Hamilton's equation and Euler-Lagrange's equation comparison

[JD_PM]

Homework Statement

I will not use summation sign: repeated pair of (upper and lower) indices are summed over: $\sum_{a} A_{a} B^{a} \equiv A_{a}B^{a} = A_{c}B^{c}$ (summed over indices are dummy indices so you can rename them as you like).

Let $V(q) = 0$. Then the Hamiltonian ($H = T + V$) and the Lagrangian ($L = T - V$) are equal:

$$H = L = 1/2 G_{ab} \dot q^a \dot q^b \ \ \ \ (1)$$

Show that you get $(1)$ either using Hamilton's equation $\dot{p}_k=\frac{\partial L}{\partial q^k}$ or Euler-Lagrange equation $\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}$

Relevant Equations

$$H = L = 1/2 G_{ab} \dot q^a \dot q^b$$

$$\dot{p}_k=\frac{\partial L}{\partial q^k}$$

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}$$

We know that the momentum for such a lagrangian is:

$$p_k = \frac{\partial L}{\partial \dot q^k} = 1/2 \Big( G_{ab} \delta_k^a \dot q^b + G_{ab} \delta_k^b \dot q^a\Big) = 1/2 \Big( G_{kb} \dot q^b + G_{ak} \dot q^a\Big) = G_{ak} \dot q^a$$

OK so using Hamilton's equation we get (and recalling that $L = H$ for this particular problem):

$$\dot p_a = -\frac{\partial H}{\partial q^a} = -\partial_a H = -1/2\partial_a G_{cb} \dot q^c \dot q^b \ \ \ \ (2)$$

By using $\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^k }\Big) = \frac{\partial L}{\partial q^k}$ we get:

$$\frac{d}{dt} \Big(\frac{\partial L}{\partial \dot q^a }\Big) = \frac{d}{dt}\Big( G_{ab} \dot q^b\Big) = \frac{\partial L}{\partial q^a} = 1/2\partial_a G_{cb} \dot q^c \dot q^b \ \ \ \ (3)$$

Mmm so there's a sign difference between $(2)$ and $(3)$. They should be equal. Why am I wrong?

 

[PeroK]

I can't figure out what you are trying to do.

All I could think of is

a) Take $L = \frac 1 2 G_{ab}\dot q^a \dot q^b$

b) Define $H = p_k\dot{q}^k - L$, where $p_k = \frac{\partial L}{\partial \dot q^k}$.

c) Show that $H = L$.

Is that the exercise?

 

[Orodruin]

You need to be very careful here. Even if you have no potential energy, the Hamiltonian and Lagrangian cannot be the same because the Hamiltonian is a function of $q$ and $p$ while the Lagrangian is a function of $q$ and $\dot q$. Since the relation between $p$ and $\dot q$ generally depends on $q$, the partial derivative of the Hamiltonian wrt q is not equal to the partial derivative of the Lagrangian wrt q. That is, generally
$$
\frac{\partial L}{\partial q} \neq \frac{\partial H}{\partial q}
$$
even if $H = L$ under the identification $p = \partial L/\partial \dot q$.

 

[JD_PM]

I can't figure out what you are trying to do.

All I could think of is

a) Take $L = \frac 1 2 G_{ab}\dot q^a \dot q^b$

b) Define $H = p_k\dot{q}^k - L$, where $p_k = \frac{\partial L}{\partial \dot q^k}$.

c) Show that $H = L$.

Is that the exercise?

What you suggest is to do the following exercise:

Show that $H = L$ given $L = \frac 1 2 G_{ab}\dot q^a \dot q^b$, $H = p_k\dot{q}^k - L$ and $p_k = \frac{\partial L}{\partial \dot q^k}$

Solution:

$$H = p_a\dot{q}^a - L = G_{ba} \dot q^b \dot q^a - \frac 1 2 G_{ab}\dot q^a \dot q^b = \frac 1 2 G_{ba} \dot q^b \dot q^a$$

So indeed

$$H = L$$

But based on what Orodruin said, this cannot be correct because 'the Hamiltonian and Lagrangian cannot be the same'...

 

[JD_PM]

Since the relation between $p$ and $\dot q$ generally depends on $q$, the partial derivative of the Hamiltonian wrt q is not equal to the partial derivative of the Lagrangian wrt q. That is, generally
$$
\frac{\partial L}{\partial q} \neq \frac{\partial H}{\partial q}
$$
even if $H = L$ under the identification $p = \partial L/\partial \dot q$.

Mmm I do not get your reasoning, could you please explain your argument based on an example?

Besides, why is the above exercise proposed by PeroK wrongly solved?

 

[PeroK]

What you suggest is to do the following exercise:

Show that $H = L$ given $L = \frac 1 2 G_{ab}\dot q^a \dot q^b$, $H = p_k\dot{q}^k - L$ and $p_k = \frac{\partial L}{\partial \dot q^k}$

Solution:

$$H = p_a\dot{q}^a - L = G_{ba} \dot q^b \dot q^a - \frac 1 2 G_{ab}\dot q^a \dot q^b = \frac 1 2 G_{ba} \dot q^b \dot q^a$$

So indeed

$$H = L$$

But based on what Orodruin said, this cannot be correct because 'the Hamiltonian and Lagrangian cannot be the same'...

Technically we have a third function, which I'll call $H'$:

$H'(q, \dot q) = H(p(q, \dot q), \dot q)$

In this case $H' = L$.

