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Travelling at the speed of light?

  1. Aug 4, 2003 #1
    If I was travelling at half the speed of light from A to B and another person travelling at half the speed of light from B to A (not directly towards each other, side-by-side), would i be travelling at the speed of light relative to the other person. If so, what would i see when i look at this other person?
     
  2. jcsd
  3. Aug 4, 2003 #2

    jcsd

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    v = (u + v')/(1 + uv'/c2)

    This is how to sum two velocites, u and v', in SR where c is the speed of light in the vacuum.

    If we input the velocities into the equation as a fraction of c, then all the c's oin the bottom of the equation cancel and we're left with:

    v = (0.5c + 0.5c)/(1 + 0.25) = 0.8c

    Therefore you will observe the other person travelling at 0.8c or 4/5 of the speed of light relative to you.
     
    Last edited: Aug 4, 2003
  4. Aug 5, 2003 #3

    jcsd

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    I'll just add for speeds alot lower that c the two velocities do add algerbraically, that is :

    v ≈ u + v'

    for values of u and v' significantly smaller than c.

    For example, taking two objects that both appear to be travelling ~650 mph (10-6 c) in oppoiste directions to a 'stationary' observer the difference in observed velocities will only be about 1 part in a trillion.
     
    Last edited: Aug 5, 2003
  5. Aug 7, 2003 #4
    So, according to your theory:

    If 2 photons were moving towards eachother, both moving at the speed of light:

    v = (c + c)/(1 + c^2/c^2) = 2c/2 = c

    The photons would observe eachother moving at the speed of light? If thats the case, how can it tell whether the other photon is moving at all of not?

    v = (c + 0)/(1 + 0/c^2) = c/1 = c
     
  6. Aug 8, 2003 #5

    jcsd

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    It's not my theory, it's Lorentz's/Einstein's (the equation is derived from the Lorentz transformations), it's called the special theory of relativity.

    According to the transformation, both photons would observe each other travelling at the speed of light, but reference frames at this speed are forbidden anyway.
     
  7. Aug 8, 2003 #6

    selfAdjoint

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    Actually the same rule holds for low speeds - it holds for all speeds, but if you try it for speeds around a billionth of c, which are the speeds we encounter (feet per second) then the Lorentz formula becomes normal addition to a high degree of accuracy. But there is no cutoff where relativity addition stops and the familiar addition begins.
     
  8. Aug 8, 2003 #7
    One photon has velocity +c and the other photon has velocity -c, both relative to some initially-given reference frame.
    The relative velocities ought to be calculated by a difference formula:

    u' = (u - v)/(1 - uv/c2)

    .

    So, for the velocity of the +c photon relative to the -c photon,

    u' = ((+c) - (-c))/(1-(+c)(-c)/c2) = (+2c)/(+2) = +c

    For the velocity of the -c photon relative to the +c photon,

    u' = ((-c) - (+c))/(1-(-c)(+c)/c2) = (-2c)/(+2) = -c

    . The relative velocities would be equal in magnitude, but oppositely directed.

    There is a problem with all this. In order to derive composition theorems like this, guaranteeing that the postulates of special relativity are enforced, it is necessary to assume implicitly that all relative velocities between observers have magnitudes less than c. So, it is not a foregone conclusion that the two photons define reference frames for which the composition formula is valid. But it does appear to give "appropriate" answers. One photon gets a relative velocity of +c and the other gets a relative velocity of -c.
     
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