Another question on Lagrangians (sorry)

[vertices]

Sorry about the endless stream of questions about Lagrangians. I am actually beginning to detest them a bit;p

Anyway, if we have a Lagrangian in three dimensional space:

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

where $$A_{i}=\epsilon_{ijk}B_{j}x_{k}$$ and B is just a constant (magnetic field).

The question is to find the equations of motion. To do this we can just use the Euler-Lagrange equations thus:

$$\frac{{\partial}L}{{\partial}q^{i}}-\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{q}}^{i}}=0$$

Now,

$$\frac{{\partial}L}{{\partial}x^{i}}=0$$

for all i.

So the EOM are given by

$$\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=0$$

Now
$$\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\dot{x}^{i}+eA_{i}$$

so d/dt of this should just be:

$$\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\ddot{x}^{i}+e\dot{A}_{i}$$

Since $$A_{i}=\epsilon_{ijk}B_{j}x_{k}\Rightarrow\dot{A}_{i}=\epsilon_{ijk}B_{j}\dot{x}_{k}$$

so we should get this:

$$\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\ddot{x}^{i}+e\epsilon_{ijk}B_{j}\dot{x}_{k}$$

but according to the solution to the problem it should be:

$$\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}}^{i}}=m\ddot{x}^{i}+2e\epsilon_{kji}B_{j}\dot{x}_{k}$$

Note the factor of 2.

I posted a similar problem and I was lead to believe that I need to differentiate wrt another superscript - I am not sure if this is at all relavent here since we are differentiating wrt time.

Would be really grateful if someone could offer me some advice...

Thanks..

 

[diazona]

Seems like
$$\frac{\partial L}{\partial x^i} \neq 0$$
because the Lagrangian depends on x through the vector potential A.

 

[vertices]

Seems like
$$\frac{\partial L}{\partial x^i} \neq 0$$
because the Lagrangian depends on x through the vector potential A.

thanks diazona. You are right:

[I will switch to subscripts, I hope that is okay]

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

where

$$e\vec{A}.\dot{\vec {x}}=[\epsilon_{123}B_{2}x_{3}+\epsilon_{132}B_{3}x_{2}]\dot{x}_{1} + [\epsilon_{213}B_{1}x_{3}+\epsilon_{231}B_{3}x_{1}]\dot{x}_{2} + [\epsilon_{312}B_{1}x_{2}+\epsilon_{321}B_{2}x_{1}]\dot{x}_{3}$$

So take

$$\frac{\partial L}{\partial x_1} = [\epsilon_{231}B_{3}\dot{x}_2+\epsilon_{321}B_{2}\dot{x}_3]$$

Now,

$$\frac{d}{dt}\frac{{\partial}L}{{\partial}{\dot{x}_1}}$$

so feeding these into the EL equation we get:

$$m\ddot{x}_{1}+e(\epsilon_{123}B_{2}\dot{x}_3+\epsilon_{132}B_{3}\dot{x}_3) - [ \epsilon_{231}B_{3}\dot{x}_2+\epsilon_{321}B_{2}\dot{x}_3 ] = 0$$

I can't see why we end up up

$$m\ddot{x}_{i}+2e\epsilon_{kji}B_{j}\dot{x}_{ k}$$

 

[diazona]

No problem with the subscripts By the way, you're missing a factor of $e$ (charge) in the last term there.

Anyway, big hint: what happens when you put in numbers for the $\epsilon_{ijk}$'s?

 

[vertices]

No problem with the subscripts By the way, you're missing a factor of $e$ (charge) in the last term there.

Anyway, big hint: what happens when you put in numbers for the $\epsilon_{ijk}$'s?

Yes, it works! (I think I made a mistake, which was confusing me somewhat!) Many thanks Diazona:)

Is decomposing into components the only way to see the levi-cevita (if that's what you call it) connection?

 

[diazona]

I'm not really sure. I guess you could try it in general tensor notation:
$$L = \frac{1}{2}m \delta_{jk} \dot x^j\dot x^k + e \epsilon_{ijk}B^i x^j \dot x^k$$
then
$$\frac{\partial L}{\partial x^j} = e\epsilon_{ijk} B^i \dot x^k = -e\epsilon_{ikj} B^i \dot x^k$$
and
$$\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot x^k} = m \delta_{jk} \ddot x^j + e\epsilon_{ijk} B^i \dot x^j$$
so when you set those equal to each other (or subtract one from the other), you wind up with
$$m \delta_{jk} \ddot x^j + 2e\epsilon_{ijk} B^i \dot x^j = 0$$

