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Your typical photon rocket is imagined as being an antimatter-powered, gamma-ray laser. A flashlight of extremely lethal ionizing radiation, capable of sterilizing a planet if pointed the wrong way.

What if you could use lower-energy photons instead? The kind that might fall in (or beyond) the Ultra Low Frequency (ULF) range. Depending on the discipline, ULF is defined by the range of 1 mHz to 100 Hz, corresponding to wavelengths 3M-3K km respectively. It's used by major military powers to communicate with submarines, penetrating both earth and seawater, and ULF EM reflections off the interior of Saturn's moon Titan point to possible subsurface water-ammonia oceans.

It seems that at lower and lower frequencies, miles of solid ice & rock look more and more translucent.

We're replacing high-energy gamma photons with low-energy ULF photons (let's not dwell on how), so we'll need vastly more of them to make up the difference.

(Caution: non-expert using math.)
Say we have an engine that generates 100 N of thrust. From the energy-momentum relationship of a photon, $E=pc$, that corresponds to 30 GJ every second. Photons with wavelength, $l$, have frequency, $f=\frac{c}{l}$, and energy-per-photon, $E_p=hf$, where $h$ is Planck's constant, 6.626E-34 J/Hz. With a wavelength of 3M km (lower bound of ULF), $E_p$ is about 6.63E-35 J. Dividing the 30 GJ requirement by the energy-per-photon, that's around 4.53E44 photons.
For comparison, the same energy from gammas would see around 20 OOM fewer photons.

Even if ULF photons are scattered/absorbed/attenuated by matter less often, having so much more of them increases the chances of interaction.

Could an ULF photon rocket be "safe" to operate around/point at Earth? Could someone stand in front of the exhaust and live to talk about it? Could a suitably powerful one lift off from Earth without annihilating everything down to bedrock? Or should Gravitational Wave Rocket Lite™ be relegated to deep space?

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    $\begingroup$ Not sure that you can ignore the "how", given that the antenna elements for even HF transmissions are multiple metres long, you'll presumably need emitters that are hundreds of metres or even kilometres long (wide) to emit ULF signals. And then you need a power source... $\endgroup$
    – KerrAvon2055
    Apr 13 at 16:49
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    $\begingroup$ Hard science tag. Answers that conform to this tag will be unintelligible to most worldbuilders. $\endgroup$
    – Zeiss Ikon
    Apr 13 at 17:44
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    $\begingroup$ at high enough intensity, even neutrinos can be lethal what-if.xkcd.com/73 $\endgroup$
    – ths
    Apr 13 at 18:03
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    $\begingroup$ That's...not typical of any photon rocket design I've come across before. What matters with a photon rocket is just output power and having the beam reasonably well collimated to reduce cosine losses. A photon rocket gains nothing from using a laser. A typical photon rocket would be something like a reactor (fission, fusion, or antimatter) producing thermal black body radiation, at the focus of a parabolic reflector. $\endgroup$
    – Christopher James Huff
    Apr 13 at 18:40
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    $\begingroup$ While you can substitute anything in a formula, I wonder: Do such "low-energy photons" exist at all? Do you consider any electromagnetic wave to be photons? The other question is: If they exist, will they follow a line of sight? $\endgroup$
    – U. Windl
    Apr 14 at 9:51

3 Answers 3

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With high intensity you walk out of the realm of linearity and non linear effects start to appear.

One of this effects is the multi-photon absorption

two-photon absorption (TPA or 2PA) or two-photon excitation or non-linear absorption is the simultaneous absorption of two photons of identical or different frequencies in order to excite a molecule from one state (usually the ground state) to a higher energy, most commonly an excited electronic state. Absorption of two photons with different frequencies is called non-degenerate two-photon absorption. Since TPA depends on the simultaneous absorption of two photons, the probability of TPA is proportional to the square of the light intensity, thus it is a nonlinear optical process.

This means that materials that at low intensity regimes are transparent, become suddenly opaque when nonlinear optical processes are involved.

And as you have already realized for gamma rays, absorbing that much energy can lead to problems.

