# The Naked Scientists Forum

### Author Topic: Red-shift of reflected light?  (Read 6700 times)

#### michi

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##### Red-shift of reflected light?
« on: 24/04/2012 12:31:58 »
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.

Thanks,

Michi.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #1 on: 24/04/2012 13:09:34 »
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.

#### Nizzle

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##### Re: Red-shift of reflected light?
« Reply #2 on: 24/04/2012 13:10:41 »
Couldn't it be that the light wave is only losing amplitude, and not frequency?

#### michi

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##### Re: Red-shift of reflected light?
« Reply #3 on: 24/04/2012 13:16:50 »
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.
« Last Edit: 24/04/2012 13:18:38 by michi »

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #4 on: 24/04/2012 13:27:33 »
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.

#### michi

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##### Re: Red-shift of reflected light?
« Reply #5 on: 24/04/2012 13:39:58 »
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.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #6 on: 24/04/2012 13:40:51 »
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.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #7 on: 24/04/2012 13:45:19 »
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?

#### Nizzle

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##### Re: Red-shift of reflected light?
« Reply #8 on: 24/04/2012 14:04:36 »
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?

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #9 on: 24/04/2012 15:28:24 »
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.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #10 on: 24/04/2012 15:30:03 »
And in fact would earn you a Nobel Prize if you can prove it without doubt.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #11 on: 24/04/2012 15:39:25 »
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?
=
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.
==

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' :)
« Last Edit: 24/04/2012 16:07:11 by yor_on »

#### JP

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##### Re: Red-shift of reflected light?
« Reply #12 on: 24/04/2012 17:29:25 »
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.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #13 on: 24/04/2012 17:45:30 »
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.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #14 on: 24/04/2012 17:52:05 »
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.
« Last Edit: 24/04/2012 18:28:49 by yor_on »

#### michi

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##### Re: Red-shift of reflected light?
« Reply #15 on: 24/04/2012 21:37:19 »
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.

Quote
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.

#### CliffordK

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##### Re: Red-shift of reflected light?
« Reply #16 on: 24/04/2012 22:30:41 »
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).

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #17 on: 24/04/2012 22:41:22 »
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.
=

and now i first wrote 'photons' meaning 'waves', it's catching :)
« Last Edit: 24/04/2012 22:44:14 by yor_on »

#### michi

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##### Re: Red-shift of reflected light?
« Reply #18 on: 24/04/2012 22:47:11 »
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.

#### yor_on

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##### Re: Red-shift of reflected light?
« Reply #19 on: 25/04/2012 00:48:07 »
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?
==

before anyone tells me that you can't mix it like that, what else have we been doing here? :)
« Last Edit: 25/04/2012 00:57:23 by yor_on »

#### MikeS

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##### Re: Red-shift of reflected light?
« Reply #20 on: 26/04/2012 06:47:46 »
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.

Isn't this the same thing?  If you change the wavelength (as c is constant) you change the frequency.

#### imatfaal

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##### Re: Red-shift of reflected light?
« Reply #21 on: 26/04/2012 15:48:05 »
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

#### MikeS

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##### Re: Red-shift of reflected light?
« Reply #22 on: 26/04/2012 16:17:10 »
imatfaal

Your quite right, I missed that.  But my point was, if you change the wavelength you also change the frequency.

#### JP

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##### Re: Red-shift of reflected light?
« Reply #23 on: 26/04/2012 22:10:51 »
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.

#### MikeS

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##### Re: Red-shift of reflected light?
« Reply #24 on: 27/04/2012 08:46:44 »
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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##### Re: Red-shift of reflected light?
« Reply #24 on: 27/04/2012 08:46:44 »