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Photons have energy, and so they are affected by gravitational lensing. More generally, they feel the force of gravity.

Gravitons have energy too, but it seems preposterous to assume they will swerve due to other gravitational fields. This should violate the inverse-square law. But how can they go free?

What is actually happening?

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  • $\begingroup$ Why do you consider that "preposterous"? Gravitons do interact with other gravitons, just like gluons, don't they? The infrared (long distance, Born approximation) inverse square law holds just fine, nonetheless. $\endgroup$ – Cosmas Zachos Nov 21 '16 at 1:22
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    $\begingroup$ See overlapping question today. $\endgroup$ – Cosmas Zachos Nov 21 '16 at 1:28
  • $\begingroup$ That question and others was my inspiration. I wanted to scale up the thought experiment to scales where the effect should not be negligible, such as the effect of a galaxy cluster on single gravitons $\endgroup$ – Ketil Tunheim Nov 21 '16 at 11:34
  • $\begingroup$ But single gravitons have not been observed, and will not, in our lifetimes... Most people assume your using the term is loose language for "gravitational waves", instead. Even the weak coupling regime of the "almost" quantized theory has daunting consistency problems. $\endgroup$ – Cosmas Zachos Nov 21 '16 at 13:19
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Gravitons are affected by gravitational fields, and viceversa.

From the classical Einstein Field Equations point of view it is straightforward. Those equations show even as an approximation that there are terms of the metric which go as 1/r at large distances, and those are the strongest surviving terms we would call gravitational radiation. Or gravitons. We can treat them linearly to determine their value (the coefficients of the 1/r terms), and it turns out that is what we are now able to detect in LIGO as they affect the distance between mirrors in the interferometers. So, gravitons indeed affect spacetime, i.e., cause gravity.

The other side is they they also are affected by it, they have to travel in the spacetime as null geodesics, i.e. Like light, to first order (I mean, they will affect that spacetime also, but that is a higher order approximation you can at first ignore). For LIGO we had to take the cosmology metric and from it calculate what redshift it would have with respect to us - Ie, the shift in freq from radiated to what we received.

Yes, it all has to be self consistent, but you can calculate often approximately right to the levels of accuracy we are able to achieve now.

As for gravitons interacting with other gravitons you'd need to do that extra higher order calculation - calculate how much the metric changes spacetime, and then the change of geodesic orbits due to that. We've not gotten to that level of accuracy. But in calculating how much gravitational wave gets produced we do go to various orders of approximation beyond the Nwtronian level, it has been parametrized to levels called eg PPN2 or PPN2.5 or higher. One of the effects we calculate is that as the grav waves get emmitted by the merging black holes it causes a recoil on the resulting leftover black hole, could be 5000 km/sec or more, just because of the conservation of momentum from emitting the gravitational energy preponderantly asymmetrically. Enough momentum to get merged black holes kicked out of their galaxies in some cases.

So, yes, all kinds of effects, and it depends on the level,of approximation you need to account for.

See the Wikipedia article, it mentions the recoil towards the end, at

https://en.m.wikipedia.org/wiki/Binary_black_hole

There's others, Google recoiling black hole mergers.

Finer or more subtle interactions we are still not able to detect

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Gravitons are definitely effected by gravitational fields (i.e. there is 'back-reaction'), this is part of why general relativity is so hard to solve in general situations (e.g. numerical solutions to the field equations).

In terms of the inverse square law:
First, and most importantly, the inverse square law comes from the symmetry and dimensionality of the problem: it's an inverse square law because 'field lines' in spherical symmetry pass through surface areas proportional to $r^2$.
Second, note that the inverse square law is only an approximation which definitely breaks down in the strong-field regime (which is also where gravitational back-reaction would be important). I think you can interpret the deviations from $r^{-2}$ as resulting from the geometry of space-time losing some of that spatial symmetry in the strong-field regime (i.e.~measuring devices will start to disagree on $r$, etc)... perhaps* this could even be interpreted as a 'back-reaction' where the existence of the gravitational field starts to cause geometric effects which then lead to deviations from $r^{-2}$.
(*Hopefully someone wiser about the field equations can comment if that's fair or not*.)

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