Does Particle Motion at Very Small Velocities Make Sense?

[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

 

[sylas]

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?

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

 

[torquil]

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

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

 

[torquil]

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

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

 

[DocZaius]

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

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?

 

[MHD93]

Thanks people..

yes it moves

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 ?

 

[sylas]

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?

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.

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 ?

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

Cheers -- sylas