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The Effect of Gravity on Photons

  1. Apr 6, 2014 #1
    I am not well versed in complex physics (I've taken honors physics in High school) so any help I can get with simpler vocabulary would be appreciated.

    I stumbled upon this problem when just thinking and have not been able to find an answer.

    We all know that gravity affects photons, as shown in black holes or even just the change in direction when passing a star. However, if a photon was emitted from a star, wouldn't it slow down? Thus you would have a photon going slower than the speed of light.

    This picture (ignoring the stuff about fusion) is what I'm trying to describe.
    PhotonSun.jpg
    (if this doesn't work, try right clicking, copying the URL and pasting it into a new tab)

    Thanks,
    jld592
     
  2. jcsd
  3. Apr 6, 2014 #2

    Drakkith

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    Staff: Mentor

    Can't see the picture. Something about not having permission to view it. I'd download the picture to your computer and then attach to the post as an attachment instead of a link.
     
  4. Apr 6, 2014 #3
    No, the photons will lose energy but will not slow down. That means its frequency, momentum, wavelength all change but not its speed.
     
  5. Apr 6, 2014 #4
    Picture link

    Here is the picture as an attachment.
     

    Attached Files:

  6. Apr 6, 2014 #5
    I just don't understand why. Gravity is usually a force, so wouldn't that affect acceleration and velocity?
     
  7. Apr 6, 2014 #6
    The force is also [itex]F=\frac{dp}{dt} = m\frac{dv}{dt}=ma[/itex], and the momentum of photon is [itex]p=h\nu[/itex]. So, the only thing which may vary for a photon is its frequency, [itex]\nu[/itex].
     
  8. Apr 6, 2014 #7

    WannabeNewton

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    Science Advisor

    Gravity is not a force in general relativity. Light rays in free fall in a gravitational field do not accelerate-nothing in free fall in the gravitational field accelerates for that matter. What happens is the light rays' frequencies change relative to observers at rest in the gravitational field (I'm trying to avoid using 'photons' because this inaccurate semi-classical conception of the photon is giving you the wrong picture).
     
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