what if information can be transmitted faster than c

[swansont]

hey rasori.

 

don’t' get intimidated.

Science is their religion.

That’s why they act that way.

Everyone who ever did anything great' date=' challenged the existing mindset.

 

you will never get support once you step outside the norm , at least until you have a hard proof.

then all the nay Sayers will be climbing over themselves trying to kiss your +++.

(sorry too crude)[/quote']

 

If science requires proof, how is it a religion?

[boxhead]

just imagine it is possible to transfer information faster than c than we can change the theory of relativity as time travel is possinle and we can change certain things in the past like birth of science forum network and can virtually delete this thread.

[Sayonara]

We can do that anyway.

[Dror]

Funny thread...

Well, classically, nothing can reach C unless it has 0 rest mass, and nothing absolutly nothing can go beyond C, without starting to travel backwards in time. There you have it..

To me that sounds kind of useless in terms if passing someone info and expecting to reach him in the future, like it should.. lol..

As for time travel to the past, enjoy.

 

Put it another way, if you pass info faster then C, the reciever won't get your message. It'll get it before it even wants it. And thats rendered useless.

[swansont]

Well' date=' classically, nothing can reach C unless it has 0 rest mass[/quote']

 

As relativity is not a classical effect, this is technically wrong. Classically, there was no speed limit.

[J.C.MacSwell]

As relativity is not a classical effect, this is technically wrong. Classically, there was no speed limit.

 

I've seen a lot of references calling "classical" pre-quantum/post relativity (even though they overlap), although I was thinking the same thing.

[swansont]

I've seen a lot of references calling "classical" pre-quantum/post relativity (even though they overlap), although I was thinking the same thing.

 

I guess it depends on your point of view. To me the dividing line is 1905. Pre- or post-Einstein's publications. You've cemented the radiation quantization with the photoelectric effect, and got relativity.

[Dror]

As relativity is not a classical effect, this is technically wrong. Classically, there was no speed limit.

 

I meant classically, as in communicating classically.

Sorry for misunderstanding.

By non classically in that respect I mean communicating with Entangled particles, which I'm still not convinced is not possible.

Contradicting it using reletivistic casualty, or lack of, is not the way to go.

As the changes to particle B caused by disturbance to particle A (regardless of entanglment being destroyed) do accur at no time at all. And it does not contradict casualty, cause it has nothing to do with reletivity.

Thats the way I see it anyway.

[alt_f13]

you dont say to a mathematician but 1 + 1 could = 3 because it doesnt (unless your counting in binary... but you know what i mean!) similarly you dont go to a physicist that you're gonna travel FTL because you cant.

 

if you think that no tricks' date=' no binary, straight forward pure maths that a single unit (1) + or added to a second single unit (1+1) could be equivelent to anything other than 2 units or 2 then i will happily stop the argument now.

 

not trying to make it sound like your dumb, just wanna make sure you're not saying like 1 and 1 = 11 or sumin![/quote']

 

I just wanted to point out that that first post made the mistake that the second post was critisizing (or whatever, seeing as neither post was really about math...)

 

In binary 1+1 = 10 (two)... not 3. Saying 1+1 in binary equals three is exactly like saying 1+1 in decimal is 11. They use the exact same digital representation too! Silly mans.

[swansont]

I meant classically' date=' as in communicating classically.

Sorry for misunderstanding.

By non classically in that respect I mean communicating with Entangled particles, which I'm still not convinced is not possible.

Contradicting it using reletivistic casualty, or lack of, is not the way to go.

As the changes to particle B caused by disturbance to particle A (regardless of entanglment being destroyed) do accur at no time at all. And it does not contradict casualty, cause it has nothing to do with reletivity.

Thats the way I see it anyway.[/quote']

 

Disturbances to A do not cause changes to B. They cause a collapse of the wave function - these are not the same things. When a measurement is made on A that collapses the wave function of B, how can either observer convey that information to the other one, other than limited by c?

[Dror]

swansont so the only "relationship" between the 2 particles is of their wave function? As in you destroyed one = you destroyed the other?

I'm trying to understand not to argue.

[swansont]

swansont so the only "relationship" between the 2 particles is of their wave function? As in you destroyed one = you destroyed the other?

I'm trying to understand not to argue.

