B Understanding Entangled Particles in Different Time Frames

BillTre
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Here is something I don't understand which I expect someone here can explain.

If one member of an entangled pair goes on a trip at relativistic speeds, there will be two different frames of observation, with two different elapsed times.
The time frames can get off-set by years, over a long trip.
If one of the pair is interacted with, determining its state, when does this "immediate" effect also determine the state of the other half of the entangled pair (in its different time frame)?

It seems like two different time frames would predict two different times for the second particle to become determined, depending on which time frame was used. Or maybe the "causal" side of the pair sets the interaction?
 
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It doesn’t matter. If they are spacelike separated then no experiment can distinguish the order
 
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BillTre said:
Here is something I don't understand which I expect someone here can explain.

If one member of an entangled pair goes on a trip at relativistic speeds, there will be two different frames of observation, with two different elapsed times.
The time frames can get off-set by years, over a long trip.
If one of the pair is interacted with, determining its state, when does this "immediate" effect also determine the state of the other half of the entangled pair (in its different time frame)?

It seems like two different time frames would predict two different times for the second particle to become determined, depending on which time frame was used. Or maybe the "causal" side of the pair sets the interaction?
This is precisely the point about quantum entanglement. It's not enough to postulate a FTL communication mechanism, since there is no absolute sense in which one measurement takes place before the other.

Postulating that the two particles communicate fails on those two points.

QM is silent on how nature achieves correlation of measurements on an entangled pair. There's a discussion of the "possibilities" here:

https://www.physicsforums.com/threads/question-about-an-entanglement-paper.966466/#post-6135402
 
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QM, or better QFT, tells us precisely, how the correlations are "achieved". It's simply, because the particles are somehow prepared in an entangled state. One example is the decay of a neutral pion ##\pi^0 \rightarrow 2 \gamma##. This creates two photons with momenta ##\vec{k}## and ##-\vec{k}## with total angular-momentum 0 (due to energy-momentum conservation and angular-momentum conservation). This makes an entangled two-photon state
$$|\Psi \rangle = \frac{1}{\sqrt{2}} [\hat{a}^{\dagger}(\vec{k},1) \hat{a}^{\dagger}(-\vec{k},-1)-\hat{a}^{\dagger}(\vec{k},-1) \hat{a}^{\dagger}(-\vec{k},1)]|\Omega \rangle.$$
Here ##\hat{a}^{\dagger}(\vec{k},\lambda)## is the creation operator for a photon with momentum ##\vec{k}## and helicity ##\lambda##. This is an entangled photon state having all the astonishing properties such states have, particularly you can perform experiments violating Bell's inequality and all that. So QT indeed explains, how the correlations come about, namely in this case simply due to entanglement following from conservation laws.
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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