Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: michi on 24/04/2012 12:31:58
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I was thinking about solar sails and reading how a perfectly reflective surface delivers twice the momentum than a perfectly absorbing surface.
This raised the question for me of what happens to the reflected light. As far as I can see, if the light imparts momentum to the solar sail, the reflected light's frequency must drop by the amount of momentum that is imparted to the sail, otherwise we have energy from nowhere.
I tried various Google searches to find more info about this, but without success. Can anyone confirm or deny whether this idea of red shift of the reflected light is correct?
Thanks,
Michi.
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What happens to light 'propagating' in glass?
Same thing, it loses momentum and, loosely described here, gets 'down-converted' in its next aproximation.
so yes, I agree, it will lose 'energy' assuming it gets reflected.
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Couldn't it be that the light wave is only losing amplitude, and not frequency?
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Couldn't it be that the light wave is only losing amplitude, and not frequency?
Hmmm… That's an interesting idea, and one I hadn't thought of. But, wouldn't that contradict the "perfectly reflective" notion? If the light loses amplitude, but not frequency, that would mean that the reflected light has the same colour, but is not as bright, meaning that some of the photons were absorbed, not reflected.
Also, seeing that photons come in quanta, is it even possible for light to lose amplitude? I would think the answer is no, because a quantum has a defined amount of energy, expressed completely by the frequency. So, I don't think it is possible for a photon to have amplitude?
Cheers,
Michi.
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A 'photon' does not have a frequency. You can if you like see it as equivalent to a wave, assuming a lot of 'photons' here, but a photon express itself as 'particle', not as a wave.
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A 'photon' does not have a frequency. You can if you like see it as equivalent to a wave, assuming a lot of 'photons' here, but a photon express itself as 'particle', not as a wave.
I'm not sure whether the particle/wave duality matters here. My understanding is that a photon has a defined amount of energy. That energy can be expressed as momentum (if I view the photon as a particle) or as frequency (if I view the photon as a wave).
For my original question, I don't think it matters. Seeing that the photons impart momentum to the light sail, I would expect the reflected light to have less energy than the incident light, otherwise we've accelerated the sail by magic. The question for me is whether that is the correct point of view, or whether I'm missing something.
Cheers,
Michi.
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You can by creating a microscopic 'cavity' of some 'wavelength' trap a photon, so maybe you can assume it has a wavelength, but it makes no real sense to me. In a 'black body radiation ' how would a 'photon' be defined then? http://en.wikipedia.org/wiki/Black-body_radiation?
To me it is a quanta of energy.
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Yes, if we're speaking photons they lose 'energy/momentum'. I've seen a lot of new formalism jumping between 'photons' and waves the last decades, without changing the original definitions? wonder why, if now a photon would have both a frequency and a wavelength?
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Yes, if we're speaking photons they lose 'energy/momentum'. I've seen a lot of new formalism jumping between 'photons' and waves the last decades, without changing the original definitions? wonder why, if now a photon would have both a frequency and a wavelength?
But how is the loss of energy/momentum observed then in a perfectly reflective surface?
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That is the original 'particle/wave' duality you're asking about. And to join those two into a 'wave definition' you also need to refute what this 'duality' is built on. Which hasn't been done so far as I know. Because then the duality is a illusion and we should be able to prove that. I'm not saying that they aren't connected, but if you want to put one before the other you better show me the proof. Which then shouldn't be directed to me, but to the original definitions and physics texts presenting them.
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And in fact would earn you a Nobel Prize if you can prove it without doubt.
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As for 'perfectly elastic collisions'. Do you know one existing? Or would this be a theoretical definition?
I don't think I've ever heard of one myself?
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Maybe? But I don't think so, as long as we're talking measurable interactions in a first hand, 'real time' experiment. Weak measurements (on the other tentacle) can be used to prove whatever you set your heart too, as it seems, as long as it follows some sort of logic causality chain. The other way might be to accept that some things by some weird reason isn't measurable as linear causality chains, as a 'photon path'.
Which one is right, or if both may be it, I don't know? But I prefer the simpler version myself.
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In a very weird way 'weak measurements' reminds me of a return to the Newtonian view of the world, in where we have 'forces' acting inside a arrow of time, defining paths and interactions? At the same time as those proposing it want the 'arrow' to be anything, except 'existing' :)
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Back to the original post: yes, the light redshifts. The problem isn't with conservation of momentum, though, it's with conservation of energy. If the photon were to reflect from the mirror with the same frequency at which it arrived, then the photon wouldn't lose any energy. But the mirror is now moving, so the mirror has gained some energy. Since total energy has to be conserved, this can't happen, and indeed the photon must lose a bit of energy in reflection so it redshifts.
The simplest physical explanation I've seen is if you imagine you fire a photon at a mirror that's at rest. Once the photon hits the mirror and reflects, the mirror is moving away from you, so the reflected photon is now being emitted by a mirror that is moving away from you which means they experience a Doppler shift towards lower frequencies. If you work out the math, you'll find that the Doppler shift does account for conservation of energy and momentum in the system.
