Don't think it can be done that way.

Come to think of it, maybe it's not about W/m² to Newtons ... but rather W/m² to a given acceleration. A watt has a unit of time involved in its definition.

 

Come to think of it, maybe it's not about W/m² to Newtons ... but rather W/m² to a given acceleration. A watt has a unit of time involved in its definition.

Clarification: the conversion works in the presence of a working medium, i.e. gas. Can it work in a vacuum?

 

"The watt is the unit of power or radiant flux in the International System of Units, equal to 1 joule per second or 1 kg⋅m²⋅s⁻³. It is used to quantify the rate of energy transfer."

If it's transferring energy, it's definitely transferring it in the form of momentum. Following the solar sail analogy.

 

A photon of wavelength \(\lambda\) (and corresponding wavenumber \(k = 2\pi\lambda^{-1}\)) has momentum \(p = \hbar k\) and energy \(E = hf = \hbar ck = cp\). If a photon strikes a surface and is absorbed, then the momentum transferred is \(p\). If it is reflected it is \(2p\) (so the total momentum remains \(p\)).

If we have a intensity of \(I\) \(\mathrm{W\cdot m^{-2}}\), and a sail with surface area \(A\) then the total power transferred is \(P = IA\). If we are ignoring heating effects and a bunch of other things and assuming all we have is a perfectly reflecting surface, then we can say:
\[\frac{dE}{dt} = P\]
\[\rightarrow c\frac{dp}{dt} = 2IA\] (the 2 for the reflection, because we gain twice as much momentum per photon as for an absorbing surface)
\[\rightarrow F = \frac{2IA}{c}\]

Thus, for a given light intensity and surface area, assuming a reflecting surface, we can estimate the amount of force being applied. For an intensity of 1 \(\mathrm{W\cdot m^{-2}}\) on a 1 square meter surface, that comes out to about 6.7 nN, and half as much if the surface is absorptive instead.

So yes: it can work in a vacuum. Just very, very slowly.

Edit: I'll add, to confirm someone's question: yes - if you have higher claimed force than what is physically possible from pure reflections, then the surface must also be emitting photons to account for the extra momentum, which means something inside must be generating energy. For an emitted photon, the emitter will have a reaction force of \(F = IAc^{-1}\), for the intensity of the light being emitted across the area emitting it.

 

A photon of wavelength \(\lambda\) (and corresponding wavenumber \(k = 2\pi\lambda^{-1}\)) has momentum \(p = \hbar k\) and energy \(E = hf = \hbar ck = cp\). If a photon strikes a surface and is absorbed, then the momentum transferred is \(p\). If it is reflected it is \(2p\) (so the total momentum remains \(p\)).

If we have a intensity of \(I\) \(\mathrm{W\cdot m^{-2}}\), and a sail with surface area \(A\) then the total power transferred is \(P = IA\). If we are ignoring heating effects and a bunch of other things and assuming all we have is a perfectly reflecting surface, then we can say:
\[\frac{dE}{dt} = P\]
\[\rightarrow c\frac{dp}{dt} = 2IA\] (the 2 for the reflection, because we gain twice as much momentum per photon as for an absorbing surface)
\[\rightarrow F = \frac{2IA}{c}\]

Thus, for a given light intensity and surface area, assuming a reflecting surface, we can estimate the amount of force being applied. For an intensity of 1 \(\mathrm{W\cdot m^{-2}}\) on a 1 square meter surface, that comes out to about 6.7 nN, and half as much if the surface is absorptive instead.

So yes: it can work in a vacuum. Just very, very slowly.

Edit: I'll add, to confirm someone's question: yes - if you have higher claimed force than what is physically possible from pure reflections, then the surface must also be emitting photons to account for the extra momentum, which means something inside must be generating energy. For an emitted photon, the emitter will have a reaction force of \(F = IAc^{-1}\), for the intensity of the light being emitted across the area emitting it.

So, we've done work on the sail by reflecting the photon, thus increasing the energy (kinetic) of the sail.

