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Where is a light year inside the earth?

  1. May 8, 2009 #1
    Raymond explains in his radically modern physics text on page
    http://physics.nmt.edu/~raymond/classes/ph13xbook/node61.html [Broken]
    that a constantly accelerated observer (acceleration a) never increases his distance from
    the point x_0-c^2/a where x_0 is the point where he started his acceleration.

    He argues that this is the reason that on earth, we are accelerated outwards,
    but nevertheless stay at constant distance from the center.

    BUT: c^2/g is about 1 light year, whereas the radius of the earth is 6400 km.
    How does his explanation fit with these numbers?

    Heinz
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 8, 2009 #2

    Dale

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    Not very well. It is a bad explanation IMO, I wouldn't waste time on it.
     
  4. May 8, 2009 #3

    russ_watters

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    I don't see anything in the link that matches anything in the OP. All he's talking about is the equivalence principe. C^2= 1 light year doesn't make any sense - the units are wrong.
     
  5. May 8, 2009 #4
    Please read the very last paragraph on the cited page. Reading helps.

    c^2/g with g=9.81 m/s^2 is a distance, about 1 light year. You better check YOUR units.

    Heinz
     
  6. May 8, 2009 #5
    Hi Heinz

    Re-read what he says as he's talking about accelerating away from the centre of the Earth. He's not talking about standing on it, feeling its gravity, but moving away from it in an accelerating reference frame.

     
    Last edited by a moderator: May 4, 2017
  7. May 8, 2009 #6

    DrGreg

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    His explanation is a little confusing, because almost all of what he says applies to an accelerating frame in flat spacetime, i.e. in the absence of gravity. Gravity complicates the issue because of the curvature of spacetime. This distorts time and space so that the distance of c2/g gets warped to a different value. (That's a bit of an oversimplification, but it gives you the general idea.)

    I would say that, really, the very last sentence of the quoted page doesn't logically follow from what precedes it (although it is true); you would need to study GR in bit more depth to come to that conclusion.
     
    Last edited by a moderator: May 4, 2017
  8. May 8, 2009 #7

    DrGreg

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    Yes he is. In general relativity, when you are standing on Earth, you are accelerating upwards (bizarre though that may sound).
     
  9. May 8, 2009 #8

    russ_watters

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    I did!
    Now that I misread - sorry.

    Anyway....still the equivalence principle.
     
  10. May 8, 2009 #9
    I'm just going off how the passage reads and it seems more like he's talking about distance of an accelerating observer moving away from the centre of the Earth, not standing on it.

    Of course you're right about the equivalence principle implying that standing stationary on the Earth means you're "accelerating" away from it, albeit at exactly the amount needed to oppose the local curvature.
     
  11. May 8, 2009 #10
    Re-reading it, now I'm really confused.

    It does sound the way you describe. What an odd way of explaining it.

    Adam

     
  12. May 9, 2009 #11
    Rymond first says:

    "General relativity says that gravity is nothing more than an inertial force. This was called the equivalence principle by Einstein. Since the gravitational force on the Earth points downward, it follows that we must be constantly accelerating upward as we stand on the surface of the Earth! The obvious problem with this interpretation of gravity is that we don't appear to be moving away from the center of the Earth, which would seem to be a natural consequence of such an acceleration. However, relativity has a surprise in store for us here."

    then he calculates and finds:

    "..., we reach the astonishing conclusion that even though the object associated with the curved world line in figure 6.5 is accelerating away from the origin, it always remains the same distance (in its own frame) from the origin. In other words, even though we are accelerating away from the center of the Earth, the distance to the center of the Earth remains constant! "

    The only problem here is that the "distance to the center/origin" in flat spacetime is c^2/g, or about a light year; in the gravity case, this distance must be 6400 km. How does curvature transform a light year into 6400 km? Is there a simple way to understand this?

    Heinz
     
  13. May 9, 2009 #12

    A.T.

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    Curvature doesn't warp a light year into 6400 km, as this formula is not valid for earth's gravitation. His derivation refers specifically to an accelerated frame or a uniform gravitational field. He uses it only to show that it is possible, to accelerate away from a position without increasing the distance to that position. But you cannot use the same simple formula for the non-uniform gravitational field of the earth, with a much more complex space-time curvature.

    The conclusion however is the same: Standing on the earth's surface you accelerate away from the earth's center, without increasing your distance to it, or without even moving in the center's frame of reference.
     
    Last edited: May 9, 2009
  14. May 9, 2009 #13
    ???
    It seems that this book is not good to learn GR
     
  15. May 9, 2009 #14

    Dale

    Staff: Mentor

    This is just a bad explanation. Don't waste time on it. Here are some reasons that it is a bad explanation:

    1) The distance to the center of the earth is wrong
    2) The spacetime around earth is Schwarzschild, not flat
    3) The Rindler coordinates' center is an event, the earth's center is a worldline
    4) The distance to all of the other points in the earth is also not increasing

    Don't waste time on trying to understand a bad explanation. That does not imply that Raymond is wrong in general or that the rest of his notes are bad. But just move on from here.
     
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