Playing with Heisenberg's uncertainity principle

[ronthepon]

I'll get to the point.

 

[math]{\Delta}x {\Delta}p > \frac {h}{4 {\pi}}[/math]

 

Or

[math]{\Delta}x {\Delta}v > \frac {h}{4 {\pi} m}[/math]

 

Where m is the mass of the particle under consideration.

 

Now let the maximum value of [math] {\Delta}v [/math] be equal to [math]c[/math] Agreed?

 

So, minimum value of [math]{\Delta}x[/math] will be given by

 

[math] {\Delta}x_{min} = \frac{h}{4 {\pi} m c}[/math]

Correct?

 

Can we take this as the... say observation radius or something?

 

And lets apply the constants.

 

I got [math] {\Delta}x_{min} = \frac{1.884 * {10^{-43}}}{m} [/math] by using Log tables.

 

There is a problem. Using the masses of neutrons and protons here, I get values at about [math]10^{-13}[/math], which is about a hundred times the size of the atomic nucleus.

 

Can somebody help me get this?

[ronthepon]

By the way, I have no idea how the uncertainity principle was derived or reached.

[UncleAl]

1) If you are hard by lightspeed the observed mass is no way near the rest mass.

 

2) Momentum is not mv except in Newtonian physics. Momentum is a conserved four vector.

 

3) Look up the Lamb shift for hydrogen and then the Lamb shift for U(91+). Relativistic corrections matter.

[ronthepon]

to number one, I'ts the uncertainity that's c, and we have no idea at all about what's the true speed anyway...

 

Meanwhile I'll be checking out the directions you gave...

[Qfwfq]

Never mind Unk's manners Ron. :) ;)

 

The expression for momentum is:

 

[math]p = \frac{mv}{\sqrt{1 - v^2/c^2}}[/math]

 

and I'm sure you can easily check that it isn't upper bound as v approaches infinity.

 

Your idea was essentially based on p having the upper limit mc, and thus also[math]\small\Delta[/math]p, but this isn't so because the denominator is infinitesimal.

[ronthepon]

Thanks a lot, Qfwfq, that clears it all up.