# What would be the force on a beam of particles from focused sunlight?

1. Jan 25, 2010

### DWM91

What would be the force imparted on a beam of particles from focused sunlight?
For the sake of argument all particles are going at the same speed.
For the sake of argument you utilize the 'full ' spectrum of the sun (or at least, most of it).

Thanks for your time and effort

DWM91

2. Jan 27, 2010

### physixlover

interesting question!!!

3. Jan 27, 2010

### chemisttree

Last edited by a moderator: Apr 24, 2017
4. Jan 28, 2010

### physixlover

Last edited by a moderator: Apr 24, 2017
5. Feb 2, 2010

### chemisttree

Yes, but photons have momentum. How would you relate momentum to force?

6. Feb 2, 2010

### Bob S

See
http://en.wikipedia.org/wiki/Thomson_scattering
Thomson scattering is inelastic scattering of photons off of free charged particles. For electrons, the cross section (probability) is about 2/3 barns (0.666 x 10-24 cm2) per particle. For protons is is about 18372 less. This is essentially low energy Compton scattering, which provides more kinematic (momentum transfer) formulas. If there were a mol (6 x 1023) electrons in 1 cm2, the probability of a photon hitting one is about 40%. If there were 1 Coulomb of electrons in 1 cm2, the probability would be about 4 parts per million.

Bob S

7. Feb 2, 2010

8. Feb 2, 2010

### Bob S

Radiation pressure is the result of electromagnetic energy (focused sunlight or photons of any energy) transferring momentum to surfaces either by absorption or reflection. The surface is a usually a solid object. The radiation-pressure momentum-transfer process can be demonstrated by a Crookes radiometer. See

For a beam of charged particles, like electrons, the incident sunlight can interact only by Thomson scattering (low energy Compton scattering). In this case, the photon energy cannot be fully absorbed due to conservation of both energy and momentum requirements, so the photon scatters off the electron and transfers some (but not all) energy and momentum to it. Sunlight can also scatter off of a beam of protons, but the probability is much lower.

For a beam of neutral atoms, like hydrogen atoms, the sunlight will not interact until the photons reach an energy of about 10.2 eV (~1215 Angstroms), which is short-wavelength UV light. For the hydrogen atom to absorb the photons, the photon energy has to match the transition energy between bound states, like the 2p-1s transition (above). The absorbed photon is re-radiated in a few nanoseconds.

Sunlight can also "reflect" off of neutral atoms, like oxygen and nitrogen atoms in the atmosphere. In this case, the radiation excites an electric dipole moment in the entire atomic electron cloud, which re-radiates in a classic radiation pattern. This was first explained by Lord Rayleigh. Like the other cases above, this photon scattering transfers momentum.

Bob S

9. Feb 3, 2010

### Lsos

I'm going to go ahead and ask the obvious question....can you attach an extremely powerful flashlight to a spaceship in order to accelerate it without any propellant?

10. Feb 3, 2010

### cragar

yes . action reaction

11. Feb 4, 2010

### lolomolo

but dude it would be sooooooooooooooooooo unefficient or you would have to have a flash light 10square miles to have eny reasonable propultion

12. Feb 6, 2010

### Lsos

Maybe so, but you wouldn't need any propellant. Just that simple fact removes many constrains, and opens the door to many possibilities...

13. Feb 7, 2010

Staff Emeritus
No, "that simple fact" does not "remove many constraints" or open any doors. You have to look at all the facts, not just one. The most relevant is that the amount of energy you need to accelerate your rocket per unit of momentum is maximized for the flashlight rocket - i.e. this is the most inefficient possible way to propel a rocket.

Numbers are important.

14. Feb 7, 2010

### Lsos

I don't claim to have made any calculations, but I doubt that you have either.

But, how efficient are our current methods of dumping out thousands of tons of mass out the tail pipe, considering that e=mc^2?

I agree that numbers are important, but I don't see any in this discussion yet. Taking your own advice then, how can we make (or dismiss) any claims at this point?

Anyways, I am too lazy and hung over, and don't have enough knowledge to attempt to run any formulas yet....

...but apparently somebody else was in a better state. As I suspected I'm not the first to have had this thought....

http://www.eetindia.co.in/ART_8800480919_1800010_NT_502b5a91.HTM

15. Feb 12, 2010

Staff Emeritus
Well, that's not just insulting, but it's also wrong.

The calculation is very simple. E = pc. dE/dt = c dp/dt, or P = F c.

The thrust of the Apollo CSM is 91.2 kN. The power needed to supply that is therefore 27 terawatts. That's about twice the total power produced by every power station on the planet.

16. Feb 12, 2010

### Lsos

Your calculations imply that absolute thrust and efficiency are the most important aspects of spacecraft propulsion.

Obviously you're not looking at "all the facts" either, since by your logic that would make currently used ion thrusters quite useless. In contrast to the CSM engine, not only does an ion thruster provide mere fractions of a newton of force, if we could even push one to 91.2 kN, it would require Gigawatts of power. Yeah it's no "tera", but it's not the pinnacle of efficiency either.

Besides that, while I admit I don't fully understand how you got 27 terawatts, I do understand that this is not a physically unattainable number. In fact, a kilogram of matter has enough energy to sustain that output for many weeks.

I am obviously looking a bit ahead of what we can currently do by burning fossil fuels, but I am by no means looking into the realm of the impossible.