# uncertainty principle

• May 4th 2006, 03:16 PM
hongster5
uncertainty principle
sry guys i know this isnt a physics forum but this forum has helped so much inteh past i bet someone will know this:
12 g bullet leaves rifle at speed of 180 m/s, if the accuracy of hte bullet was determined by only the uncertainty principle, by how much might the bullet miss a pinpoint target 200m away? thx alot
• May 4th 2006, 04:43 PM
topsquark
Quote:

Originally Posted by hongster5
sry guys i know this isnt a physics forum but this forum has helped so much inteh past i bet someone will know this:
12 g bullet leaves rifle at speed of 180 m/s, if the accuracy of hte bullet was determined by only the uncertainty principle, by how much might the bullet miss a pinpoint target 200m away? thx alot

Interesting question. Note that we are considering a Classical trajectory in this problem, which means that we know exactly what the different coordinate positions are. Specifically, let the positive z direction be the direction from the gun to the pinpoint target. Using a typical (that is to say left-handed) coordinate system we may define the positive y direction to be up and then the x direction sticks out horizontally from the yz plane. We know without error that the x component of momentum is zero (barring wind, etc.) This requires that the x component of displacement is completely unpredictable. Thus we have absolutely no ability to predict if the bullet hits the target.

Ignoring this point we can focus on the y components of momentum and displacement. I am assuming the gun and target are at the same height for the sake of simplicity, so we can use Classical Physics to predict the angle the gun has to be at to hit the target. What will happen is that the magnitude of the y component of the final momentum (when it hits the target) will be the same as the magnitude of the y component of the initial momentum. The problem is, of course, that we know the magnitude of the momentum, not the root-mean-square value of the error (more or less simply called the "error") in the momentum.

There are various and sundry ways to estimate an error in the momentum. One argument (which I like because it tremendously simplifies things) is to assume a momentum measurement will take not a Gaussian distribution, but a Poisson distribution, in which case the error in the measurement will simply be $\Delta p \approx \sqrt p$. (This is justified by considering the total momentum as the sum of a large quanta of momentum.) I honestly don't know the range of validity of this assumption, but we have to have some way to calculate an error for this problem. You may wish to check with your professor about how to get an error for the momentum.

Assuming the error estimate above is valid, we can now calculate an error for the displacement in the y direction.
$\Delta y \, \Delta p_y \approx \hbar/2$
$\Delta y \approx \frac{\hbar}{2 \sqrt p_y}$

This number should be fairly small, probably undetectable, as makes sense for the result of the uncertainty in a Classical measurement, so the bullet will be fairly close to the target as far as the y direction is concerned.

We can do a similar analysis using the x component of the final momentum.

(Of course, these results are completely overshadowed by the result in the x direction!)

-Dan
• May 4th 2006, 05:01 PM
hongster5
hmm thanks that makes sense, cept btw its not h bar/2 its just delta x delta p=h bar... so you're saying that the 200m thing doesnt matter at all?
• May 5th 2006, 04:41 AM
topsquark
Quote:

Originally Posted by hongster5
hmm thanks that makes sense, cept btw its not h bar/2 its just delta x delta p=h bar... so you're saying that the 200m thing doesnt matter at all?

The constant in the Uncertainty Principle equation has many values depending on which book you read. In reality it should go something like:
$\Delta x \, \Delta p \ge \hbar$
and most people use some sort of approximation with their favorite constant.

Since you are concerned with hitting the target, you are interested in the error in the displacement (ie. position) when it reaches the target. The error in the displacement has nothing to do with the actual position of the projectile so yes, I would say the 200 m doesn't mean anything in regard to the problem.

-Dan
• May 5th 2006, 05:16 AM
CaptainBlack
Quote:

Originally Posted by hongster5
sry guys i know this isnt a physics forum but this forum has helped so much inteh past i bet someone will know this:
12 g bullet leaves rifle at speed of 180 m/s, if the accuracy of hte bullet was determined by only the uncertainty principle, by how much might the bullet miss a pinpoint target 200m away? thx alot

But the uncertainty principle isn't physics its mathematics, it spills over
into physics and engineering because of the form QM and EM take.

It is in fact a property of the Fourier transform, that when these things are
appropriately defined and scaled that:

$
\Delta t \Delta \omega \ge 1
$
.

This is most easily seen with the Gaussian density, which with appropriate
scaling of the two domains and juggling with the constants involved is
its own FT, then $\Delta t$ can be taken as the "standard deviation" in the $t$-domain,
and $\Delta \omega$ the same in the $\omega$ domain.

RonL
• May 5th 2006, 10:22 AM
topsquark
Quote:

Originally Posted by CaptainBlack
But the uncertainty principle isn't physics its mathematics, it spills over
into physics and engineering because of the form QM and EM take.

It is in fact a property of the Fourier transform, that when these things are
appropriately defined and scaled that:

$
\Delta t \Delta \omega \ge 1
$
.

This is most easily seen with the Gaussian density, which with appropriate
scaling of the two domains and juggling with the constants involved is
its own FT, then $\Delta t$ can be taken as the "standard deviation" in the $t$-domain,
and $\Delta \omega$ the same in the $\omega$ domain.

RonL

That's pretty much exactly where it comes from. The Uncertainty Principle is the only example I can think of for a principle that comes from wave mechanics that isn't also derivable from matrix mechanics. In matrix mechanics we pretty much have to assume the commutator form of the Uncertainty Principle, though technically we can make a logical inference that leads to it, whereas in wave mechanics it pretty much falls out on its own.

-Dan
• May 5th 2006, 11:01 AM
hongster5
thx for all the help guys, appreciate it