Momentum Kick From Flashlight
Most recent answer: 06/21/2011
Q:
1. Since light has momentum, does that mean that my flashlight has a (very small) kickback? If so, how powerful would a flashlight/laser/whatever need to be in order to have a kickback comparable to that of a gun?
2. When a photon hits a solar sail, it imparts energy to the sail and its frequency decreases slightly. What would happen to the photon if you kept repeating this ad infinitum? Is there an absolute low-energy state for light?
- Nick (age 37)
Charlotte, NC, USA
- Nick (age 37)
Charlotte, NC, USA
A:
Comparing the flashlight kickback to a bullet kickback is a little tricky because the bullet kickback happens in a quick burst, but the flashlight kickback occurs gradually. Let's say, however, that you have some light source that dumps its energy out quickly, more or less like a bullet. Then we can make a very crude calculation without knowing much.
Except for a factor of two for the bullet (way inside our error bars), both the light and the bullet have momentum p=E/v, where E is the energy and v is the speed. The bullet speed is somewhere around the 300 m/s, (again, not worrying about x2) compared to 3x108 m/s for the light. So for a given E, the light only has 10-6 times as much p.
Let's say that the batteries have about the same energy density as the gunpowder. (That's not true for real batteries, but maybe we have some super-duper new fuel cells.) Then to get the same momentum you'd need batteries with about 106 times a bullet's worth of volume. My memory of shooting 22's in junior high is that they have a cm3 or so of volume. So you'd need about a cubic meter of high-density batteries dumping their energy quickly into light to get a kickback comparable to that from a small bullet.
I think that the total energy per pulse of the collection of lasers used in inertial confinement fusion research is somewhere around the right magnitude. () The energy is dumped from a huge bank of capacitors. Of course these lasers are designed to give zero net momentum, so they'd have to be aimed the same direction to give a net kick.
On your second question, electrodynamics has no known lowest energy scale or longest wavelength. However, in our universe there are going to be some problems describing things with wavelengths greater than about 10 billion light years, because different parts would be outside each others horizon.
Mike W.
Except for a factor of two for the bullet (way inside our error bars), both the light and the bullet have momentum p=E/v, where E is the energy and v is the speed. The bullet speed is somewhere around the 300 m/s, (again, not worrying about x2) compared to 3x108 m/s for the light. So for a given E, the light only has 10-6 times as much p.
Let's say that the batteries have about the same energy density as the gunpowder. (That's not true for real batteries, but maybe we have some super-duper new fuel cells.) Then to get the same momentum you'd need batteries with about 106 times a bullet's worth of volume. My memory of shooting 22's in junior high is that they have a cm3 or so of volume. So you'd need about a cubic meter of high-density batteries dumping their energy quickly into light to get a kickback comparable to that from a small bullet.
I think that the total energy per pulse of the collection of lasers used in inertial confinement fusion research is somewhere around the right magnitude. () The energy is dumped from a huge bank of capacitors. Of course these lasers are designed to give zero net momentum, so they'd have to be aimed the same direction to give a net kick.
On your second question, electrodynamics has no known lowest energy scale or longest wavelength. However, in our universe there are going to be some problems describing things with wavelengths greater than about 10 billion light years, because different parts would be outside each others horizon.
Mike W.
(published on 06/21/2011)
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