Lagrange's equations to Newton's equations

[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 .

 

[George Jones]

Actually,

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

What do you get when you substitute $T$ and [itex]V[/tex] into Lagrange's equations?<br /> <br /> Regards,<br /> George[/itex]

 

[Reshma]

Sorry for the typo and thank you replying. I made the substitution in Lagrange's equation.
$$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q_i}}\right) - \frac{\partial L}{\partial \dot{q_i}}= Q_i'$$

Setting q₁=x,q₂=y,q₃=z

I get the Lagrange's equation for variable x as:
$$m\ddot{x} + \frac{\partial V}{\partial x} = Q_1'$$

How does this equation show Newton's second law of motion?

 

[George Jones]

What is

$$-\frac{\partial V}{\partial x}?$$

Regards,
George

 

[Reshma]

Umm...is it part of the net force?

 

[George Jones]

Umm...is it part of the net force?

Yes. What is the force that results from a potential $V=V\left(x,y,z\right)$, and what type of force is it?

Regards,
George

 

[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?

 

[George Jones]

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?

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

 

[Reshma]

Thank you very much for all your replies.

Yes. What is the force that results from a potential $V=V\left(x,y,z\right)$, and what type of force is it?

Regards,
George

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?

 

[Reshma]

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?

Be easy on the Caps !

 

[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.