Spaceship drive by light

1. Aug 21, 2014

Helios

Would a spaceship that is half reflective ( the stern ) and half flat-black ( the bow) be propelled through space just by the ambient radiation from stars and the CMBR?

2. Aug 21, 2014

Ryan_m_b

Staff Emeritus
Technically yes, you're essentially describing a solar sail. They're incredibly low thrust though and I doubt you could practically exit the solar system with one.

3. Aug 21, 2014

A.T.

The black side will heat up and radiate.

4. Aug 21, 2014

A.T.

Solar sails work fine with particles coming from one direction. But the OP seems to ask about approximately isotropic radiation from all sides. It seems to me, that converting such ambient radiation continuously into a directed movement of a ship would reduce entropy, and thus violate the laws of thermodynamics.

5. Aug 21, 2014

Staff: Mentor

Right. Of course if we were willing to supply power to an on-board refrigeration system, we could transfer heat to the shiny side to maintain the thrust... Which just goes to show that we aren't going to get any useful work out of an isotropic energy bath without putting some work in ourselves, just as you said.

6. Aug 21, 2014

D H

Staff Emeritus
Very little, and not in the direction you think! Details follow.

Heat up? It will cool down. In fact, the black side will radiate a lot until it cools down to T0=2.72548 K, the CMBR temperature. There will be no net thrust if the black side cools to the CMBR temperature. Ignoring thrust from starlight, the only way to get thrust is if the black side has a temperature above that of the CMBR.

Note: I'm dismissing a device temperature below that of the CMBR as a physical impossibility. That would require cooling, and that takes work, and that's at best perfect. The work will generate heat, the amount of which will at best keep the device on average at the CMBR temperature. Any inefficiencies and the temperature will be greater than that of the CMBR.

I'll assume the spacecraft is trying to make its across one of those huge voids in space (wiki article: http://en.wikipedia.org/wiki/Void_(astronomy)). That means there's no pesky starlight to make a mess of my hemispherical cow thrust assumptions.

My hemispherical cow assumptions: The reflective side is perfectly flat perfect mirror, while the black side is a perfectly flat perfect black body at a uniform temperature TT0.

The reflective side specularly reflects all incoming CMB radiation over the half-sphere visible to that side of the spacecraft. The CMB radiation is isotropic, but the spacecraft only gets thrust from the normal component of the radiation, effectively halving the intensity. But since the reflection is specular, we get to double the thrust. The net effect is a factor of 1. The total thrust from the reflective side is thus $F_{\text{reflective}} = \frac {\sigma A {T_0}^4} c$, where the numerator is the Stefan-Boltzmann law for radiative power and the numerator reflects the fact that photons travel at c. This thrust is in the direction of the black side.

The black side absorbs all incoming CMB radiation over the half-sphere visible to that side of the spacecraft. Integrating over the half-sphere, the only component that survives is the normal component. This is half the thrust from the reflective side, but this thrust is directed toward the reflective side. The black side also radiates per the Stefan-Boltzmann law over the half-sphere visible to that side of the spacecraft. Assuming this is isotropic over the half-sphere, the net thrust from this outgoing radiation is $\frac 1 2 \frac{\sigma A {T}^4} c$. The total thrust from the black side is thus $F_{\text{reflective}} = \frac 1 2 \frac {\sigma A {T_0}^4} c + \frac 1 2 \frac {\sigma A {T}^4} c = \frac 1 2 \frac {\sigma A} c \left({T_0}^4 + T^4 \right)$. This thrust is in the direction of the reflective side.

Summing the thrusts from both sides vectorially yields $F_{\text{total}} = \frac 1 2 \frac {\sigma A} c \left(T^4 - {T_0}^4\right)$, toward the reflective side. There is no thrust if the spacecraft does cool to the CMBR temperature.

Suppose on the other hand the spacecraft has some power source that keeps the spacecraft alive, powers computations, and powers the spacecraft's rotational control. The equilibrium temperature T given some power P is $T= \left({T_0}^4+\frac {2P}{\sigma A}\right)^{1/4}$. The net thrust will be $\frac P c$, in the direction of the reflective side. Note that the area of the spacecraft and the CMBR temperature have dropped out of the picture.

