[MannoFr]

.
Is there something that connects expansion to having a direction in time for you to discount any other cause?
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Time is the distance between events.

The further the distance the slower the time.

If the expansion rate is greater than the time rate, the events will receed.

Distance is gathered by the gravitational force. The resulting equal but opposite reaction is expansion. So events remain local.

For a short period of time...

Hope that helps.

This may or may not be written by something slightly coherent

[Slonopotam845]

.
Is there something that connects expansion to having a direction in time for you to discount any other cause?
-------------------------

Time is the distance between events.

The further the distance the slower the time.

If the expansion rate is greater than the time rate, the events will receed.

Distance is gathered by the gravitational force. The resulting equal but opposite reaction is expansion. So events remain local.

For a short period of time...

Hope that helps.

This may or may not be written by something slightly coherent

No, that provides some very good points to address. Will get on to it shortly. Thanks for your patience and contribution.

[HedgeYourBets]

Heheheheh. He plays with us allowing us to even believe we can live through the debate, but just when we think we are getting on top of things he crushes us in his jaws with a deft statement such as above. :-))

it actually says a lot about the idea of this thread, of which the summarising looks like a bit monumental with some consideration.

[tgs]

it actually says a lot about the idea of this thread, of which the summarising looks like a bit monumental with some consideration.

Don't be so hard on yourself Postpoc. It happens to me all the time on this forum. The lesson I get from it is that it really doesn't need to be that complex, but it does require you to understand the maths and to be able to drill deep down into a particular interpretation. For those of us who don't have these skills (like me), we just have to rely on the accurate interpretation of the maths (and that's where I pin my trust in KJW). For those maths guru's however, it still doesn't come easy and upon making a choice of which road they wish to take, they just have to hope that they have made the right choice.

For a generalist and a poor mathematician, who likes to keep the bridges open, it is almost impossible to get a deep understanding of all the theoretical alternatives and that's where the confusion lies. :-))

[Drugmachine]

was mostly being dramatic, but I appreciate the sentiments. Have a good night. I need some sleep.

[Peptobismol]

Night Postpoc. One day we will trip him up *sniggers* :-))

[Slonopotam845]

I thought decoherence or wave function collapse was a symptom of measurement in our classical universe and that in the Many Worlds Interpretation an 'entangled' superposition persists. Need an enlightened one to help me out on this? :-))

Entanglement results from an interaction between objects. Superposition is the notion that a quantum state is a vector sum of orthogonal eigenstates. Entanglement itself isn't anything remarkable, but when there is a superposition of entangled states, then you have the non-classical behaviour that Bell's theorem is about.

One thing that may be a point of confusion in this thread is that if A and B are entangled, then measurement of A (or B) generally destroys the entanglement between A and B. In other words, entanglements generally don't persist.

[S.T.D.]

Entanglement itself isn't anything remarkable, but when there is a superposition of entangled states, then you have the non-classical behaviour that Bell's theorem is about.

Although often the term "entanglement" is referring to the superposition of entangled states.

[Peptobismol]

I was just wondering how you could 'in a laboratory' in principle ever achieve a superposition that was not entangled in some way with another object. :-))

[Beerinkol]

Would it be permissable to speculate that for any physical change to occur, a wave function collapse might be necessary, but the moment of collapse might not have to be present during the moment of change?

I don't think so. Wavefunction collapse (or the appearence of wavefunction collapse) occurs when a microscopic system interacts with a macoscopic system. However, microscopic systems can still interact with each other. The result is a superposition of the entanglement that results from the interaction. Because the total system is still microscopic, there is no wavefunction collapse (or the appearence of wavefunction collapse).

[Slonopotam845]

I was just wondering how you could 'in a laboratory' in principle ever achieve a superposition that was not entangled in some way with another object. :-))

A single particle in a pure quantum state is a superposition, but there is no entanglement because there is only one particle. However, when the particle is measured, it becomes entangled with the measuring device (and us when we look at the measuring device).

[Slonopotam845]

A single particle in a pure quantum state is a superposition, but there is no entanglement because there is only one particle. However, when the particle is measured, it becomes entangled with the measuring device (and us when we look at the measuring device).

Hmmm. I just realised that one is going to have entanglement with the source of the particle.

[Lt_Apple]

>>Hmmm. I just realised that one is going to have entanglement with the source of the particle.

Haha :-))

[LottiFurmann]

Hmmm. I just realised that one is going to have entanglement with the source of the particle.

So....what is an electron? :-))

[LottiFurmann]

Haha :-))

Create a pair a entangled photons, then measure one of the photons. The other photon is no longer entangled to anything.

[Raj_Copi_Jin]

Create a pair a entangled photons, then measure one of the photons. The other photon is no longer entangled to anything.

Surely it is just not entangled with respect to the basis vector that was being measured. What about the rest of the superposition? :-))

[PhillipHer]

Surely it is just not entangled with respect to the basis vector that was being measured. What about the rest of the superposition? :-))

Once you measure one of the photons, there is no longer a superposition and the two-particle state is just the tensor product of the two single-particle states, which means that the two photons are now independent.

[tgs]

You win. I'm off to bed. Night KJW :-))

[tgs]

Once you measure one of the photons, there is no longer a superposition and the two-particle state is just the tensor product of the two single-particle states, which means that the two photons are now independent.

On the other hand, the unmeasured photon is still entangled with the measured photon + measuring device system.

[Paul Bunyan]

Originally Posted by KJW (Maths)
Once you measure one of the photons, there is no longer a superposition and the two-particle state is just the tensor product of the two single-particle states, which means that the two photons are now independent.

On the other hand, the unmeasured photon is still entangled with the measured photon + measuring device system. This illustrates an aspect about the maths of quantum mechanics: one normally doesn't include the measuring device into the description of a quantum mechanical system being measured. In the case of quantum decoherence, the irreversible nature of wavefunction collapse is the result of not including the measuring device in the description, while including the measuring device leads to unitary evolution of the total system wavefunction.