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A clock rate equal to c?

  1. May 1, 2012 #1


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    Hello All,

    in 4-space, the position is x=(x0,x1,x2,x3), where x0=ct

    The question is -
    Can one define the clock rate in the frame of the object at rest as
    local clock rate = [itex]\frac{x_{0}}{t}[/itex] = c

    Then all objects, moving with respect to this frame will have clock rates less than c.

    E.g., a position within another frame of reference is x'=(x'0,x'1,x'2,x'3), where x'0=ct';
    Then the clock rate of the 'other' (primed) frame as perceived from the laboratory frame is = [itex]\frac{x'_{0}}{t}[/itex] = c(t'/t) < c
    Last edited: May 1, 2012
  2. jcsd
  3. May 1, 2012 #2


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    Gold Member

    You have a misunderstanding here. The position vector (ct,x,y,z) ( I've dropped the spacetime indexes) is a point in 4-D spacetime. 'ct' is not a rate it is a position on an axis, like x, y and z. Note that ct,x,y,z have the same units.

    Be aware that [itex]x_0=ct[/itex] is a definition and should read [itex]x_0 \equiv ct[/itex]. Your x/t = c is not meaningful.
  4. May 2, 2012 #3


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    Hi jpo,

    I think you're talking about four velocity. But what you really want to express is the derivative of proper time versus coordinate time [itex]\frac{d\tau}{dt}[/itex], which works the way you're describing it.
  5. May 2, 2012 #4


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    An observer is at rest. A mass m is at rest too. There is no relative motion, thus the spatial velocity of m in the reference frame of the observer is (0,0,0).

    Now, for the time travel (or clock rate):
    Both the observer and m travel in the time direction with velocity c.
    This statement is from Wikipedia (Four-velocity):
    "Thus all rest-massive objects can be thought of as moving through spacetime at the speed of light."
    In what reference frame is this statement valid?
  6. May 2, 2012 #5


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    Ok, let me also ask this:

    If an observer is at rest, is he in any way aware of his temporal motion? Because by convention, an observer in a frame is someone, who is not aware of the motion of his own frame of reference.
  7. May 2, 2012 #6
    This makes no sense. Velocity is movement in spacetime. A stationary observer has velocity 0, not c. Do not confuse clock rates with velocities.
  8. May 2, 2012 #7


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    At rest: Their spatial velocity is (0,0,0) with a norm of 0; their four-velocity is (c,0,0,0) with a norm of c
  9. May 2, 2012 #8
    Yes, he notes he is getting older. I can vouch for this personally.
  10. May 2, 2012 #9


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    An observer in a rotating frame is getting dizzy... saying this from experience

    But on a serious note - how does one avoid being tripped by semantics? If the 4-velocity in a rest frame is (c,0,0,0), can one express the clock rate in this frame by using the value of c? Just for the sake of comparing with other frames, where the four velocity, seen from a stationary frame will be (c',v1,v2,v3) and their perceived clock rate will be expressible through c'
  11. May 3, 2012 #10

    The norm of the four-velocity is always c, no matter what the actual velocity of the object it. In fact the norm of any four-vector is a constant.
  12. May 4, 2012 #11
    What do you mean by "this frame"? You just defined a 4-velocity, not a frame.

    To me, "clock rate" means the number of ticks per second it produces so you can't say what the rate is without knowing something about the clock itself. In relativity, the proper time between consecutive ticks for a perfect clock would be exactly uniform and independent of the frame.
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