Energy, mass and Noether’s theorem

[md2perpe]

Mentors note: split from https://www.physicsforums.com/threads/energy-mass-equivalence.1003564/

> Mass is defined as energy measured in COM frame of reference where total momentum is zero.

How do you measure energy?

 

[Dale]

How do you measure energy?

Energy is the conserved quantity associated with time translation. In general it will be a composite term including multiple measured quantities.

It is often easier to measure mass than to measure energy.

 

[md2perpe]

It is often easier to measure mass than to measure energy.

Do you mean that you use mass to compute energy? Is it through $E=mc^2$?

In $(mc^2)^2 = E^2 - (pc)^2$, how are $m$, $E$ and $p$ defined?

 

[Dale]

In $(mc^2)^2 = E^2 - (pc)^2$, how are $m$, $E$ and $p$ defined?

They are all defined in their usual ways. $E$ is defined as the conserved quantity associated with the time translation symmetry of the Lagrangian. Similarly $p$ is defined as the conserved quantity associated with spatial translation symmetry. Then $m$ is defined as the invariant norm of the four momentum $(E,p)$.

Since $m$ is invariant, it is often easiest to measure. E.g. with a balance.

 

[md2perpe]

So then $E$ is the total energy and $p$ is the total linear momentum, since those are the conserved quantities?

 

[Dale]

Yes. That is correct

 

[Sagittarius A-Star]

So then $E$ is the total energy and $p$ is the total linear momentum, since those are the conserved quantities?

Yes. They are components of the four-momentum, which is a conserved quantity.

 

[md2perpe]

If so, why is it used on individual particles in a multi-particle interaction? According to what you write such use is incorrect since it should only be used on the total energy and momentum.

 

[Sagittarius A-Star]

The four-momentum is additive. If you have an isolated system of free particles, then you can add the individual four-momenta of the particles to get the four-momentum of the system.

 

[vanhees71]

One has to be careful. If it comes to scattering processes one should rather think in terms of quantum field theory rather than classical point-particle mechanics (which is anyway a problematic topic in relativity when it comes to interactions). Here it comes clear that a particle interpretation of the quantum fields, which are the fundamental mathematical entities in the description of particles in realtivistic QFT, is only possible for asymptotic free states (and even this is a more complicated notion than often thought). In a naive sense you can define them as Fock states of the various quantum fields describing particles in a situation, where the particles are located so far distant from each other that all interactions are negligible. A one-particle basis of these states is then defined by the momentum-spin eigenstates, which are automatically also energy eigenstates, where the relation between energy and momentum is
$$E^2=m^2 c^2 + \vec{p}^2.$$
That's the so-called mass shell in momentum space.

Scattering processes are then described by the socalled S-matrix. The S-matrix elements (or rather their modulus squared) describe a transition rate for a process described by a reaction, where (usually) two asymptotic free particles are shot at each other, then interact and then the reaction products fly appart and what's measured are then the asymptotic free particle states (usually many particles, which can be different from the original particles) corresponding to the reaction you are interested in. That's usually done with help of perturbation theory using Feynman diagrams, which are mainly just a very clever notation of the corresponding mathematical formulae but also provide, if correctly interpreted, a quite intuitive picture of the scattering process. For these scattering processes the energies and momenta (or in relativistic lingo their four-momenta) of the incoming and outgoing asymptotic free particles always (a) obey their corresponding on-shell conditions and (b) overall energy and momentum conservation, i.e.,
$$p_1+p_2=p_1'+p_2'+p_3'+\cdots,$$
where $p_1$ and $p_2$ are the on-shell four-momenta of the incoming asymptotic free particles and the $p_1'$, $p_2',\ldots$ those of the outgoing asymptotic free particles.

It should be understood that a particle interpretation of the quantum state of the involved quantum fields during the scattering process itself, where the states are not asymptotically free, no particle interpretation is possible. Often one refers to the corresponding internal lines of Feynman diagrams of perturbation theory as "virtual particles", but that's just the usual physicists' slang which has to be understood in the right way. What's physically observable are the asymptotic free states and, through making measurements of very many scattering processes, the probabilities (encoded in terms of cross sections) for certain scattering processes calculated from the S-matrix elements.

 

[md2perpe]

The four-momentum is additive. If you have an isolated system of free particles, then you can add the individual four-momenta of the particles to get the four-momentum of the system.

