# Lagrange's equations to Newton's equations

1. Oct 1, 2005

### Reshma

I want to show the Lagrange's equations reduces to Newton's equation of motion if we take the Cartesian coordinates as the generalised coordinates.

So let T be the K.E. of the system and V be the P.E. of the system. So the Lagrangian is L=T-V.

So $$T = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$$

& $$V = V(x,y,z)$$
Help me proceed with the proof .

Last edited: Oct 1, 2005
2. Oct 1, 2005

### George Jones

Staff Emeritus
Actually,

$$T = \frac{1}{2}m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right).$$

Regards,
George

7. Oct 2, 2005

### samalkhaiat

YOUR LAGRANGE EQUATION IS NOT RIGHT. THERE SOULD BE ZERO ON THE WRITE HAND SIDE NOT Q. AGAIN IF YOU PUT Q=0 IN YOUR SECOND EQUATION YOU GET;
MASS X ACCELERATION = - THE DIREVATIVE OF THE POTENTIAL WITH RESPECT TO x.
= FORCE.
ISN'T THIS THE SECOND LAW OF NEWTON?

8. Oct 2, 2005

### George Jones

Staff Emeritus
The terms involving the potential and the terms involving the Q' s represent different types of forces. This is what I was trying to point the way towards in my last post.

Regards,
George

9. Oct 3, 2005

### Reshma

Thank you very much for all your replies.
There is part of the force derivable from PE and the other part independent of the PE. So one has to be the conservative part and the other the non-conservative part. So Q's represent the non-conservative parts and V(x,y,z) represent the conservative part, right? Can you elaborate more on these forces?

10. Oct 3, 2005

### Reshma

Be easy on the Caps !!

11. Oct 5, 2005

### samalkhaiat

so you wrote down Lagrange equation with a source on the right hand side of it. Then you ask about deriving newton second law from it. But you actually did derive it.
The forces as you said conservative (derivable from potential) and non-conservative (which are not) like friction forces etc.