 

[PeroK]

PS The following appears a lot in physics texts:

You have some function $\psi(x)$, say, and a change of variable $x = f(y)$. Which leads to $\psi(y)$. But, in fact, what we really have is a new function:

$\psi'(y) = \psi(f(y))$

To take an example. If $\psi(x) = \sin(x)$ and $x = 2y$, then:

$\psi(y) = \sin(y)$ (technically, as $\psi$ is simply the sine function)

But:

$\psi'(y) = \sin(2y)$

Is what's really meant. If we are taking about a real wave amplitude, say, then $y = \pi/4$ means $x = \pi/2$ and the amplitude should be $1$ when $y = \pi/4$ . Which it is for $\psi'$ but not for $\psi$.

 

[Orodruin]

Solution:

H=pa˙qa−L=Gba˙qb˙qa−12Gab˙qa˙qb=12Gba˙qb˙qa​

This is incorrect because, as already mentioned, the Hamiltonian is a function of q and p, not q and $\dot q$. You obtain the Hamiltonian from the Lagrangian by insering $\dot q$ as a function of q and p where $p = \partial L/\partial \dot q$.

Take the simplest example $L = m \dot q^2/2$. You would find that $p = m\dot q$ and therefore $\dot q = p/m$ leading to $H = p^2/(2m)$. Here, neither L or H depend on q so the partial derivatives both vanish. However, this would not be the case if you let m depend on q.

 

[JD_PM]

Technically we have a third function, which I'll call $H'$:

$H'(q, \dot q) = H(p(q, \dot q), \dot q)$

In this case $H' = L$.

Oh I think I see it now. The issue is that we get a function (which you labeled $H'$) which is a function of $q$ and $\dot q$. By definition this cannot be the Hamiltonian, as $H(q, p)$ is certainly not $H'(q, \dot q)$. Thus we need to apply a change of variables to get the Hamiltonian out of $H'(q, \dot q)$; in this case we need $\dot q = p(q, \dot q)$.

By the way, I think there is a typo in #6;

I get:

$$H'(q, \dot q) = H(q, p(q, \dot q))$$

Where I've used the change of variables $\dot q = p(q, \dot q)$

Instead of $H'(q, \dot q) = H(p(q, \dot q), \dot q)$

Do you agree?

 

[PeroK]

Oh I think I see it now. The issue is that we get a function (which you labeled $H'$) which is a function of $q$ and $\dot q$. By definition this cannot be the Hamiltonian, as $H(q, p)$ is certainly not $H'(q, \dot q)$. Thus we need to apply a change of variables to get the Hamiltonian out of $H'(q, \dot q)$; in this case we need $\dot q = p(q, \dot q)$.

By the way, I think there is a typo in #6;

I get:

$$H'(q, \dot q) = H(q, p(q, \dot q))$$

Where I've used the change of variables $\dot q = p(q, \dot q)$

Instead of $H'(q, \dot q) = H(p(q, \dot q), \dot q)$

Do you agree?

Hmm ... I guess it depends what we mean by the "Hamiltonian". Like the Lagrangian, it takes a different form in different coordinates.

If we take $L, H, H'$ to be specific functions (of specific variables), then we need to stay consistent in what these mean.

The way I defined $H$ in post #3 I think is consistent with post #6.

 

[JD_PM]

Hmm ... I guess it depends what we mean by the "Hamiltonian". Like the Lagrangian, it takes a different form in different coordinates.

Your Hamiltonian is a function of $p$ and $\dot q$. I think that the Hamiltonian has to be $H(q, p)$ (by definition) instead.

 

[PeroK]

Your Hamiltonian is a function of $p$ and $\dot q$. I think that the Hamiltonian has to be $H(q, p)$ (by definition) instead.

How do you define it?

 

[Orodruin]

Your Hamiltonian is a function of $p$ and $\dot q$. I think that the Hamiltonian has to be $H(q, p)$ (by definition) instead.

This is correct.

How do you define it?

What he is saying is that you wrote that the Hamiltonian is a function of $p$ and $\dot q$. It is a function of $p$ and $q$.

 

[JD_PM]

...he...

He/she/bot ...

 

[JD_PM]

OK So my answer to #1 would be:

This exercise is faulty; let's argue why

It asks to show that

$$H = L = 1/2 G_{ab} \dot q^a \dot q^b \ \ \ \ (1) $$

We know that $L(q, \dot q) = 1/2 G_{ab} \dot q^a \dot q^b$

But

$$H'(q, \dot q) = p_a\dot{q}^a - L = G_{ba} \dot q^b \dot q^a - \frac 1 2 G_{ab}\dot q^a \dot q^b = \frac 1 2 G_{ba} \dot q^b \dot q^a$$

And of course

$$H'(q, \dot q) \neq H(q, p)$$

Thus $(1)$ should be:

$$H(q, p) \neq H'(q, \dot q) = L(q, \dot q) = 1/2 G_{ab} \dot q^a \dot q^b \ \ \ \ (1) $$

I think I got it! Thank you both :)

WoW I am learning a lot these days...

 

[PeroK]

Oh I think I see it now. The issue is that we get a function (which you labeled $H'$) which is a function of $q$ and $\dot q$. By definition this cannot be the Hamiltonian, as $H(q, p)$ is certainly not $H'(q, \dot q)$. Thus we need to apply a change of variables to get the Hamiltonian out of $H'(q, \dot q)$; in this case we need $\dot q = p(q, \dot q)$.

By the way, I think there is a typo in #6;

I get:

$$H'(q, \dot q) = H(q, p(q, \dot q))$$

Where I've used the change of variables $\dot q = p(q, \dot q)$

Instead of $H'(q, \dot q) = H(p(q, \dot q), \dot q)$

Do you agree?

Yes, you're right. The correct functional form of $H$ is as a function of $q, p$.