## Solar panels and pesky photons

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### Solar panels and pesky photons

Hi.

Today in the physics lesson in my school (no, this is not a homework question) we essentially spoke about solar panels and their use in travelling to distant galaxies to get to know some cool extraterrestical dudes. We were supposed to calculate the force that the sunlight applies to a one square meter solar panel in average earth orbit's distance to the sun. For simplicity's sake, we pretended that the whole light the sun radiates has a wavelength of 500nm.

[The following paragraph is not interesting]
What I therefore calculated was the surface area of an imaginary sphere around the sun containing the solar panel, which (as obviously all the light the sun radiates will hit this surface) allowed me to calculate the accumulated output power of all the photons hitting a one square meter area on this surface, which then divided by the energy of each single photon (E=h\cdot f, c=\lambda\cdot f) gives me the number of photons that hit this one square meter surface. I can also calculate the impulse of the photons by E=m\cdot c^{2}. If I derive impulse over time, I get force, which is what I wanted to have (well, I still get force/area, which is pressure, but I can just apply that to my one square meter solar panel and get force), and it could be that simple.

But it isn't. The problem that I had a long discussion with my teacher about is the actual force that each single photon applies to the surface. The general formula for all those photons should be like:
F=n\cdot k\cdot p_{Photon}
where k is the magic factor the discussion was about.
First, let's pretend the body was perfectly black: in this case, k would be around 1, right? After all, the photons just hit it and get annihilated, and even if they do somewhat warm up the body, they should mostly make it move. This was the part my teacher and me agreed about, even though now I'm not even sure anymore (as I don't know how much warmth we actually get).
Second example, and this was the difficult one: What if the body was a perfect mirror? My teacher said that k would have to be 2, but that, I think, is just totally impossible as it would violate the law of the conversation of energy. His idea was that each photon would not only hit the mirror and give it kinetical energy, but would also bounce back, in this again accelerating the solar panel (like if you jumped of a boat, he said). I didn't agree on this (pleading on the law of conversation of energy), and my teacher made some points about Compton effect and energy-impulse-invariance, but I do still not agree with him: IMO the value for k should be very, very low, in fact only affected by the Compton effect: A photon goes into the atom, then goes out again (in direction of angle of reflection and stuff), somewhat redshifted, with all the energy that it lost by this redshift now turned into kinetic energy for the solar panel (E=h\cdot f again) just because that's the only place the energy could possibly go (warmth aside again, but that's just kinetic energy, too, strictly speaking).

Could someone clarify on both these points? I'd be very grateful for any answer.
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cemper93

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### Re: Solar panels and pesky photons

cemper93 wrote: His idea was that each photon would not only hit the mirror and give it kinetical energy, but would also bounce back, in this again accelerating the solar panel (like if you jumped of a boat, he said). I didn't agree on this (pleading on the law of conversation of energy)

Your teacher is right (I'm not sure about the k value, but his explanation). If we're in a vacuum, I throw you a ball an you catch it, to conserve momentum you'll gain velocity in the direction of the ball's initial travel. If you then throw it back to me, to conserve momentum you'll travel in the opposite direction of the ball, meaning you've received two kicks.
The thing that I think you're missing is that when you reflect the photon, is loses energy (since h and c don't change, this means you redshift). Essentially, you just need to solve two simultaneous equations:
pbefore=pafter and Ebefore=Eafter

We can solve this since we know:

pbefore = h/λ1
pafter = m v - (h/λ2)

Ebefore = (h c)/λ1
Eafter = ((h c)/λ2) + (1/2) m v2

Actual solving this simultaneous equation is left as an exercise to the reader. Should probably use relativistic momentum and energy, but I'm both lazy and tired.

Edit: And I have a feeling I screwed up something special relativity, but I can't spot it.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

yurell

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### Re: Solar panels and pesky photons

k would be a bit smaller than 2 for a perfect mirror. The reflected photons lose a tiny amount of energy, as they give a part of their energy to the ship and the ship moves after that. Both the absorption and the emission accelerate the ship, so k is not small, but close to 2.
If the ship is not at rest, the description is a bit different for the system of the sun (photons lose a relevant amount of energy, as P/F is higher) and the ship (photons lose nearly no energy).
mfb

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### Re: Solar panels and pesky photons

If we're in a vacuum, I throw you a ball an you catch it, to conserve momentum you'll gain velocity in the direction of the ball's initial travel. If you then throw it back to me, to conserve momentum you'll travel in the opposite direction of the ball, meaning you've received two kicks.
Doesn't that only make sense if you don't think too hard about it? Okay, I see it's just an analogy, but in classical physics, if I throw something, I generally loose some chemical energy that was stored in my arms or something and turn it into kinetic energy for the ball. If I get hit by an object in space, I receive only one kick, even if it bounces of me (say, a baseball or something). These laws also hold if I throw back a lighter object (whether I do only throw back half the baseball or a redshifted version of the photon). But I guess you just wanted to leave out some quantum weirdness effect, which is very kind of you, but giving me a rough idea of what's happening would be nice, too.
Or is the thing that I don't get the flipped sign of the photon's momentum due to its way of travel now being the other way round? Yes, that would actually make a lot of sense: I do first slow down the photon to zero speed, then accelerate it again to -c speed, and have exactly two kicks. Uah, my mind sometimes just doesn't work right (but my teacher really explained it badly, too). Thank you two, anyway.

