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Silly idea?

  1. Jun 10, 2010 #1
    how about this: silly idea, but think about it:

    the square root of 1 = {+1,-1}

    and we all know: t'=t/(1-(v/c)^1/2)

    in essance, t' must simultaneously have values of {+t',-t'} for all velocities not equal to zero and approaching c.

    this means that for a relativistic frame of reference, (i.e., twin paradox) there are not 2, but four (4) solutions. and, the sum of all 4 solutions = zero.

    does anyone have a good idea why the denominator in t'=t/(1-(v/c)^1/2) can only have a positive value?
  2. jcsd
  3. Jun 10, 2010 #2
    Re: Does a photon experience time?

    Gamma is always positive because no object can excel to the speed of light, or beyond. Also there's a square root in there, and the second solution (the negative) would imply negative energy, which we have not observed.
    Last edited by a moderator: Jun 10, 2010
  4. Jun 10, 2010 #3
    Re: Does a photon experience time?

    sorry, gamma MUST be always both positive and negative at ANY velocity other than c.

    however, zero in the denominator has neither positive or negative values. In fact, it is only at v = c that there is 1 unique solution.
  5. Jun 10, 2010 #4
    Re: Does a photon experience time?

    p.s. negative energy has been observed - it in fact, is proposed to make up greater than 70% of the energy in the known universe....
  6. Jun 10, 2010 #5
    Negative energy is not the same as dark energy.
  7. Jun 10, 2010 #6
    When we have the equation (x^2)-1=0, and we solve for x, we find that x has two solutions, 1 and -1. Either value will work. It is not both 1 and -1 at the same time. In this sense, what we say is that the value of gamma could be either negative or positive, however a negative value implies an object having negative energy. This has not been observed.
  8. Jun 10, 2010 #7
    Re: Does a photon experience time?

    At v=c there is no solution. Undefined is not a solution.
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