Proving K+U is constant with tensor notation.

  • #1
461
16

Main Question or Discussion Point

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:

Answers and Replies

  • #2
461
16
I've noticed ##m_i \ddot{r_i} = -\frac{\partial \Phi}{\partial r_i}## is wrong too. As the LHS is a scalar value.
 

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