[Reading Notes] Mechanics by Landau and Lifshitz

The following is an excerpt from [\S 43 The action as a function of coordinate]. I would like to comment on the formal derivation of Hamilton’s equations and the problems of independence of variations.

Aside. Interpretation of the action in classical mechanics

捕获

 

Note that in the derivation above, the variations \delta p and \delta q are regarded as independent.* Actually, \delta q is arbitrary but \delta p is not, even though p, q are both independent variables. Since p in connected with \dot{q} and  \delta p and \delta \dot{q} are not independent.

Notice that before (43.8) is derived, we have applied Ledrendre  Transformation which requires that

\dot{q}=\frac{\partial H}{\partial p}.                                               (1)

So the coefficient of \delta p is  0,  and \delta q is arbitrary, so its coefficient must be 0, Hence we get another Hamilton’s  equation

\dot{p}=-\frac{\partial H}{\partial q}.                                              (2)

Notice that we only derive half of Hamilton’s equations from the procedure above.

Since we can not say that we derive Hamilton’s equations by applying Hamilton’s equations. In order to make this induction above complete, we have to give the proof of another half of Hamilton’s equations, that is (1).

(1) is related to the definition of  p=\frac{\partial L}{\partial \dot{q}}. From the definition of Hamiltonian

H=\Sigma \dot{q}p-L

and \dot{q}=\dot{q}(p,q,t), then

捕获

With the definition of p, p=\frac{\partial L}{\partial \dot{q}}, we have

\dot{q}=\frac{\partial H}{\partial p}.

Hence the half part of Hamilton’sequations is derived.

 

In this way of deriving Hamilton’s equation, strictly, we first derive (1) from the definition of p, and then by applying (1) in \delta S, (2) can be derived.

*We should notice that variations here are simultaneous variations and it’s for a complete system.

 

 

 

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s