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About the accelerating spaceship

  1. Jul 28, 2008 #1
    I have some questions about one of the old elevator/spaceship thought experiments.

    1. If I am on a spaceship that is far from any massive bodies and is accelerating upward (from my point of view) uniformly at g, is the spacetime in my ship curved or flat?

    2. If a beam of light enters the ship from a hole in one of the walls and is initially moving in a direction parallel to the floor, does its path curve downward?

    3. If the path is parabolic, does the horizontal velocity decrease so that the overall speed remains at c?

    I would appreciate any responses.
     
  2. jcsd
  3. Jul 29, 2008 #2

    Dale

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    1) Flat
    2) Yes
    3) That's a great question, I hadn't thought of it before. Don't forget that time also dilates from top to bottom in the spaceship. I think that alone is sufficient to keep light moving at c everywhere locally, but I haven't actually run through the calculations myself.
     
  4. Jul 29, 2008 #3
    It is not that the photon enters the hole using an initial parallel or horizontal path. Due to the increasing speed of the ship the photon path appears parabolic to the ship. You are catching it at an instant when the tangent happens to be parallel. The proper vector speed "c" is measured along the tangent.
    When you shine a flashlight across a room it crosses at "c" unless you vector the path diagonally across the room then of course measuring only the horizontal speed of crossing will be less than “c”.
     
  5. Jul 29, 2008 #4
    This may be a non sequitur, but if the spacetime is flat why is the light path curved?

    Edit: oh, wait; someone may have just answered that...

    RandallB, I don't quite follow you. Are you saying that from the perspective of the observer on the ship, the overall path of the light beam is curved, but while it is inside the ship it is a straight line even to him?
     
    Last edited: Jul 29, 2008
  6. Jul 29, 2008 #5
    The path is curved because the frame attached to the accelerated spaceship is not inertial, I think.
     
  7. Jul 29, 2008 #6
    The spacetime is flat because if an observer in the nose of the rocket lets go and falls towards the tail of the rocket the light path will be exactly straight to him. Any experiments he does while falling will be identical to those done if was not moving and far away from gravitational sources. The spacetime around a spherical planet is not flat because a free falling observer will not see light traveling exactly in straight lines unless he confines his measurements to a very small volume. So curved spacetime is spacetime that cannot be completely straightened out by a free falling inertial observer.

    From the perspective of the non inertial accelerating observer in the rocket the light path is always curved (unless it is exactly vertical). The path might aproximate a straight line to him if he looks at a very small segment of it. I think all RandallB is saying is that when the light path is curved its horizontal velocity might be getting slower, buts its vertical velocity is getting faster and overall the speed is always c as measured by a local accelerating observer. You could imagine lots of local accelerating observers each measuring the velocity tangential to a small segment of the light beem local to them. It is only the direction of the photon that is changing. To an inertial observer outside the rocket the light path is always a straight line (assuming there is no gravity of the curved variety around ;)
     
  8. Jul 29, 2008 #7
    Is the local speed of light is always c whether the spacetime is flat or curved?
     
  9. Jul 29, 2008 #8
    Yes, because locally, the principle of equivalence says that spacetime is lorentzian, which means that physics experiments obey the principles of special relativity. One of them tells you that the speed of light is always c.
     
  10. Jul 29, 2008 #9
    Well, thank you all for answering my questions. I have a follow-up if anyone is interested: Is all this consistent with response #85 in the thread called "rest length in gr"?

    https://www.physicsforums.com/showthread.php?t=242014&page=6

    #85 is a response to #82, which is a response to #50, etc. The subject matter of that thread is rather deep and I cannot claim to understand it, but I got the impression from my limited participation in it that the speed of light is only constant in inertial frames of reference.
     
  11. Jul 30, 2008 #10

    Fredrik

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    #85 is one of mine, so I guess I should answer this. :smile: There's actually nothing strange with a speed of light that's different from c in non-inertial frames (even in special relativity). The reason is that a coordinate system is just a function that maps a subset of spacetime onto a subset of [itex]\mathbb R^4[/itex], and you can pick almost any such function to be your coordinate system. (There are some technical requirements mentioned in the definition of a manifold, e.g. if x and y are coordinate systems, [itex]x\circ y^{-1}[/itex] has to be [itex]C^\infty[/itex], i.e. it has to be differentiable an arbitrary number of times).
     
