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Would the one accelerating please stand up?

  1. Feb 6, 2015 #1
    Ok. I'm trying to get this straight in my mind. I am familiar with Einstein's equivalence of acceleration and gravity. But what I am trying to figure out is: who exactly is accelerating? If someone is falling to the earth and I am looking at his pitiful circumstance, he is clearly - according to my observation and calculations - accumulating distance geometrically...and therefore accelerating. Oh, but wait a minute. He feels weightless. According to him - if he hasn't suffered an infarction already - all the subjective signs of one accelerating are missing: he does not feel any force pressing against him (we'll discount the atmosphere). However, I in my stationary position do feel a force against my feet. This leads me to conclude that I am the one accelerating and not him. After all, when I am in a car, I know when I am accelerating when I am pushed back into my seat.

    Furthermore, the time dilation, the Lorentzian contraction, and all that mess.....does that apply to people in a free fall toward a gravitational body or only to people that are on the ground (or a platform or what have you) and who are feeling the "acceleration" of the ground toward their feet?
     
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  3. Feb 6, 2015 #2

    phinds

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    Yep, you got it. Everyone is motionless in their own frame of reference but for you to stay motionless relative to the Earth, the Earth has to push up against you while spacetime curvature wants you to move toward the center of the Earth. The guy in free-fall is happily following spacetime curvature.

    They apply to ALL objects that are moving when viewed from a specific frame of reference in which they are not at rest, it's just that the effects are SO tiny at human-scale speeds that they might as well not be happening (except where incredible precision is needed, such as in the GPS system). Just to be sure I'm clear, all motion is relative. NOTHING is ever moving in its own frame of reference and so nothing ever "experiences" time dilation, for example. It is something that is observed/calculated by someone in a different frame of reference. You, for example, are MASSIVELY time dilated right now from the frame of reference of an "accelerated" particle at CERN (because it sees itself as motionless and you moving at .999999c)
     
  4. Feb 6, 2015 #3
    Yeah, I realize about the frame of reference part. What I mean is: if someone flew down from outer space, placed a watch on the falling guy, stole it back during mid-fall and brought it to flying spaghetti monster or what have you and flying spaghetti monster looked at the watch, wouldn't he see that the watch read the same time as his (after correcting for the slight acceleration from and to the earth), whereas if he picked up a clock from the Dead Sea he would see the clock show an earlier time than his own (however slight a difference it would be)? Because when the falling guy was in free-fall, he was not effected by gravitational time/length/mass warps (according to spaghetti monster's frame of reference). Or am I way off?
     
  5. Feb 6, 2015 #4

    phinds

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    I say again, time dilation is an observation/calculation from a frame of reference in which an object is not moving. It is utterly irrelevant WHY the object is moving in that FoR, it only matters that it is moving. Your guy in free-fall is moving relative to the spaceship (unless it performs acceleration to exactly match, outside the atmosphere, the spacetime path of the guy in free-fall. The space-ship can't free-fall at the same rate at the guy, because it is farther outside the Earth's gravity well.

    There are two kinds of time dilation. One is cause by relative motion and the other is caused by gravitational fields. The GPS system, for example, has to compensate for both. The GPS satellites are all moving relative to the surface of the Earth AND they are farther outside the Earth's gravity well than a clock on the surface so they have to compensate for both. If the GPS satellites were geosynchronous (which many people mistakenly believe them to be) the speed issue would not be there.
     
  6. Feb 6, 2015 #5

    Dale

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    There are two separate but related concepts of acceleration. One is called "coordinate acceleration", it is the second time derivative of the position and it depends on the reference frame. The other is called "proper acceleration", it is the acceleration measured by an accelerometer.

    In your frame: You have a coordinate acceleration of 0 and a proper acceleration of g upwards. The free fall guy has a coordinate acceleration of g downwards and a proper acceleration of 0.

    In the free fall guys frame: You have a coordinate acceleration of g upwards and a proper acceleration of g upwards. He has a coordinate acceleration of 0 and a proper acceleration of 0.

    Note that the coordinate acceleration changes depending on which frame you use. Note that the proper acceleration does not change depending on which frame you use.
     
