Yeah, I know, but that's a relation (them relative me) changing, as I see it. from the frames of the photons themselves it won't matter from where I study them, as I think about it :)They have only one defined final path, relative each other, ignoring the observer. At least that was what I meant

Physics is hard

It all goes back to 'energy' doesn't it, and as 'gravity' is coupled to mass and then those photons can be defined to have a mass, we now have a circumstantial chain pointing to 'energy' as the origin of 'gravity' too, which makes sense. With the exception of me thinking of it as coupled, not generated as in 'created' by mass. to me it's more of something 'sticking' to 'it all' SpaceTime that is, including 'energy'

As a geometry defining energy relative the observer sort of, and always local. Even though intrinsically invariant those light quanta's then can be seen as a summation of 'gravity', changing relative the observers position in SpaceTime.

But what do you mean by "If you compute the invariant mass including the atoms which will emit the photons, then invariant mass has one value at all times, no matter what reference frame you're in."?

That if we redefine the closed system to also in-cooperate the 'photons' sources, the atoms, you will find a same invariant mass, no matter the 'rotations' you apply on them by your position relative them? And then it is the observer too, he must always be included right?

So in that case, there is no difference between that system and a planet?

The mass (of the two photon system) will now be invariant, no matter what rotations you apply on it.

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Had to clean it up so I could understand what I wrote