Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Age of Photons

  1. Nov 6, 2005 #1
    I'm a real noob in physics. After reading some things about special and general relativity I came up with a question based on a statement i saw on a physics book (don't remember wich). If my memory is not failing me i remember having read that photons have exactely the same age now than when the universe begins. This because they travel at c speed and according to special relativity time doesn't affect u if u travel at c speed.
    However correct me if i'm wrong light only travels at c speed in vacuum cause in other medium it travels at inferior speeds, so if
    t = t'/sqroot(1-(v^2)/(c^2)) , and v < c photons have an age.
    Is this correct???
    BTW, if a photon has only traveled trough vacuum since Big Bang how do u solve the equation???
    t = t'/sqroot(1-(c^2)/(c^2)) <=> t = t'/0 ???
    Sorry but the maths i know don't allow me to solve this.
    PS: BTW what force makes the speed of light drop below c in other medium than vacuum. I don't know much about forces, only that there are 4 so plz tell me. My guess would be on gravity as i heard it affects photons because of G. relativity.
    Sorry for my english and for my noobish questions
  2. jcsd
  3. Nov 6, 2005 #2


    User Avatar

    Staff: Mentor

    That is correct. Nevertheless, individual photons always travel at speed c. The photons that leave a piece of glass are not the same photons that enter it. The photons that enter are absorbed, and the atoms of the glass radiate new photons.
  4. Nov 6, 2005 #3


    User Avatar
    Gold Member

    jtbell is correct. I have nothing to add.
    Last edited: Nov 7, 2005
  5. Nov 7, 2005 #4

    The result t = t'/0 just tells you that you are exactly over the lightlike worlline, i.e., the particle is traveling at speed of light (be a photon or graviton, for example ).
    Last edited: Nov 7, 2005
  6. Nov 7, 2005 #5
    As jtbell pointed out, individual photons always travel at speed c. What slows a light ray down in a material is the absorption/emission process as it interacts with atoms in the material. But the individual photons are always travelly at c (think of it this way: the space between atoms is a vacuum). Because of this the individual photons do not age, as per SR.
  7. Nov 8, 2005 #6
    Last edited by a moderator: Apr 21, 2017
  8. Nov 8, 2005 #7
    Last edited by a moderator: Apr 21, 2017
  9. Nov 8, 2005 #8
    The problem is that you're trying to find a relationship between the rate at which a clock ticks when it is traveling at the speed of light in the inertial frame S as measured by an observer at rest in S. This is an impossible situation and that is the reason for the division by zero. The faster a clock moves then the slower you will measure the rate of the clock ticking. As it approaches the speed of light the time between ticks, i.e, "t", increases. The division by zero means that the clock stopped. But since a real clock can never travel at the speed of light this will not happen. That's one of the reasons we can't place a clock at rest near a photon so as to measure its age.

  10. Nov 8, 2005 #9
    But is it still corrent to say that photons haven't aged, using the concepts of limits.. say as v approaches c, gamma just gets bigger and bigger, so the time period experienced by the photon as observed by us (and all other inertial frames) approaches zero... or is the concept of time to a photon seen as meaningless?

    Also, using the velocity transformations to see how fast one photon travels as observed by another moving in the opposite direction, you find that it only moves at a speed of c.

    One more question, can you even use the concept of a limit, given that the left hand limit (considering only v<c) is "infinity", what is the right hand limit?

    gamma = 1/sqrt(1 - vv/cc), so as v>c but v tends to c, vv/cc will get close to 1 from above 1, and 1-vv/cc will get close to 0 from negative, and then what happens? sqrt(1-x), as x gets small but x<0, gets close to zero but ... hmm. would the right hand limit be infinity but a different kind?

    Just some random thoughts ...
  11. Nov 8, 2005 #10
    Both really, its the same thing.
    The right hand limit is the same as the left hand limit, its just imaginary approaching zero, rather than real and approaching zero.
  12. Nov 8, 2005 #11
    Hello, sorry to butt in, but I have a question relevant to this topic. What would the outside world appear to a photon, knowing that it is impossible for a photon to see, but just hypothetically what would it look like. Would the photon see everywhere it would be and was at the same time? I have no clue. I asked this question earlier using electrons but I got tied up in red tape. Apparently electrons and photons are different. Hopefully this is a less retarded version of the question.
  13. Nov 8, 2005 #12
    The entire universe would appear contracted into an infinite plane.
  14. Nov 9, 2005 #13
    cool.. do you mean a plane perpendicular to its direction? So only other photons would be outside the plane. Would the plane also appear frozen?

    What about when the photon moves? Does it keep its view of the plane so that the point of the photon lies on the plane all the time? Or does it see the plane move away behind it?
  15. Nov 9, 2005 #14
    Unless you specify what the term "age" means then there can be no answer to your question. To an observer at rest in an inertial frame he can start his clocks when the photon is created/emitted. Then so long as the photon exists we can keep the clock running. When it has been destroyed/absorbed then we stop our clocks and call that the total age of that photon. It has no meaning to refer to the time as measured in the photons frame since we can't place clocks in that frame. All we can do is to look at a limiting process with a particle with finite proper mass and let its speed approach, but never equal, c.

  16. Nov 9, 2005 #15


    User Avatar

    Staff: Mentor

    Frankly, neither dies anybody else, in the context of special relativity, anyway. When you try to work through the equations, you end up trying to divide by zero and things like that. It's impossible to make meaningful, consistent statements about the reference frame of a photon because there simply is no such thing.
  17. Nov 9, 2005 #16
    Another way of looking at the ‘speed’ of light slowing down through a medium:
    Think of the light going though a ‘hall of mirrors’ as the absorb & re-radiate process of going through the medium. As it reflects up&down, back&forth, left&right though the non-vacuum it never slows down – it just travels farther. How far, & if it gets deflected & by how much depends on the material and its shape.

    As to photons aging, the CMB started out very hot in the Big Bang traveling for some 13 Billion years of our time to reach us cooled down to a Microwave Frequency. Some look at that as “tired light” getting weaker and less energetic rather than slowing down. For light to get that tired from aging I’d guess it to be maybe 87 k-Yrs old by its own ‘clock’. But by accounting for the dramatic Red-Shift with the Hubble Expansion there is no need to account for the energy change in light with photon aging or getting tired, thus no need or ablity to give a photon a ‘clock’ to read.
  18. Nov 9, 2005 #17
    yes a plane perpendicular to its direction of motion. Essentially all points along the photons trajectory would appear contracted into a single point. The photon would see itself as existing at all points on its trajectory simultaneously.

    Again though, this is all based on taking the limit of the Lorentz transformations as v approaches c. The results may or may not be physically valid (or for that matter meaningful) but that is what the limits of the lorentz transformations predict.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook