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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.

heat map of CMB

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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Feb 24 '17 at 15:43
  • $\begingroup$ Sorry for the confusion, everyone. I made a mod flub and converted an answer to comment instead of migrating comments to chat. It is fixed now, but there are duplicate comments across the chat rooms. $\endgroup$ – called2voyage Feb 24 '17 at 15:44
  • $\begingroup$ I think the ±0.00057 K "variance" is not the range of temperatures, but the total uncertainty in the average temperature value, combining data from many experiments and including all uncertainties in each, AND correcting for the 600 km/s doppler shift from our galaxy's motion. See last sentence of section 4 in arxiv.org/abs/0911.1955 According to the caption from your image and here also the range of the plot is +/- 0.0002 K. But this plot has been doppler shifted. $\endgroup$ – uhoh Feb 28 '17 at 6:10
  • $\begingroup$ The main dipole moment of +/-0.0035 K peak-to-peak is from the 600 km/s motion of our group of galaxies, and has been subtracted before your plot was made. As I've mentioned here and thereabouts, the theoretical maximum efficiency for a Carnot cycle with such a small temperature difference is about 0.26%, or a few nano-watts per square meter theoretical thermodynamic maximum. Your milage will vary. $\endgroup$ – uhoh Feb 28 '17 at 6:18
  • $\begingroup$ Aw you gave away the bounty right as I answered this, complete with numbers. $\endgroup$ – DrZ214 Mar 5 '17 at 0:49
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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.


enter image description here

above: Spectrum of the CMB from here.

enter image description here

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

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – called2voyage Feb 24 '17 at 15:43
  • $\begingroup$ You seem to be doing analysis for "naked" antennas, which is fine, but don't forget we could focus a large area of incoming rays to a spot, like with a parabolic reflector. See my answer for the details. $\endgroup$ – DrZ214 Mar 5 '17 at 0:52
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    $\begingroup$ @DrZ214 For an essentially isotropic source, focusing provides no help. I know that this is hard to believe at first, but remember while increasing the area by a factor X you decrease the solid angle (cone of light) by X at the same time. The simplest way to demonstrate this is to go outside during the day with a magnifying glass and try to "focus" blue sky on to a piece of paper. A collimated source like the sun will give you a dangerous blindingly bright hot spot, but there is barely any change in blue sky at all. $\endgroup$ – uhoh Mar 5 '17 at 1:01
  • $\begingroup$ @uhoh nonsense. particles moving parallel can be reflected with the parabola, which has the exact geometric property we want. It is just a way of collecting all the photons in one area so we can have a smaller receiver. Either the receiver itself has to be very large or the dish has to be very large. Are you thinking of the odd photovoltaic property of photon intensity (amplitude) has no effect on the electricity produced? That much is true, but we are still focusing/collecting more photons, so more watts. $\endgroup$ – DrZ214 Mar 5 '17 at 1:09
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    $\begingroup$ @DrZ214 Please do the experiment with the magnifying glass. If you get a bright blue spot, photograph it and post it here. The principle can be called called conservation of phase space or Liuville's theorem or conservation of etendu to get around it would violate the 2nd law Thermodynamics so it's pretty much the law $\endgroup$ – uhoh Mar 5 '17 at 1:19
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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

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  • $\begingroup$ that would awfully increase the dry mass of the craft though :) $\endgroup$ – SF. Mar 1 '17 at 21:12
  • $\begingroup$ It's fine. Just leave then engine where it is and beam power to your spacecraft. $\endgroup$ – Schlusstein Mar 1 '17 at 21:13
  • $\begingroup$ Brilliant! +n! In this case, a reflector might in fact help. While one can't focus the diffuse CMB hot source, one could use a reflector to give the radiator a much larger solid angle "view" of the cold source. That would allow a larger distance from the black hole. It would have to be a low emissivity reflector, but in this case that turns out to be OK (good reflectors are indeed low emissivity at their high reflectivity wavelength). $\endgroup$ – uhoh Mar 5 '17 at 1:23
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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)

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    $\begingroup$ I was more thinking in terms of antenna - a detector radio uses no power supply; all its power is gathered through the (sizable) antenna and sufficient to move an earphone - but tune it to no radio station and you hear the white noise of cosmic background radiation; its energy moving the earphone membrane. It requires good grounding besides the antenna though, and considering it's to be on a probe in space, this approach doesn't really work. $\endgroup$ – SF. Feb 23 '17 at 21:24
  • $\begingroup$ if anything, I'm more interested in electric power than thermal. $\endgroup$ – SF. Feb 23 '17 at 22:32
  • $\begingroup$ @SF. Electrical power is governed by thermodynamics after all, like all other forms of power. There is no electric perpetuum mobile, because electricity is governed by the thermodynamic laws. But your response gave me new insight, I was never aware, that the spectrum of radiation would play a role. Quasar radiation can be harvested, because it is narrow-band. $\endgroup$ – Andreas Feb 23 '17 at 22:49
  • $\begingroup$ Can you explain what is mean (in reality) by "exposing" a surface to one hemisphere? What temperature is the other side of that surface exposed to - in reality? There is a dipole moment in the distribution of 0.0035K, so do you mean the intermediate surface facing the "insides" of both surfaces would be spot-on average, and the CMB non-uniformity thermodynamic extractor would be using this difference of +/- 0.0035K? What is the theoretical efficiency of that Carnot Cycle? $\endgroup$ – uhoh Feb 28 '17 at 0:53
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    $\begingroup$ @uhoh Assuming a dipole moment of +/- 0.0035K and a mean of 2.726K, the theoretical upper limit for efficiency would be 0.0026 . This means that only 1/380 of the energy you receive on the hot end can be used to do work (= to drive irreversible processes). Formula from here. (Corrected factor 10 error, blush...) $\endgroup$ – Andreas Feb 28 '17 at 6:40
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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.

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    $\begingroup$ I'm meaning any kind of radiation / energy available in interstellar/intergalactic space, other than light/radiation of distant stars/galaxies. $\endgroup$ – SF. Feb 24 '17 at 10:08
  • $\begingroup$ Can you add some numbers, and links to support your statements? Weak , strong, tiny, big, not much, significant, exceedingly... are subjective. I'm not necessarily disagreeing with you, but this many 'value judgements' without any sourcing, and actually without any numerical values makes this an "exceedingly" weak answer. Let's see if we can nail down how much power is available from cosmic rays outside the Earth's magnetic field's influence. $\endgroup$ – uhoh Feb 24 '17 at 10:09
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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.

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