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Thanks a lot :)

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Thanks a lot :)

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Vanadium 50

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Hold on here, the lagrangian isL = EK-EP. It is supposed to be conserved in particle interactions.

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

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

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

But it is

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 [tex]\mu\in[x,y,z,t][/tex] for instance, and [tex]\partial_{\mu}S[/tex] would be a 4-vector containing the partial derivative of the scalar [tex]S[/tex].Also, what does this symbol mean; [tex]\partial[/tex] with a small [tex]\mu[/tex] symbol either at top or bottom next to it?

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

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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 isnotwhat 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 [tex]\mu\in[x,y,z,t][/tex] for instance, and [tex]\partial_{\mu}S[/tex] would be a 4-vector containing the partial derivative of the scalar [tex]S[/tex].

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

Thanks, your explanations help :). So just to be sure i understood partial derivative right; [tex]\partial_{\mu}S[/tex] 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?

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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.

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Thanks, your explanations help :). So just to be sure i understood partial derivative right; [tex]\partial_{\mu}S[/tex] 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?

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.

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malawi_glenn

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lower that index!

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malawi_glenn

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[tex] \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}) [/tex]

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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 togethernot according to my lecture notes.

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

[tex]\partial_{\mu}=\frac{\partial}{\partial x^{\mu}}[/tex] and [tex]\partial^{\mu}=\frac{\partial}{\partial x_{\mu}}[/tex]

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

but[tex]\partial_{\mu}V^{\mu}=g_{\mu}^{\,\,\,\,\nu}\partial^{\nu}V^{\mu}[/tex]

[tex]\partial_{\mu}V^{\mu}=g_{\mu\nu}\partial^{\nu}V^{\mu}[/tex]

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malawi_glenn

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

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