Is Uncertainty in Velocity Equal to the Velocity Itself?

  • Context: Graduate 
  • Thread starter Thread starter darkfall
  • Start date Start date
  • Tags Tags
    Velocity
Click For Summary

Discussion Overview

The discussion revolves around the relationship between uncertainty in position and uncertainty in velocity, specifically questioning whether the uncertainty in velocity can be equated to the velocity itself. The scope includes theoretical considerations related to quantum mechanics and the implications of the uncertainty principle.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant proposes that if the uncertainty in position is equal to de Broglie's wavelength, then the uncertainty in velocity might also be equal to the velocity itself, seeking clarification on this idea.
  • Another participant references the uncertainty principle, stating that if the uncertainty in position is set to the wavelength, it leads to a derived expression for the uncertainty in momentum and subsequently in velocity, indicating a lower limit on the uncertainty in velocity.
  • A subsequent post seeks confirmation on whether the derived expression for uncertainty in velocity is indeed v/(4π).
  • A later reply acknowledges the previous contributions without further elaboration.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the uncertainty in velocity is equal to the velocity itself, and the discussion remains unresolved with differing interpretations of the uncertainty principle.

Contextual Notes

The discussion includes assumptions about the non-relativistic motion and the application of the uncertainty principle, which may not be universally applicable without further context.

darkfall
Messages
4
Reaction score
0
If the uncertainity in position is equal to de broglies wavelength ..den is the uncertainity in velocity equal to the velocity itself?
If yes how?
 
Physics news on Phys.org
##\displaystyle \Delta x \Delta p \geq \frac{\hbar}{2}##. Let ##\displaystyle \Delta x = \lambda = \frac{h}{p}##, then ##\displaystyle \Delta p \geq \frac{\hbar}{2 \Delta x} = \frac{\hbar p}{2 h} = \frac{p}{4\pi}##.
If the motion is non-relativistic, p=mv and ##\displaystyle \Delta v \geq \frac{v}{4\pi}##.
Note that this is a lower limit on the uncertainty.
 
So its v/(4*pi) ??
 
All rite...thanks
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 70 ·
3
Replies
70
Views
6K