Energy balance after reflection of light - Help wanted

[h2ouniverse]

Hi all,<br /><br />I am desperately trying to find a CLEAR and physical statement about energy balance for the phenomenon of radiation pressure.<br />When facing the problem of radiation pressure, people consider balance of momentum. So for a fully absorbed photon you go from:<br />one photon with h.nu1/c momentum + one impacted object with 0 momentum <br />to:<br />one impacted object with m.v momentum<br />and you have mv=h.nu1/c<br /><br />For fully reflected photon<br />you go from:<br />one photon with h.nu1/c momentum + one impacted object with 0 momentum <br />to:<br />one impacted object with m.v momentum + one reflected photon with -h.nu2/c momentum<br />and you have mv=h.nu1/c + h.nu2/c<br /><br />Does anybody know whether nu2 = nu1 or not?<br /><br />If (as I could see written) there is no energy loss of the photon, then h.nu1 = h.nu2, and mv=2h.nu but we have a problem of conservation of the energy of the system...<br />We then even have an even weirder issue of if you consider two 45° perfect mirrors attached to each other<br />. . . . . . . .//--------- /> outcoming photon<br />mirror1 . // |<br />. . . . . . .// .| .//<br />. . . . . . . . .|.// mirror 2 <br />------------- />//<br />incoming photon<br /><br />because although you will have 0 momentum in the end for the pair of mirrors, you will have had a net displacement of the mirrors horizontally after the two reflections. (mirrors set in motion after first bounce, then stopped at second bounce).<br />That would mean you can get a mere displacement either from:<br />1) loss of energy of photon if nu2 < /> nu1<br />2) no energy at all if nu1=nu2 !!!<br />(reminds you of something?)<br /><br />Even if 2 is false, 1 would still be fantastic.<br />Where am I wrong?<br /><br />Note: things are even worse if you do not attach the mirrors. Then you end up with still 0 global momentum but with two objects with a non-zero kinetic energy...<br />

 

[emperor_of_localgroup]

Two things before I attempt to answer your questions.<br /><br />1)You must be a physics person, we engineers don't use nu anymore, we use 'f' for frequency. ,br /><br />2) It has been long since I studied this stuff in physics class, so I may be wrong. But I'll take a shot at it. <br /><br />your problem is in the statement <br /><br /><font color="cyan">one photon with h.nu1/c momentum + one impacted object with 0 momentum<br />to:<br />one impacted object with m.v momentum + one reflected photon with -h.nu2/c momentum<br />and you have mv=h.nu1/c + h.nu2/c </font><br /><br />If the photon is completely absorbed, h.nu1/c term shouldn't appear in the reflection part, only mv. <br /><br />For absorption: mv=h.nu1/c <br /><br />For radiation: mv=h.nu2/c <br /><br />If emitted photon has less energy, then mv=h.nu2/c+mv2 <br /><br />mv2 comes from the remaining speed of impacted object. Anyone who is taking a modern physics class can correct me. <br /> <div class="Discussion_UserSignature"> <font size="2" color="#ff0000"><strong>Earth is Boring</strong></font> </div>

 

[h2ouniverse]

Actually I am an engineer. But we still use nu when dealing with photons energy (that is to say once every five years on average <img src="/images/icons/smile.gif" />)<br /><br />My equations are correct for we use them for calculating solar disturbance torques on satellite (and correlating them from in-orbit measurement!).<br />Of course we have intermediate case with reflectivity rho between 0 and 1 (moreover on Solar Arrays, it slightly depends too on the fact that the section is electrically connected or in open circuit).<br />But of course given the ridiculously low value of the radiation presssure, and the large mass of the satellite, we never bother about nu2 (or f2), i.e. the frequency of the photons after reflection. <br />We generally assume nu1=nu2=nu and so have mv=h.nu/c for rho=0, and mv=2.h.nu/c for rho=1.<br /><br />I have just written constancy of total momentum: the sum must be h.nu1/c before and after the reflection in a Galilean referential where the object was motion less before the impact.<br />I have assumed a refletion perpendicular to a perfect mirror. (no angle, so momentum of reflected photon is negative vs motion of the object)<br />m = mass of the object<br />v= speed of object after the impact<br /><br />So I cannot figure out what is v2 in your post. Sorry I may not have been clear enough in the wording of the first post.<br /><br />

 

[jaxtraw]

