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Falling off a building

  1. Feb 1, 2016 #1
    In the spirit of Einsteins happiest moment, I'm working on a new theory of gravity and would like to test a prediction of mine against the experts here at PF. We know that Bob standing on top of the empire state building close to Earth's gravity is aging slower than Major Tom floating out at some random Lagrange point, but...

    What if Bob jumps off the building? On his way down in free fall, my prediction is that Bob would be aging at the maximum rate possible, maybe even faster than major Tom. Am I right?
     
  2. jcsd
  3. Feb 1, 2016 #2

    Someone in free fall would be in an inertial state, so it's as if they are far from any gravitational bodies that influence their passage of time. Folks at the top and bottom of the building are in accelerated states. I wouldn't say Bob would be aging at the *maximum rate* possible, because it's relativity after all. But I think you have the right idea. Hate to burst your bubble though, but this isn't *new*.
     
  4. Feb 1, 2016 #3

    bcrowell

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    Your first two examples are static observers, so it makes sense to compare their time dilations. (E.g., in mountain-valley experiments with atomic clocks, we can make sure that the driving time is negligible compared to the time the clock spends on the mountain.) But the free-falling observer is not static, and there is no meaningful way to compare his time dilation with the static ones.

    What you could do is to compare two non-static observers, both of whom start at event A and both of whom end up at event B. A common way of axiomatizing SR is to say that in this situation, a free-falling observer has maximal proper time.
     
  5. Feb 1, 2016 #4

    A.T.

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    So he ages at the maximum rate possible, but only maybe faster than major Tom?
     
  6. Feb 1, 2016 #5

    PeterDonis

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    You might want to first try a warmup exercise. Bob is standing next to Charlie on top of the building. Alice is standing at the bottom of the building. Alice is aging at a slightly slower rate than Bob and Charlie, because of her lower altitude. Everyone's clocks are set so that they all read zero at some instant. (Note that we have implicitly adopted a simultaneity convention--the one that is natural for observers who are static, i.e., at rest relative to the Earth, which is the source of the gravitational field.)

    Now, at the instant when everyone's clocks read zero, Bob jumps off the building. When he reaches the bottom and meets Alice (we assume he has some way of stopping himself at the bottom without injury), Charlie's clock reading will be slightly greater than Alice's (because he is aging slightly faster--note, once again, that we are implicitly adopting a simultaneity convention to make this comparison). What about Bob's? How will it compare to Alice's? To Charlie's?
     
    Last edited: Feb 1, 2016
  7. Feb 1, 2016 #6

    Janus

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    If you mean that he will start aging faster at the moment he leaves the top of the building, then no.

    GR predicts that the combination of velocity and gravity results in a time dilation factor of

    [tex]\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2}- \left ( \frac{rc^2}{2GM}-1 \right )^{-1} \frac{v_{\|}^2}{c^2}}[/tex]

    where [itex]v_{\|}[/itex] is the radial velocity.

    Since at the moment Bob leaves the building his velocity is zero, this reduces to the gravitational time dilation formula alone. Meaning that Bob, at the moment he starts his free fall is aging at exactly the same rate as he was when standing on the building (at the same altitude) and at the instant he reaches the ground, he will be aging slower. ( combination of lower gravitational potential and relative motion to when he was standing on the building).
     
  8. Feb 1, 2016 #7

    PeterDonis

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    Why is the extra term in the radial velocity there? The ##v^2 / c^2## term already captures the effects of velocity.
     
  9. Feb 1, 2016 #8

    bcrowell

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    Re the formula posted in Janus's #6, I would claim, for the reasons given in #3, that it's not even well defined what one is being asked to calculate in this situation, so giving a formula without specifying a definition doesn't mean much.
     