You could also make an attempt at it in vector notation,
$$L = \frac{1}{2}m (\dot{\vec{x}})^2 + e(\vec{B}\times\vec{x})\cdot\dot{\vec{x}}$$
but you'd have to use a vector identity,
$$(\vec{A}\times\vec{B})\cdot\vec{C} = (\vec{B}\times\vec{C})\cdot\vec{A} = (\vec{C}\times\vec{A})\cdot\vec{B}$$
so you could write
$$L = \frac{1}{2}m (\dot{\vec{x}})^2 + e(\dot{\vec{x}}\times\vec{B})\cdot\vec{x}$$
Then you get
$$\frac{\partial L}{\partial \vec{x}} = e \dot{\vec{x}}\times\vec{B} = -e \vec{B}\times\dot{\vec{x}}$$
and
$$\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{\vec{x}}} = \frac{\mathrm{d}}{\mathrm{d}t}\left[m \dot{\vec{x}} + e \vec{B}\times\vec{x}\right] = m \ddot{\vec{x}} + e \vec{B}\times\dot{\vec{x}}$$
and again, when you put them together, you get
$$m \ddot{\vec{x}} + 2e \vec{B}\times\dot{\vec{x}} = 0$$

It's the same math going on underneath, it's just that there are several different ways to write it out. If I didn't already have a lot of experience with using vector or tensor notation, I'd probably do it the way you did it, manually, one component at a time.

 

[vertices]

Hi Diazona - this is really, really helpful:)

Can I ask a really stupid question though:

How do you get from this (I've omitted the term in B):

$$L = \frac{1}{2}m \delta_{jk} \dot x^j\dot x^k$$

to this (this what the third line implies, ie. before d/dt'ing it):

$$\frac{\partial L}{\partial \dot x^k} = m \delta_{jk} \dot x^j$$

Where has the factor of 1/2 gone?

Thanks again:)

 

[diazona]

Ah, well in that one you have to remember that j and k are dummy indices, so that $\dot{x}^j$ and $\dot{x}^k$ are really the same thing. So the derivative applies to both of them equally.

If you want to show it explicitly, you can apply the product rule. I'm going to relabel the index of the $\dot{x}^k$ in the derivative so there's no confusion (of course, the label makes no difference because it is a dummy index):
$$\frac{\partial}{\partial \dot{x}^a}\biggl[\frac{1}{2}m\delta_{jk}\dot{x}^j\dot{x}^k\biggr] = \frac{1}{2}m\delta_{jk}\frac{\partial \dot{x}^j}{\partial \dot{x}^a}\dot{x}^k + \frac{1}{2}m\delta_{jk}\dot{x}^j\frac{\partial \dot{x}^k}{\partial \dot{x}^a}$$
Now, do you accept that $\partial \dot{x}^j/\partial \dot{x}^a = \delta^{j}_{a}$? Then
$$\begin{align*} \frac{\partial}{\partial \dot{x}^a}\biggl[\frac{1}{2}m\delta_{jk}\dot{x}^j\dot{x}^k\biggr] &= \frac{1}{2}m\delta_{jk}\delta^{j}_{a}\dot{x}^k + \frac{1}{2}m\delta_{jk}\dot{x}^j\delta^{k}_{a} \\ &= \frac{1}{2}m\delta_{ak}\dot{x}^k + \frac{1}{2}m\delta_{ja}\dot{x}^j \\ &= \frac{1}{2}m\delta_{ja}\dot{x}^j + \frac{1}{2}m\delta_{ja}\dot{x}^j \\ &= m\delta_{ja}\dot{x}^j \end{align*}$$
I went from $\delta_{ja}\dot{x}^j$ to $\delta_{ak}\dot{x}^k$ just by relabeling that index from j to k, and then switching the order of indices in the Kronecker delta because it's a symmetric tensor.

 

[vertices]

Aahh that is clear now - it's so much better knowing how things work, than just accepting them as facts... thanks ever so much diazona:)

I have one final question (I promise I won't bother you anymore after this!) - Infact, it's in this thread, but I'll reproduce it here:

If we have the following Lagrangian:

$$L={\frac{m}{2}}{g_{ij}(x)}.{\dot{x^{i}}{\dot{x^{j} }$$

I want to show the momentum is this:

$$(\frac{{\partial}L}{{\partial}\dot{x^{i}}})=mg_ {ij}x^{j}$$

So I differentiate wrt another superscript (as per George's advice):

$$\frac{{\delta}L}{d \dot{x}^k}=\frac{m}{2}g_{ij}[\frac{d \dot{x}^i}{d \dot{x}^k}{\dot{x}^j+\frac{d \dot{x}^j}{d \dot{x}^k}{\dot{x}^i]$$

This is equal to:

$$\frac{m}{2}g_{ij}[{\delta}_{ik}\dot{x}^{j}+{\delta}_{jk}\dot{x}^{i}]$$

Now replacing k with i we find

$$\frac{{\delta}L}{d \dot{x}^i}=\frac{m}{2}g_{ij}\dot{x}^{j}$$

ie. still have a factor of 1/2. My concern is that we can't relabel indices (as you did above), as there are no repeated indices.

Can you shed any light on where I have gone wrong?

Thanks..