25 Jul 16
Originally posted by humyhttps://en.wikipedia.org/wiki/Photon
Both of those things is now exactly what I strongly suspect. I now plan to later dig up all the relevant equations (from my university courses and I have forgotten these relevant equations ) and see if I can do all the calculations and check that that is all exactly right.
If that is exactly right, then you were right in suggesting the velocity imparted by the photons may be greater than what I would think.
Energy, frequency and momentum are essentially all the same equation. In fact you could say that its energy entirely consists of momentum. They are just different units for the same thing.
So, if you have an object moving in the opposite direction to a photon, it is impacted with greater momentum just as it would with a more conventional particle.
25 Jul 16
Originally posted by twhiteheadYour first line is correct due to quantum effects, in this kind of scattering soft photons will be emitted, so there is an extra channel for momentum to be carried away. However, I'm concerned about your second line as you seem to be assuming that momentum can be transferred to the mirror, but have the wavelength changing by an amount corresponding to twice its momentum. The incoming photon has momentum p1 and the outgoing one -p2 (note its a vector quantity). If p1 = p2 then E1 = E2 and no energy has been transferred to the mirror. In that case the momentum of the photon changes by twice it's initial value (p -> -p). If any energy is transferred to the initially stationary mirror then its energy is initially just equal to its mass (let's set the speed of light to 1 to avoid annoying constants). Its final energy will be E^2 = (p1 + p2)^2 + m^2. This lets us use energy and momentum conservation to solve for p2. We use deBroglie's equation (p = h/wavelength) to get the new wavelength of the outmoving photon. But if no energy is transferred (in the idealized case of a fixed mirror) then no wavelength change can occur.
I believe so. And I believe you will find it is close to, if not identical to, the amount of momentum lost by the moving mirror.
I believe you will find that even a stationary mirror changes the wavelength of a photon by an energy amount equal to twice its momentum.
25 Jul 16
7 edits
Originally posted by twhiteheadI am probably just being pedantic here but; energy has units of measurement of the same units as momentum multiplied by velocity (as opposed to just "momentum" ) and that "multiplied by velocity" part changes the v part of the units for momentum to the v^2 part for the units for energy. So, as you just said, they consist of different units.
In fact you could say that its energy entirely consists of momentum.
So I doubt that it is verbally accurate to say "energy entirely consists of momentum" although not really sure how one should interpret that statement.
25 Jul 16
Originally posted by DeepThoughtYes, I got it all wrong. Essentially we can say that the wavelength is negative for the outgoing photon as it goes in the other direction. So without taking direction into account, the photons wavelength doesn't change but it does impart twice its momentum to the mirror, seemingly without loosing any energy.
However, I'm concerned about your second line ....
But if you think about it, the same would apply if you throw a bouncy ball at a wall. It comes back with about the same velocity having imparted twice its momentum to the wall.
If you move the wall towards the ball, then the ball comes back a bit faster, but also imparts more momentum to the wall.
25 Jul 16
Originally posted by humyI think twhitehead meant that there is no mass energy or potential energy component. For a photon E = pc, so the energy is due solely to its momentum. Besides, in Natural Units one sets the reduced Planck's constant, the speed of light, and Bolzmann's constant to 1, so then E = p and they do have the same units.
I am probably just being pedantic here but; energy has units of measurement of the same units as momentum multiplied by velocity (as opposed to just "momentum" ) and that "multiplied by velocity" part changes the v part of the units for momentum to the v^2 part for the units for energy. So, as you just said, they consist of different units.
So I ...[text shortened]... entirely consists of momentum" although not really sure how one should interpret that statement.
25 Jul 16
Originally posted by DeepThoughtIt says that in the title, why would this perpet not work.
I was not claiming it was perpetual, the work done on the piston cannot exceed the energy that the photon has to start with, which eliminates the possibility of a perpetual motion machine from the start. However, within my idealization, the photon is contained inside the cylinder which is perfectly reflective. I'm assuming the piston can be push ...[text shortened]... fectly reflective across all possible wavelengths and infinitely large cylinders can't be built.
25 Jul 16
3 edits
Originally posted by humyOK, I have got the relevant equations which are;
Both of those things is now exactly what I strongly suspect. I now plan to later dig up all the relevant equations (from my university courses and I have forgotten these relevant equations ) and see if I can do all the calculations and check that that is all exactly right.
If that is exactly right, then you were right in suggesting the velocity imparted by the photons may be greater than what I would think.
E = hc/λ = energy of a photon
p = h/λ = momentum of a photon
KE = 1/2 m v^2 = kinetic energy of moving mass
( not sure this one above is needed )
and the effect of the Doppler shift on the change of wavelength of photon for the mirror moving towards the light source is:
λ = λ' √ ( ( 1 – v/c ) / ( 1 + v/c ) )
where λ' is the wavelength of the photon from the perspective of the light source and λ is the wavelength of the photon from the perspective of the mirror moving towards the light source.