Yes, the four-momentum is additive. But in the formula $(mc^2)^2 = E^2 - (pc)^2$ you just confirmed that $E$ is the total energy and $p$ the total momentum, so the formula shouldn't be used for the energy and momentum of individual bodies, however such things are defined.

 

[Sagittarius A-Star]

energy and momentum of individual bodies, however such things are defined.

As I wrote, this is defined for a system of free (classical) particles. Of course, the phone battery in the OP is not a system of free particles, because the particles interact via the electromagnetic field. In this case it is not possible to break down, which part of the total energy belongs to which particle.

 

[md2perpe]

As I wrote, this is defined for a system of free (classical) particles.

So it wasn't true that $E$ and $p$ are the total energy and total momentum?

 

[Dale]

If so, why is it used on individual particles in a multi-particle interaction? According to what you write such use is incorrect since it should only be used on the total energy and momentum.

You can always redraw your “system” boundaries to include only an individual particle if you like. When you do so then again, it is the total energy and total momentum of that particle that is four momentum.

In other words, for a single free particle you do not exclude its KE or its rest energy. You use its total energy.

 

[Sagittarius A-Star]

So it wasn't true that $E$ and $p$ are the total energy and total momentum?

It is true, that $E$ and $p$ are the total energy and total momentum. In a battery, energy is also stored in the electric field.

 

[md2perpe]

They are all defined in their usual ways. $E$ is defined as the conserved quantity associated with the time translation symmetry of the Lagrangian.

Does Noether's theorem really specify the value of the energy? Doesn't it just give an expression $h(t,x,\dot{x})$ that is conserved? I mean, if $h$ is conserved, then so are $h+C$ and $Ch$ where $C$ is a constant. Can you take $E$ to be any of these?

 

[Dale]

Does Noether's theorem really specify the value of the energy? Doesn't it just give an expression $h(t,x,\dot{x})$ that is conserved? I mean, if $h$ is conserved, then so are $h+C$ and $Ch$ where $C$ is a constant. Can you take $E$ to be any of these?

That is a good question. I don’t know about that.

The $Ch$ is not an issue since that is just different units. But the $h+C$ is more interesting.

 

[md2perpe]

The $Ch$ is not an issue since that is just different units.

So if I multiply $E$ with one constant and $p$ with another constant that will not be a problem, you think?

 

[Dale]

So if I multiply $E$ with one constant and $p$ with another constant that will not be a problem, you think?

Correct, that is not a problem. That is what the various factors of c do already for SI units.

You would get $m^2=k^2_{m/E }E^2 - k^2_{m/p }p^2$ where the $k$ factors are the ratio of a unit mass to a unit energy or momentum.

 

[Herman Trivilino]

a) Can we convert energy to mass (matter) in every day life?

b) When we charge a phone battery, its mass (weight) increases according to E=mc2 . Does it mean we convert energy to matter? If not, how its mass increases?

Mass is not a measure of the quantity of matter.

 

[ergospherical]

So if I multiply $E$ with one constant and $p$ with another constant that will not be a problem, you think?

No, you cannot do this. Whilst it is true that $E_* = \alpha E$ and $\mathbf{p}_* = \alpha' \mathbf{p}$, for $\alpha,\alpha' \in \mathbf{R}$, are also conserved, it's not true generally that $(E_*, \mathbf{p}_*)$ is any longer a 4-vector. Because if\begin{align*}
p'^{\mu} = {\Lambda^{\mu}}_{\nu} p^{\nu}
\end{align*}then, for example, you have for $E_*'$ the following\begin{align*}
E_*' = \alpha E' &= \alpha\left[ {\Lambda^0}_0 E + {\Lambda^0}_i p^i \right] \\
&= {\Lambda^0}_0 E_* + {\Lambda^0}_i \alpha p^i \\
&\neq {\Lambda^0}_{\nu} p_*^{\nu}
\end{align*}unless $\alpha = \alpha'$; i.e. you would have to scale energy and momentum by the same factor in order to preserve the 4-vector character.

