Recent content by ebrattr

Yes, I mean \dfrac{\partial q}{\partial \mathbf{r}} As you know if you do \dfrac{\partial}{\partial \mathbf{r}} both sides you get \dfrac{\partial q}{\partial \mathbf{r}} -\sin q = \dfrac{\hat{\mathbf{k}}}{L} . Solving for \dfrac{\partial q}{\partial \mathbf{r}} you get \dfrac{\partial...

Hi ! I'm trying to inverse a mass matrix so I need to do something like this \dfrac{q}{\partial \mathbf{r}} where \cos q = \dfrac{\mathbf{r}\cdot \hat{\mathbf{k}}}{r} However, when \mathbf{r} = \hat{\mathbf{k}} \text{ or } -\hat{\mathbf{k}} I have problems. ¿What can I do...

Yeahhh... its cero. Thanks.

Hi everyone, I would like to know if this stament is true or not. I have two variables u,v both of them distributed as normal distribution with mean 0 and variance a^2. Is it true that the expected value of uv is a^2 ? Thanks

Thanks ! I have found the problem !, when I compute the Total energy, in the internal cycle, it is not k \neq j , it is instead k > j .

Suppose I have mass of 1 kg. This not real important.

Mass. Magnitude of the force due to Lennard Jones potential.

F_i^j is the resultant force over the particle i in the j-chain. f_{LJ} is the corresponding magnitude.

Yeah I am tired changin the time step. And I'm not using a cutoff for the LJ potential.

Yes. I do not have an angular potential. And I calculate Lennard Jones potential over all particle pairs, except those that have an angle or a bond. In fact, in the most simple case, with M chains and 2 particles for every chain, I get the same thing. In fact, can we only discuss this...

By setting every positiion in terms of those coordinates I mentioned before? I don't understand the question.

Yeah, off corse. But what part ? There many lines of code.

No. It is not rotating.

Yeah, I compute all the velocities to one frame, my global frame.

No, I don't have velocity dependent forces, because my only forces are due to a Lennard-Jones Potential. I was a little confused. What I do really have is a rotating frame. Got it. I was not including the potential energy due to inertial centrifugal force. I was just including, kinetick energies...