  7. Feb 6, 2015 #6
    So, will I observe the same degree of Lorentz contraction of the falling dude as he does of me? (I'm talking over my own head here.)
     
  8. Feb 6, 2015 #7

    Dale

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    No. The time dilation formula only applies to inertial frames, which are frames where objects with 0 proper acceleration move in straight lines at constant speed. Your frame and his frame are not equivalent to each other.
     
  9. Feb 6, 2015 #8

    collinsmark

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    [Edit: On further consideration, the contents of this post are wrong (more than just the 42 vs 84 corrections; rather the whole thing is wrong). Sorry for the confusion.]

    Allow me to replace the scenario with a new one that hopefully demonstrates the important concepts, without the need to bring in the flying spaghetti monster and its delivery person.

    Consider a simple, non-rotating Earth. Drill a hole through the Earth from one side to the other. (If you wish, you can also remove the atmosphere from the Earth so it doesn't leak into the hole. If not, just neglect air resistance).

    Now drop a clock down the hole. Also keep another clock (initially synchronized) on the surface of the Earth near one of the ends of the hole. After around 42 minutes [Edit: should be 84 minutes, not 42], the clock that you dropped down will come back up the same hole (the clock will oscillate back and forth through the Earth, very similar to a mass on a spring, with a period of roughly 42 minutes [Edit: should be 84 minutes, not 42]). At that moment, you can easily compare the two clocks.

    Which clock has ticked slower?

    Before you answer, here are a couple of things to note:
    • At the moment you dropped the clock, both clocks were synchronized, and existed in the same place (well close enough) at the same time and had zero relative velocity.
    • When the clock comes back up through the hole (after about 42 [Edit: should be 84, not 42] minutes or so), once again both clocks exist in the same place and time and have zero relative velocity; although now they are no longer synchronized. (The clocks are easy to compare though.)
    The important point with the above notes is that both clocks share a single point in spacetime at the beginning, and they also share another point in spacetime at the end.

    So which path through spacetime extremalizes proper time, and which does not?

    Be careful here. A little knowledge about relativity can actually lead you to the wrong answer. Many would follow the following logic and reach the wrong conclusion:
    • Well, the clock falling through the Earth is moving back and forth, and thus, like the standard twin paradox, it would tick slower, right?
    • Also, the falling clock goes deeper into Earth's gravity well, so that's another reason why the falling clock should tick slower, due to gravitational time dilation, right?
    With that, someone might conclude that the clock falling through Earth would tick slower, for multiple reasons. But that someone would be wrong.

    The correct answer is that the falling clock will, overall, tick faster! The clock sitting there at rest by the end of the hole will experience more time dilation than the falling clock.

    There is a little truth to the logic about the gravity-well deepness part, but the comparison to the standard twin paradox is very wrong here.

    The falling clock is in freefall, meaning it is tracing a path through spacetime on a geodesic. And that causes its proper time to be extremal. And it will always tick faster than any other clock that went though spacetime taking a different path. (Recall that both clocks share the same beginning and ending points in spacetime.)
     
    Last edited: Feb 7, 2015
  10. Feb 6, 2015 #9
    Thank you, collinsmark. That was very helpful, enlightening,.....and somewhat humorous. What you said is basically what I thought was the case (though my bumbling language above might have indicated otherwise): That someone accelerating, or as Dale Spam named it, accelerating in the proper way, would have a slower clock than someone in free-fall.
     
  11. Feb 6, 2015 #10

    collinsmark

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    Yes, you are correct. That is always the case.

    If lazy Bob synchronizes his clock with Alice when the floating freely in space (together, at the same place and time, and same relative velocities), and then Alice goes off and explores the galaxy, leaving Bob effortlessly floating the whole while (Bob never has any proper acceleration), then later comes back to Bob, such that they again share the same space, time and velocity, Bob will always age more than Alice. It doesn't matter what Alice does -- travels the various stars, rests on the surface of heavy planets -- anything, she will never age more than Bob during her exciting trip.

    [Edit: this not necessarily true. There are situations where it is decidedly false. Sorry for the confusion. :oops:]

    To be fair though, in order to compare Alice's and Bob's aging, they need to be in the same place at the same time at the time of comparison. Making the comparison half way through Alice's trip doesn't make sense.
     