    Last edited: Jul 30, 2008
  12. Jul 30, 2008 #11
    Hi Snoopies,

    As I understand it, all observers, inertial AND non inertial see the LOCAL speed of light as constant. For example, imagine a series of observers on different levels of a high tower in strongly curved spacetime. An observer mid way up the tower would measure a light signal sent from the the top of the tower as moving faster than c while it is aproaching him from above and getting progressively slower as it falls below him. However, he will measure the speed of the light signal as it passes him in his local vicinity as c. In fact all observers on the tower, wherever they are, will measure the speed of the light signal in their local vicinity as c as the signal passes them. None of the tower observers are inertial observers. If all the observers got together and compared notes they would form a global picture of the motion of light and conclude light slows down as it falls but it appears to be constant to all local observers due to gravitational time dilation and length contraction of their measuring devices. This global picture is what is generally referred to as coordinate measurements. The Schwarzchild metric shows the coordinate velocity c' of a radially falling photon varies according to the distance R from the gravitational source as c' = c(1-2GM/R/c^2).

    An inertial observer in flat spacetime will measure the speed of light as constant everywhere and not just in his local vicinity. What I am not sure about and maybe one of the experts here can clarify, is whether a free falling inertial observer in strongly curved spacetime will see the speed of light as constant at any distance from them. I suspect not.
     
  13. Jul 30, 2008 #12

    Fredrik

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    Kev is talking about the most general kind of physical observers and their associated coordinate systems, but there are other coordinate systems. The physical observers (accelerating or not) correspond to a certain class of coordinate systems called "local inertial frames". When a physical observer measures a speed, the result is the value of [itex]|\sqrt{dx^2+dy^2+dz^2}/dt|[/itex] in a local inertial frame. (I think we have to consider this a postulate of both SR and GR). The speed of light is 1 (i.e. =c) in all local inertial frames, so it's true that both theories say that all observers will measure the same speed of light.

    However, you can easily imagine coordinate systems that aren't local inertial frames. For example, start with a (global) inertial frame on Minkowski space and make the coordinate change t'=2x, x'=x, y'=y, z'=z. The function that assigns (t',x',y',z') coordinates to events is a coordinate system that isn't a local inertial frame.
     
  14. Jul 30, 2008 #13
    Fredrik I may be misreading you, but you seem to me to be saying that the speed of light is whatever we choose it to be by our choice of coordinate system. Is this correct?
     
  15. Jul 30, 2008 #14

    Fredrik

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    Yes, that's what I'm saying.
     
  16. Jul 30, 2008 #15
    accelerating frames

    But you already know that “the speed of light AS YOU OBSERVE OR MEASURE IT is whatever we choose it to be by our choice of RELATIVE coordinate systemS” don’t you?
    I’m assuming you know of the Shapiro Effect where light passing near the sun slows in our view – observes on the surface of the sun would measure light passing by far away near us as traveling ‘fast’ due to the differences in reference frames. Essentially ‘equivalent’ to what you are seeing in your accelerated frame example.

    What is important to see is that while ordinary particles are viewed as traveling different speeds by observers in different inertial frames where the diff is based on speed. That kind of frame difference between observers does not change how they compare the speed of light between them, they both get the same result. Only acceleration differences between frames, where at least one cannot be inertial, cause them to measure light differently.
     
  17. Jul 30, 2008 #16
    I don't know what you mean by "relative coordinate systems".

    If I measure the speed of a beam of light that is not close to me, I can understand that I may not get c if a gravitational field is involved. But I don't understand how it would be legitimate for me to then use the same data with a different coordinate system, arrive at a different speed, and conclude that both speeds are equally correct. Shouldn't the proper coordinate transformation keep the speed the same?
     
  18. Jul 30, 2008 #17
    If you are moving relative to me at a relative speed or acceleration our coordinate system are moving relative to each other and they may or may not be inertial what can be confusing here?
    No one said you should.
    Keep what speed the same as what with respect to what, your question has no context. Are you somehow not getting that a non-inertial reference frame does not measure observe light the same as a inertial one?
    Reread the posts by KEV.
     
  19. Jul 30, 2008 #18
    To me the phrase "coordinate system" is not synonymous with "frame of reference". That may have been the only point of confusion here.
     
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