# Energy, mass and Noether’s theorem

• A
As @vanhees71 said, the four momentum is a vector, which is a single unchanging geometrical object. What you are doing when you say ##E’=2E## is simply to represent that single geometric object in a different basis. I.e. as a larger multiple of smaller vectors. This change of size of your basis is a change of units. It doesn’t change the geometrical vector, just the numbers you use to represent it.
You seem to think that I am just changing for example 1 Joule to 2 half-Joules. That is *not* what I am doing. I am changing 1 Joule to 2 Joules. Noether's theorem just gives an expression ##h## (with a physical dimension!) that is conserved and doesn't say whether the correct energy to use is ##E=h## or ##E=2h##.

ergospherical
What you cannot do is build a 4-vector out of, say, ##\left(\alpha p^{0}, \alpha' \mathbf{p} \right)## where the scalings ##\alpha \neq \alpha'## are not equal, because as I already showed in post #28 this set of quantities does not transform as a 4-vector.
Thus we can not just, as Dale did, say that ##E## and ##p## are the entities given by Noether's theorem by invariance under time and spatial translations. There is another condition that restricts the possible choices.

Dale
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You seem to think that I am just changing for example 1 Joule to 2 half-Joules. That is *not* what I am doing.
Why would you not do that? It doesn’t work the other way. I had assumed that you were intending to do correct things.

Thus we can not just, as Dale did, say that ##E## and ##p## are the entities given by Noether's theorem by invariance under time and spatial translations. There is another condition that restricts the possible choices.
Right, it is what @vanhees71 already pointed out. The resulting quantity needs to also be a four-vector. Thus the four-momentum ##(E,p)## is the conserved four-vector associated with the symmetry of the Lagrangian under spacetime translations.

vanhees71
What you cannot do is build a 4-vector out of, say, ##\left(\alpha p^{0}, \alpha' \mathbf{p} \right)## where the scalings ##\alpha \neq \alpha'## are not equal, because as I already showed in post #28 this set of quantities does not transform as a 4-vector.

Is'nt that what 't Hooft is doing?

't Hooft equ. (1.12) page 6 said:
##p^\mu = \begin{pmatrix}
p_x \\
p_y \\
p_z \\
iE \end{pmatrix}##
Source:
https://webspace.science.uu.nl/~hooft101/lectures/genrel_2013.pdf

ergospherical
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No T’Hooft uses the “##ict##” convention so that you can use the canonical inner product on ##\mathbf{R}^4## instead of the Lorentz metric.

vanhees71
vanhees71
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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.

Dale
PeterDonis
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But it has been written that using different factors for ##E## and ##p## is just about units. It has also been suggested to put the unit conversion factors into the metric. That would change the metric.
It changes the mathematical representation of the metric (and other things, such as ##E## and ##p##). It does not change any actual physical observables.

vanhees71
PeterDonis
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as I already showed in post #28
You didn't show what you claim in post #28. All you showed is that, if you are using different units for different components of a 4-vector, you need to change the transformation equations accordingly.

You appear to have forgotten the obvious fact that we can represent 4-vectors in any coordinate chart we like, and in many commonly used charts, different components do have different units. The simplest examples are cylindrical or spherical coordinates, where the angular components of 4-vectors have different units from the other components. (And of course there is the choice of whether to use units where ##c = 1## or not, which affects the relative scaling of timelike and spacelike coordinates.) The transformation equations between coordinate charts automatically take care of all this.

vanhees71
ergospherical
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As far as I am concerned both Dale and yourself are not addressing the question. @md2perpe asks why one cannot form a 4-vector out of, for example, 2E and 3p - the answer is that this does not have the correct transformation properties. We are not changing, or even considering, the space time coordinates.

PeterDonis
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@md2perpe asks why one cannot form a 4-vector out of, for example, 2E and 3p
And the answer is that you can, if you adjust the coordinates appropriately.

We are not changing, or even considering, the space time coordinates.
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.

Thus we can not just, as Dale did, say that ##E## and ##p## are the entities given by Noether's theorem by invariance under time and spatial translations.
Yes, you can. Remember that Noether's theorem starts with the Lagrangian, and the Lagrangian is a function of the generalized coordinates and their derivatives. As commented just above, when you say you can switch from a conserved quantity ##K##, derived from the Lagrangian by Noether's theorem, to ##hK + C##, what you are actually doing is changing the Lagrangian by changing the generalized coordinates. There is no freedom in the derivation of Noether's theorem to insert arbitrary factors or constants into the conserved quantity without changing the Lagrangian.

To put this another way, as I said in post #64, when you say you can switch from ##E## and ##p## to ##2E## and ##3p##, you can only mean that you are changing the mathematical representation of the physics. You can't change any actual physical observables just by saying so. The physically observed energy and momentum are what they are. If you want to represent those physically observed quantities by ##2E## and ##3p## instead of ##E## and ##p##, you have to make other corresponding changes in the mathematical representation.

Dale
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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

vanhees71
ergospherical
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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
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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.

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ergospherical
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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
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That is a bit of a stretch, don’t you think?
No. That is why I answered this way from the beginning.

ergospherical
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.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
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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.

Dale
ergospherical
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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...

docnet
PeterDonis
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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.

Dale
ergospherical
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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
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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
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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
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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
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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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