implied observer. An event is specified by one time coordinate and three spatial coordinates. The distance measured by the spatial coordinates can be referred to as coordinate distance to distinguish it from proper distance.

    Interpreting all of those is the (relatively) difficult part.

    Proper time - say you have an inertial clock, elapsed time on that clock is proper time. (t')

    Proper length (distance) - say you have an inertial rod, the rest length of the rod (ie where the ends of the rod are simultaneous) is proper length (or proper distance between the ends of the rod). (L)

    Coordinate time - we have an implied observer, the time on the observer's clock when events take place is coordinate time. (t)

    Coordinate distance - we have an implied observer, the distance between the observer and an event is coordinate distance. A coordinate length would be the delta between two events, for a rod that would mean the ends of that rod. (L')

    The relationship between coordinate time and proper time is given by:


    The relationship between coordinate length and proper length is given by:


    These equations might be familiar :smile:


    Last edited by a moderator: May 4, 2017
  15. May 27, 2009 #14
    No, Alice's clock will show an elapsed time of ~1/7 yrs, since the tape measure was ~1/7 light yrs long in her frame. The "length contracted" distance divided by Alice's elapsed time will equal .99c. The two factors don't combine, they're complementary to result in an invariant light speed.
  16. May 27, 2009 #15
    OK, thanks everyone for the helpful answers. Which I should have known, but it's easy to confuse oneself on this. The mistake always seems to be in trying to create a hybrid of two different reference frames, rather than maintaining the discipline of examining results in each frame separately.
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