 

Basically, yes. The particles are in a superposition. Measuring one particle puts both of them in definite states. The collapse of both wave functions happens simultaneously - this is what appears to violate relativity, but because you can't communicate the information about the collapse faster than c, it really isn't. (this is the "spooky action at a distance" that bothered Einstein. But that was on philosophical grounds, not physics ones, so it really doesn't matter)

[Dror]

swansont,

What you CAN do is check B after the colapse of both, to see if it's current state could have possibly accur without tinkering with A.

Put it even more simply, check for B and see if it's in a state of 0 probablity.

 

Correct me if I'm wrong.

[swansont]

swansont' date='

What you CAN do is check B after the colapse of both, to see if it's current state could have possibly accur without tinkering with A.

Put it even more simply, check for B and see if it's in a state of 0 probablity.

 

Correct me if I'm wrong.[/quote']

 

How does B's observer know if A's observer made a measurement?

 

And B is in a superposition of 2 states. It's going to be in one or the other.

[blike]

How does B's observer know if A's observer made a measurement?

Can't you perform an experiment to determine whether particle B is behaving as a classical particle, or according to the wave function? For example, when a classical particle is subjected to the two-slit experiment, it will show a classical banding pattern.

 

For instance, assume two billion particles are entangled. Spaceship A takes one half of the entagled particles, and spaceship B takes the other half. Spaceship B will then travel to the edge of the galaxy. When it arrives, it will collapse the wavefunction of all the particles (by measurement of some sort). Meanwhile, spaceship A is running tests continuously to determine whether its half of the entangled particles are behaving classically. If they are, then spaceship B has arrived at the edge of the galaxy. If they aren't, then spaceship B hasn't arrived.

 

However, this really isn't FTL communication, because to separate the particles you still have to give them a ride on the spaceship.

 

Edit: Also, the two-slit experiment was just an example. You'd probably have to find a way to determine whether or not the property you've entangled is still in a superposition.

[5614]

In binary 1+1 = 10 (two)... not 3. Saying 1+1 in binary equals three is exactly like saying 1+1 in decimal is 11. They use the exact same digital representation too! Silly mans.

I was referring to when 1+1=11 (two ones, one after the other) which in binary is 3.... silly man

[swansont]

Can't you perform an experiment to determine whether particle B is behaving as a classical particle' date=' or according to the wave function? For example, when a classical particle is subjected to the two-slit experiment, it will show a classical banding pattern.

 

For instance, assume two billion particles are entangled. Spaceship A takes one half of the entagled particles, and spaceship B takes the other half. Spaceship B will then travel to the edge of the galaxy. When it arrives, it will collapse the wavefunction of all the particles (by measurement of some sort). Meanwhile, spaceship A is running tests continuously to determine whether its half of the entangled particles are behaving classically. If they are, then spaceship B has arrived at the edge of the galaxy. If they aren't, then spaceship B hasn't arrived.

 

However, this really isn't FTL communication, because to separate the particles you still have to give them a ride on the spaceship.

 

Edit: Also, the two-slit experiment was just an example. You'd probably have to find a way to determine whether or not the property you've entangled is still in a superposition.[/quote']

 

I'm not sure what you mean by the "classical banding pattern." Classically it's a wave, and it shows interference. In QM you can get the pattern one particle at a time, but it's the same pattern.

 

You detect the interference by detecting the particles at a particular location. Photons are absorbed in a detector. That gets rid of the photon and with it, the entanglement.

[Janus]

swansont' date='

What you CAN do is check B after the colapse of both, to see if it's current state could have possibly accur without tinkering with A.

Put it even more simply, check for B and see if it's in a state of 0 probablity.

 

Correct me if I'm wrong.[/quote']

 

The instant you check B, you cause it's waveform to collapse, if it hasn't already. The thing is that you cannot know whether the state you now measure was caused by "tinkering" with A or by your own measurement of B.

[Dror]

Janus yes you can.

Using quantum mechanics you can forsee all the possible states you might find the particle in. By tinkering with A you essensialy colapse the wavefunction of B and hence you will find B in a state you did not forsee - 0 probability. And take that as communication.

However as swansont said, at the first time you check B you colapse both wave functions. This calls for sending a classical signal of "check now" to B just after you tinkered with A. And that's a really funny way of communicating - useless.

 

Now I get it.

[swansont]

The particle is in a superposition of two states. When you collapse the wavefunction, it will go to one of those two states, not one with 0 probability.