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Does Doppler effect change the energy of emitted light?

  1. Sep 28, 2009 #1
    A very basic question (I hope this is the right place for it).

    A distant star emits, say, a high-energy x-ray. Due to expansion of space, this reaches us as visible light (say, red part of the spectrum).

    The light that reaches us has less energy per photon (E=h*nu) then the light emitted from the star.

    What happened to the difference? Where did the energy go?
     
  2. jcsd
  3. Sep 29, 2009 #2

    Ich

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    From the emitter's point of view (in his standard coordinates), the photon didn't lose energy, but we can use and measure only a fraction of it; the rest is used to increase our kinetic energy.
    In standard cosmological coordinates, where this expanding space picture comes from, energy is not conserved, that's how these coordinates are defined. You know, "energy" is a coordinate-dependent value.

    With gravitation, everything becomes more complex. It's difficult to account for gravitational energy in GR.
     
  4. Sep 29, 2009 #3

    Wallace

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    As Ich said, energy is not conserved in GR. That being said, the reason is beacause energy in a Newtonian sense is a 3D concept. In the 4D theory (GR) there are still conservation laws that apply to 4D quantities, but any 3D 'projection' will not neccessarily be conserved. I only point this out to make it clear that the non-conservation of energy is not a problem or a mystery, which it can seem at first, particularly as we are taught in basic physics classes that energy is always conserved.

    To put it another way, imagine you are standing on a tall tower. Someone on the ground throws a ball up to you. As you catch is you can measure its energy and you find that it is less than the energy that the thrower imparted it with. Where did the energy 'go'? We would say that some of the kinetic energy has been converted in gravitational potential energy (and in a realistic calculation some is also lost in air resistance, but lets not worry about that). Now, in the case of a photon in an expanding universe, it turns out that in an analogous way, the conserved qauntity originally given to the photon has been somehow converted, but is none the less conserved (as long as we consider the whole universe in our calculation). The details are more complicated in the GR case (including gravity) as Ich suggested, but in an analogous way we are able to account for everything that was given to that photon.
     
  5. Sep 29, 2009 #4
    Thank you both, the general explanation makes sense.

    I do have a sub-question, although feel free to ignore it if the answer is too complex for forum explanations: what is the energy converted into?

    In the analogy with the ball, the potential energy is still available (I can let go off the ball and release it upon impact when the ball hits the ground). Is the converted energy from the photon also available, or is it lost?

    Or, in quasi-thermodynamic terms, does expansion of space lower the amount of free energy in the universe?
     
  6. Sep 29, 2009 #5

    Wallace

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    I fear I may have lead you astray, let me backtrack a little. I didn't mean to imply the analogy could be taken so literally, I was just trying to point to a more familiar example of where energy might appear to have 'gone missing' if not everything is taken into account.

    Again, don't take my little analogy literally in the case of a photon. There is no conversion of energy from one form to another, it really is 'lost'. On the other hand, the 4D conserved quantity that you can describe a photon with is, as the name suggests, conserved.

    Maybe think about it like this, the unit of energy are mass, length^2 per time^2 (usually expressed in Newtons which is [tex] Kg m^2 s^{-2}[/tex]). Now, when it comes down to it, you can always express the difference in energy as perceived by the emitter and observer of a photon by the differences in their rods and clocks, in other words the different length and time calibrations between them. In relativity we are familiar with the concept of length and time dilation, and it is these effects that give rise to redshift.

    Put simply, the redshift is effectively the change in frequency of the photon. Think of the emmitter as an oscillator and if there is a relative time dilation between the emmitter and observer then clearly they will disagree about the frequency, and hence energy, of the photon.

    The reasons for the time dilation can be either due to relative motion of the two things or gravity, however once you make that statement how much each contribute turns out to depend on how you define co-ordinates, so this is not universal and needs to be done with care. In the end the result is the same though, redshift is due to the difference in rods and clocks between the observer and emmitter. Since the 4D conserved quantity takes these into account it remains unchanged, while the 3D energy does not, hence we see that 'energy is not conserved'.

    I hope that helps, let me know if that is just confusing the issue!
     
  7. Sep 30, 2009 #6
    It kind of makes sense, but I think I'm reaching the boundaries of my understanding. To go further on this, I'll need to crack open a few math and physics textbooks.

    Oh well, that is what libraries are for... :)

    Thanks again for the explanations. :)
     
  8. Sep 30, 2009 #7

    Wallace

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    The classic simple example that might be relevant is the rotation of a vector. Imagine a vector in 2D (x-y). If you rotate it, the length doesn't change, but the components of the vector (the length in the x and y directions) do change. So, the length is conserved, but if you are only looking at the length of a single component you would conclude that the rotation didn't conserve the length.
     
  9. Oct 1, 2009 #8

    Chronos

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    Redshifted photons do not lose energy. They are time dilated, hence the net energy transmitted over time is conserved.
     
  10. Oct 4, 2009 #9
    Exactly, Chronos. the reduced energy/frequency wavetrain is longer. Energy x time remains a constant.
     
  11. Oct 5, 2009 #10
    I'm not familiar with the properties of photons, but I do know that a high energy EM burst(say X-ray) from a point of origin moving away from you will 'spread out' upon impact.

    By this I mean that the speed it releases from the origin(light speed) and the speed of impact(light speed minus drift speed) will cause the burst to lengthen(red shift) and the energy will be impacting the target over a slightly longer period of time. This makes it look like it is less intense, but the total energy remains the same.

    If by 'energy' you mean the impact velocity of the photon on target, then that 'energy' is used up in 'acceleration' from point of origin to accomodate for the point of origin moving away. Like a baseball thrown at you out the back of a moving bus. If it is thrown at 50 mph and the bus is moving 40 mph it only hits you at 10 mph, making it seem weaker. The energy was used to 'accelerate' from the bus, then the excess is what hits you.
     
  12. Oct 5, 2009 #11
    Thank you all for the answers. It now makes sense.

    I'll still have to open a book, though. The first thought that came to mind was "but wouldn't that require a photon to split into multiple photons - i.e. if a single photon of 200 nm light is detected as 400 nm, we would need to have two photons to maintain E=h*nu".

    Then I realized that I was thinking about photons in a completely wrong way. :smile:
     
  13. Oct 5, 2009 #12
    Aer-ki no! Completely wrong.
    "speed it releases from the origin(light speed) and the speed of impact(light speed minus drift speed) "
    there's no minus. Electromagnetic waves travel at C. Nomatter what the speed of the emitter or receiver. That's an axiom of relativity.
    There is a gravitational effect on the energy of a photon, but that's not relevant here.
     
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