7. Aug 21, 2014

A.T.

Sorry, bad choice of words. I meant that you cannot prevent it from radiating, by cooling it down to 0K.

8. Aug 21, 2014

D H

Staff Emeritus
There are two problems with your line of thinking.

Problem #1: Transferring heat to the shiny side accomplishes nothing because the shiny side doesn't emit radiation. At any given frequency, the emissivity, transmissivity, and reflectivity of some object sum to one. Our object presumably is a perfect reflector at all frequencies, so it's emissivity is zero at all frequencies.

Problem #2: This is the thermodynamical equivalent of lifting yourself by your own bootstraps. In fact, its worse. Nothing much happens I try to lift myself by my own bootstraps. You can cool down a part of the black side, but that takes work, and that generates heat, and that in turn raises the temperature somewhere else on the black side. The net effect is always more thrust from the black side than if you hadn't mucked with trying to cool things down.

The only way to have the black side cooler than T0 is to bring some coolant that has been cooled to below the CMBR temperature before the craft departs. That coolant will eventually run out, and then we're stuck. The thrust is zero once the black side's temperature reaches T0.

The only way out is to take advantage of the fact that the net thrust will be in the direction of the reflective side so long as the vehicle still has power. It will need some power, a trickle charge to keep memory intact, a bit of power for computing where the vehicle is, a bit of power for keeping the vehicle pointing in the right direction. Of course, the power source will eventually run out, and once again we're stuck.

9. Aug 21, 2014

A.T.

I assume that is just the initial thrust. As soon it starts moving, the incoming radiation stops being isotropic due to Doppler effect and aberration.

10. Aug 21, 2014

Helios

The cosmic background radiation looks bluer to the bow of the ship and redder off the stern. It would seem that this anisotropy would effect things.

11. Aug 21, 2014

D H

Staff Emeritus
P/c is an incredibly small value. Even at Mr. Fusion levels (1.21 gigawatts), P/c amounts to a whopping 4 newtons. Presumably the spacecraft won't consume power at that rate. There are no garbage cans with which one could replenish Mr. Fusion in those voids in space. We need to be miserly with our power consumption to cross that void.

Let's assume the spacecraft manufacturers have learned the art of miniaturization to a tee, and have also learned to minimize power consumption. If the vehicle uses 1 milliwatt of power per kilogram of spacecraft to keep things alive, perform computations, and control the vehicle, the tiny value of P/c means it will take 3 billion years to accelerate to 1/1000 of the speed of light. At that tiny speed, the anisotropy of the CMBR is a tiny, tiny drag on the vehicle.

Here's the calculation of that 3 billion year figure at WolframAlpha: http://www.wolframalpha.com/input/?i=(0.001+c)/((1+milliwatt)/c/(1+kg)). Note that I've ignored drag from the dipole anisotropy of the CMBR. It's a very small amount of drag, and all that accounting for this will do is to make it take even longer than 3 billion years to reach 1/1000 of the speed of light.

That exists, but it's even smaller than the minuscule acceleration described above. The Doppler shift is a rather small effect for a vehicle moving much, much slower than the speed of light. For smallish velocities (v<<c), this will result in a very, very small drag term that grows linearly with velocity.

12. Aug 22, 2014

256bits

This is the wiki on Crookes radiometer, which may be of interest.

Of special interest is that is will not rotate in pure vacuum, but it needs a certain amount of rarefied gas to make it spin.
Under a partial vacuum, it spins shiny side leading when subjected to light or heat.
When subjected to a cooler outside temperature, it spins black side leading.

The explanation is that the rarefied gas provides the momentum transfer to produce the spinning.

13. Aug 22, 2014

sophiecentaur

Those are my thoughts, too. There has to be a gradient (of 'something') out there to supply the power needed for propulsion. Any talk of on board refrigeration systems implies using an on board power source, so you no longer have free energy.

14. Aug 22, 2014

D H

Staff Emeritus
That's not applicable here because radiation pressure most certainly is a real effect. Just because the rotation of the Crooke's radiometer does not result from radiation pressure does not mean that radiation pressure doesn't exist. It most certainly does. It is the basis for solar sails, and this concept has been successfully demonstrated with the IKAROS mission.