# Proving K+U is constant with tensor notation.

1. Sep 5, 2015

### davidbenari

Suppose we have a system of particles that interact via conservative forces. I wish to prove that $K+U$ is a constant of the system with tensor analysis. Here is my procedure:

The Lagrangian is $L=\frac{1}{2}m_i\dot{ r_i}^2-\Phi$

Lagrange's equations $\frac{d}{dt}(\frac{\partial L}{\partial \dot{r_i}}) = \frac{\partial L}{\partial r_i}$ yield

$m_i\ddot{r_i}=-\frac{\partial \Phi}{\partial r_i}$
If the RHS is multiplied by $\dot{r_i}$ we get the chain rule for partial derivatives and get $\frac{d}{dt}\Phi$ which would be a crucial step in my derivation. However the LHS ends up with three indices which is not allowed. And that is the crux of my problem. How do I avoid this three indices to prove this property of classical mechanics?
Namely
$m_i\dot{r_i}\ddot{r_i}=-\dot{r_i}\frac{\partial \Phi}{\partial r_i}$

is not allowed.

Thanks.

I'm not terribly familiar with Lagrangian mechanics so I could've screwed up in this respect somewhere.

In case its not obvious what I'm trying to prove, it is that $\frac{d}{dt}(K+U)=0$ for n-particles interacting via conservative forces. I'm specifically looking for a derivation with tensor analysis. I know of other derivations.

Last edited: Sep 5, 2015
2. Sep 5, 2015

### davidbenari

I've noticed $m_i \ddot{r_i} = -\frac{\partial \Phi}{\partial r_i}$ is wrong too. As the LHS is a scalar value.