Just in general, I think in your "2 mirror problem", the energy used to displace the mirrors will have been extracted from the photon, which will be slightly redshifted as a result. That's obvious with the first mirror- after the photon has hit it, it's moving away from the photon, so the photon emitted will be redshifted. The second mirror looks like it ought to blue shift the photon, since it's moving towards it when struck. However, I think I can get around this with some SR-- I'll choose my inertial frame to be the mirror, which in this inertial frame can be considered stationary (while the universe travels past it, heh). Since it's guaranteed to see a photon travelling at c, as the photon hits it, in this inertial frame it is again pushed away from the direction of photon travel-- so the photon is once more redshifted.<br /><br />The above may be nonsense, but it works for me <img src="/images/icons/smile.gif" />

 

[emperor_of_localgroup]

Solar cells on a satellite in space - that is very good combination of physics and engineering <br /><br /><font color="cyan">We generally assume nu1=nu2=nu and so have mv=h.nu/c for rho=0, and mv=2.h.nu/c for rho=1.</font><br /><br />I have no problem with zero reflection (rho=0) mv=h.nu/c <br /><br />But for a perfect reflection (rho=1), 2h.nu/c is the change in momentum, I'm not sure under which conditions we can equate that to mv. For example, when a tennis ball with momentum mv hits a wall, it bounces back with momentum mv, we say its change in momentum, (delta)p=2mv. From which we can compute force applied to the wall as (delta)p/(delta)t, (delta)t is the time of contact between the ball and the wall <br /><br />In your case, does the impacted particle really double its speed when rho=1? That'd be an interesting experiment in space. <br /><br />v2 comes in my equation when rho is between 0 and 1, assuming the impacted particle moves with momentum mv, then emits a photon of momentum h.nu2/c (which is lower than h.nu1/c), and the particle will have the remaining momentum mv2 (v2 is lower than v). <br /><br />Now I remember reading somewhere that proposes use of photon impacts as 'thrust" to move spaceships in zero gravity.<br /> <div class="Discussion_UserSignature"> <font size="2" color="#ff0000"><strong>Earth is Boring</strong></font> </div>

 

[h2ouniverse]

Emp,<br /><br />Yes it really doubles. In your tennis ball experiment, the 2p change is for the ball which goes from +p to -p, but by impacting the wall, it has communicated 2p to the wall (hence to Earth, and given Earth's mass, you are entitled to neglect v but you cannot neglect momentum 2p).<br />So you go from +p in ball plus zero on wall/Earth to -p in ball plus 2p on wall/Earth, conserving a total momentum of +p.<br />the equations I put above are with assumption that v=0 before impact (motion less object). So v means speed after impact. The impacted satellite does really double (or almost double if there is a redshift of the photon) its speed versus the rho=0 case.

 

[h2ouniverse]

Jaxtraw,<br />Thanks for your reply. I would assume too that there is a redshift but I have not succeeded in finding a formula to compute it!<br />Actually in my 2 mirror problem, another complication comes from the fact that the second impact spacetime point B is just marginally in the causality cone of the first impact spacetime point A. So if A and B are connected by matter, the cohesion forces that prevent the two impacts from quartering the set of mirrors will have a hard time reaching B before the photon does! So the photon will exert a force in B before the cohesion forces propagate from A to B to transmit the "order" of motion due to the first impact. A part of the two impact forces will be stored in internal deformation in the matter that connects the two mirrors. But even this way, there should be a resulting displacement along the photon's direction.<br />Even if momentum is conserved, that means it makes a way to transform light energy into a displacement with no final kinetic energy...<br /><br />Jaxtraw, do you know how to compute the redshift?<br /><br /><br />

 

[jaxtraw]

I'm too ignorant and lazy to offer a formula here, but if I were trying to calculate it, I'd work out the velocity of the mirror from the momentum-- which would of course mean the mass of the mirror has to be known.<br /><br />Then I'd work out the redshift by considering the mirror to be simply an object of a certain velocity, emitting a photon of the initial frequency of the photon, so it's a straightforward Doppler shift, I think. I don't think you can do it without knowing the mirror's mass, because then you'd be asking an undefined question...<br /><br />Mind you, I'm still quite thrilled at discovering that Google can do astronomical calculations. You can type in "1 megasparsec in angstroms" and it tells you. It's 3.08568025 × 10^32 by the way <img src="/images/icons/smile.gif" />

 

[h2ouniverse]

Thanks Jaxtraw.<br />Actually I was not thinking about the Doppler shift. But may be this is what makes the energy balance. I need to think about it!

