# Derivative of Lagrangian with respect to momentum

1. Jun 12, 2013

### geoduck

I'm trying to figure out when is $$\frac{\delta L}{\delta p}=\dot{q}$$.

From $$L=p\frac{\delta H}{\delta p}-H$$

I get that $$\frac{\delta L}{\delta p}=p\frac{\delta^2 H}{\delta p^2}=p \frac{d}{dp}\dot{q}$$.

For this to equal just $$\dot{q}$$, it must obey the dfe $$\frac{d}{d ln p}\dot{q}=\dot{q}$$

which implies that $$\dot{q}=cp$$ for some constant c.

However, I have verified that for the relativistic Lagrangian $$L=\sqrt{p^2+m^2}$$ it is still true that $$\frac{\delta L}{\delta p}=\dot{q}$$. However, the relation between p and $$\dot{q}$$ is not linear:

$$p=\frac{m \dot{q}}{1-\dot{q}^2}$$

How can this be? What did I do wrong?

2. Jun 12, 2013

### Avodyne

The lagrangian depends on $q$ and $\dot q$, and not on $p$. So $dL/dp=0$. Also, when taking derivatives, $\dot q$ should be treated as independent of $p$, so $d\dot q/dp=0$.

3. Jun 12, 2013

### geoduck

Well, the reason I'm considering the Lagrangian as a function of momentum is because in the Dirac Lagrangian, psi^dagger is the conjugate momentum to psi, so I want to treat the psi^dagger not as a field but a conjugate momenta. So say in a path integral formulation you have [dx][dp] as your measure, and usually you can integrate out [dp] to get a Lagrangian formulation. But I want to say that [dp]=[d psi^dagger], and the Dirac Lagrangian looks like:

psi^dagger psi^dot - Hamiltonian.

Is this right?

4. Jun 13, 2013

### bhobba

Not so sure about that. By definition H = pV - L so L = pv - H which, since v can be a function of p, L can be a function of x and p. Indeed since v can be a function of p then L can depend on p and x directly. That's what you are ignoring - v can be an implicit function of p.

I will have a bit of think about the original question and see what I can come up with - but the derivative of L wrt p is not in general zero.

OK - had a bit of a muck around with the equations.

δL/δp = δL/δv δv/δp = p δv/δp

Now classically δv/δp = 1/m so p δv/δp = v as you suppose.

Relativistically I worked through your math and cant see the error either - maybe someone else can spot it.

Thanks
Bill

Last edited: Jun 13, 2013
5. Jun 13, 2013

### stevendaryl

Staff Emeritus
That isn't the actual relativistic Lagrangian. The actual relativistic Lagrangian is:

$L = -mc^2 \sqrt{1-\frac{v^2}{c^2}}$

or in terms of momenta

$L = -\dfrac{m^2 c^4}{\sqrt{p^2 c^2 + m^2 c^4}}$

Last edited: Jun 13, 2013
6. Jun 13, 2013

### dextercioby

The Dirac field is first oder in derivatives, hence automatically a constrained system. The Hamiltonian (density) as exception can be written in terms of fields only, but generaly it has the generalized momenta and coordinates as functional vartiables (as is should, based on the geometrical formulation in terms of manifolds).

7. Jun 13, 2013

### geoduck

ah, that's the stupid thing I did wrong! Of course the derivative of the expression I gave is equal to the velocity, because $$\sqrt{p^2+m^2}$$ is the Hamiltonian, and derivative of Hamiltonian like that is the velocity by definition pretty much. I think I took note of the fact that in non-relativistic mechanics, the Lagrangian and Hamiltonian are the same, and just assumed it to be true for relativistic case.

it's interesting how in the non-relativistic case the Lagrangian and Hamiltonian are the same in the sense of scalars: L(v[p])=H(p), which is not true for the relativistic case.