 

[Dale]

No, you cannot do this. Whilst it is true that $E_* = \alpha E$ and $\mathbf{p}_* = \alpha' \mathbf{p}$, for $\alpha,\alpha' \in \mathbf{R}$, are also conserved, it's not true generally that $(E_*, \mathbf{p}_*)$ is any longer a 4-vector. Because if\begin{align*}
p'^{\mu} = {\Lambda^{\mu}}_{\nu} p^{\nu}
\end{align*}then, for example, you have for $E_*'$ the following\begin{align*}
E_*' = \alpha E' &= \alpha\left[ {\Lambda^0}_0 E + {\Lambda^0}_i p^i \right] \\
&= {\Lambda^0}_0 E_* + {\Lambda^0}_i \alpha p^i \\
&\neq {\Lambda^0}_{\nu} p_*^{\nu}
\end{align*}unless $\alpha = \alpha'$; i.e. you would have to scale energy and momentum by the same factor in order to preserve the 4-vector character.

Not necessarily. You can simply put the unit conversion factors into the metric. This is the difference between $ds^2=-c^2 dt^2+ dx^2 + dy^2 + dz^2$ in SI units and $ds^2 = dt^2 + dx^2 + dy^2 + dz^2$ in natural units. You can even scale different directions of space differently, like fathoms for vertical distances and nautical miles for horizontal ones.

 

[ergospherical]

The metric isn't relevant to the point I was making, which was that a $4$-vector by definition must transform like $A' = \Lambda A$, and the vector $(c E, c'\mathbf{p})$ does not transform like that.

 

[Dale]

The metric isn't relevant to the point I was making, which was that a $4$-vector by definition must transform like $A' = \Lambda A$, and the vector $(cE, c'\mathbf{p})$ does not transform like that.

Sure it does, you just have to write $\Lambda$ appropriately.

 

[ergospherical]

Sure it does, you just have to write $\Lambda$ appropriately.

No! $\Lambda$ is a fixed Lorentz transformation (between two given co-ordinate systems), you cannot touch it.

 

[PeterDonis]

The metric isn't relevant to the point I was making

Yes, it is, because the form of the line element depends on the coordinates. See below.

No! $\Lambda$ is a fixed Lorentz transformation (between two given co-ordinate systems), you cannot touch it.

Changing the units in the way @Dale describes changes the coordinates. Therefore it also changes the components of the Lorentz transformation.

 

[Dale]

No! $\Lambda$ is a fixed Lorentz transformation (between two given co-ordinate systems), you cannot touch it.

Agreed. But changing the units is changing the coordinate systems. If I draw one chart where the integer coordinate lines are separated by meters and another where the integer coordinate lines are separated by miles it is a simple coordinate transform between the two. The expression for the metric changes accordingly as does the expression for the Lorentz transform.

 

[ergospherical]

I don't understand what either of you mean. You rescale $E \rightarrow \alpha E$ and $\mathbf{p} \rightarrow \alpha' \mathbf{p}$. On the other hand, you do not touch the co-ordinates $x^{\mu}$. The components of $\Lambda$ are clearly unchanged, ${\Lambda^{\mu}}_{\nu} = \dfrac{\partial x'^{\mu}}{\partial x^{\nu}}$. It is clear that $(\alpha E, \alpha' \mathbf{p})$ does not transform like a 4-vector unless $\alpha = \alpha'$.

Also, none of this bears direct relation to the metric, line element, etc., since I merely discuss the transformation properties of the vectors.

 

[Dale]

I don't understand what either of you mean. You rescale $E \rightarrow \alpha E$ and $\mathbf{p} \rightarrow \alpha' \mathbf{p}$. On the other hand, you do not touch the co-ordinates $x^{\mu}$.

If $\alpha’\ne \alpha$ then you will have to change the coordinates or the metric.

 

[ergospherical]

If $\alpha’\ne \alpha$ then you will have to change the coordinates or the metric.

Something has gotten badly lost in translation; the original premise of this discussion was that from Noether's theorem you get four conserved quantities $\dfrac{\partial L}{\partial \dot{x}^{\mu}} \delta^{\mu}_{\mu_0}$ due to invariance of the Lagrangian under spacetime translations, and the question was whether one can choose to define each of $E$ and $p_i$ as any multiples of that. The coordinates are of no relevance to this question, and are presumed as fixed.

As far as I can tell, you are instead suggesting scaling the co-ordinates $x^{\mu} \rightarrow \alpha x^{\mu}$ and considering the effect of this change of units on the components of the 4-momentum, but one must understand that this is an entirely different question!