But the photon still has exactly the same energy as it started with, just in the opposite direction.

How is this not an overall increase in energy of the system, and also a violation of conservation of momentum?

For example, why can I not have another solar sail traveling in the opposite direction, all the while bouncing the photon between them? Each bounce will impart additional energy alternately to the corresponding sail indefinitely.

 

But the photon still has exactly the same energy as it started with, just in the opposite direction.

I doubt that is the case. The photon most likely loses energy in the form of a diminished frequency... but I'm jus guessing here

 

I doubt that is the case. The photon mos likely loses energy in the form of a diminished frequency... but I'm jus guessing here

No losses are indicated in @ZCochran98's computations.

 

No losses are indicated in @ZCochran98's computations.

It was assuming perfect reflections, though I DID forget to incorporate conservation of energy. Total energy WOULD need to be conserved, so through a combo of conservation of momentum and energy you'd end up with some smaller overall amount of momentum gained. So the light reflected off of the reflecting surface would experience a redshift, but if you assume the surface is moving at non-relativistic speeds and the energy per photon was significantly less than the relativistic energy (\(E = mc^2\) for the reflecting surface), the "real" answer will approximate out to what I have. The error between what I had and what would be the "real" value is, for something starting at rest, very small. The full relativistic equations are a pain. This article has the full calculations for such a situation. Equation 4 in that paper lets us get the change in frequency for a given initial velocity (I'm going to use 0, based on my earlier calculations) with a mirror of mass \(m\):
\[\omega_f = \frac{\omega_i}{1+\frac{2\hbar\omega_i}{mc^2}}\]
\[\Delta\omega = \omega_i - \omega_f = \omega_i\left(1 - \frac{1}{1+\frac{2\hbar\omega_i}{mc^2}}\right) = \frac{\omega_i}{1+\frac{mc^2}{2\hbar\omega_i}}\]
So, for a mirror starting at rest with mass of 1 kg and an initial frequency of 400 THz (red light at 750 nm if I did my math right) the frequency shift is \(\Delta\omega \approx 6\times10^{-23}\) Hz (if, again, I did my math right).

 

It was assuming perfect reflections, though I DID forget to incorporate conservation of energy. Total energy WOULD need to be conserved, so through a combo of conservation of momentum and energy you'd end up with some smaller overall amount of momentum gained. So the light reflected off of the reflecting surface would experience a redshift, but if you assume the surface is moving at non-relativistic speeds and the energy per photon was significantly less than the relativistic energy (\(E = mc^2\) for the reflecting surface), the "real" answer will approximate out to what I have. The error between what I had and what would be the "real" value is, for something starting at rest, very small. The full relativistic equations are a pain. This article has the full calculations for such a situation. Equation 4 in that paper lets us get the change in frequency for a given initial velocity (I'm going to use 0, based on my earlier calculations) with a mirror of mass \(m\):
\[\omega_f = \frac{\omega_i}{1+\frac{2\hbar\omega_i}{mc^2}}\]
\[\Delta\omega = \omega_i - \omega_f = \omega_i\left(1 - \frac{1}{1+\frac{2\hbar\omega_i}{mc^2}}\right) = \frac{\omega_i}{1+\frac{mc^2}{2\hbar\omega_i}}\]
So, for a mirror starting at rest with mass of 1 kg and an initial frequency of 400 THz (red light at 750 nm if I did my math right) the frequency shift is \(\Delta\omega \approx 6\times10^{-23}\) Hz (if, again, I did my math right).

So, again to my original point: only a very small portion of the 1 watt per square meter is converted to thrust, at least for a solar sail.

My question was: is their a law that prohibits conversion of 1 watt per meter square to the equivalent newton thrust (at some reasonable efficiency) in a vacuum?

 

So, again to my original point: only a very small portion of the 1 watt per square meter is converted to thrust, at least for a solar sail.