  10. Feb 1, 2016 #9

    PeterDonis

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    I was assuming that he was giving a formula for ##d\tau / dt##, where ##\tau## is proper time along a worldline that, at some event, has radial coordinate ##r## and coordinate velocity ##v##, and ##t## is coordinate time, "coordinate" meaning Schwarzschild exterior coordinates. But you're right, this should be stated explicitly and not assumed.
     
  11. Feb 1, 2016 #10

    bcrowell

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    Sure, that's one possible definition, but it's a coordinate-dependent one. There's no obvious reason (to me) why we'd care about the Schwarzschild ##t## coordinate in particular.
     
  12. Feb 1, 2016 #11

    PeterDonis

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    Because it's the natural coordinate we are led to by the timelike Killing vector field of Schwarzschild spacetime: the metric is independent of ##t## in these coordinates, and it is also diagonal, meaning that surfaces of constant ##t## are orthogonal to the integral curves of the KVF, so they are natural surfaces of simultaneity for static observers (whose worldlines are those integral curves). The normalization of ##t## is given by the normalization of the KVF; i.e., the "gravitational time dilation factor" goes to 1 at infinity.
     
  13. Feb 1, 2016 #12

    Dale

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    Closed pending moderation
     
  14. Feb 1, 2016 #13

    PeterDonis

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    The actual scenario under discussion (unlike the brief mention of a personal theory in the OP) is within bounds for PF, so the thread is reopened.
     
  15. Feb 2, 2016 #14
    Yeah, I was being a little tongue and cheek with the "new theory of gravity" comment, sorry. I was just playing off of Einstein's revelation with the "painter falling off the roof" legend. I know better than to wax any personal theories here on PF. That said, I was trying to "channel" Albert's "Gedankenexperiment" after watching the infamous water-pitcher demo of Brain Greene:



    That is an interesting Gedankenexperiment in itself right there. My guess is that, at the instant he hits the ground next to Alice, Bob's clock will read the fastest time of the three. Charlie's clock will read a bit slower than Bob's, and Alice's will read the slowest. My reasoning for this is that, during the epoch of Bob's free-fall, he was not subject to any influence of a gravitational potential, insofar as the definition for it that I know of. I'm assuming here that the time dilation effect of being in a gravitational potential means that you (i.e., Alice, Bob, and Charlie) actually have to be experiencing the acceleration effect of the potential. Maybe I'm wrong. If I'm not, then in the epoch from where Bob jumps off the roof until he hits the ground, he is not undergoing the effects of time dilation as is Alice and Charlie, who continue to experience acceleration due to the gravitational potential through the epoch.

    The above discussion brings up a related and equally interesting question which I'm guessing there's already a "stock" answer to. What if we took Bob, put him in an elevator, and accelerated him at a steady "one g" through deep intergalactic space free from any gravitational potential or tidal forces. Would he age at the same rate as he would standing at the bottom of the Empire State Building? In other words, does the equivalence principle apply here? Do tidal forces make a difference? More generally, is it the acceleration aspect of the gravitational potential that accounts for the time dilation, or is it some other aspect or some combination thereof?
     
  16. Feb 2, 2016 #15

    PeterDonis

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    Bad guess. To see the correct answer, consider the following:

    Suppose that, instead of jumping off the building from a standing start, Bob had started off on the ground with Alice, and had jumped vertically upward just fast enough to come to rest for an instant at the top of the building next to Charlie, and then had just free-fallen back down to the ground next to Alice. For this case, it is easily seen that Bob's clock has more elapsed time than Alice's, because the two of them are spatially co-located at the start and end of Bob's jump, and Bob is in free fall, so Bob's proper time must be the longest, since his worldline, as a geodesic, must have maximal proper time between those two events. (Note that there are actually complications to this in curved spacetime, but we can ignore them for this scenario.)

