Mass falling to ground (uncertainty principle)

In summary: You should get a value of \Delta x that minimizes the uncertainty.In summary, the problem is to determine the total error in the final location of a ball dropped from a height L over ground with mass m = 1 g, taking into account the uncertainty in initial position (\Delta x) and initial momentum (\Delta p). This is done by minimizing the function \Delta x + (\Delta p / m) \sqrt{2L/g}, under the constraint that \Delta x \Delta p = \frac{\hbar}{2} and using the uncertainty principle to eliminate \Delta p in favor of \Delta x. The resulting value of \Delta x will minimize the uncertainty in the final position.
  • #1
naggy
60
0
This is a 2d problem.

If I drop a ball with mass m from a height L over ground. How large is the interval that the ball can possibly be on if m = 1 g and L = 2m. Use the uncertainty principle [tex]\Delta[/tex]x[tex]\Delta[/tex]p ~h to determine the interval.

Am I supposed to think of this as a classical mechanics problem and just use the uncertainty principle? I'm not sure what [tex]\Delta[/tex]x or [tex]\Delta[/tex]p would be? Or am I supposed to construct a solution to the Schrodinger equation and work with that?
 
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  • #2
I guess the idea is this: you're trying to drop the ball exactly on a given point. So you try to precisely position the ball right over the point, and let it go with an initial momentum of 0. However there will be some error [tex]\Delta x[/tex] in your initial position and some error [tex]\Delta p[/tex] in your initial horizontal momentum, both of which will result in the ball landing off target. You could make [tex]\Delta x[/tex] very small but then [tex]\Delta p[/tex] grows large and the ball flies off very far from the target. You could make [tex]\Delta p[/tex] small but then [tex]\Delta x[/tex] grows large and you end up dropping the ball from an initial position far from the target. So you're supposed to calculate (classically) the total error in the final location of the ball resulting from an initial position error of [tex]\Delta x[/tex] and an initial momentum error of [tex]\Delta p[/tex] and then minimize the error in the final position, under the constraint that [tex]\Delta x \Delta p = \frac{\hbar}{2}[/tex].
 
  • #3
The_Duck said:
I guess the idea is this: you're trying to drop the ball exactly on a given point. So you try to precisely position the ball right over the point, and let it go with an initial momentum of 0. However there will be some error [tex]\Delta x[/tex] in your initial position and some error [tex]\Delta p[/tex] in your initial horizontal momentum, both of which will result in the ball landing off target. You could make [tex]\Delta x[/tex] very small but then [tex]\Delta p[/tex] grows large and the ball flies off very far from the target. You could make [tex]\Delta p[/tex] small but then [tex]\Delta x[/tex] grows large and you end up dropping the ball from an initial position far from the target. So you're supposed to calculate (classically) the total error in the final location of the ball resulting from an initial position error of [tex]\Delta x[/tex] and an initial momentum error of [tex]\Delta p[/tex] and then minimize the error in the final position, under the constraint that [tex]\Delta x \Delta p = \frac{\hbar}{2}[/tex].

I have that the velocity is [tex]\sqrt{2gL}[/tex] at landing and momentum at landing is [tex]m\sqrt{2gL}[/tex]. How do I connect this to my original [tex]\Delta p[/tex] and [tex]\Delta x[/tex]
 
  • #4
The vertical velocity is unimportant, I think. However if you have an uncertainty in your initial horizontal momentum of [tex]\Delta p[/tex], and a fall time of [tex]\sqrt{2L/g}[/tex] then this translates into an uncertainty in the landing position of [tex](\Delta p / m) \sqrt{2L/g}[/tex], which adds to the uncertainty in initial horizontal position, [tex]\Delta x[/tex]. So you need to minimize [tex]\Delta x + (\Delta p / m) \sqrt{2L/g}[/tex]
 
  • #5
The_Duck said:
The vertical velocity is unimportant, I think. However if you have an uncertainty in your initial horizontal momentum of [tex]\Delta p[/tex], and a fall time of [tex]\sqrt{2L/g}[/tex] then this translates into an uncertainty in the landing position of [tex](\Delta p / m) \sqrt{2L/g}[/tex], which adds to the uncertainty in initial horizontal position, [tex]\Delta x[/tex]. So you need to minimize [tex]\Delta x + (\Delta p / m) \sqrt{2L/g}[/tex]

How do you get that uncertainty from the fall time??

Also, how am I to minimize that function. Is it a derivative that is set to zero? It can´t be that, if it would be then derivative with respect to what?
 
  • #6
Suppose you have an object with horizontal velocity [tex]v[/tex], which falls for time [tex]t[/tex]. It will cover a horizontal distance [tex]vt[/tex] during that time. Or equivalently, if you have an object with uncertainty in velocity [tex]\Delta v[/tex], after time [tex]t[/tex] this translates to an uncertainly in position of [tex](\Delta v) t[/tex]. Velocity is related to momentum by [tex]v = p/m[/tex], so [tex]\Delta v = (\Delta p) / m[/tex]. So the final uncertainty in position arising from the initial uncertainty in velocity is [tex](\Delta v) t = t (\Delta p) / m = (\Delta p / m) \sqrt{2L/g}[/tex], which you can just add on to the initial uncertainty in position [tex]\Delta x[/tex].

So you get uncertainty as a function of two variables, [tex]\Delta x[/tex] and [tex]\Delta p[/tex]. But you can eliminate [tex]\Delta p[/tex] in favor of [tex]\Delta x[/tex], using the uncertainty principle, so you get uncertainty as a function of one variable, [tex]\Delta x[/tex]. Then, as you say, you can find the minimum as you do any function, by taking the derivative with respect to [tex]\Delta x[/tex] and setting it equal to zero.
 

1. What is the uncertainty principle?

The uncertainty principle, also known as Heisenberg's uncertainty principle, is a fundamental principle in quantum mechanics that states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa.

2. How does the uncertainty principle relate to the mass falling to ground?

The uncertainty principle applies to all particles, including those that fall to the ground. This means that when we measure the position of a falling mass, there will always be some uncertainty in its momentum, and therefore its speed and trajectory.

3. Can the uncertainty principle be observed in everyday life?

Yes, the uncertainty principle is a fundamental aspect of the behavior of all particles, including those that make up everyday objects. However, its effects are usually only noticeable at the microscopic level.

4. How is the uncertainty principle related to the concept of probability in quantum mechanics?

The uncertainty principle is closely linked to the probabilistic nature of quantum mechanics. It states that we can only know the probability of a particle's position and momentum, rather than their exact values.

5. Is there a way to overcome the uncertainty principle?

No, the uncertainty principle is a fundamental aspect of quantum mechanics and cannot be overcome. However, scientists can use mathematical techniques to minimize the uncertainty and make more precise measurements of particles.

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