Since light beams carry momentum, why would a person holding a flashlight not feel a recoil similar to that of a rifle being fired?
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ Because they have extremely little momentum when compared to a bullet. $\endgroup$– JavierCommented May 22, 2014 at 0:10
-
$\begingroup$ Thank you for your answer, but I am looking for one that is more in-depth. I believe I should talk about photons and such, I cannot formulate a strong answer, however. $\endgroup$– yankeejman1042Commented May 22, 2014 at 0:11
-
$\begingroup$ To a significant extent, @JavierBadia's comment is the "more in-depth" version. That really is all there is to it. $\endgroup$– dmckee --- ex-moderator kittenCommented May 22, 2014 at 1:05
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
One can do a simple calculation. Suppose you hold a $60\text{W}$ lightbulb and you turn it on. Each second it emits $60 \text{J}$ of energy. Using the energy-momentum relation for photons, $E=pc$, we get a momentum of $p = 60\text{J}/c = 2\cdot 10^{-6}\ \text{kg}\cdot\text{m}/\text{s}$, or a force $F = 2\cdot 10^{-6}\ \text{N}$. In contrast, a Wikipedia page about firearms says that each bullet of a .44 Remington Magnum comes out with a momentum of roughly $5\ \text{kg}\cdot\text{m}/\text{s}$. That's quite a bit more than the flashlight.
(I should make a disclaimer that I know nothing about guns, I just pulled the numbers from Wikipedia. When we have a difference of six orders of magnitude, you don't really care about these things.)
-
$\begingroup$ $(60\ \mathrm{J})/(3\times10^8\ \mathrm{m/s}) = 2\times10^{-7}\ \mathrm{m/s}$ $\endgroup$– user10851Commented May 22, 2014 at 6:21
-
$\begingroup$ @user10851 what is that quantity supposed to be? $\endgroup$– Mark CCommented Jun 22, 2017 at 15:41