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Do photons have kinetic energy?

If a photon of radio wave and a photon of gamma wave moves at the same speed, how can both of them have different energies?

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  • $\begingroup$ Essentially a dupe of physics.stackexchange.com/q/180977/25301 $\endgroup$ – Kyle Kanos Dec 14 '19 at 4:02
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    $\begingroup$ @KyleKanos not really, did not see a good answer there $\endgroup$ – anna v Dec 14 '19 at 5:00
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    $\begingroup$ @annav dupes are proposed not necessarily because of (the entirely subjective basis of) "good answer" but on what the question actually asks. In this case, the questions are fairly similar; however I did not vote to close it as a duplicate. $\endgroup$ – Kyle Kanos Dec 14 '19 at 12:12
  • $\begingroup$ If an elephant and a flea move at the same speed, do they have the same kinetic energies? $\endgroup$ – Eric Towers Dec 14 '19 at 22:57
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    $\begingroup$ @Eric Towers Do photons have mass? $\endgroup$ – user248881 Dec 15 '19 at 2:48
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It is called special relativity, and it is the kinematics ruling at the level of particles and of large velocities generaly, close to the speed of light.

The concept of Energy in special relativity includes the energy inherent in the rest mass of the system .

$$\sqrt{P\cdot P}=\sqrt{E^2-(pc)^2}=m_0c^2$$

Here p is the momentum vector of the particle, and one can say the $(pc)$ is the kinetic energy term of the particle in special relativity. When mass equals zero, as with the photon, the total energy is kinetic energy. For photons the $E=hν$ holds, where $h$ is Planck's constant and $ν$ the frequency of light.

Thus, it is the difference in the frequency that differentiates a gamma ray photon and a radio wave photon.

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    $\begingroup$ one can say the $(pc)^2$ is the kinetic energy of the particle Did you mean $pc$? The quantity $(pc)^2$ has the dimensions of energy squared. $\endgroup$ – G. Smith Dec 14 '19 at 6:39
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    $\begingroup$ @anna v Isn’t it very odd that, referring a light wave, energy increases as the frequency increases; while referring a sound wave, energy decreases as the frequency increases. $\endgroup$ – user248881 Dec 15 '19 at 2:32
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    $\begingroup$ @user248881 hyperphysics.phy-astr.gsu.edu/hbase/mod6.html Actually for light only in the wrong solution of the classical em wave the energy goes up with frequency ; that is why quantization of black body radiation was discovered. Due to quantization the number of modes falls with frequency. The ν in the individual quantum energy goes up with frequency, not the light wave energy $\endgroup$ – anna v Dec 15 '19 at 8:10
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    $\begingroup$ @user248881 In the standard model of particle physics all photons are point particles, i,e, have no size. They do not have extent in space, the only difference between a gamma and an infrared photon is in the energy. The E=hν. The Planck constant transforms energy to a number that miraculously is the frequency of the light that will be built up by zillion such photons. All this can be shown with mathematics. The tiny describes energy, not extent. $\endgroup$ – anna v Dec 16 '19 at 5:30
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    $\begingroup$ Look at E=hν .That is the energy of a photon, and h is a constant.It is the ν that differentiates red and blue. No, their energy comes from the initial interaction that created them, either scattering or deexcitation of atoms/molecules/lattices. They do not vibrate. The wave nature comes in the probability of detectting them, i.e. many of them with the same frequency show a wave nature.See sps.ch/en/articles/progresses/… $\endgroup$ – anna v Dec 17 '19 at 5:28
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Yes. The relativistic definition of kinetic energy $K$ for a particle of mass $m$ is

$$K=E-mc^2=\sqrt{(mc^2)^2+(pc)^2}-mc^2\approx\frac{p^2}{2m}+…$$

where $E$ is the relativistic energy and $p$ the relativistic momentum.

Set $m=0$ and you get

$$K=E=pc$$

for a photon. This relation involves only Special Relativity.

In addition, quantum mechanics tells us that the energy is related to the angular frequency $\omega$ by

$$E=\hbar\omega$$

and the momentum is related to the wavenumber $k$ by

$$p=\hbar k$$

so we get the expected relation between angular frequency and wavenumber for an electromagnetic wave,

$$\omega=kc.$$

The photons of a radio wave and a gamma wave have different frequencies and thus different energies, and also different wavenumbers and thus different momenta. They may have the same speed $c$ but they have different $\omega$, $k$, $E$, and $p$ and this makes them interact differently with other particles.

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Photons behave a little like mechanical objects and a little like their own thing in this regard.

Suppose you have a squirt gun that shoots a series of water droplets. If you sit still with respect to the gun, the droplets hit you with a certain frequency, momentum, and energy. If you run toward the gun, the frequency, momentum, and energy all go up.

The energy of a photon is proportional to its frequency. $E = h \nu$.

If you run upstream into a beam of light. The frequency increases because of the Doppler shift. And so do the momentum and energy of the photons. If you ran at a suitable relativistic velocity, you could turn a radio wave into a gamma ray wave.

At the same time, you do not increase the speed the photons travel with respect to you by running. They always travel at the speed of light.

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It is very confusing when you read the definition of kinetic energy (the energy of a particle that it possesses due to its motion), and since both photons (with different wavelength) travel at the same speed c, you could think they should have the same kinetic energy.

It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

https://en.wikipedia.org/wiki/Kinetic_energy

But the photon is massless, and it does not have a rest frame, it by definition travels at speed c.

Photons always travel at speed c in vacuum, when measured locally. Its energy and momentum are related by E=p*c (where p is the magnitude of the momentum vector).

$E^{2}=p^{2} c^{2} + m^{2} c^{4}$

Since the photon is massless, this will reduce to E=pc.

The energy and momentum of the photon only depend on its frequency or inversely on its wavelength. $E=h\nu=\frac{hc}{\lambda}$

$p=\frac{h\nu}{c}=\frac{h}{\lambda}$

The photons energy (E) is kinetic energy (p*c) and in your case the different wavelength photons have different energy and momentum. Still, they both travel at speed c in vacuum when measured locally.

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  • $\begingroup$ Why the downvote? $\endgroup$ – Árpád Szendrei Dec 14 '19 at 16:38
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Yes.

Moreover, you could say the energy of a photon is purely kinetic energy.

In relativity theory, massive particles have both kinetic energy and a potential energy which is proportional to their mass. Photons have no mass, hence their energy is purely, and wholly, kinetic.

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