# Lagrangian is defined as the difference between kinetic and potential

1. Jun 14, 2008

### faen

So after reading a bit particle physics, the lagrangian is defined as the difference between kinetic and potential energy: L = EK-EP. It is supposed to be conserved in particle interactions. However i dont know why it is conserved, so can anyone please explain to me why?

Thanks a lot :)

2. Jun 14, 2008

Staff Emeritus
The Lagrangian is a part of a technique used in advanced classical mechanics called, not surprisingly, Lagrangian mechanics. The same basic idea can be applied to quantum mechanics. It will be difficult to understand the QM version before understanding the classical version.

3. Jun 14, 2008

### faen

I just read noethers theorem, so this is what i learned: The lagrangian leads to the equations of motions. So any symmetry observation in the equations of motion means that the symmetry must exist in the lagrangian of a wavefunction also because it is related to the equations of motions. And visa versa.. Am i right or did i get it wrong? :p

4. Jun 14, 2008

### humanino

Hold on here, the lagrangian is not conserved ! A conserved quantity would be the energy, I mean total E= Ek + Ep. If L and E were both conserved, then Ek and Ep would be both conserved as well. There would be nothing happening whatsoever, no motion, no force.

If you integrate L along a physical trajectory, you will obtain a number (the action) which is less than for any other possible (but unphysical) trajectory. So L is not conserved but is (in some sens) minimum. Technically, this is nothing more than the principle of least action.

The way it is formulated, the action is not necessarilly least, it could be most or it even could just be stationnary, that is a local inflection point in the space of all possible trajectories. I state that to try to make it clearer, not to claim that "least action" is a bad name.

5. Jun 14, 2008

### faen

Thanks, i understand a bit more now. But was what i said in my previous post right or wrong?

Also, what does this symbol mean; $$\partial$$ with a small $$\mu$$ symbol either at top or bottom next to it?

6. Jun 14, 2008

### humanino

In your previous post, you said that any symmetry of the equation of motion is also a symmetry of the lagrangian. I have not put too much thought in this statement, and I do not want to confuse you, but I would say that apart from symetry breaking mechanisms (spontaneous, dynamical), your statement seems all right.

But it is not what Noether's theorem is about. Noether's theorem is about the fact that, to any symmetry is associated a conserved quantity.

Energy conservation <-> time translation invariance
Momentum conservation <-> space translation invariance
Angular momentum conservation <-> space rotation invariance
This refers to the partial derivative with respect to space-time coordinates. So $$\mu\in[x,y,z,t]$$ for instance, and $$\partial_{\mu}S$$ would be a 4-vector containing the partial derivative of the scalar $$S$$.

You need to check you metric $$g_{\mu\nu}$$ which could be (+,-,-,-) or (-,+,+,+) to define $$\partial_{\mu}V^{\mu}=g_{\mu}^{\,\,\,\,\nu}\partial^{\nu}V^{\mu}$$ (well really it defines the inner product of 4-vectors...)

7. Jun 14, 2008

### faen

Thanks, your explanations help :). So just to be sure i understood partial derivative right; $$\partial_{\mu}S$$ means that i get a vector where the first component is the partial derivative of x, then the next vector component is the partial derivative of y, and so on?

8. Jun 14, 2008

### Haelfix

"you said that any symmetry of the equation of motion is also a symmetry of the lagrangian"

Nitpick, is that true necessarily, I haven't thought about that for awhile either. You gotta be a little careful here I think and distinguish between the action, lagrangian and lagrangian density.

9. Jun 14, 2008

### K.J.Healey

first term is time, then the three spatial.
Anytime you see V^mu it's a 4-vector {t,x,y,z} or rather {x0,x1,x2,x3}.
so basically what you said but with the first term is derivative wrt time.

edit: actually I think when learning classical mechanics they did something like {x,y,z,t}, but reagardless, yes, its just the notation for a derivative vector.

10. Jun 15, 2008

### malawi_glenn

$$\partial ^{\mu} = (\frac{\partial }{\partial x^0},-\frac{\partial }{\partial x^1},-\frac{\partial }{\partial x^2},-\frac{\partial }{\partial x^3}) = (\frac{\partial }{\partial x^0},-\vec{\nabla})$$

11. Jun 15, 2008

### K.J.Healey

lower that index!

12. Jun 15, 2008

### malawi_glenn

not according to my lecture notes.

$$\partial _{\mu} = (\frac{\partial }{\partial x^0},\frac{\partial }{\partial x^1},\frac{\partial }{\partial x^2},\frac{\partial }{\partial x^3}) = (\frac{\partial }{\partial x^0},\vec{\nabla})$$

13. Jun 15, 2008

### humanino

In QFT in flat spacetime, many people simply ignore upper and lower indices. This is because covariant and contravariant indices transform identically (numerically) when the basis is (pseudo-)orthogonal. Actually, when I say pseudo, I mean when you have a lorentzian metric, and it is not quite true that the components are equal : the are actually opposite for those dimensions corresponding to the negative part of the signature. If say you take (+,-,-,-), the time component is really unchanged, but the space components flip signs. However, in the end we do not really care because any time we want to evaluate the inner product we already know we must take a minus sign for the space components, so we usually forget those technicalities all together

Remember how we can canonically identify a tangent vector to the dual form which gives the projection on this vector ? In principle a tangent vectors and its (dual) forms are just the contravariant and covariant representation of the same (1-)tensor. But in euclidean space, the coordinates are the same if you choose an orthogonal basis. This basically is just the fact that orthogonal projection on a basis vector will give the same result as the projection along (dim-1) hyperplane spanned by the other basis vectors if (and only if) the basis is orthogonal.

Covariance and contravariance of vectors on wiki.

But if you want to keep track of indices, you should have
$$\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}$$ and $$\partial^{\mu}=\frac{\partial}{\partial x_{\mu}}$$
In the end, this just stems for the fact that we are doing symbolic manipulations : if you put a contravariant in the denominator, then you will pick up the inverse of the jacobian instead of the jacobian as would have been if the contravariant were in the numerator. There is not much more to these notations.

What I wrote earlier was wrong BTW, it should not have been
but
$$\partial_{\mu}V^{\mu}=g_{\mu\nu}\partial^{\nu}V^{\mu}$$

14. Jun 15, 2008

### malawi_glenn

this is great, my professor will give me a golden star =D