Can cosmic background be harvested for energy?

Question

I know it would be awfully weak, but it's present even where all other power sources would be unavailable - could cosmic level radiation be used to power minimal set of systems of a probe, say, a clock to "wake it up" after reaching destination (in a couple hundred years) and activate more efficient energy sources "hibernated" for duration of the travel? What orders of magnitude of power input could be expected, per some unit (mass, surface?) of the "harvesting" apparatus?

Specifically, it appears the cosmic background - specifically, CMB - is not uniform.

The variance is only ±0.00057 K but the features (hot and cold spots) seem to remain mostly persistent over time, so the device, while needing to pivot to maintain orientation relative to CMB, doesn't have to reconfigure 'absorber' and 'radiator' layout on the fly.

The question remains, how can one use such tiny variance, especially with total input of 400–500 photons/cm$^3$ - and how much (...how little) energy could be gained that way?

As clarification what I'm asking for - any external energy source; it doesn't need to be CMB specifically - that is available during intergalactic travel, where you can't depend on solar wind, light of nearby stars or similar; you must depend on what's available there: thermal cosmic background radiation, cosmic microwave background, high-energy cosmic rays, whatever means are available in intergalactic space, other than what you brought with the craft.

Answer 1

I'm guessing you mean cosmic microwave background radiation.

A radio tuned to an empty frequency hears "static" noise. A small fraction of the static received by an analog television is in fact the CMB. Could an antenna system collect some of this, and with a diode or something similar, convert some of it to a small amount of useful electricity? CMB detectors are usually bolometers - antennas + resistors + temperature sensors, but in this case they could be diodes, as you've pointed out here, which can provide some DC power from the original AC.

tl;dr: nope! Thermodynamics always wins, unless your spacecraft and electronics are below 2.7 K.

But let's see what we lost:

To get a rough idea of the amount of power potentially available, without considering the actual functioning circuit which would need to be fairly broadband to make efficient use of the spectrum, here's a quick calculation of the power available incident microwave power. It is only an upper limit to what would be possible to collect.:

Instead of integrating the thermal distribution, let's just use the full width of 200GHz and a heigh of 400 MJy/sr (see below). The data is available here also.

MJy is 1E+06 Jansy. One Jansky is 1E-26 Watts per square meter per Hertz.

Let's say the antenna has some directionality (they always do unless they play tricks with polarization) and call the acceptance 1/2 of a sphere, or 2$\pi$ sr.

400E+06 Jy/sr * 2$\pi$ sr * 200E+09 Hz * 1E-26 W/m^2 ~ 1E-06 W/m^2

So that's 1 micro Watt per square meter possible. Your milage may vary.

To double check that, use the Stephan-Boltzman equation $P=\sigma T^4$. With $\sigma$=5.67E-08 watts per square meter per degree Kelvin^4, that gives 3E-06 W/m^2. Remembering the factor 0.5 for the antenna acceptance, it works out nicely.

But can a hot (say 273 K) crystal radio (diode rectification) actually extract power from a cold thermal distribution? No, heat flows from hot to cold unless you are using energy to drive a heat pump. From a radio point of view, random electrical fluctuations in your radio's electronics will also be broadcasting microwaves back into the cosmos.

So you'd have to cool your radio down below 2.7K to get any power. And at that point, you've just built a thermal collector and a cold, radio-black plate could do roughly as well as the radio. But you wouldn't be gaining energy because you'd be doing more work cooling it.

So in a word, nope! Thermodynamics always wins, unless your spacecraft and electronics are below 2.7 K.

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above: Spectrum of the CMB from here.

above: Spectrum of the CMB plotted with x-axis units converted from 1/cm to GHz, data from here.

Answer 2

Yes! If we assume that you have access to a large black hole and materials that probably can't exist, you can run a heat engine off of the cosmic microwave background radiation.

Reasonably sized black holes are cold. Very cold. Billionths of a Kelvin for a stellar mass black hole, which is numerous orders of magnitude colder than the CMB. The temperature of a black hole is called the Hawking Temperature.

That means you could take a heat engine using helium as working fluid. Your cold sink will be a large radiator facing the black hole that you're orbiting very closely. Your heat source is a similar plate facing away from the black hole. You will need some sort of multi-layer insulation between them for optimal performance. At this point it should be noted that if any matter is falling into the black hole it will be producing radiation and you won't be able to use it as a cold sink.

How much energy will you get? Not much, but since the helium would not be a gas I don't know how to calculate it. As an upper limit though, the Stefan-Boltzmann Law limits it to less than 3.1 microwatts per square meter.

Honestly though it would probably be easier to use the Penrose Process if you can do real engineering that close to a black hole. https://en.wikipedia.org/wiki/Penrose_process

Answer 3

You probably want to harvest usable energy, a physicist would rather call that low entropy. Harvesting energy would mean you heat up indefinitely. Accepting that thermodynamics drives deterministic technical processes (as well as life on earth) is is the key to understanding this problem.