    However, the same reasoning does not hold when comparing Bob's proper time to Charlie's, because Bob and Charlie are only spatially co-located once, not twice (at the event where Bob reaches the top of the building and comes to rest for an instant). We could imagine Bob jumping off the building, bouncing when he hits the ground, and flying back up to the top of the building in free fall, so that he and Charlie would start and end spatially co-located; but then Bob's worldline will not be geodesic everywhere--he undergoes a large acceleration at the bounce. So we have to use another line of reasoning.

    The correct line of reasoning to compare Bob and Charlie is as follows: Bob's altitude is never greater than Charlie's, and is almost always lower; also, Bob's velocity relative to Earth is never less than Charlie's (since Charlie is at rest relative to Earth) and is almost always larger. Both of these effects make Bob's "rate of time flow" slower than Charlie's, so we can safely conclude that Bob's elapsed proper time is less than Charlie's. So the correct ordering of proper times, from shortest to longest, is Alice -> Bob -> Charlie.

    Of course, the above reasoning is verbal, not mathematical; but the math is straightforward and confirms what I said above. Your original scenario is just the second half of the above one, so the same reasoning applies to it as well.

    Yes, you're wrong, as you can see from the above. Here is how "gravitational potential" works.

    First of all, the concept of "gravitational potential" only makes sense in a stationary spacetime. The technical definition of "stationary" is "has a timelike Killing vector field"; but another way of stating it, which is easier to conceptualize, is that a stationary spacetime has a family of observers who all see an unchanging spacetime geometry along their worldlines. The geometry can differ from one observer to another, but each observer never sees any change. We then say that each member of the family of stationary observers is "at rest" in the stationary spacetime, and therefore each one marks out a unique "position in space" in an invariant sense. In the scenario under discussion, the stationary observers are the ones who are "hovering" at a constant altitude above the Earth and constant angular coordinates. So Alice and Charlie are stationary, but Bob is not.

    The "gravitational potential" in a stationary spacetime is a function of spatial position, and can be interpreted physically as a "gravitational time dilation factor" associated with that position, relative to some "standard" position. In the scenario under discussion, the "standard" position is at spatial infinity--that is where the gravitational time dilation factor is 1. (Because of the spherical symmetry, only the radial coordinate ##r## actually affects the value of the time dilation factor in this scenario; in more general stationary spacetimes, all three spatial coordinates can affect it.) For a stationary observer, the gravitational time dilation factor is the only one involved, so comparing their rates of time flow is simple if you know their positions.

    For moving observers, the time dilation factor is a combination of two things: the gravitational time dilation factor at the position they are currently at, and the rate at which their position is changing; the latter can be thought of, conceptually, as working like time dilation due to motion does in SR, but only with respect to stationary observers--the faster an observer is moving relative to stationary observers, the slower his rate of time flow becomes, over and above the gravitational time dilation effect due to his position (which itself can be changing as he moves). There are technical points here to do with how we define "rate", which are important when we look at the details of the math; but for the simple reasoning I gave above, just knowing that an observer like Bob is moving relative to stationary observers is enough.

    There's no meaningful way to make such a comparison. In order to compare the "rate of time flow" of two observers, you have to have one of two things:

    (1) The observers are spatially co-located at the start and end of some process, so the elapsed times on their clocks can be directly compared. For example, in the standard twin paradox, the twins are spatially co-located at the start and end of the traveling twin's trip. In the scenario described above, Bob and Alice are spatially co-located at the start and end of Bob's jump.

    (2) There is some way to determine "corresponding" events on the two observer's worldlines, so the elapsed times on their clocks between successive "corresponding" events can be compared. In general, this will be a matter of convention (simultaneity convention), but in some cases there are reasonable invariants that can be used to pick out a "natural" simultaneity convention to use. For example, in the scenario described above, there is a natural simultaneity convention associated with the "stationary" observers, Alice and Charlie, and we can use that to pick out events on Charlie's worldline that "correspond" to the events where Alice and Bob are spatially co-located, at the start and end of Bob's jump.

    In the case you're proposing now, however, neither of the above conditions hold, so no meaningful comparison can be made.
     
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