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Wgat would a light clock measure in free fall?

  1. Nov 6, 2011 #1
    Wgat would a "light clock" measure in free fall?

    I've read in a couple of different books that the similarities between acceleration from gravity and rockets or whatever is only local. Both books said one of the reasons is because with acceleration from gravity two objects in free fall starting from the same place and a distance between them will slowly converge as they free fall towards the gravitational body. (oddly one of the books said the objects would meet at the centre of the body?)

    Are the objects in free fall following "straight lines" (geodesic)?

    With a light clock in free fall horizontally what happens to the time measurement if the mirrors are attached (distance between mirrors maintained)? (from the perspective of an observer from where the light clock was released) An easy guess is the clock ticks more slowly.

    What happens to the time measurement if the mirrors are not attached and slowly coverage with the higher gravitational potential? (from the perspective of an observer from where the light clock was released) I guess the observer sees the clock tick at the same rate. Said differently that the reduced distance between the mirrors makes up for the time dilation.

    Not the point of my question, but my guesses to my questions tells me the book that says the objects meet at the centre of the body is wrong. And also that no matter how close the objects were at the start of their free fall they never meet. Seems like that would define a singularity if they did.
     
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  3. Nov 6, 2011 #2

    Dale

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    Re: Wgat would a "light clock" measure in free fall?

    Yes.

    Yes, this is the standard gravitational time dilation formula.

    I don't know, that is a good question.

    They do meet at the center. That is a simple consequence of the spherical symmetry. Since the solution is spherically symmetric, the two objects will have the same radial position as a function of time. When r=0 then they meet.
     
  4. Nov 6, 2011 #3
    Re: Wgat would a "light clock" measure in free fall?

    When you discuss that light-clock I lose you somewhat. Are you discussing the Lorentz contraction? This one gives a good explanation I think http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_basics/index.html#Moving1 [Broken]

    As for them meeting at the centre, you can find a similar definition thinking of 'Newtons shells'. In it he defined mass as always situated at the centre, of whatever mass you define, like earth. So Earth's 'gravity' points into the exact middle, well maybe not exactly in the middle according to GR, but in Newtons definition it does.
     
    Last edited by a moderator: May 5, 2017
  5. Nov 6, 2011 #4
    Re: Wgat would a "light clock" measure in free fall?

    I think I see what you are saying Dalespam, however wouldn't the geodesic paths look like this?
    Geodesicpathsv1.jpg

    Of course there is no scale here, it is just to illustrate that the paths narrow most where the gravitational potential is highest (iirc near the surface for Earth) and returns to "normal" (from perspective of where the light clock was released) at the centre of the body (iirc Earth has net zero gravity at centre).

    Are you saying the geodesic paths look like this image below?

    Geodesicpaths2.jpg
     
    Last edited: Nov 6, 2011
  6. Nov 6, 2011 #5

    Dale

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    Re: Wgat would a "light clock" measure in free fall?

    The geodesics you describe in your OP are radial geodesics. Radial geodesics are, well, radial, so they would never look like your first drawing.

    Think about it: there is spherical symmetry, since there is no initial velocity in the tangential direction then, by symmetry, there cannot ever be any tangential acceleration. (which direction would it accelerate and why?)
     
  7. Nov 6, 2011 #6
    Re: Wgat would a "light clock" measure in free fall?

    The problem is when I think about it :smile:

    I thought the gravitational potential is highest at the surface, and net zero at the centre. (a quick wiki search says that gravity is strongest at the outer edge of the outer core, which is about halfway to the centre, this is where the paths in my picture should be closest)

    So as the objects magically pass the surface the paths diverge as they get towards the centre of the body because of the reduced gravitational potential below the surface, and the net zero potential at the centre. (less gravitational acceleration, more distance between paths / no gravitational acceleration, no change in distance between paths, all as compared to where the objects were released which was nearly zero gravity).

    I don't think the paths below the surface would be linear as I drew them, but the effect I was trying to convey is the same-ish.

    Another way I think of it is a string of clocks from well above (almost zero gravity) the Earth straight through centre to the other side and out again. The clocks nearest the surface tick most slowly, the clocks nearest the centre tick similarly to the the clocks far from the surface.

    lol does the Earth juice flow up to the crust?
     
    Last edited: Nov 6, 2011
  8. Nov 6, 2011 #7

    PAllen

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    Re: Wgat would a "light clock" measure in free fall?

    Try to think deeper about what Dalespam said. Imagine tunnels through the planet so the objects could continue falling passed the surface. Your belief is that if these tunnels led to the center of the earth, the objects, on passing the surface, would veer off and hit the side of the tunnel. How? Why? There is absolutely no feature of the situation (in the ideal, where the planet isn't rotating) that is not spherically symmetric. What could cause the motion to deviate from radial?

    Possibly you are thinking the speed of fall relative to earth slows after the falling objects pass the surface. This is true, but doesn't change the directions.
     