My question was: is their a law that prohibits conversion of 1 watt per meter square to the equivalent newton thrust (at some reasonable efficiency) in a vacuum?

There is no law that prohibits conversion of 1 W/m^2 to the equivalent newton thrust in a vacuum, no; the thrust is just very, very small. IKAROS is an example of a real solar sail. You WILL get a frequency shift (and corresponding energy shift) in the reflected light, but, unless you're going at appreciable fractions of the speed of light, the shift will be very small.

 

So, again to my original point: only a very small portion of the 1 watt per square meter is converted to thrust, at least for a solar sail.

My question was: is their a law that prohibits conversion of 1 watt per meter square to the equivalent newton thrust (Cat some reasonable efficiency) in a vacuum?

Conservation of energy. The kinetic energy added by the thrust per second cannot exceed the power in. 1W can add no more than 1 Joule each second.

 

Conservation of energy. The kinetic energy added by the thrust per second cannot exceed the power in. 1W can add no more than 1 Joule each second.

You think?

 

Don't think it can be done that way.

EM has momentum.

 

EM has momentum.

No, I mean you cannot efficiently extract thrust from photonic momentum. As I described above, the reflected photon retains almost all the energy it had prior to reflection. Very little was transferred to the sail.

I'm asking if there is a law that says we cannot impart physical force through annihilation of the photon, i.e efficiently converting its total energy to thrust.

 

I'm asking if there is a law that says we cannot impart physical force through annihilation of the photon, i.e efficiently converting its total energy to thrust.

Wow ... I hadn't understood your question before ... and it's a very interesting one.

E=mc² and nobody disputes that mass and energy are interchangeable. But I've always thought about E in terms of energy particles, i.e. photons. It hadn't occurred to me that said E could also be interpreted in terms of pure simple momentum.

 

I just googled this:



It makes the premise of the sci-fi series "The Expanse" quite believable.

 

Photons do not have an antiparticle, so there is no way to "annihilate" a photon (though you can convert it into a particle and its corresponding antiparticle, each with mass \(m = \frac{\hbar\omega}{2c^2}\) (assuming the particles remain at rest; otherwise, the actual particle mass will be dependent on how much momentum it also has because of relativistic effects). However, if you absorb a photon you do gain its energy...just split amongst a bunch of different things (kinetic energy, thermal energy, electron orbitals, etc). But yeah: to convert a photon purely into kinetic energy (or its equivalent momentum, as I understand what's being asked now) is an interesting idea, but I have no idea how it could be done, if it can be done.

Edit: I just looked it up, and I was wrong: photons are technically their own antiparticle. Not sure how that works without having to go dig a little deeper.

 

No, I mean you cannot efficiently extract thrust from photonic momentum. As I described above, the reflected photon retains almost all the energy it had prior to reflection. Very little was transferred to the sail.

I'm asking if there is a law that says we cannot impart physical force through annihilation of the photon, i.e efficiently converting its total energy to thrust.

Ok, I see what you mean. The photonic momentum is expressed in the duality as a particle while the photonic energy is carried as a wave. These thrust devices claim, by some novel physics invention, to transform EM wave energy (RF , that RF wave energy can be generated, transfered and converted in wireless systems) into in effect, particle momentum by 'pushing' on nothing. The basic law violated IMO is conservation of momentum in the closed system of these types of drives.
https://www.google.com/url?sa=t&sou...UQFnoECBAQAQ&usg=AOvVaw3paFsOO6czhbQURGgj-L1L

https://phys.libretexts.org/Courses/Tuskegee_University/Algebra_Based_Physics_I/07:_Linear_Momentum_and_Collisions/7.04:_Conservation_of_Momentum

 

You think?

Yes, I do, do think. The alternative is that conservation of energy is violated. Of course, that is not to say that is the tightest possible limit, conservation of momentum is also in play. I was not aware that your question was specifically related to a light powered sail. If that is the case the momentum gained by reflecting a photon back in the same direction is exactly twice the momentum of the original photon.