Most of your field of view, the cosmic microwave background, is almost perfect black body radiation with a temperature of 3.7K. Once your spacecraft completely cools to this point, your entropy balance will be even.

You can however use small fluctuations in the spatial distribution to harvest usable energy by exposing the "hot-end" of a harvesting apparatus to the hotter-than-average parts of the sky, the "cold-end" to the other parts. A temperature gradient would build up and can be used.

Any electromagnetic radiation that deviates from a planck spectrum can also be used by reradiating it as thermal radiation with higher entropy (solar panels do exactly that, if you think about it). Visible light and quasar radiation would be a good source. Unfortunately, point sources and sources of narrowband radiation only take up a very small part of your field of view. Quasars are rare and power decays with inverse square law, so visible light from stars may be easier to obtain.

Can a clock oscillator be powered from solar panels in outer space? Depends on surface area. Low entropy reservoirs, like RTGs, are far easier solutions.

(some of the comments below may refer to the previous version of my answer , which was quite different)

Answer 4

Yes, you can theoretically harvest the CMBR. However...

The other answers seem to be describing the (tiny) limits for "naked" antennas. But don't forget you can gather a large area of incoming photons and focus them, say, via a parabolic dish.

How big of a dish? That's up to you and your spaceship. You quoted about 450 photons per cubic centimeter so I think you can figure out how much you want. For reference, the CMBR frequency is about 160 GHz, in the microwave range of the spectrum. It is definitely possible to build a smooth enough dish that will reflect microwaves. (We have such dishes on Earth.)

Edit: Nevermind that. It appears the CMBR is actually a diffuse source of photons (not parallel photons), so a parabolic dish will not focus them.

However...

You mentioned:

any external energy source; it doesn't need to be CMB specifically - that is available during intergalactic travel, where you can't depend on solar wind, light of nearby stars or similar; you must depend on what's available there: thermal cosmic background radiation, cosmic microwave background, high-energy cosmic rays, whatever means are available in intergalactic space, other than what you brought with the craft.

The external energy source I think would be best is nearby starlight. Imagine you have a gigantic parabolic dish on your spacecraft, trying to collect the tiny CMBR. Instead, it would collect a lot more starlight from nearby stars.

How much starlight? Let's just simplify with average stars at average distances. The sunlight near Earth, but not yet hitting the atmosphere, comes in at about 1.36 kW per square meter. The nearest star to our Sun is about 4 lightyears away.

So let's pretend your spaceship, during its journey, is always within 2 LY of a star. 1 LY is about 63,241.1 AU (1 AU = distance from Sun to Earth). Therefore, the starlight is coming in at 1.36 / 63,241.1^2 = 34.005 nanowatts per square meter. But there are probably several nearby stars. Let's say on average, your ship is surrounded by 8 stars each 2 LY away. So you have about 272 nanowatts per square meter.

Not a lot but I'm pretty sure it's orders of magnitude above what you could get with the CMBR. It also does not include light from more distant stars. Just from knowing the mathematical series of 1/2 + 1/4 + 1/8 + 1/16 ..., I know that sums to 1. So I'll just guess that all the other stars farther away add up to another 272 nanowatts per sq meter. Total is 544.

So if you want 1 watt, you need a solar panel that's about 1.85 million square meters big...or a parabolic reflector taking 1.85 M sq m and focusing it onto a solar panel of just 1 sq m.

Keep in mind, tho, that you'll hafta point the solar panel or its parabolic focuses at the nearest star. You could have many such things surrounding the ship, each capable of swiveling. There have been some creative ideas about large lightweight reflectors for this purpose, like the unfolding solar sail that reflects and focuses light back to the towed capsule.

Answer 5

To be clear: do you actually mean "cosmic microwave background radiation" or simply "cosmic radiation"? Because the two are two very different things (although the answer is the same for both, but the reason different). Cosmic radiation consists of GCR (galactic cosmic radiation), a weak flux of very high energy particles, and solar, which is a stronger flux of weaker particles.

While such radiation is very damaging to the body, the flux doesn't represent much total power. Or to put it another way, it can kill you, but you're not going to feel any warmth in the process. While the power (watts) of the flux is tiny (and no, your immediate reaction of "what about a big...".... I'll just stop you right there, the answer is "no"), it's damaging because it releases the energy inside of your cells, not on the protective epidermis layer. By the time a single particle of GCR, some insignificant number of joules, is full decelerated, the energy has generally been split into many other particles, and each has left trails of destruction through your cells like microscopic bullet paths.

There are some GCR particles that can have significant energy - see the "Oh My God Particle" (a single particle measured with the energy of a fast-pitch baseball) for an example. But such things are exceedingly rare. Plus it takes basically a whole atmosphere to stop them.

Answer 6

Actually if you could go up toward relativistic speeds, that radiation would be charged and very strong and hot. You could use the microwave radiation as energy if you had a way to harvest it. Unless you could be at near relativistic speeds, the background radiation would be too weak but let’s see if at 99.999% of the speed of light and it would be overpowering,