    Last edited: Feb 7, 2015
  12. Feb 6, 2015 #11

    collinsmark

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    [Once again I'm temporarily deleting this post until I can double check something about a claim I made.]
     
    Last edited: Feb 6, 2015
  13. Feb 6, 2015 #12

    collinsmark

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    I'll leave my last post deleted. I had made a claim about very low-Earth-Orbit satellite clocks (lower than GPS -- so low they're just skimming the Earth's surface) still ticking faster.

    My claim was incorrect. Although the effects of being in freefall would tend to speed up the orbiting clocks (relative to a terrestrial clock), it's not enough to compensate for the ticking slowdown due to the changing of reference frames during the initial acceleration.

    My mistake was that even though the satellite and terrestrial clock share the same points in spacetime at the comparison events, they still do not share the same frame of reference (their instantaneous, relative velocities are different). And that's enough to keep the very low orbit satellite clocks ticking slower.

    (The hole through the Earth scenario is still correct though, as-is.)
     
    Last edited: Feb 7, 2015
  14. Feb 6, 2015 #13
    Is free-fall measurably continuous? Can something be in "more of a free-fall" than another thing? Or is it a go/no-go sort of thing?
     
  15. Feb 6, 2015 #14

    collinsmark

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    Interesting question. :)

    One thing you can say for sure: Check your accelerometer. If your accelerometer reads 0, you are in freefall.

    But what happens if you experience a little air resistance when falling out of a tree? You're still accelerating toward Earth. But your proper acceleration is not quite 0, but it's almost 0. Is that "almost freefall"?

    I won't comment on the terminology, but I can say that the effects of time dilation are continuous, in this respect.
     
  16. Feb 6, 2015 #15

    PeterDonis

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    Are you sure? I strongly suggest checking your intuition here with math.

    In flat spacetime, you would be correct without qualification. In curved spacetime, however, there are complications. A free-falling worldline is always locally extremal; that is, it always has more elapsed proper time between two given events than non-free-falling worldlines that are "close" to the free-falling one. But the worldline of the person who stays at one end of the hole is not "close" to the free-falling one in this sense.

    Globally, however, a given free-falling worldline is not necessarily the one with the absolutely longest proper time between two given events; there may be other worldlines, not free-falling, which have more elapsed proper time. You have, I believe, (apparently unwittingly) given one example of such a case.

    It is true that, given a pair of events in curved spacetime, the worldline between them which does have the globally maximal proper time will be a freely falling worldline. But the statement that any free-falling worldline must always have more elapsed proper time than any non-free-falling worldline between the same two events, is much stronger than that (and is false).

    [Edit: I see you've retreated from that very strong version, since you agree that the free-falling clock in the "skimming the surface" orbit has less elapsed proper time than the clock at rest on the surface. So I've edited my statement above to no longer attribute that strong version to you; but I think it's still worth stressing that it's false.]

    Have you checked the math? If so, can you show it?
     
    Last edited: Feb 6, 2015
  17. Feb 6, 2015 #16

    Dale

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    An example which is closely related to the hole-in-the-earth scenario would be the "thrown clock" scenario.

    If you have two clocks, one at rest on the table, and one thrown briefly upwards so that it passes by the table clock twice, then the thrown clock will accumulate more time. This is basically just a twins scenario with the thrown clock being like the stay-at-home twin.

    Here "briefly" means that the time and distance are small enough that tidal effects are negligible.
     
  18. Feb 6, 2015 #17

    PeterDonis

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    "Closely related" in the sense that it illustrates different proper times for different worldlines, yes. But in this scenario, as you say, the free-falling clock (the thrown one) has greater elapsed time. (In fact, the thrown clock's worldline is the globally maximal one among all worldlines between the same start and end events.)