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When it comes to a 'wave' crossing a border/density, as in glass, it will be diverted. That's referred to as 'refraction' and it will change the waves wavelength, but not its frequency. A frequency is defined as the 'number of cycles per unit time' that you measure whereas a wavelength is defined as the distance between one peak, or crest, of that same wave of light.
The frequency is what defines the color we see, so if we assume that the frequency change with a wave getting reflected then all mirrors should change the color of what they reflect as the light travels through glass first to then interact with some silvery medium before getting reflected. Only in space, assuming a perfect elastic collision, could you expect the color to stay the same for a beam of light.
The index of refraction is 'n' ( n = c / v ) where 'c' is lights speed in a vacuum, 'v' is your new speed inside that medium, or density. And to get to the density inside something?
The index of refraction can also be stated in terms of wavelength (sorry, tried to upload the equation but? The upload won't work?) Ah well This one makes sense, referring to ocean waves. (http://dev.physicslab.org/Document.aspx?doctype=3&filename=GeometricOptics_refractionLight.xml)
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And that's how one intuitively might assume that a wavelength has to do with a photon. Because any idea of a wavelength is like a 'frozen snapshot' of that small part of a waves frequency, as just describing it from crest to crest frozen in time. And a 'photon' is a corpuscle of light although 'dimensionless', something with 'borders' of a sort.
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Back to the original post: yes, the light redshifts. […] If the photon were to reflect from the mirror with the same frequency at which it arrived, then the photon wouldn't lose any energy. But the mirror is now moving, so the mirror has gained some energy. Since total energy has to be conserved, this can't happen, and indeed the photon must lose a bit of energy in reflection so it redshifts.
Yes, that was my reasoning too: the energy has to have come from somewhere.
The simplest physical explanation I've seen is if you imagine you fire a photon at a mirror that's at rest. Once the photon hits the mirror and reflects, the mirror is moving away from you, so the reflected photon is now being emitted by a mirror that is moving away from you which means they experience a Doppler shift towards lower frequencies.
Thanks for that, that makes perfect sense!
Michi.
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If a mirror is moving towards the light source, you get blue-shift.
Stationary - no color change.
Moving away from source - red shift.
So, if you think of your bathroom mirror, essentially fixed, then you should not get any redshift/blueshift.
With your solar sail, logically you would get a very slight redshift, if there is actually an acceleration, or a positive change in velocity.
So, if you have a billiard ball hitting a fixed bumper, theoretically it bounces back at the same velocity it struck the bumper. However, if it strikes another ball, it imparts energy to the ball it struck, and bounces back with much less force (generally stopped, but one could vary this with different masses of balls).
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Well, in a refraction you can't speak about a red/blue shift as that would assume that your window pane would red shift 'waves', but JP was talking about reflection as I understood, not a refraction. And the idea of frequencies getting stretched or compressed depending on 'relative motion' is workable, as we know from our every day life.
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and now i first wrote 'photons' meaning 'waves', it's catching :)
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So, if you think of your bathroom mirror, essentially fixed, then you should not get any redshift/blueshift.
OK, so follow-up question:
There is no such thing as a "perfectly fixed mirror" in practice, I would think. If light hits my bathroom mirror, it still imparts momentum to the mirror, and the mirror will move ever so slightly, probably converting the imparted momentum to heat.
So, wouldn't it follow that light reflected from my bathroom mirror would also be (ever so slightly) red-shifted?
Cheers,
Michi.
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Heh :)
That has to do with definitions, but treating it as photons you definitely should impart a momentum to what you 'hit'. But then you have 'elastic collisions' to consider too, on a theoretical plane. And the wave/particle duality as well. A wave with a wavelength larger than the bumps and 'grooves' on the mirror-surface it hits, a 'smooth' surface, is defined as reflecting all light back as I understands it. I'm not sure at all about that one? Is specular reflection defined as a 'perfect' elastic collision?
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before anyone tells me that you can't mix it like that, what else have we been doing here? :)
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When it comes to a 'wave' crossing a border/density, as in glass, it will be diverted. That's referred to as 'refraction' and it will change the waves wavelength, but not its frequency. A frequency is defined as the 'number of cycles per unit time' that you measure whereas a wavelength is defined as the distance between one peak, or crest, of that same wave of light.
The frequency is what defines the color we see, so if we assume that the frequency change with a wave getting reflected then all mirrors should change the color of what they reflect as the light travels through glass first to then interact with some silvery medium before getting reflected. Only in space, assuming a perfect elastic collision, could you expect the color to stay the same for a beam of light.
The index of refraction is 'n' ( n = c / v ) where 'c' is lights speed in a vacuum, 'v' is your new speed inside that medium, or density. And to get to the density inside something?
The index of refraction can also be stated in terms of wavelength (sorry, tried to upload the equation but? The upload won't work?) Ah well This one makes sense, referring to ocean waves. (http://dev.physicslab.org/Document.aspx?doctype=3&filename=GeometricOptics_refractionLight.xml)
Isn't this the same thing? If you change the wavelength (as c is constant) you change the frequency.