 

[jaxtraw]

Well, I think the redshift <i>is</i> the Doppler shift. It's the same thing as solar sails- a photon hits it, now the sail is moving a little, so the emitted (reflected) photon is emitted from a moving source, so it has a red (Doppler) shift. That's where the sail gets its momentum from <img src="/images/icons/smile.gif" />

 

[h2ouniverse]

Then, Houston, we have a problem with my 2 mirror thought experiment. Because the photon is redshifted twice, and the set of mirrors has shifted toward right by distance d and ends speedless. Momentum is no longer conserved... (one photon hnu1/c in plus 0 momentum on set of mirrors; becomes => one photon hnu2/c emitted plus still 0 momentum on mirrors; where has h(nu1-nu2)/c gone?).<br />Part of energy is absorbed in internal deformation, then re-emitted as heat ultimately. But there is no reason to impose directionality to this heat emission. You should assume the photons of the heat emission have a sum of momentum that is statistically 0. <br /><br />On the opposite for energy that can still be OK. You can say photon energy loss = internal deformation energy + work of the radiation pressure force over distance d.

 

[jaxtraw]

Then I withdraw my previous gobbledegook and propose that the photon is blueshifted at the second mirror, thus momentum is conserved... <img src="/images/icons/smile.gif" /><br /><br />I was trying to avoid doing any work (moving the mirrors) because I couldn't see where the energy comes from to do it if the photon ends up back as it was. But maybe that doesn't matter. But...<br /><br />If I attach a magnet to one of the mirrors, and put it in a coil, I can extract energy from the system, and I don't see where that energy's coming from if the photon emerges with as much energy as it went in. So now I'm very confused, but I think that's because I'm not very bright...

 

[h2ouniverse]

No no that is OK for you to speculate! . I think this is a very difficult problem.<br />I do not think you can have a blueshift with the second bounce because it occurs at the limit of the causality cone of the first impact A. Coordinates in spacetime of second impact B are not in the causality cone of the mediating particles ensuring the cohesion forces between the two mirrosr: the matter connecting them has to be longer than A-B distance. So matter at B cannot be moving yet when hit by the photon (the signal to set it in motion is still on its way from A). You cannot have a blueshift IMO. And even if you had, you would have to substract a loss to account for the bounce.<br />I maintain the paradox. May be it is even EPR paradox-related. <br /><br />You have a variant that is as mind-boggling. Imagine the two mirrors are free to move vs each other but inside a box, with limited gap. Then after the two bounces the mirrors shift from each other in opposite directions, but after few µs they both hit the box internal walls, canceling their kinetic energy and momentum. You still end up with a zero momentum box+mirrors, and a photon momentum that has decreased due to two redshifts..

 

[jaxtraw]

Thinking about this in the wee small hours, I think the answer is there isn't a paradox, at least from the perspective I've approached it from, and the answer is a bit dull <img src="/images/icons/frown.gif" /><br /><br />Calling the first mirror the photon hits A, and the second B, and assuming that A and B are connected by a rod I think the scenario is this--<br /><br />The photon hits A and is reflected. A starts moving, and the photon has been redshifted- which means the photon is now less energetic, which I think is important. Now, as you've pointed out, forces in the rod won't travel fast enough to start B moving before the reflected photon hits it. So, the photon hits B, and it starts moving away from A- but with less momentum (velocity) since the photon is less energetic. The photon is reflected again and redshifted again, so it's less energetic again.<br /><br />Some time after the second reflection, the pull from A reaches B, stops it moving and starts pulling it back towards A, a little (since the momentum of A is greater than B). The rod has been strained slightly, which eventually stops A moving and pulls it back most of the way towards where it started. The net result is a small movement of A, which has simply been lessened by the presence of B. The energy extracted from the photon will appear as heat in the rod.<br /><br />So really it's a grossly inefficient solar sail. All that B achieves is to rob A of most of its motion, and to convert that motion into heat in the rod.<br /><br />That's my take on it, anyway <img src="/images/icons/smile.gif" /><br /><br />

 

[h2ouniverse]

That is my view also. But the heat will ultimately be radiated isotropically, so with 0 resulting global momentum. And we have lost the balance of momentum!!!<br />

 

[jaxtraw]

Well, our tie-rod and mirror system looks to me like it'll oscillate and ultimately the momentum is lost due to frictional forces(?) of that oscillation. I have no idea TBH what stops a tuning fork humming eventually. I presume it's internal friction. So far as I can see, the momentum has been transferred to the atoms of the system, which will eventually radiate it as heat.<br /><br />? <br /><br /><br />Okay, here's another thought experiment. I'm going to build a reactionless space drive! I take a wooden plank, and screw a battery powered light bulb to one end. At the other end ("the front") I attach a light sail. Will it propel me (slowly) to the stars? Can I get some momentum for free?