  9. Nov 6, 2011 #8

    pervect

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    Re: Wgat would a "light clock" measure in free fall?

    The good news is that there is a paper that gives you a physical model to look at of what the r-t plane of the Schwarzschild metric looks like as an emedding in a 3-d space, so you can draw radial geodesics on it.

    GR ultimately boils down to drawing space-time diagrams on a curved surface. Usually we use the metric functions to deal with curvature. But you can think of drawing your space-time diagrams on an actual curved surface if you want, just as you can think of drawing maps of the Earth's surface on a globe.

    Note that the r-t plane has a Lorentzian signature, so the model you get this way will be something you can draw space-time diagrams on, but being purely a spatial surface it won't have the negative signature for the time axis. So you have to have some idea of special relativity to start with, if you can't draw an SR space-time diagram at all, it won't help you much to know the curved surface that you'd have to draw it on to do GR. Of course, the Lorentzian signature really only matters when you want to transform reference frames.

    The other bad news is that having the actual surface model to draw your space-time diagrams on doesn't directly tell you how to draw geodesics on the surface. So it's not particularly good for doiing calculations. You need to know enough either about GR or about curved surfaces to construct the geodesics , which generally means solving the goedesic equations, to get actual geodesics out of the model. ANd it's pretty much easier to solve the geodesic techniques using GR than it is to try and get them from the surface model - it's defiitely better documented.

    On the plus side, you can visualize what's going on a bit better with an actual model that represents a comonly occuring physical condition - a spacetime dominated by one large mass.

    The paper that describes this wasn't really the easiest read, either, but you can find it at http://arxiv.org/abs/gr-qc/9806123, "Space Time Emebedding Diagrams for Black Holes" by Marolf.
     
  10. Nov 6, 2011 #9

    Dale

    Staff: Mentor

    Re: Wgat would a "light clock" measure in free fall?

    the reduced gravitational potential would cause it to accelerate less, but the acceleration would still be directed toward the center. Given the symmetry, what could possibly make it accelerate left or right?

    See PAllens comments above.
     
  11. Nov 6, 2011 #10
    Re: Wgat would a "light clock" measure in free fall?

    I was thinking of how the acceleration slows and thinking that is a consequence of changes in the path through spacetime.

    That being said I have read a bit more and am more confused then before.

    Seems I am asking questions I don't yet understand. :smile:

    One more question I don't understand, would a photon traveling through that tunnel follow a straight path to the centre as well.

    Wow GR has alot more jargon then SR. Maybe out of reach for casual interest.
     
    Last edited: Nov 6, 2011
  12. Nov 6, 2011 #11
    Re: Wgat would a "light clock" measure in free fall?

    Yea I think I need to have a better understanding of sr. time like space like intervals specifically. and what it means for an interval to be calculated as +--- or -+++. Maybe that will help me understand geodesics. looks like I gotta play with math to understand it :frown: And if GR is as mathematically ridiculous as I see on wiki then Im s.o.l. as far as understanding GR lol.
     
  13. Nov 6, 2011 #12

    pervect

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    Re: Wgat would a "light clock" measure in free fall?

    The way with the least math that I know of to really calculate a geodesic path requires some familiarity with the calculus of variations.

    [add]I was looking at EF Taylor's website, and they have an approach to deriving equations of motion using Lagrange's principle that only requires ordinary calculus. But it's oriented more towards classical mechanics than GR. The URL is http://www.eftaylor.com/pub/lagrange.html. You might be able to extend the ideas there to the GR case, but they mostly discuss using the variational principle for classical Newtonian mechanics. Which is still not a bad place to start.

    You sort of have to take it as a given that the path a object takes is a path of extremal aging, though there might be some clever way to motivate this (I'm drawing a blank though as far as clever motivation goes). The extremal proper time will be maximum proper time a lot of the time, but you can't really rely on it being a true maximum always - there's some papers on that aspect of things, "When Action is not Least", on Taylor's website (see above).

    By writing down the formula for the proper time to go from point A to point B, you can then write down the differential equations that the "extremal aging " condition imposes, which will be the geodesic equations.

    If you follow a formal derivation, (I was looking for one, I'm pretty sure I've seen one, perhaps in MTW, but I ddidn't find any on the WWW), you'll see the Christoffel symbols in the geodesic equation, but at this stage you won't have to worry about their significance much.

    Another issue is that said equations are rather annoying to solve, if you don't take advantage of the symmetries. The Killing vectors give you a few conserved quantities that allow you to rewrite the geodesic equations in a manner that are much easier to solve, (d/dtau of conserved quantity = 0), but this is usually an advanced topic.

    I've heard Taylor's "Exploring Black Holes" is a good basic treatment of GR, unfortunately I don't have it. The first few chapters are available on the www - they cover more basic introductory material rather than getting the equations of motion, so I'm not quite sure about how Taylor chose to address the equations of motion problem, I"m pretty sure the text does address it though.
     
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