# Very small velocity

1. Jan 27, 2010

### MHD93

Hello everybody..

If a particle has a very small velocity.. (about 10 ^ -40 m/s) .. does it really move ?
or my question makes no sense?

thanks

2. Jan 27, 2010

### sylas

When you know the velocity to that accuracy, it might be hard to tell where the particle is, especially for a lightweight particle.

Consider a proton, for example. Mass is 1.67*10-27 kg. The momentum "p" is... tiny.

By Heisenburg's uncertainity principle, the location of the particle is only defined to within
$$\Delta x \geq \hbar/2/\Delta p = 3.15 \times 10^{35} \; m$$​
That size is many times larger than the observable universe.

Or how about me. I'm around 75 kg. If I was at rest to that accuracy, my location could only be defined to within 7km or more. Of course, it makes no sense to have a large composite object like me with motions defined to that accuracy.

In other words particles don't really have a velocity defined sufficiently strongly to make any difference between being at rest and having such a small velocity, by the uncertainty principle.

Cheers -- sylas

Last edited: Jan 27, 2010
3. Jan 27, 2010

### torquil

A localized wave packet can be prepared at x=0 at t=0. If it is later measured to be at x=1m at t=10^40s, then it makes sense to say that its average velocity has been 10^-40m/s during that period. And it has obviously moved. The average velocity of a particle during a finite time period can be found to be any number.

On the other hand, if your particle is in a momentum eigenstate corresponding this velocity around the number you mention (with p=mv), then it is completely delocalized. But that doesn't mean that is doesn't move, In that case the particle is a wave that moves. It just happens to have a non-zero amplitude all over the universe.

Torquil

4. Jan 27, 2010

### torquil

Sorry, I forgot that this is the classical physics section. In that case I could have just said: yes it moves, since v is not equal to zero :-)

Torquil

5. Jan 27, 2010

### DocZaius

How were you able to use Heisenberg's uncertainty principle for this, when the OP gave a velocity without any +/- uncertainty? I was under the (perhaps wrong) impression that that equation can only be used in regards to uncertainties, and when a velocity is declared without any, that equation cannot be used. What am I missing?

6. Jan 27, 2010

### MHD93

Thanks people..

Ok.. when a crate of 10 kg mass freely falls to the earth, suppose that the only force exerted on the earth is the weight of this crate, therefore, the earth's acceleration is
a = F / m
a = 100N / (6 * 10 ^ 24 kg)
a is about 1.66 * 10 ^ -23 m/s^2

now that makes the earth move.. (with ignoring every particle except the earth and the crate)
that's right ?

7. Jan 27, 2010

### sylas

The question is whether there's any difference between such a velocity and being at rest. The difference is so small that it is less than the intrinsic uncertainty in velocity for any reasonable level of certainty on position.

Yes, that's right. How far are you proposing this crate should fall?

Cheers -- sylas

Last edited: Jan 27, 2010