1
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – mmesser314
    Mar 10 at 1:18
  • 1
    $\begingroup$ "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. $\endgroup$
    – Solomon Slow
    Mar 10 at 1:18

3 Answers 3

Reset to default
2
$\begingroup$

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 $\Delta t$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you. This was helpful. $\endgroup$
    – trading
    Mar 23 at 2:08
0
$\begingroup$

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?

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you! Yes, that does look better. $\endgroup$
    – trading
    Mar 23 at 2:09
-2
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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