Uncertainty principal: mean square momentum

Originally posted by @metal-brain
You have contributed nothing to this thread. That is because you can't. You know much less than me.

Don't count on it. Where did you get your Phd?

Originally posted by @sonhouse
Don't count on it. Where did you get your Phd?

Tell me, what has he contributed? Trolling isn't a contribution.

Originally posted by @metal-brain
You have contributed nothing to this thread.

and so haven't you.

Originally posted by @humy
and so haven't you.

I created this thread.
That is a contribution.

Originally posted by @metal-brain
I created this thread.
That is a contribution.

well done.
You created this thread and your only contribution to it was your comment of "I don't understand" because, for you, that is an extreme rare example of a comment that is both totally honest and totally devoid of delusional arrogant overtones that hint that you think you know better than someone else (which is rarely the case).

Originally posted by @sonhouse
You mean one of the three could be negative but the other 2 are positive? What does it mean in the physical world if all three components are negative?

It means it is moving in a direction opposite to the directions your arbitrarily selected axes point in. Momentum is a vector quantity and so it is measured relative to some set of orthogonal axes. It's like putting your car in reverse, relative to forwards gears your car has negative velocity.

Originally posted by @humy
well done.
You created this thread and your only contribution to it was your comment of "I don't understand" because, for you, that is an extreme rare example of a comment that is both totally honest and totally devoid of delusional arrogant overtones that hint that you think you know better than someone else (which is rarely the case).

You have done nothing but troll in this thread. I created this thread so I would learn something, not necessarily teach anything. The only thing you have taught is your apparent uncontrollable desire to troll.

Originally posted by @metal-brain
I have always been curious as to why electrons do not crash into protons given such a powerful attraction.Durac's wave function was the best explanation I had found until reading the Feynman lecture today.

Here is an excerpt from the link below:

"You know, of course, that atoms are made with positive protons in the nucleus and with electrons outsi ...[text shortened]... m. I did an internet search but had a difficult time finding info about it in the right context.

Kazet's explanation is correct, but a little terse. I've been thinking about how to explain this, as the quantity arises quite naturally. Let's forget about the proton for a minute and just consider a particle confined to a box. Classically this is just the ideal gas model but with only one particle. The ideal gas model has a collection of particles which do not interact with each other but do bounce off the walls of the container perfectly elastically. Suppose there are N helium atoms of mass m, the iᵗʰ atom has momentum pᵢ and velocity vᵢ, where I'm using bold face to indicate a vector quantity, for simplicity we'll assume the box is a cube of side length L. Let's focus on the component of velocity in the x direction, vᵢₓ the atom takes time T = L/vᵢₓ to traverse the container and so will hit the right hand wall every 2L/vᵢₓ seconds. The force exerted on that wall of the container is equal to the momentum change of the atom per collision, 2pᵢₓ, divided by the number of times per second the atom hits it:

Fᵢ = pᵢₓvᵢₓ/L = mvᵢₓ²/L

The pressure exerted on the wall is the force divided by the area, so we have:

PᵢL³ = mvᵢₓ²

L³ is just the volume V. We can repeat this analysis for the other components of velocity and get:

3PᵢV = mvᵢ²

Pᵢ is the pressure attributable to atom i. The total pressure is the sum of these:

3PV = m Σᵢvᵢ²

So we have the pressure in terms of the sum over the square of the particle velocities. I'll use <x> to mean average of x. <v²> is the mean square velocity, and since there are N atoms we have:

<v²> = Σᵢvᵢ² / N

and so:

3PV = Nm <v²>

or in terms of momentum:

PV = (N/3m)<p²>

Now, return to the quantum mechanical problem of a particle in a box. We can picture a quantum particle in a box as being the same as a collection of classical particles in boxes, each in a separate box. When we measure the momentum of the quantum particle it is as if we selected one of the classical boxes at random and measured the momentum of that classical particle. We can work out our averages from that.

Let's focus on the component of momentum in the x direction again. By symmetry the mean momentum is zero and is given by:

<p> = Σᵢpᵢ / N = 0

We can work out the standard deviation, whose square is the average square deviation from the mean. This is given by:

Δpₓ² = Σᵢ (pₓᵢ - <pₓ> )² /N = <pₓ²> - <pₓ>² = <pₓ²>

Where the last equality follows because the mean momentum is zero. So the uncertainly in the momentum is just the root mean square momentum (assuming a symmetric box).

The uncertainty relation tells us that ΔpₓΔx > h. Since the uncertainty of the position of the electron is just the size of the box and the uncertainly in the momentum is the root mean square momentum we can write:

<pₓ²>L² > h²

and so

<p²> > 3h²/L²

But we can relate this to the pressure equation we got earlier and have that

3mPV = <[b]p[b]²> > 3h²/L²

so:

P ~ h²/L⁵

or, in terms of total force on the container walls:

F ~ h²/L² ~ h²/A

Returning to an atom, although the electrostatic force increases as 1/r² as the proton tries to confine the electron in a smaller space the uncertainty relation produces a kinematic force proportional to the surface area of the volume it is being confined to. But the "uncertainty force" (my term) goes as 1/r² so they tend to cancel out.