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Everybody can buy a single photon counter. Why are there no single graviton counters?

Obviously, graviton sources are rare. But why are graviton detectors so hard to make? Is it just because the typical gravitons (say, from black hole mergers) have such low frequency, and thus very low energy?

In other words, is it a noise issue, in the sense that any graviton detector has intrinsic difficulties to distinguish a signal from noise, because the signal has such a low level?

Or is there still another reason?

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    $\begingroup$ Since gravitons have never been detected, it's a little premature to start marketing detectors for them. $\endgroup$
    – D. Halsey
    Oct 13, 2019 at 13:42
  • $\begingroup$ Of course, and therefore the question asks about exactly why such detectors are so hard to make. $\endgroup$
    – frauke
    Oct 13, 2019 at 19:55
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    $\begingroup$ "Hard to make" seems like you are saying they have been made before $\endgroup$ Oct 14, 2019 at 5:57

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The best reference on this topic that I'm aware of is Rothman and Boughn, "Can Gravitons Be Detected?," http://arxiv.org/abs/gr-qc/0601043 . They argue that the cross-section for a graviton to interact with pretty much any target is on the order of the square of the Planck length. This is not obvious, and they refer to a lot of previous authors who came up with other estimates, and claim that with hindsight those estimates were wrong. So basically we can't detect single gravitons because the Planck length is very small.

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  • $\begingroup$ That paper is wonderful! Thank you for pointing it out. $\endgroup$
    – frauke
    Oct 14, 2019 at 4:37
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Look at the difference in coupling constants between electromagnetism and gravity:

electrmag

grav

The coupling constants are what define the ballpark of the probability calculations in quantum mechanics. Supposing that gravitons exist ( it is an effective, not decisive, quantization of gravity that is used) the probability of their interacting with the atoms and molecules of a detector is smaller than for photons at least by $10^{35}$ times.

Since Avogadro's number( molecules in a mole of matter) is of order $10^{23}$ you can imagine how big a detector would have to be in order for a graviton to interact with one of its molecules. Then there is the probability of the interaction being non trivial in energy so some photons are produced in order to see that an interaction happened.

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Photons interact readily with individual charged particles and have a relatively high energy compared with the energy required to trigger a chain of events that leads to a count in a detector.

Gravitons don't interact with individual particles directly, they have the effect of warping spacetime. The effect an individual graviton is expected to have is incredibly tiny compared with the impact of a photon on an atom, so a vastly more-sensitive detector would be required.

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  • $\begingroup$ All this makes sense, except one statement: "Gravitons don't interact with individual particles directly." How do we know this? $\endgroup$
    – frauke
    Oct 13, 2019 at 10:13
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    $\begingroup$ Contrary to what this answer says, it is common to do quantum gravity calculations in which individual gravitons interact with individual particles such as electrons. The interaction is simply too weak to detect, since gravity is 40 orders of magnitude ($10^{40}$) weaker than electromagnetism. $\endgroup$
    – G. Smith
    Oct 13, 2019 at 16:13
  • $\begingroup$ @frauke That's good point, sorry. What I meant to say was that the macroscopic effects of gravity are the cumulative effect of very (and I mean very) large numbers of gravitons- the effect of a single graviton is tiny by comparison with the impact a photon of visible light might have, which is enough to excite an electron. Thanks to G.smith too. $\endgroup$ Oct 13, 2019 at 17:45
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    $\begingroup$ Well, gravity is 10^40 times weaker than electrodynamics. But a one eV graviton has the same momentum as a 1 eV photon. So detection should be just as easy, when it hits an electron. I suppose the interaction is not too weak to be detectable, but too improbable to happen. $\endgroup$
    – frauke
    Oct 13, 2019 at 19:58
  • $\begingroup$ It might be instructive to calculate how much energy would be carried by a single graviton whose frequency is in the range of gravity waves detectable by LIGO, and how high the frequency would have to be to carry enough energy to be electronically detectable, and then think about what kind of physical system could conceivably oscillate with that frequency. $\endgroup$
    – S. McGrew
    Oct 14, 2019 at 2:17
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It is very important to differentiate between virtual gravitons (in QFT calculations of interactions) and real gravitons (GW quanta). Now there is no consensus on this site whether gravitons are the quanta of GWs or not.

https://physics.stackexchange.com/a/215180/132371

There are two ways we can detect gravitons, either build detectors as you suggest, or create them at the LHC.

Now detecting single gravitons is very unlikely, because gravitons interact with matter very very weakly.

$$\alpha_\mathrm{G} = \frac{G m_\mathrm{e}^2}{\hbar c} = \left( \frac{m_\mathrm{e}}{m_\mathrm{P}} \right)^2 \approx 1.7518 \times 10^{-45} $$

The gravitational coupling constant characterizes the gravitatonal attraction between elementary particles.

$$\alpha_\mathrm{G}$$ is 42 orders of magnitude smaller then $$\alpha$$

Now if you use the energies at LHC, you will see that it is similarly orders of magnitude less likely to produce a single graviton at the LHC then a single photon.

https://en.wikipedia.org/wiki/Gravitational_coupling_constant

Thus, the correct answer to your question is that in the foreseeable future we will not be able to build detectors that can detect single gravitons nor will we be able to produce them at the LHC.

You are correct, it is the cross section.

Gravitons have a much lower cross section then photons to interact with matter.

https://arxiv.org/pdf/gr-qc/0607045.pdf

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  • $\begingroup$ why the downvote? $\endgroup$ Oct 14, 2019 at 14:56

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