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Homework Help: Uncertain on a few things photons

  1. Jul 19, 2006 #1
    right, so im supposed to draw a free body diagram of the avg photon impact on a spacecraft, as well as approximate a value for the force vector those photons may create. It's a really rough estimation, but im having trouble with a few things:

    a. the fundamnetal frequency of a photon
    b. calculating the force, since photons travel at a constant c, and dont accelerate.
    c. do photons have mass or not? My teacher told me it wasnt entirely true that photons were massless, however everything I find online contradicts him.

    due tmrw morning, thanks
  2. jcsd
  3. Jul 20, 2006 #2

    Andrew Mason

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    a) The energy of a photon of wavelength [itex]\lambda[/itex] or frequency [itex]\nu[/itex] is:

    [tex]E = hc/\lambda = h\nu[/tex]

    So the frequency of a photon is its energy divided by h (Planck's constant).

    b). The momentum of a photon is:

    [tex]p = E/c = h/\lambda = h\nu/c[/tex]

    While it is true that photons do not accelerate, (they always travel at c, the speed of light) they carry momentum. So they impart a force to the absorbing object when absorbed. The force is difficult to measure because the time over which it acts is so small. But you are interested in the force x time or impulse imparted to the spacecraft. That impulse is the photon's momentum, E/c

    c) photons do not have rest mass or inertia. But they transport mass/inertia across space, as Einstein first noticed when discovering the famous relationship, [itex]E = mc^2[/itex]. So when the photon is absorbed, there is a small increase in mass of the spaceship (so small it is practically immeasureable) in the amount [itex]m = E/c^2 = h/c\lambda = h\nu/c^2[/itex].

    Last edited: Jul 20, 2006
  4. Jul 20, 2006 #3
    For a), a photon can have any frequency, take the EM spectrum for eg, from X-ray to microwave, any frequency is possible. For b), photons still exert a momentum, hence a force, so v = c. For c), the mass of a photon is very very small, so normally we assume to be close to zero or negligible.
  5. Jul 24, 2006 #4
    That is incorrect. Photons are completely massless, at least according to current theory.

    The weird thing is that while photons do not have mass, they do have momentum. The relation given above that p = h/lambda is correct, it was one of the formulas which gave birth to quantum mechanics.

    Also, photons can accelerate. Remember that velocity is a vector quantity. While the speed of photons is always c, the direction of motion can change, which is an acceleration.

    [tex] \vec{p} = \hbar \vec{k} [/tex]

    [tex] |\vec{k}| = \frac{2\pi}{\lambda} [/tex]

    and the [itex] \vec{k} [/itex] points in the direction of motion of the photon. So for this problem, the change in momentum of the photon will be the difference in the k vectors times hbar. Since the wavelengths don't change, you can pull the hbar2pi/lamdas out and subtract the unit vectors. That's the momentum added to the spaceship.
  6. Jul 25, 2006 #5

    Andrew Mason

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    Hmm. An accelerating photon. Are you sure it is the same photon that is changing direction?

    I would suggest that a photon changes velocity only due to a gravitational field. And I would be reluctant to call that acceleration.

  7. Jul 25, 2006 #6
    Well that depends on the coordinate system.
  8. Jul 25, 2006 #7

    Andrew Mason

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    What, exactly, depends on the coordinate system?

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