1. Feb 23, 2015

### Nikitin

Hi!

Two exerts from my lecture notes:

"Assume we have a system of point masses in position $\vec{r_i}$ which are influenced by forces $\vec{F_i}$."

"Let's say you have a system where $\vec{F_i} = - \nabla_i V$"

In the second line, what does the notation $\nabla_i$ mean? Why is that sub $i$ there?

They use similar notation in here "$\nabla_{\vec{r_k}}$" http://en.wikipedia.org/wiki/Virial_theorem#Connection_with_the_potential_energy_between_particles

What does the r_k mean?

2. Feb 23, 2015

### Orodruin

Staff Emeritus
The force on particle $i$ is equal to the gradient of the potential with respect to the coordinates of particle $i$.

3. Feb 24, 2015

### Nikitin

Shouldn't it instead be $\vec{F_i} = \nabla V(\vec{r_i})$ ?

4. Feb 24, 2015

### Orodruin

Staff Emeritus
No. The minus sign is conventional so that you decrease the potential when the force does work. Compare with gravitational potential in a homogeneous field. It increases with height and so has positive z derivative, yet the force points down.

5. Feb 24, 2015

### Nikitin

Oh I forgot to add the minus sign. never mind that. What I was asking about were the indexes.

6. Feb 24, 2015

### Orodruin

Staff Emeritus
No, the potential is a function that involve all positions, which is why you need to specify which position you take the gradient with respect to. For example, if you have two charged particles, the potential will be a function of the distance between them, which is a function of both positions.

I think you are imagining the case when all of the particles move in an external potential with no inter-particle interactions. Then it will be possible to divide the potential into several contributions V(r1,r2,...) = U(r1) + U(r2) + ... Note that I here introduced the external potential U, which is a priori a different function than V - the total potential. Inserting this into the expression you were asking about will give you something like what you quoted, but as I said, this is not the general situation with interactions within the system.

7. Feb 25, 2015

### Nikitin

OK, I see. Thank you!