Energy, mass and Noether’s theorem

[Dale]

No, it doesn’t mean I have rescaled the coordinates, it means I have translated by $k \epsilon$…

If you have a Lagrangian $L(q,\dot q,t)$ and $q$ is cyclic then there is one and only one conserved quantity which is called the canonical momentum conjugate to $q$: $p_q=\partial L/\partial \dot q$. This is the only quantity referred to when we talk about the conserved quantity from the $q$ symmetry of the Lagrangian.

Yes, $kp_q$ or indeed any given $f(p_q)$ is also conserved. But those are not the quantities referred to with $L(q,\dot q,t)$.

If you want to make one of those other conserved quantities the specific conserved quantity referred to then you do in fact need to do a change of coordinates $L(q’,\dot q’, t’)$.

 

[ergospherical]

If you have a Lagrangian $L(q,\dot q,t)$ and $q$ is cyclic then there is one and only one conserved quantity which is called the canonical momentum conjugate to $q$: $p_q=\partial L/\partial \dot q$. This is the only quantity referred to when we talk about the conserved quantity from the $q$ symmetry of the Lagrangian.

It's actually not. You're talking about the canonical momentum, but this is one special case of the general Noether theorem.

For Noether's theorem in full, you have the freedom to pick any symmetry generator $G^{\alpha}$ and of course any dilation of the generator by a constant also dilates the conserved quantity.

 

[Dale]

For Noether's theorem in full, you have the freedom to pick any symmetry generator Gα and of course any dilation of the generator by a constant also dilates the conserved quantity.

But not all generators are spacetime translations. Twice a translation generates a conserved quantity but it is not the conserved quantity associated with spacetime translations, unless you are also changing the coordinates.

 

[ergospherical]

Twice a translation generates a conserved quantity but it is not the conserved quantity associated with spacetime translations, unless you are also changing the coordinates.

I'm afraid this is non-sense because there is not a unique conserved quantity associated with any given spacetime translation (there are an infinity of them).

There is scaling freedom in the choice of generator (which you could absorb into the parameter $\epsilon$ when doing the functional differentiation of $L$). You are not in any sense changing the coordinates (I don't even know why you think it would be doing that).

 

[PeterDonis]

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