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Why does light move at

  1. Jun 29, 2008 #1
    a constant velocity whether or not you are traveling toward the source, or away from the source?
  2. jcsd
  3. Jun 29, 2008 #2
    If I'm not mistaken, someone please correct me if I'm wrong, but if you're going 1000mph west and you fire a bullet at the speed of light towards the east, the bullet travels at the speed of light - 1000mph. I vaguely recall the topic of "if you had a car traveling at the speed of light and you turn on you're headlights, what would happen?" My understanding from that would be the light from the headlights would be 2x the speed of light.

    I may be in error and if I am I would enjoy clarification. Just contributing to the discussion to understand it myself.
  4. Jun 29, 2008 #3
    Nobody knows. That's why Einstein postulated it rather than proved it.

  5. Jun 29, 2008 #4

    There are two errors in your assumptions. First, nothing with rest mass (eg cars and bullets) can move at the speed of light even in theory. Second, velocities do not add in the simple way you imagine, but according to the relativistic velocity addition equation. See http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/headlights.html
  6. Jun 29, 2008 #5


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    In the usual formulations, the constancy of the speed of light is built right into the setup. The answer to "why" is, therefore, obvious and trivial -- e.g. it might be given by the trivial proof "the speed of light is constant because the speed of light is constant".

    If you seek a nontrivial explanation, you need to give some indication of what alternative formulation of physics you want to use.
  7. Jun 29, 2008 #6

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    Actually, the speed of electromagnetic waves in free space can be derived from Gauss', Ampere's and Faraday's laws. Start here:

    [tex] \nabla \times (\nabla \times E) = \nabla \times (\nabla \times E) [/tex]

    Apply a vector calc identity to the left hand side.

    [tex]\nabla (\nabla \ldot E) - \nabla^2 E = \nabla \times (\nabla \times E) [/tex]

    Apply Gauss' Law to the left hand side to kill the first term and Faraday's Law to the right hand side.

    [tex]- \nabla^2 E = \nabla \times ( - \frac{dB}{dt}) [/tex]

    Now, switch the order of space and time derivatives on the right hand side.

    [tex]- \nabla^2 E = - \frac{d} {dt}(\nabla \times B)} [/tex]

    Apply Ampere's Law,

    [tex] - \nabla^2 E = - \frac{d}{dt} (\mu_0 \epsilon_0 \frac{dE}{dt}) [/tex]

    Bring the constants out front

    [tex] \nabla^2 E = \mu_0 \epsilon_0 \frac{d^2E}{dt^2} [/tex]

    And you have a wave equation with wave velocity equal to [tex]\sqrt{\mu_0 \epsilon_0}[/tex], which one will see numerically is equal to c.

    If you like, you can repeat this showing magnetic waves travel at the same speed. There is little additional insight to be gained by doing this.
  8. Jun 29, 2008 #7
    Other than that Maxwell's laws or supposed to invariant under a change in coordinates, so the wave equation with that particular velocity should also be expected to hold true in all coordinate systems.

    As a sort of proof for what the OP was asking, notice that the simple addition of velocities formula cannot hold in any universe with a finite upper bound on speed.
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