And I have a feeling I screwed up something special relativity, but I can't spot it.
Maybe that in special relativity, impulse isn't p = m \cdot v, but this weird stuff like \frac{m\cdot v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}? (Yeah, I just wanted to act as if I had understood anything, too.)
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cemper93

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### Re: Solar panels and pesky photons

cemper93 wrote:For simplicity's sake, we pretended that the whole light the sun radiates has a wavelength of 500nm.

The wavelength doesn't matter. Light alway carries a momentum of E/c.

cemper93 wrote:After all, the photons just hit it and get annihilated, and even if they do somewhat warm up the body, they should mostly make it move.

No, mostly to heat and very little to make it move. Of corse the actual energy going into increasing its momentum by E/c depends on it velocity. If it is standing still it takes no energy (for differential increase) if it is moving towards the sun it even loses energy. This energy is balanced by the apparent red/blue shift from the ships POV)

cemper93 wrote:My teacher said that k would have to be 2, but that, I think, is just totally impossible as it would violate the law of the conversation of energy. His idea was that each photon would not only hit the mirror and give it kinetical energy, but would also bounce back, in this again accelerating the solar panel (like if you jumped of a boat, he said). I didn't agree on this (pleading on the law of conversation of energy)

k is more than two when moving towards the sun and less than 2 when moving away (because of conservation of energy as you say), but at realistic speeds this is a very small effect. If your movement is slow enough that the photon does not change wavelength appreciably, then k is two, and a negligible amount of the photons energy has gone to the ship. c is very large, and photon drives/solar sails are very inefficient when compared to the amount of light in use. Solar sails only work because there is such an abundance of light.

Tass

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### Re: Solar panels and pesky photons

The wavelength doesn't matter. Light alway carries a momentum of E/c.

But then what is E? The only formula I know to calculate a photon's energy is the de Broglie Formula E=h\cdot f, which involves wavelength.

In fact, I can totally not understand how light's momentum can be completely independent of everything at all. Because if it is, redshifting doesn't change a photon's momentum and I could construct some kind of perpetuum mobile like this:
As if F=n\cdot k\cdot p_{Photon} would hold, and p_{Photon} were constant (n and k being constant because n has to and it would blow my mind if k weren't), I could hit the same surface as often as I wanted with the same photon and would apply the same force every time, which just doesn't make any sense whatsoover. (Yes, I know there aren't any perpetuum mobiles, I'm just too stupid to understand this.)

No, mostly to heat and very little to make it move

Okay, yes, makes sense. But then, why do black bodies move at all if I hit them with photons? Yeah, they irradiate warmth, but AFAIK that doesn't make them move, or does it?

If your movement is slow enough that the photon does not change wavelength appreciably, then k is two, and a negligible amount of the photons energy has gone to the ship.

:O My mind! Shouldn't I expect that larger values of k make the photon loose more energy to the ship, not less? As, after all, k is a factor that I multiply with the photon's momentum to determine how much momentum the ship gains from the impact? And if the ship gains more momentum, the photon should loose more and therefore loose lots of energy, too?
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cemper93

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### Re: Solar panels and pesky photons

cemper93 wrote:
The wavelength doesn't matter. Light alway carries a momentum of E/c.

But then what is E? The only formula I know to calculate a photon's energy is the de Broglie Formula E=h\cdot f, which involves wavelength.

I believe Tass is saying that you can run your calculations without needing to know the wavelength of the photon, just set the photon's momentum to E/c before and E '/c after ... a photon's energy is wavelength dependent.

cemper93 wrote:
No, mostly to heat and very little to make it move

Okay, yes, makes sense. But then, why do black bodies move at all if I hit them with photons? Yeah, they irradiate warmth, but AFAIK that doesn't make them move, or does it?

To conserve momentum. If you stop the photon, you've suddenly gained all its momentum and you heat up to conserve energy (note that you can't think of your ship as just one perfect solid object at this point — if you consider an electron in a vacuum, it can't actually absorb an electron because there's no way to conserve both energy and momentum, and so it always reflects it).

The warmth itself can make them move, if it's not distributed evenly. If I have a hot rear and a cold front, I will be spitting more photons out my rear than my front, accelerating me forward until the temperature difference is removed and I begin emitting isotropically.
cemper93 wrote:Dude, I just presented an elaborate multiple fraction in Comic Sans. Who are you to question me?

yurell

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### Re: Solar panels and pesky photons

yurell wrote: if you consider an electron in a vacuum, it can't actually absorb an electron a photon because there's no way to conserve both energy and momentum, and so it always reflects it or ignores it.

FTFY

PM 2Ring

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### Re: Solar panels and pesky photons

cemper93 wrote:
The wavelength doesn't matter. Light alway carries a momentum of E/c.

But then what is E? The only formula I know to calculate a photon's energy is the de Broglie Formula E=h\cdot f, which involves wavelength.

You know the power. Whether the suns output power is divided into fewer or more photons doesn't matter to a solar sail. F=P/c. (Possibly with a factor k).

Basically going around wavelength and number of photons back to radiation pressure is roundabout. The quantization of light does not matter here, and you don't need Planck's constant.

Tass

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