But now I have got stuck; how to use the above equations to show that the energy of the system with the mirror moving towards the light source (sliding on a frictionless surface ) is conserved because the increase in the energy of the photon from the Doppler shift is exactly the same amount as the reduced kinetic energy of the mirror because of the momentum of the photon transferred to the mirror?
Anyone?
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30 Jul 16
2 edits
Originally posted by humy"KE = 1/2 m v^2 = kinetic energy of moving mass
OK, I have got the relevant equations which are;
E = hc/λ = energy of a photon
p = h/λ = momentum of a photon
KE = 1/2 m v^2 = kinetic energy of moving mass
( not sure this one above is needed )
and the effect of the Doppler shift on the change of wavelength of photon for the mirror moving towards the light source is:
λ = λ' √ ( ( ...[text shortened]... energy of the mirror because of the momentum of the photon transferred to the mirror?
Anyone?
( not sure this one above is needed )"
I don't solve relitivistic problems like this, but two things: mass is zero for a photon, and "v" is "c"
As you can see this relationship is of the classical variety (v<<c and m > 0). You will undoubtedly be needing relitivistic momentum and energy relationships.
EDIT: Nevermind, I see that was in reference for the mirror
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30 Jul 16
2 edits
Originally posted by joe shmoAgain... This may not be precise, but classically if you were to apply conservation of momentum assuming a perfectly elastic collision (total momentum and kinetic energy of the system is conserved).
"KE = 1/2 m v^2 = kinetic energy of moving mass
( not sure this one above is needed )"
I don't solve relitivistic problems like this, but two things: mass is zero for a photon, and "v" is "c"
As you can see this relationship is of the classical variety (v<<c and m > 0). You will undoubtedly be needing relitivistic momentum and energy relationships.
EDIT: Nevermind, I see that was in reference for the mirror
m•v + h/λ = m•v' + h/λ'
h/λ - h/λ' = m•v'- m•v
h( 1/λ - 1/λ' ) = m( v'-v )
h[( λ'-λ ) / ( λ'•λ )] = m( v'-v )
δv = ( v'-v ) δλ = ( λ'-λ )
h • δλ / ( λ'•λ ) = m•δv
Which indidcates the Doppler Shift is a maifestaion of the change in momentum of the mirror. Also, kinetic energy is conserved by assumption of perfectly elastic collision. However, I feel a bit foggy today, and I'm not a Physicist; I most likely am out in left field.
31 Jul 16
5 edits
Originally posted by joe shmoI don't quite follow but I still suspect you probably got it correct as your conclusion agrees what I now strongly suspect but don't get how to algebraically prove.
Again... This may not be precise, but classically if you were to apply conservation of momentum assuming a perfectly elastic collision (total momentum and kinetic energy of the system is conserved).
m•v + h/λ = m•v' + h/λ'
h/λ - h/λ' = m•v'- m•v
h( 1/λ - 1/λ' ) = m( v'-v )
h[( λ'-λ ) / ( λ'•λ )] = m( v'-v )
δv = ( v'-v ) δλ = ( λ'-λ )
h ...[text shortened]... However, I feel a bit foggy today, and I'm not a Physicist; I most likely am out in left field.
01 Aug 16
2 edits
Originally posted by vivifyyes; but, although that would be perpetual motion, that would be no more a perpetual motion machine than an electron spinning around an atom i.e. it wouldn't be a perpetual motion machine.
Question: regarding perpetual motion, let's say there's an object spinning that's light-years away from the nearest thing to it (in other words, far enough that gravity from something else has a negligible effect on it), would that object continue spinning indefinitely?
01 Aug 16
Originally posted by humySuppose in this same space that I described, there was a spinning turbine, that generated electricity; couldn't this be used to make a perpetual motion machine?
yes; but, although that would be perpetual motion, that would be no more a perpetual motion machine than an electron spinning around an atom i.e. it wouldn't be a perpetual motion machine.
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01 Aug 16
1 edit
Originally posted by vivifyThat spinning turbine cannot do any Work while maintaining its angular momentum. That is, as soon as you try to use power from the turbine to drive a load, the turbine will slow down. It can theoretically be perpetual motion, but perpetual motion in itself is worthless. The idea is behind perpetual motion machines is they somehow violate the law of physics and provide work at no cost to the supply. That is "Work" just magically and endlessly appears out of nothing.
Suppose in this same space that I described, there was a spinning turbine, that generated electricity; couldn't this be used to make a perpetual motion machine?
01 Aug 16
Originally posted by joe shmoAnd for some reason this appeals to people more than putting a solar panel on the roof.
The idea is behind perpetual motion machines is they somehow violate the law of physics and provide work at no cost to the supply. That is "Work" just magically and endlessly appears out of nothing.