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I haven't really read the thread Mike - but c is constant in a vacuum, light does not propagate at c through a medium. Apologies if this is not the problem
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imatfaal
Your quite right, I missed that. But my point was, if you change the wavelength you also change the frequency.
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When you enter a medium where light propagates slower, you do not change its frequency. Both the speed and the wavelength decrease, however. If you changed it's frequency, the oscillation in time wouldn't be continuous across the interface, which would be non-physical.
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JP
"Any electromagnetic wave's frequency multiplied by its wavelength equals the speed of light."
http://www.qrg.northwestern.edu/projects/vss/docs/communications/2-how-are-frequency-and-wavelength-related.html
Your quite right I was forgetting that changing the speed of light alters the wavelength.
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Presumably this is a consequence of energy conservation.
Something to do with density and wiggle rate?
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Yeah, that's one way of looking at it. The frequency of the radiation doesn't change (or its photon energy would) but light moves slower and bunch closer together, so the energy density goes up, and as a consequence, the speed of its flow has to go down.
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It seems you can't (easily) beat the oldies. If energy suddenly disappeared, it probably didn't.
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A 'photon' does not have a frequency. You can if you like see it as equivalent to a wave, assuming a lot of 'photons' here, but a photon express itself as 'particle', not as a wave.
If a photon strikes a prism, it will come out somewhere on the other side. Depending on where this photon comes out, you can tell if this photon was originally emitted by an atom of sodium, neon or helium.
- Similar for a diffraction grating, only the photon can end up at one of several points... The beauty of the diffraction grating is that you can tell the wavelength from the physical structure.
So I say that individual photons do have a frequency and wavelength, and you can measure it.
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So I take it that you see the beam being reflected back , significantly redder and low-energy , than when first projected ?
P.
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you see the beam being reflected back , significantly redder and low-energy
In current experiments within the solar system, you would have to say "insignificantly redder".
- For a mirror fixed to the wall (which is fixed to the floor, which is fixed to your laser), you would have to say "not red shifted at all".
You really need to get up to a significant fraction of the speed of light to see a significant red-shift.
- The Breakthrough Starshot project is hoping to achieve interstellar travel by accelerating a solar sail up to 20% of the speed of light - now that's significant!
See: https://en.wikipedia.org/wiki/Breakthrough_Starshot
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So I say that individual photons do have a frequency and wavelength, and you can measure it.
Do they have other properties that we find in electromagnetic wave, such as amplitude (in magnetic and electric field), and polarity (either linear, circular, or elliptical)?
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Evan won't know that you're addressing him , due to the space between your answer and his post . You might want to message him .
P.M.
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Do (photons) have other properties that we find in electromagnetic wave, such as amplitude (in magnetic and electric field), and polarity (either linear, circular, or elliptical)?
Yes, like most quantum particles, a photon has more than one quantum property/quantum number.
- It has a spin of ±1
- As I understand it, this gives rise to a helicity of ±h, which indicates circular polarization
- If you have a source of photons which can emit vertically (↑) or horizontally polarized photons (→), you can distinguish these with a polarizing mirror
- But if you have a source which can emits photons having all possible polarizations, you can't determine this with a polarizing mirror; the act of measuring the polarization forces it to be one or the other.
See: https://en.wikipedia.org/wiki/Photon#Physical_properties
As an isolated particle, a single photon:
- Does have an energy, usually measured in electron-Volts
- Does not have a constant intensity, which you would usually measure in Watts =Joules/second. This is more a characteristic of a laser beam, which emits huge numbers of photons every second, for as long as it remains powered.
- Does not have a constant electric field strength, which you would measure in volts/m (or a constant magnetic field, either).
I set up a Radio Frequency Interference lab many years ago, and we had antennas to measure electric and magnetic field strength emitted by electronic equipment (which could interfere with nearby radio receivers). These fields were somewhat random, as they depended on what the software was doing at the time. But they did represent the aggregate power of zillions of radio-frequency photons emitted over long periods of time (a zillion being a highly technical term...).
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As an isolated particle, a single photon:
- Does have an energy, usually measured in electron-Volts
- Does not have a constant intensity, which you would usually measure in Watts =Joules/second. This is more a characteristic of a laser beam, which emits huge numbers of photons every second, for as long as it remains powered.
- Does not have a constant electric field strength, which you would measure in volts/m (or a constant magnetic field, either).
If we have a 1 second long light pulse containing 1 million photons, each having 1 eV energy, what would be the power of each photon? What if the same amount of photon is spent in 2 seconds?
Each photon has fluctuating electric and magnetic field strength with period inversely proportional to its frequency.
Since its energy is constant when not disturbed, its amplitude of electric and magnetic field strength should be constant.
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1 eV = 1.6 x 10-19J so the power of 106 eV/second = 1.6 x 10-13 W. If you deliver the same total energy over 2 seconds, it's 0.8 x 10-13W.
"Power per photon" is a bit meaningless.