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Does an observer on a carousel see a horizon?

  1. Jun 19, 2009 #1
    Special relativity shows that any accelerated observer
    sees an event horizon. In fact, if an observer is accelerated
    by a, the horizon is at distance l=c^2/a in the direction
    opposite to a.

    If an observer is on a carousel or merry-go-round,
    he is accelerated inwards. Does he then see a horizon
    on the outside?

    If so, does a horizon also appear for an observer in orbit,
    thus when circling the earth or the sun?

  2. jcsd
  3. Jun 19, 2009 #2

    George Jones

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    For an observer with constant 4-acceleration, there is a region of spacetime from which no signal reaches the accelerated observer. The boundary of this region is the horizon.

    For the observer on a carousel, there is no such region of spacetime, and thus no horizon (boundary). A spacetime diagram that has two space dimensions and one time dimension shows this clearly.
  4. Jun 20, 2009 #3
    Thank you! Can I read this somewhere, maybe with a picture of the diagram?

  5. Jun 20, 2009 #4
    An observer in orbit does not accelerate. Such an observer would travel on a "straight line".
  6. Jun 21, 2009 #5

    George Jones

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    I haven't tried to find this anywhere.

    Suppose that the centre of the carousel is in an inertial reference frame, that the plane of the carousel is the x-y plane, and that a person on the edge of the carousel moves with constant speed 1/2 (c=1).

    Then, the coordinates of the person on the edge are [itex]x = \cos \left( t/2 \right)[/itex] and [itex]y = \sin \left( t/2 \right)[/itex]. Plotting this worldline on a t-x-y spacetime diagram gives a helix about the t-axis (worldline of the centre).

    Now pick an arbitrary event in spacetime. The attached spacetime diagram shows that there is a lightlike path from the event to the worldline of the person on the edge of the carousel.

    Attached Files:

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