How does the potential and kinetic energy of a photon relate? Do they mean the same thing?

Also how does De broglie wavelength and Potential relate?

  • $\begingroup$ Potential energy is a property of the potential that something moves in. For instance, photons will gain and lose energy in a gravitational field (we'll forget general relativity for now and go with a simple 1/r potential picture here). That potential energy comes out of the field and it increases or decreases the energy of the photons, which, of course, translates into a change of frequency and wavelength. Does that "relate" to kinetic energy? Only trough energy conservation, since kinetic energy is an observer dependent quantity, to begin with. $\endgroup$ – CuriousOne May 4 '15 at 11:20
  • $\begingroup$ Ok now, say a photon with kinetic energy K enters a place where the potential is V . What is the total energy of the photon now? $\endgroup$ – slhulk May 4 '15 at 13:29
  • $\begingroup$ Nor the velocity nor the energy content of a photon could be involved by gravity. The source and the receiver are located in points with different gravitational potential and that is the reason they capable to emit and receive photons at different frequencies. $\endgroup$ – HolgerFiedler May 4 '15 at 14:09
  • $\begingroup$ possibly helpful: hyperphysics.phy-astr.gsu.edu/hbase/relativ/blahol.html $\endgroup$ – Carl Witthoft May 4 '15 at 15:40

I'm going to have to give an answer that's very different to Jimmy360's. Apologies.

How does the potential and kinetic energy of a photon relate?

They don't. The photon is all kinetic energy.

Do they mean the same thing?

No. When you drop a brick, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic energy as radiation, you're left with a mass deficit, see Wikipedia. From this you know that the potential energy was rest-mass energy. You also know that a photon doesn't have any rest mass, so you ought to know that potential energy doesn't apply.

Also how does De broglie wavelength and potential relate?

An electron has a de Broglie wavelength. When you drop the electron, its gravitational potential energy is converted into kinetic energy. When you dissipate this kinetic energy as radiation, the electron then has a mass deficit, and its de Broglie wavelength is increased.

People say that a descending photon is blueshifted, and that it gains energy. But I'm afraid it doesn't. Gravity is not a force in the Newtonian sense. If you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c². Conservation of energy applies. You're like the electron writ large. When you descend potential energy is converted into kinetic energy, which gets dissipated. So your total energy is reduced. So you measure the selfsame photon energy as increased. The frequency doesn't actually change, but gravitational time dilation means you and your clocks are going slower, so you measure the frequency as being increased.


Imagine a beam of light, going towards a massive object. It has potential energy in the gravitational field. Of course, the potential energy has to become kinetic energy. This is done by shifting frequency. The energy of a photon is given by $E = hf$ so to increase kinetic energy we must increase frequency. If the beam of light was red, it will be a higher frequency light beam such as violet. This also works for a beam goon away from a gravitational field, except that the frequency will decrease.

  • 1
    $\begingroup$ A photon with less kinetic energy upon entering a position with high potential changes it's kinetic energy to a higher value?? $\endgroup$ – slhulk May 4 '15 at 13:30
  • $\begingroup$ @slhulk no, it is just like dropping a ball. At the top all energy is potential. As the ball falls, the potential energy becomes kinetic. It is the same for light, except the light doesn't change speed to increase kinetic energy, it changes frequency. $\endgroup$ – Jimmy360 May 4 '15 at 13:33
  • 1
    $\begingroup$ Oh. Okay now say you have a photon with kinetic energy K with De brogie wavelength L. Now if it enter a position with potential V. What it's DB wavelength now? $\endgroup$ – slhulk May 4 '15 at 13:39
  • $\begingroup$ Perhaps related physics.stackexchange.com/questions/149604/… and this too physics.stackexchange.com/questions/172854/… $\endgroup$ – HolgerFiedler May 4 '15 at 14:06
  • $\begingroup$ @slhulk It isn't a De Broglie wavelength. It is just the wavelength of light discussed in classical mechanics. $\endgroup$ – Jimmy360 May 4 '15 at 20:43

The kinetic energy of a photon is hf, where f is the photon's frequency.

