# Pressure due to a single photon

I'm interested in the pressure exerted on a material when a single photon is absorbed.

I have written the following expression for pressure:

$$P = \frac{hf}{cA\Delta t},$$ where $$A$$ is the area over which the force is exerted, $$\Delta t$$ is the time over which the force is exerted, $$h$$ is Planck's constant, $$f$$ is the frequency of the photon, and $$c$$ is the speed of light.

Something seems fishy.

• This is like asking what is the pressure when a single atom is adsorbed onto a surface. Pressure comes from the average force when many such interactions are averaged over an area. Mar 10 at 1:18
• "Pressure" is a statistical phenomenon. What you get from a single impact is called impulse—change in momentum—of which the unit is Newton seconds. Mar 10 at 1:18

I don't think that $$A$$ is well-defined concept for a photon, and I'm not too sure about $$\Delta t$$ either. What does make sense is to interpret your expression as the mean pressure exerted per photon when a decent number of randomly distributed photons lands normally on area $$A$$ of the absorber in time $$\Delta t$$.

• Thank you. This was helpful. Mar 23 at 2:08

Too many variables:

$$f = c/\lambda$$

and by dimensional analysis:

$$\Delta t \propto \lambda/c$$

and

$$A \propto \lambda^2$$

so

$$P\propto \frac{hc/\lambda}{c\lambda^2\lambda/c} = \frac{hc}{\lambda^4}$$

Does that look better?

• Thank you! Yes, that does look better. Mar 23 at 2:09

It's incorrect to measure energy for a time interval without knowing rate of energy transference, power. So Pressure can be easily known from intensity of photons and rate of transportation of photons, thus $$p=\frac{\varepsilon}{c}$$

Answer given by JEB is incorrect because wavelength of light is in direction of propogation not along the dimension of incident area just because dimensionally it is equal.