thinking out of the cat's box

Andre

A simple one, Schrödingers cat.

A cat is placed in a box, together with a radioactive atom. If the atom decays, and the geiger-counter detects an alpha particle, the hammer hits a flask of prussic acid (HCN), killing the cat.

Now why is the cat alive when the observer opens the box?

Jimmy Snyder

Answer:

The experiment wouldn't make sense if the cat were not alive at the time that the observer opened the box to put the cat in.

eom

Andre

okay let me rephrase, opening the box after N minutes after the start of the experiment. But good thinking.

Jimmy Snyder

Answer:

Because the observer is in the box with the cat. If the cat dies, so does the observer, and the box is never opened. If the observer opens the box, it means the observer is alive, and that means the cat is alive.

eom

Andre

Incredible how many way you can think out of the cat's box, It's starting to look like 'why did the duck cross the street'

Think proverbal.

Jimmy Snyder

I did early on, but I didn't think much of it. Answer:

The cat will only lose at most one of its nine lives. However, the cat may have lost eight lives before being confined, so this answer doesn't cover all cases.

eom

Andre

Right, So did Schrödinger think in of out the box?

Anyway, in another forum yesterday, I met an old forum buddy who inquired if I was still thinking out of the box while busy with a complex elaboration. So I replied that the quartenary box was way too small to think in so somebody else posted this picture



asking if this box was big enough. So I needed some ammo to dismiss the joke and carry on with the thread. So I explained why not even Schrödinger thought out of the box.

lanfearxt

it's alive because that's the first time he opened the box to put the cat in!

DaveC426913

Is there really a trick to this?

The cat is alive because the atom didn't decay in N minutes. The atom is U238, which has a half-life of 4.5 billion years.

Personally, I think the setup is shaky. One atom? How are you going to detect one alpha particle?

ƒ(x)

Are we to assume that the geiger-counter works?

DaveC426913

Is there a reason to assume otherwise?

ƒ(x)

In that case, wouldn't the obvious answer be that the atom didn't decay?

DaveC426913

You read post #9, right?

ƒ(x)

Of course, but does the atom have to be U238?