# B Heisenberg applied to accelerated particles

1. Jul 26, 2017

### Chorlton

I may be getting these Thread Levels wrong. I get the impression that they apply to my level of knowledge such that someone else might be able to give an answer that I might be able to understand rather than the level of knowledge required to answer the question.

Rest assured I am quite stupid.

Anyway. Assume I produce a bunch or gas of nominally stationary, at rest, protons but, ignoring the fact that they are going to fly apart, they have random X,Y,Z motions.

Taking the X axis assume there is an uncertainty in their momentum along that axis expressed in electron volts, ueV.. Uncertainty in electron volts.

I think this may be some sort of Maxwell-Boltzmann kind of thing.

Let's say that they are at room temperature so about 25meV.. Milli Electron Volts. Ignore my rounding down error. Take that 25meV figure as being the uncertainty in their energy and, possibly incorrectly, extrapolated to the uncertainty in their momentum along the X axis.

I am waving hands.

What happens if I accelerate them to 100KeV along the X axis using a low, effectively zero noise, voltage source?

Specifically have I reduced the uncertainty in their Energy/Momentum along the X axis by a factor of 100KeV/25meV or 4E6?

Again I have no clue but Frames of Reference might apply.

Besides being completely wrong I might also be a factor of three and a square root out.

Last edited by a moderator: Jul 26, 2017
2. Jul 26, 2017

### Jilang

Consider the momentum first. Any uncertainty in that will remain. The spread will still be be roughly constant.

3. Jul 26, 2017

### Chorlton

This is why I mentioned Frame of Reference. If you sit amongst the accelerated particles then... 25meV

If you try to observe them outside of their accelerated frame then 100KeV/25meV.. You are, about, 4E6 time more certain as to what there dE/dP is and therefore about 4E6 times less certain what their dT/dX is.

4. Jul 26, 2017

### Staff: Mentor

It's a combination of both. Many questions can be answered at different levels of knowledge; the answers just get more complicated and comprehensive as the knowledge level increases. So the thread level gives an indication of the level of knowledge at which you want the answer to be given. But it also requires that the question can be answered at the required level; some questions can't really be answered at all if you don't have enough background knowledge.

The level you gave to the question in this thread doesn't preclude an answer, but there won't be a lot that can be said about it at the "B" level.

5. Jul 27, 2017

### Chorlton

Thanks for the replies and clarification as to 'grade'.

I think I may have resolved my own confusion by referring to the Wikipedia pages on the Uncertainty Principle and Standard Deviation. The page for Uncertainty mentions that the [in]equality is based on the standard deviations of position and momentum while the page for Deviation mentions that it is based on variance about a mean. As such I would conclude that when stationary at room temperature the particles have a mean equivalent to 0eV, 0K, with a deviation that might be approximated as being 25eV. If I now accelerate the particles through 100KeV they now have a mean of 100KeV but the deviation remains the same at 25eV. As such the acceleration has no effect on the original [in]equality. I had hoped that by accelerating the particles along the X axis the higher momentum would result in less knowledge about position, the particles would appear to be stretched along the X axis. Apparently that is not the case. Perhaps someone can confirm my somewhat convoluted reasoning.