16. Photon Momentum

Now that a photon model has been deduced, the next sections (16-22) will see how the model corresponds to various light and photon observations. In this section, the momentum of the photon is derived, giving the expected result.

The photon has a magnetic field based on the wave solution from Eq.(15-1). The electric and magnetic fields are, in relation to the magnitude of the electric field¹:

(1)

(2)

The momentum density can be derived in the usual way² from Eq.(1-5). This yields

(3)

Substituting Eq.(1)&(2) into Eq.(3) gives

(4)

Note that the momentum density is all in the same direction. If the entire photon is absorbed, the momentum transferred can thus be found by summing over the total volume of the photon. We then obtain:

(5)

By substituting in the energy of the electric field, this becomes

(6)

The energy in the magnetic field is given by (from Eq.(1-5)):

(7)

Substituting Eq.(2) here yields

(8)

Thus we can see that the energy in the magnetic field of a photon is equal to the energy in the electric field. We can thus show the magnitude of momentum of a photon in relationship to the total energy of a photon (x) by:

(9)

Thus, if a photon is absorbed, it will transfer this amount of momentum³.

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1. The photon, even though having a cylindrical field can be treated as a plane wave in the direction of propagation. The relationship then is like what can be found in Griffiths¹ 357.

2. This can be found in Griffiths¹ 331 or Jackson² 239.

3. This is the expected amount (see for example Griffiths¹ 481) supported by observations such as the photo-electric effect. This also relates to radiation pressure which can be calculated macroscopically (Griffiths¹ 359, Jackson² 265) or by summing over the number of photons.

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