 

[h2ouniverse]

In your thought experiment, it is like blowing a boat sail with a fan on the boat. You will not get momentum.<br />The photons emitted from the bulb will produce an opposite momentum -p. So you get -p on bulb +p on photon= /> -p+p=0. Then when the photon hits the sail you get a -p reflected photon, still a -p bulb, and a +2p momentum on the sail. So -p-p+2p=0 overall momentum. <br /><br />Well tried! <img src="/images/icons/smile.gif" />. <br /><br />Actually, if you want a thrust, you should have no sail. Just the bulb.

 

[jaxtraw]

Hehe, that's precisely what's been suggested as the cause of the Pioneer effect. The sail isn't pulling the plank forward- but it prevents photons leaving the front of the plank, allowing those leaving the bulb unmolested via the rear to push it forward. It's been suggested as the cause of the Pioneer effect (photons from the RTG bouncing off the communications dish) <img src="/images/icons/smile.gif" />

 

[emperor_of_localgroup]

For your mirror problem where photons are hitting the mirrors at 45 deg, you have to use x and y components of v or momentum. If you take components into consideration, there'll be a motion along the y direction also. That may be your missing energy. I'm assuming they are not tied together.<br /><br />I didn't try a math equation because I'm not sure if photons can have any directional components. Because our science God (Einstein) says photons are sacred, they dont have to follow any rules of the land, I mean reference frame. <br /><br />Btw, I learned quite a bit from this thread. <div class="Discussion_UserSignature"> <font size="2" color="#ff0000"><strong>Earth is Boring</strong></font> </div>

 

[h2ouniverse]

Emp,<br />if x is the direction of the incoming (and outcoming!) photon, momentum along y is 0 at beginning and 0 in the end: it will be canceled in the same way after two bounces. You have either a net displacement along x and -y plus heat dissipation of internal forces, or just the heat dissipation.<br />That balances energy for sure.<br />But the momentum corresponding to the two redshifts of the photon is still unbalanced and that is a big issue for conservatives... <br /><br />At stake is the credibility of any new prop based on energy transfers between momentum, net displacement, heat.

 

[jaxtraw]

Two quick points;<br /><br />firstly, I goofed. The photon transfers is redshifted at the first bounce, so it imparts less energy at the second bounce (mirror B) which can't quite couteract the momentum transferred to mirror A- so the (mirrors+rod) end up with a net momentum i.e. velocity which should equal the (momentum lost at A minus momentum lost at B). (?)<br /><br />The rest will go as heat I guess. Friction allows you to not conserve momentum...

 

[h2ouniverse]

The delta between the two redshifts is of second order.<br />In reality, the energy communicated to the impacted object is very low (momentum multiplied by half-speed, or in other words momentum squared divided by twice the mass). So if the redshift is deltanu, the ratio of energy loss for the photon is deltanu/nu after one reflection, which is far far lower than 1.<br />After two redshifts, the total loss ratio is <br />1-(1-deltanu/nu)(1-deltanu/nu)<br />= 1 - 1 + 2 deltanu/nu - deltanu²/nu²<br />=2deltanu/nu - deltanu²/nu²<br />As deltanu<<nu, this is very close to the 2deltanu/nu unbalance you have if you neglect the effect you describe.<br />This term deltanu²/nu² is very weak in practice to compensate for the momentum unbalance. Note that even if deltanu/nu is non negligible, you still have deltanu²/nu² < 2deltanu/nu.<br /><br />Well tried again <img src="/images/icons/smile.gif" /> but I maintain that if there is a redshift, there is loss of momentum in the two-mirror problem.<br />

 

[managalar]

I might be way off here, I'm just thinking out loud. I have my BS in Physics, and I've always had a problem treating Photons like balls. Balls, by nature, don’t always exist a fixed speed. Are we 100% sure the impulse is twice the momentum of the photon? I've derailed my Optics Prof on this question a few times (might be lucky that I passed) and I realize the question is pretty much nonsense (the answer is so obvious with the ball analogy), but I’m still looking for the evidence (that’s how I found this thread). I think it is pretty close to the question you’re asking.

 

[h2ouniverse]

I share your perplexity about implications of fixed-speed bounce. The proponents of emdrive btw use that to justify their device and I am still wondering whether they can be right after all despite all the suspicion they generate.<br /><br />About the factor of 2 though, I can testify, as a satellite engineer having observed (and re-correlated with specialists) solar disturbance torques on in-orbit satellites attitudes (via telemetry!!) that this is physical.<br />In reality of course reflectivity is between 0 and 1 so you have a factor between 1 and 2 depending on the surface.<br /><br />Best regards.

 

[h2ouniverse]

Btw, I am interested in your opinion (and the one of your professor if you communicate with him/her) about the 45 degree mirrors thought experiment and the issue of momentum conservation...<br /><br />Thank you in advance for your light in that matter...