# Yet another question on Lagrangians

1. Jan 9, 2010

### 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...

2. Jan 9, 2010

### 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.

3. Jan 9, 2010

### vertices

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!).

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

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.

4. Jan 9, 2010

### diazona

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.
Yep, that all seems reasonable to me...

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.

5. Jan 9, 2010

### vertices

Thanks for reassuring me of the basics there!

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).

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.

6. Jan 12, 2010

### diazona

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!)