# [Ask a Geek] Applying uncertainty principle for a macroscopic object

Curiosity Science Physics [Ask a Geek] Applying uncertainty principle for a macroscopic object

• Hello Geekswipe! I have a problem involving quantum physics, or more specifically, the uncertainty principle:
If ΔpΔx >= h/4pi where h = planck’s constant

Then if Δp = m*v*Δ
and if m = .4kg, v = 40ms, and Δ = 6.625*10^-34

Then Δx = .498M

Now assuming the object is a football, how can you be uncertain where it is to a meter? You can observe it, and to my understanding, prove that it is both there and traveling at roughly that speed, yet when you narrow in on it’s p, the Δx seems to be bigger than possible.

Could you help me understand how an object can be that you can see have that much uncertainty?

Thank you for the help.

Sincerely,

Appreciate your curiosity. There are few things I don’t get in your question.

You have mentioned Δ = 6.625*10-34. I don’t get it. The value is Planck’s constant (h). It should be Δv.

As far as the answer goes, Δx (uncertainty in position) will be very very very very very :) little for macroscopic objects. And you should not use the uncertainty principle to macroscopic objects like football as you would not be able to observe the teeny tiny variation (a little greater than the extremely small 6.626 x 10-34 Js).

Said that… The wave-particle duality cannot be observed (as it is too small) for macroscopic objects. But it does not mean that the uncertainty principle only applies to quantum mechanics.

If we consider the football to behave as a wave, the de Broglie wavelength λ can be found.

λ = h/p (where p is the momentum)

λ = h/p = 4.141 x 10-35 m/s (This is a very small value!)

This is the reason why the football seems deterministic

If you were to apply HUP to the football, then the uncertainty in position and velocity will both be a minuscule non-zero number. I think your confusion arises from assuming the change in velocity too close to Planck’s constant for ease of calculation. It is evident that the Δx value had to go up to satisfy the relation. When you substitute realistic values (which will still be too small), you will get some agreeable results.