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  • $\begingroup$ A minimum energy is required to push electrons into higher energy levels, right? Yet we're dealing with extremely low-energy photons. If my math is right, it takes 2.18E-18 J to ionize a hydrogen atom. Each ULF photon carries 6.63E-35 J, which means we'll need around 3.29E16 of them arriving at one atom at the same time. $\endgroup$
    – BMF
    Apr 13 at 18:02
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    $\begingroup$ Hydrogen atoms have a radius 0.05 nm (double Bohr's radius) and cross-section of about 8E-21 m^2. If the exhaust stream has a radius 1 m, and the ULF photons are evenly distributed, each photon can expect about 7E-45 m^2 all to itself. Assuming that's all correct, each hydrogen atom can expect to share its cross-section with about 10^24 ULF photons, 8 OOM more than needed to ionize it. $\endgroup$
    – BMF
    Apr 13 at 18:22
  • $\begingroup$ Seeing as how ionization energy increases towards hydrogen, other atoms might be worse-off. $\endgroup$
    – BMF
    Apr 13 at 18:25
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    $\begingroup$ Correction: I forgot that the beam having such a long wavelength will diverge very rapidly. It isn't so much a beam as a preferential hemisphere. Unsure whether Airy wave can be trusted at that point. All the photons in a 1 meter radius is an implausible setup. The aperture size should be on the order of the wavelength (from 1/4 to 1/3, I think) which drastically changes the picture. Only millions of ULF photons per hydrogen atom, many OOM below ionization. $\endgroup$
    – BMF
    Apr 13 at 22:00
  • $\begingroup$ @BMF You're basing that on your figure of 4.53e44 photons? But that's 4.53e44 photons per second. I'd imagine that the timescale for multiple photon ionization is many orders of magnitude less than a second, which would mean the number of photons for this purpose should be proportionally many orders of magnitude smaller, causing a far smaller chance of ionization. $\endgroup$
    – causative
    Apr 14 at 16:03
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Those ULF photons still carry the same total energy (less per photon, many more photons, as noted in the question), so anything that absorbs that energy will be affected.

Obviously, the beam area will be much, much larger than your gamma ray flashlight -- the smallest possible focus spot approximates the wavelength (as does the minimum emitter size, BTW, and it'll be many times that large to produce the highly directional beam that doesn't waste most of the beam energy by radiating in non-thrusting directions), so the total heating where the beam impinges will be more diffuse -- but that just means you may cook everything on the surface of an entire planet (albeit, as noted in comments, very, very slowly), rather than punch a hole all the way through (or at least into the mantle or core) -- but gamma can pass through many objects without much absorption and is still quite lethal with its low interaction cross section. ULF at very high power might well have the same issue; we simply don't know.

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  • $\begingroup$ With ULF of a certain frequency the ULF ray could just diffrant off Earth as the wave length would be more than twice larger than Earth's diameter. Yet with the required energy even if 99% of ULF photons would miss Earth because of particle-wave effects, the remainder could cause various things. Not sure which, and not sure if the photon of that frequency would not just, say, reflect off anything macroscopic. $\endgroup$
    – Vesper
    Apr 14 at 4:38
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    $\begingroup$ "Cook everything on the surface of an entire planet"? The OP states 30 GW. That is similar to the sunlight on a plot of land 30 km^2. If the beam is spread across the whole Earth, that is a fifth of a milliwatt per square meter, a thousand times less than moonlight. Nothing is getting cooked. $\endgroup$
    – causative
    Apr 14 at 5:40
  • $\begingroup$ Yep, @causative, you're right -- I hadn't done the math, even at BOTE levels. $\endgroup$
    – Zeiss Ikon
    Apr 15 at 17:15
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The first issue is that to generate thrust, you need to send more energy backwards than forwards. This requires an antenna of the same order of magnitude size as the wavelength or larger- clever antennas that are much smaller than the wavelength are forced to have symmetrical transmissions, for extremely complicated reasons. (eg, https://physicsopenlab.org/2020/05/03/loop-antenna-for-very-low-frequency/)

So, the spaceship you propose in your question is the size of the earth and generating 100 newtons of thrust- it isn't going anywhere fast.

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  • $\begingroup$ I see, thanks for the reference! $\endgroup$
    – BMF
    Apr 14 at 15:46

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