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I need to study the relativistic lagrangian of a free particle. It's $\ L= - m c^2 \sqrt[2]{1- \frac{|u|^2}{c^2}} $ I need to study the translation, boost and rotation symmetry. I say it doesn't depend of the position, so it has translation symmetry and the momentum will conserve. It's rotation invariant because depends only of the modulus of the speed $|u|$(What is the conserved quantity derived by this symmetry?) What can I say about the boost transformation, I can use the boost of the speeed $u$, but I can't see any conclusion about that.

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Possible duplicate: physics.stackexchange.com/q/12559 –  Nathaniel May 22 '13 at 0:59
I'm asking in a particular case, without all the formalism of the tensors. –  NArgento May 22 '13 at 3:01

2 Answers 2

Time and space translation invariance imply conservation of energy and momentum.

Rotation invariance implies conservation of angular momentum.

As I recall: Boost (a 4-d rotation) invariance tells you that all inertial reference frames have constant velocity wrt each other.

Caveat: I haven't thought about the last one for years.

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Only the action S is a relativist invariant : invariant under translations, rotations, and boosts. The Lagrangian itself is not invariant under boosts.

The action $S$ is : $S = -mc \int ds = - mc \int \sqrt {ds^2} = - mc \int \sqrt {c^2 dt^2 - \vec {dx}^2} = - mc^2 \int \sqrt {1 - (\frac{\vec {dx}}{c \, dt})^2} \,\,\,dt$

$= \int L \, dt$

where L is the Lagrangian : $L= - mc^2 \sqrt {1 - (\frac{\vec {dx}}{c \, dt})^2}$

$ds^2$ is clearly a relativist invariant, and so the action S is too.

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