29.4 Photon momentum (Page 3/5)

College physics Page 3 / 5

(a) A payload having an umbrella-shaped solar sail attached to it is shown. The direction of movement of payload and direction of incident photons are shown using arrows. (b) A photograph of the top view of a silvery space sail. — (a) Space sails have been proposed that use the momentum of sunlight reflecting from gigantic low-mass sails to propel spacecraft about the solar system. A Russian test model of this (the Cosmos 1) was launched in 2005, but did not make it into orbit due to a rocket failure. (b) A U.S. version of this, labeled LightSail-1, is scheduled for trial launches in the first part of this decade. It will have a 40-m ² sail. (credit: Kim Newton/NASA)

Relativistic photon momentum

There is a relationship between photon momentum $p$ and photon energy $E$ that is consistent with the relation given previously for the relativistic total energy of a particle as $E^{2} = (pc)^{2} + (mc)^{2}$ . We know $m$ is zero for a photon, but $p$ is not, so that $E^{2} = (pc)^{2} + (mc)^{2}$ becomes

E = pc,

p = \frac{E}{c} (photons).

To check the validity of this relation, note that $E = hc / λ$ for a photon. Substituting this into $p = E / c$ yields

p = (hc / λ) / c = \frac{h}{λ},

as determined experimentally and discussed above. Thus, $p = E / c$ is equivalent to Compton’s result $p = h / λ$ . For a further verification of the relationship between photon energy and momentum, see [link] .

Photon detectors

Almost all detection systems talked about thus far—eyes, photographic plates, photomultiplier tubes in microscopes, and CCD cameras—rely on particle-like properties of photons interacting with a sensitive area. A change is caused and either the change is cascaded or zillions of points are recorded to form an image we detect. These detectors are used in biomedical imaging systems, and there is ongoing research into improving the efficiency of receiving photons, particularly by cooling detection systems and reducing thermal effects.

Photon energy and momentum

Show that $p = E / c$ for the photon considered in the [link] .

Strategy

We will take the energy $E$ found in [link] , divide it by the speed of light, and see if the same momentum is obtained as before.

Solution

Given that the energy of the photon is 2.48 eV and converting this to joules, we get

p = \frac{E}{c} = \frac{(2.48 eV) (1 . 60 \times 10^{–19} J/eV)}{3 . 00 \times 10^{8} m/s} = 1 . 33 \times 10^{–27} kg \cdot m/s .

Discussion

This value for momentum is the same as found before (note that unrounded values are used in all calculations to avoid even small rounding errors), an expected verification of the relationship $p = E / c$ . This also means the relationship between energy, momentum, and mass given by $E^{2} = (pc)^{2} + (mc)^{2}$ applies to both matter and photons. Once again, note that $p$ is not zero, even when $m$ is.

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29.4 Photon momentum (Page 3/5)

Relativistic photon momentum

Photon detectors

Photon energy and momentum

Problem-solving suggestion

Section summary

Conceptual questions

Problems&Exercises

Questions & Answers

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