Legendre Transformation of the Hamiltonian

[Simfish]

It's given as this

[tex]H\left(q_i,p_j,t\right) = \sum_m \dot{q}_m p_m - L(q_i,\dot q_j(q_h, p_k),t) \,.[/tex]

But if it's a Legendre transformation, then couldn't you also do this?

[tex]H\left(q_i,p_j,t\right) = \sum_m \dot{p}_m q_m - L(p_i,\dot p_j(p_h, q_k),t) \,.[/tex]

 

[Dickfore]

No, you can not.

 

[Simfish]

Why not? Is it something we physically cannot do (perhaps due to our assumptions), or mathematically cannot do? The Wikipedia article for Legendre transformations only showed these transformations for functions with single-variable arguments, not multivariate arguments

 

[Dickfore]

If you have a function:

[tex]
z = f(x, y)
[/tex]

and the derivative of [itex]z[/itex] w.r.t. [itex]x[/itex] is denoted by:

[tex]
X(x, y) \equiv \frac{\partial z}{\partial x}
[/tex]

then, the function:

[tex]
w \equiv z - X \, x
[/tex]

is a Legendre transform of [itex]z[/itex], w.r.t. [itex]x[/itex] only! Its total differential is:

[tex]
dw = dz - X \, dx - x \, dX = X \, dx + \frac{\partial z}{\partial y} \, dy - X \, dx - x \, dX
[/tex]

[tex]
dw =\frac{\partial z}{\partial y} \, dy - x \, dX
[/tex]

i.e. w is to be treated as a function of X and y:

[tex]
\frac{\partial w}{\partial X} = -x, \ \frac{\partial w}{\partial y} = \frac{\partial z}{\partial y}
[/tex]

Of course, you need to eliminate [itex]x[/itex] from the equation:

[tex]
X = X(x, y) \Rightarrow x = f(X, y)
[/tex]

Notice that the variables w.r.t. which we have not performed a Legendre transform still remain arguments of the transforme dfunction.

Similarly, [itex]-H[/itex] is a Leg. trans. of [itex]L[/itex] w.r.t. [itex]\dot{q}[/itex] and the respective partial derivative:

[tex]
p \equiv \frac{\partial L(t, q, \dot{q})}{\partial \dot{q}}
[/tex]

is the generalized momentum. Therefore, [itex]H[/itex] is to be treated as a function of [itex]q[/itex] (a variable over which we had not performed a Legendre transform). [itex]p[/itex] (the derivative w.r.t. the variable that we had transformed, namely the generalized velocity) and, possibly time [itex]t[/itex] for open systems.

 

[Simfish]

Oh okay I see. Thanks!

 

[Dickfore]

[tex]
H\left(q_i,p_j,t\right) = \sum_m \dot{p}_m q_m - \mathbf{L(p_i,\dot p_j(p_h, q_k),t)} \,.
[/tex]

The bolded part does not make any sense at all because the Lagrangian is defined as a function of generalized coordinates, velocities and, possibly, time.

 

[vanhees71]

Of course the equation should read

[tex]H(q,p,t)=\sum_{m} p_m \dot{q}_m(q,p,t)-L[q,\dot{q}(q,p,t)].[/tex]

It is very important to remember that the Hamiltonian depends on position variables, canonical momenta and sometimes explicitly on time, while the Lagrangian depends on position variables, their time derivatives (generalized velocities) and (sometimes) explicitly on time.