    However, if we set up the scenario so that the thrown clock goes up and comes back down in 84 minutes, the same time it takes for the clock in the hole to come back and for the clock in the "skimming the surface" orbit to come back, the proper time elapsed on the clock in the hole will not be the same as the proper time elapsed on the clock thrown upwards; it will be less. If I've done my calculations correctly, the ordering of proper times will be:

    clock in hole < clock in orbit < clock at rest on surface < clock thrown upwards
     
  19. Feb 6, 2015 #18

    collinsmark

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    Here is a link to an pdf document, compete with the mathematics, that explains the principle of extremal aging, that (as far as I can tell) does a better job than I could do (particularly in short order):

    http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCAQFjAA&url=http://ocw.mit.edu/courses/physics/8-224-exploring-black-holes-general-relativity-astrophysics-spring-2003/assignments/gravitationalforcesnotes3a.pdf&ei=x5TVVJftL4irogTz9oHwCw&usg=AFQjCNG0njnplKtWXXZcKeHwTNNOA_S-ew&sig2=lR5FIGqQk-iflesM0G8w6Q&bvm=bv.85464276,d.cGU

    Here is a relevant quote:

    "There is no way to distinguish between the effects on an observer of a uniform gravitational field and of constant acceleration.

    "The meaning of this statement is shown by a famous thought experiment proposed by Einstein. Consider a person in an elevator whose cable has broken and is accelerating at 9.8 ms-2 toward the center of the earth. The person is weightless and floats in the middle of the elevator with no acceleration relative to the elevator. (This follows from the Newtonian equivalence of gravitational and inertial mass discussed in the notes Coordinates and Proper Time -- recall that all bodies fall the same way in a gravitational field.) Locally, that is over sufficiently small intervals of time and space, there is no way for the person in the elevator to determine whether she is in an elevator falling in a gravitational field or floating freely in an unaccelerated spaceship far from all sources of gravity.

    "We know already that, in the absence of gravity, unaccelerated bodies take the path that maximizes proper time. Arguing from the Equivalence Principle, Einstein proposed that the same "extremal aging" principle holds in curved spacetime. Locally there is no difference between free-fall and unaccelerated motion. Locally, bodies must therefore follow paths that maximize proper time. If a long path is divided into many short segments, each of which maximizes proper time, then the total path maximizes proper time. The effects of gravity must be represented not by local acceleration but rather by global curvature."
    This applies to the "hole-in-the-Earth" scenario, by the way.

    The "thrown clock" scenario is different, and I don't think it necessarily applies. The thrown clock undergoes acceleration before it's thrown and after it lands. That needs to be considered before comparing it to a clock just sitting on the surface. [Edit: Oh, wait. I didn't read that "passing the tabletop twice" part. It still might not apply though since the two clocks have different velocities at the comparison events.]

    On the other hand, the "hole-in-the Earth" scenario presents the principle of extremal aging because
    • The falling clock is always in a state of freefall -- from beginning to end
    • At the comparison events, not only do the two clocks share the same points in spacetime, but also their relative velocity is zero.
     
    Last edited: Feb 6, 2015
  20. Feb 6, 2015 #19

    PeterDonis

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    Once again, I strongly suggest that you actually do the math. You will find that the principle of extremal aging doesn't mean quite what you appear to think it does. The word "locally" is very important in the quote you gave: a free-falling worldline has locally maximal proper time, but as I said in an earlier post, that does not necessarily mean it has globally maximal proper time. But the way you are applying the principle to the hole in the Earth scenario assumes that the proper time must be globally maximal, not just locally maximal.

    That doesn't matter. The clock at rest on the surface and the clock in the "skimming the surface" free-fall orbit also have different velocities at the comparison events; but you can still compare their proper times between successive events where they meet. Comparison of proper times does not require that relative velocities are zero at the start and end events. It only requires that those events are common to both worldlines.
     
    Last edited: Feb 6, 2015
  21. Feb 7, 2015 #20

    PeterDonis

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    I forgot to comment on this. The math given in that document only applies in vacuum; it does not apply in the interior of the Earth, because the interior of the Earth is not vacuum and the metric there is not the metric given in the pdf. The general principle as given in equation (2) of the pdf is correct, but to apply it to the worldline of the clock falling through the hole in the Earth, you have to know the metric in the Earth's interior (which is not the one given later on in the pdf).

    However, you don't have to know the full metric in the Earth's interior to prove the statement I made in a previous post about the ordering of the proper times. You only have to know certain general features of it, as compared to the metric in the vacuum region outside the Earth.

    (Note that even in vacuum, there can still be free-falling worldlines between two events which are not globally maximal; the worldline of the clock in the "skimming the surface" orbit is an example.)
     
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