If a photon is fired straight up from the surface of a planet it is said to "lose energy" to "the (gravitational) field". The loss of energy is visible to an observer overhead as a lower frequency, and therefore a lower energy, photon. This is the so-called "red shift".

Saying energy is lost to the field amounts to saying that there is greater potential energy and less kinetic in the configuration of planet and photon when the photon is higher than when it is lower.

That is the same basic idea as when a cannonball is fired straight up. There is greater potential energy and less kinetic energy in the configuration of planet and cannonball when the cannonball is higher than when it is lower.

The major difference between cannonball and photon is that the cannon ball slows down until its kinetic energy is exhausted, then turns around and falls until all its kinetic energy is restored, reaching the point from which it was fired, whereas the photon loses ever more kinetic energy and never returns unless it runs into a mirror (for example), but if it runs into a mirror at a sufficiently high point, the photon will have an extremely low kinetic energy, which it will surrender to the mirror before falling back to the planet, recovering all but the extremely low amount of energy it lost to the mirror.

To me these two cases are similar enough that I don't have a problem saying that the photon gains potential energy as it rises and regains kinetic energy as it falls except for the tiny loss to the mirror.

In both cases it is a little odd to speak of the potential energy as being in the field (what does that mean?) or in the photon/cannonball (what does that mean?) rather than being in the configuration of photon (or cannonball) and planet, but those are the traditional ways of saying it. I don't fight it.

The justification for saying the potential energy is in the photon/cannonball is the gross asymmetry of the situation. A photon or a cannonball is almost nothing in comparison with a planet. The planet will not be detected as having shifted in its orbit as a result of the firing, but the photon/cannonball has obviously moved relative to the seemingly immovable planet.

As regards the potential energy of a particle and its deBroglie wavelength: The potential energy is (from Wikipedia)

U = -G (m M)/r

where r is the distance between two masses with mass m (of the particle) and mass M (of the planet), and G is the gravitational constant.

The deBroglie wavelength is w = h/(lmv), where h is Planck's constant, l is the Lorentz factor, and v is the velocity (a scalar here) of the particle.

There doesn't seem to be much of a relationship except that m appears in both. By multiplication we get

Uw = -hGM/(lvr)


Uwlvr = -hGM.

Hmm. Beyond this, there seems nothing to get rid of.

  • 1
    $\begingroup$ Welcome on Physics SE :) You might want to see this part of our help section for typesetting formulas :) $\endgroup$ – Sanya Nov 28 '16 at 18:53

Quote from a webpage a bit over my head :-) greatians.com

Photon has linear momentum. Photon travels in vacuum space at the ultimate speed of light. Photon has the quantized energy of hf as given by eq. WD.1.2.
E = hf                                                                                                  … eq. WD.1.1
where   h = Plank’s constant
            f = frequency of photon

The energy of photon can be further sub-divided into two portions. There are the kinetic and potential energy of photon. The energy equation of photon is described below,
E = hf = pv + tf                                                                                … eq. WD.1.2
where   p =  momentum of photon
            v = traveling speed of photon = the Kong vector
            τ = torque, angular force between electric and magnetic component
            f = twisting or deform angle of M&E components

When the M&E and Kong vectors are not perpendicular, the photon travels at the lower speed. The kinetic energy of photon is
Kp = pv                                                                                               … eq. WD.1.3

And the potential energy of photon is
PP = τf                                                                                                … eq. WD.1.4

When the M&E and Kong vectors of photon are perpendicular, the photon is traveling at the speed of light. The deform angle of M&E vectors is zero. Therefore, the total energy of photon becomes the kinetic energy, where eq. WD.1.2 becomes

E = Kp = hf = pc                                                                                … eq. WD.1.5
where   c = the speed of photon

Not the answer you're looking for? Browse other questions tagged or ask your own question.