Energy, mass and Noether’s theorem

[Dale]

Argh. That's just an oldfashioned convention to avoid the Minkowski-metric components $\eta_{\mu \nu}$ for the fundamental form of Minkowski space in pseudo-Cartesian coordinates and make everything look like Eulidean Cartesian coordinates. I'd not recommend to use this convention anymore, because it's inconvenient and cannot be generalized to arbitrary bases of Minkowski space nor, of course, to general relativity.

I would like to echo this. It is akin to relativistic mass. Something that was indeed published in the past, but has since been recognized as being a bad idea and justifiably abandoned by the mainstream scientific community

 

[ergospherical]

Yes, you are. $E$ and $p$ are derived from the Lagrangian by Noether's theorem. Switching to $2E$ and $3p$ means you are changing the generalized coordinates in the Lagrangian accordingly.

No, not necessarily. It’s just by observation that a single conserved quantity gives rise to an infinite of other possible conserved quantities. No need to start interpreting this as changing the coordinates.

 

[Dale]

No, not necessarily. It’s just by observation that a single conserved quantity gives rise to an infinite of other possible conserved quantities. No need to start interpreting this as changing the coordinates.

That is not correct. While those other quantities are indeed also conserved, they are not the quantities obtained by Noether’s theorem unless you change the coordinates.

In other words, if $a$ is the conserved quantity associated with a specific symmetry of a given Lagrangian, per Noether’s theorem, then it is true that any given $f(a)$ is also conserved. But the given $f(a)$ is not obtained from the Lagrangian using Noether’s theorem. To obtain it from the Lagrangian using Noether’s theorem requires a coordinate transform.

 

[ergospherical]

That is not correct. While those other quantities are indeed also conserved, they are not the quantities obtained by Noether’s theorem unless you change the coordinates.

That is a bit of a stretch, don’t you think? In any case I might have a think about this some other time, it’s a fiddly little question but I’ve sort of lost interest. Some good points all around, though, I think.

 

[Dale]

That is a bit of a stretch, don’t you think?

No. That is why I answered this way from the beginning.

 

[ergospherical]

.In other words, if $a$ is the conserved quantity associated with a specific symmetry of a given Lagrangian, per Noether’s theorem, then it is true that any given $f(a)$ is also conserved. But the given $f(a)$ is not obtained from the Lagrangian using Noether’s theorem. To obtain it from the Lagrangian using Noether’s theorem requires a coordinate transform.

I would add, though, that I don’t think this is right. If $G^{\alpha}$ generates a symmetry then any multiple of $G^{\alpha}$ generates the same symmetry. So you can get different conserved quantities corresponding to the same underlying symmetries by playing with the generators rather than the coordinates.

 

[PeterDonis]

If $G^{\alpha}$ generates a symmetry then any multiple of $G^{\alpha}$ generates the same symmetry.

Write down the math for this explicitly. You will see that, in order to use a multiple of $G^\alpha$ as the generator of the symmetry, you have to modify the generalized coordinates in the Lagrangian. The symmetry generator is determined by the Lagrangian; there is no freedom of choice to insert an arbitrary factor.

 

[ergospherical]

Write down the math for this explicitly. You will see that, in order to use a multiple of $G^\alpha$ as the generator of the symmetry, you have to modify the generalized coordinates in the Lagrangian.

You most definitely do not...

If $G^{\alpha}$ generates a symmetry then so does $kG^{\alpha}$, i.e. $x'^{\mu} = x^{\mu} + \epsilon k G^{\mu}$,
\begin{align*}
\delta L &= \dfrac{\partial L}{\partial x^{\mu}} \epsilon k G^{\mu} + \dfrac{\partial L}{\partial \dot{x}^{\mu}} \epsilon k \dot{G}^{\mu} \\
&= \epsilon k G^{\mu} \dfrac{d}{d\lambda} \left( \dfrac{\partial L}{\partial \dot{x}^{\mu}} \right) + \dfrac{\partial L}{\partial \dot{x}^{\mu}} \epsilon k \dot{G}^{\mu} \\
&= \epsilon \dfrac{d}{d\lambda} \left( k\dfrac{\partial L}{\partial \dot{x}^{\mu}} {G}^{\mu} \right) = 0
\end{align*}Which implies that $k\dfrac{\partial L}{\partial \dot{x}^{\mu}} {G}^{\mu}$ is conserved. It's very simple, and one does not touch the coordinates, of course...

 

[PeterDonis]

If $G^{\alpha}$ generates a symmetry then so does $kG^{\alpha}$

Using $k G^\alpha$ instead of $G^\alpha$ as your generator means, reading directly off your formula for $x'^\mu$, that you are translating by $k \epsilon$ instead of $\epsilon$, which means you have rescaled your coordinates.

 

[ergospherical]

Using $k G^\alpha$ instead of $G^\alpha$ as your generator means, reading directly off your formula for $x'^\mu$, that you are translating by $k \epsilon$ instead of $\epsilon$, which means you have rescaled your coordinates.

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

 

[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]

Thread closed for moderation.