Hamiltonian Formalism: Equations of Motion for Lagrangians | Explained

[vertices]

I know this is getting really ridiculous but I have yet another question on Lagrangians...

This is our Lagrangian:

$$L=\frac{1}{2}m\dot{\vec{x}}^{2}+e\vec{A}.\dot{\vec {x}}$$

Using the fact that:

$$\vec P= \frac{\partial L}{\partial \dot{\vec{x}}}=m \dot{\vec{x}} + e\vec A$$

and substituting P for $$\dot{\vec{x}}$$ *

We get this Hamiltonian:

$$H=\frac{1}{2m}(\vec P - e\vec A )^2$$

(From * $$\dot{\vec{x}} = \frac{1}{m}(\vec P - e\vec A )$$, so presumably, the above Hamiltonian is equal to $$\frac{m}{2} \dot{\vec{x}}^2$$)

The question is to find the Equations of Motion in the Hamiltonian Formalism, ie. we need to determine $$\dot{\vec{x}}$$ and $$\dot{P}_i$$:


We know $$\dot{\vec{x}} = \frac{1}{m}(\vec P - e\vec A )$$ from above.

Now:

$$\dot{P}_i = \frac{\partial H}{\partial x^i} = \frac{1}{2m}\frac{\partial [(\vec P - e\vec A )^2]}{\partial x^i} = \frac{m}{2}\frac{\partial [(\dot{\vec{x}} ^2]}{\partial x^i}=0$$

HOWEVER the answer seems to be this:

$$\dot{P}_i = \frac{e}{m}(P-eA)_j \epsilon_{jki} B_k$$

Can someone please explain, why P_dot is non zero?

Thanks...

 

[diazona]

The short answer would be that it's because of the vector potential... if this is anything like your previous questions,
$$A_{i}=\epsilon_{ijk}B_{j}x_{k}$$
It's actually $x$ and $p$ that are the independent variables, not $x$ and $\dot{x}$. In the simple case, when there's no vector potential, both statements are equivalent, but that's not the case when the momentum has that spatial dependence in it.

 

[vertices]

The short answer would be that it's because of the vector potential... if this is anything like your previous questions,
$$A_{i}=\epsilon_{ijk}B_{j}x_{k}$$
It's actually $x$ and $p$ that are the independent variables, not $x$ and $\dot{x}$. In the simple case, when there's no vector potential, both statements are equivalent, but that's not the case when the momentum has that spatial dependence in it.

Thanks for the reply diazona. I must admit, I don't know very much about vector potentials - I will try reading up on this. I guess the key thing to take away from your post is that x_dot and x aren't independent in the case of vector potentials (a revelation to me!).

Can I ask whether:

$$\dot{\vec{x}}^{2} = \sum_{i,j}\delta_{ij} \dot{\vec{x}}_i \dot{\vec{x}}_j$$

and whether we can write:

$$(\vec P - e\vec A )^2=(\vec P.\vec P - 2e\vec P.\vec A + e^2 \vec A.\vec A)$$

Where $$\vec P.\vec P$$

and $$\vec P.\vec A$$
and $$\vec A.\vec A$$

are just scalar products?

Because if I assume this, I get that:

$$\frac{\partial H}{\partial x^i}=-2e\epsilon_{kji}B_j(P_k-eA_i)$$

-not the expression I should get.

Thanks.

 

[diazona]

Thanks for the reply diazona. I must admit, I don't know very much about vector potentials - I will try reading up on this. I guess the key thing to take away from your post is that x_dot and x aren't independent in the case of vector potentials (a revelation to me!).

It's nothing specific about vector potentials - the vector potential $\vec{A}$ is just one of many possible things that can be in a Lagrangian or Hamiltonian. It might be better to say that when you switch from the Lagrangian to the Hamiltonian formalism, $x$ and $\dot{x}$ stop being completely independent of each other, and $x$ and $p$ become the independent variables.

If you find this a little confusing, you're not alone... I'll see if I can come up with a better explanation.

Can I ask whether:

$$\dot{\vec{x}}^{2} = \sum_{i,j}\delta_{ij} \dot{\vec{x}}_i \dot{\vec{x}}_j$$

and whether we can write:

$$(\vec P - e\vec A )^2=(\vec P.\vec P - 2e\vec P.\vec A + e^2 \vec A.\vec A)$$

Where $$\vec P.\vec P$$

and $$\vec P.\vec A$$
and $$\vec A.\vec A$$

are just scalar products?

Yep, that all seems reasonable to me...

Because if I assume this, I get that:

$$\frac{\partial H}{\partial x^i}=-2e\epsilon_{kji}B_j(P_k-eA_i)$$

-not the expression I should get.

hmm, how exactly did you get that? There's definitely something wrong if you're getting a sum that involves different unsummed indices in different terms.

 

[vertices]

Thanks for reassuring me of the basics there!

It's nothing specific about vector potentials - the vector potential $\vec{A}$ is just one of many possible things that can be in a Lagrangian or Hamiltonian. It might be better to say that when you switch from the Lagrangian to the Hamiltonian formalism, $x$ and $\dot{x}$ stop being completely independent of each other, and $x$ and $p$ become the independent variables.

I kind of see this. Ofcourse the first thing I was taught in this crash course was that H=H(p,q) and L=L(q,q_dot).

hmm, how exactly did you get that? There's definitely something wrong if you're getting a sum that involves different unsummed indices in different terms.

I just spent ages typing out my solution, but as per usual I made I tiny mistake. And indeed, I got the right answer in the end.

 

[diazona]

Cool, glad you got it.

Here's something you might be interested in: hopefully you know that the state of a physical system is described by a point in phase space. For a 1D system, you have a 2D phase space, so the state of the system is described by two coordinates. But you can choose those coordinates to be either $x$ and $\dot{x}$ (the Lagrangian way) or $x$ and $p$ (the Hamiltonian way). Or you could choose some set of generalized coordinates that are combinations of those.

The point is, basically what you're doing by picking coordinates is choosing a basis for a vector space. A particular state of a system corresponds to a vector in the space, which you can express as either
$$x\hat{x} + \dot{x}\hat{\dot{x}}$$
(Lagrangian) or
$$x\hat{x} + p\hat{p}$$
(Hamiltonian). I'm using letters with hats to denote basis vectors and letters without hats to denote the coordinates. Visually, it looks like this:

(okay, well that's kind of small, but the full-size version is clear)

When you take a partial derivative, like $\frac{\partial}{\partial x}$, physically that corresponds to shifting the state of the system by some infinitesimal amount $\mathrm{d}x$ in the direction of increasing $x$ - that is, the $\hat{x}$ direction, which is to the right in these diagrams. You'll notice that if you take the Lagrangian view (right), $\hat{x}$ and $$\hat{\dot{x}}$$ are orthogonal, which means that when you adjust the state by that infinitesimal amount in the $\hat{x}$ direction, the $\dot{x}$ coordinate doesn't change at all. But the $p$ coordinate does. This is what it means to say that $x$ and $\dot{x}$ (but not $x$ and $p$) are the independent variables in the Lagrangian formulation. Conversely, in the Hamiltonian view (on the left), when you adjust the state by $\mathrm{d}x$, the $p$ coordinate doesn't change, but $\dot{x}$ does. Thus $x$ and $p$ are the independent variables in the Hamiltonian formulation. (There are in fact two different meanings of "independent" being thrown around here, so it can easily get confusing!)