Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: chris on 10/11/2018 20:19:22

A visitor to a page on the site about the force exerted on objects by light (https://www.thenakedscientists.com/articles/questions/canlightexertforcemoveobject) has suggested that a verticallyangled torch can push particles of dust along.
I've not done the maths, but my instincts are that the force on a dust particle from the flimsy beam of a torch will be tiny  but I wondered if anyone could help me to do the calculation?
So, the question I want to solve is, how much force does the beam from a torch apply to the average dust particle passing through the beam?

Around 1 micronewton per square meter.

Thanks  but can you explain how I calculate that to prove the value?

The pressure is going to fall off the further away from the light you are, so I'll assume a best case scenario where we are looking at the pressure the light is producing right at the face of the flashlight.
The equation for calculating radiation pressure is P = I_{f}/c, where:
P is pressure in pascals
I_{f} is the irradiance in watts per square meter
c is the speed of light in meters per second
If I have a 5 watt flashlight with a lens radius of 2.5 centimeters (0.025 meters, 0.0019634954 square meters), then the irradiance will be about 2,546.48 watts per square meter. This assumes that 100% of those watts are going into the creation of electromagnetic radiation, which we know it wouldn't be. The actual efficiency might run anywhere from 10% to 90% depending on the kind of bulb used.
So now we put the relevant numbers into the equation:
P = 2,546.48 watts per square meter/299,792,458 meters per second
P = 0.00000849414 pascals (newtons per square meter)
Multiply that number by a value from 0.1 to 0.9 depending on how efficient the lightbulb is to get a better estimate.
EDIT: According to Engineering Toolbox, household dust particles range in size from 0.05 to 100 microns across. For the sake of convenience, I'll assume they are spherical, resulting in crosssectional areas of 0.0019634954 square microns (1.9634954 x 10^{15} square meters) and 7,853.98163397 square microns (7.85398163397 x 10^{8} square meters) respectively. Now to calculate the force on these particles:
F = 0.00000849414 newtons per square meter x 1.9634954 x 10^{15} square meters
F = 1.6678205 x 10^{20} newtons
F = 0.00000849414 newtons per square meter x 7.85398163397 x 10^{8} square meters
F = 6.671282 x 10^{13} newtons
To find the acceleration that the light imparts on these particles, their mass will need to be known. If I assume that they are made of organic matter, then a density similar to water might not be far off the mark (if it is, we can modify this later). A sphere 0.05 microns across has a volume of 6.54 x 10^{5} cubic microns (6.54 x 10^{23} cubic meters). A sphere 100 microns across has a volume of 5.24 x 10[/sup]5[/sup] cubic microns (5.24 x 10^{13} cubic meters). The density of water is 1,000 kilograms per cubic meter, so:
m = 6.54 x 10^{23} cubic meters x 1,000 kilograms per cubic meter
m = 6.54 x 10^{20} kilograms
m = 5.24 x 10^{13} cubic meters x 1,000 kilograms per cubic meters
m = 5.24 x 10^{10} kilograms
Now for the acceleration:
A = F/m
A = 1.6678205 x 10^{20} newtons / 6.54 x 10^{20} kilograms
A = 0.255 meters per second squared (for 100% efficient light)
A = 0.23 meters per second squared (for 90% efficient light)
A = 0.0255 meters per second squared for (for 10% efficient light)
A = F/m
A = 6.671282 x 10^{13} newtons / 5.24 x 10^{10} kilograms
A = 0.00127 meters per second squared (for 100% efficient light)
A = 0.00114 meters per second squared (for 90% efficient light)
A = 0.000127 meters per second squared (for 10% efficient light)

...of course, a dust particle is continually being knocked by air molecules, which absorb any accumulated velocity, and impart their own random velocity to the dust particle.
See: https://en.wikipedia.org/wiki/Brownian_motion

Interestingly, that means that (for some dust particles and a really good torch) you you could "levitate" dust on the Moon.
https://www.universetoday.com/35565/gravityonotherplanets/
For a given material, reducing the radius 10 fold would reduce the area (and, therefore the force) 100 fold and the mass 1000 fold so the acceleration would increase 10 fold.
That means that very small dust particles would be blown away from the moon's gravity by sunlight.

...of course, a dust particle is continually being knocked by air molecules, which absorb any accumulated velocity, and impart their own random velocity to the dust particle.
See: https://en.wikipedia.org/wiki/Brownian_motion
Yes, that's true. The drag forces imposed on a dust particle would be relatively large due to its tiny size.
Another potential complicating factor if one were to try to observe this at home would be convection currents generated by the heat of the light bulb. For all I know, those might impose more force on dust than the light beam.

Thanks @Kryptid  one question  how is this equation  P = I_{f}/c  derived in the first place. I'm trying to do the dimensions / units on it to understand it, but I'm struggling.

Thanks @Kryptid  one question  how is this equation  P = I_{f}/c  derived in the first place. I'm trying to do the dimensions / units on it to understand it, but I'm struggling.
It seems to have something to do with the Poynting vector: https://en.wikipedia.org/wiki/Poynting_vector
That's a little above my understand, unfortunately.

lol, not a single mention of thermal "heat" transfer being the primary driving force! lol

Around 1 micronewton per square meter.
This suffices but a square meter is more of a flood light! lol

lol, not a single mention of thermal "heat" transfer being the primary driving force! lol
Possibly because. in principle, the light might all be reflected which gives no rise in temperature.
Why did you find that funny?

Why did you find that funny?
It's an indictment on pretentiousness! lol

Why did you find that funny?
It's an indictment on pretentiousness! lol
I must be missing something.
What pretentiousness?

I must be missing something.What pretentiousness?
yawn! lol

I must be missing something.What pretentiousness?
yawn! lol
That didn't really address the issue.

This suffices but a square meter is more of a flood light! lol
The value he stated was a measure of radiation pressure. It does not imply that the light in question necessarily covers an area of one square meter.
I'll now attempt to calculate the maximum speed of a dust particle due to this force. I'll do this by using the terminal velocity equation:
V_{t} = sqrt((2mg)/(ρAC_{d})), where
V_{t} is terminal velocity in meters per second (which is just maximum speed in this case)
m is the mass of the dust particle in kilograms
g is the acceleration due to the gravity in meters per second squared (but will be substituted for the accelerations from earlier)
ρ is the density of the air in kilograms per cubic meter
A is the crosssectional area in square meters
C_{d} is the drag coefficient (0.47 for a sphere)
V_{t} = sqrt((2(6.54 x 10^{20})(0.0255))/((1.225)(1.9634954 x 10^{15})(0.47)))
V_{t} = sqrt((3.3354 x 10^{21})/(1.13048247655 x 10^{15}))
V_{t} = sqrt(2.95042167 x 10^{6})
V_{t} = 0.0017177 meters per second
V_{t} = sqrt((2(5.24 x 10^{10})(0.000127))/((1.225)(7.85398163397 x 10^{8})(0.47)))
V_{t} = sqrt((1.33096 x 10^{13})/(4.5219299257582275 x 10^{8}))
V_{t} = sqrt(2.943345 x 10^{6})
V_{t} = 0.0017156 meters per second