I The Energy - Momentum Equation vs the Energy - Mass Equation

[SiennaTheGr8]

Leaving aside the non-rotational rotational [edited] case—how would you define the kinetic energy of the two-electron system I mentioned? Would you just define it as the sum of the particles' KE?

 

[Sagittarius A-Star]

According to Wikipedia, both definitions are possible.

Definition 1:

A system of bodies ... The kinetic energy of the system is the sum of the kinetic energies of the bodies it contains.

Definition 2:

When discussing movements of a macroscopic body, the kinetic energy referred to is usually that of the macroscopic movement only. However all internal energies of all types contribute to body's mass, inertia, and total energy.

Source:
https://en.wikipedia.org/wiki/Kinetic_energy#Kinetic_energy_of_systems

 

[vanhees71]

In manybody physics Definition 2 is the usual one. Kinetic energy is a very inconvenient quantity, because it's not nicely transforming under Lorentz transformations. That's why one always includes the "rest energies" so that together with momentum one gets a four-vector.

 

[SiennaTheGr8]

I might call Definition 1 "the kinetic energy in a system" for short in some contexts, but never "the kinetic energy of a system"—otherwise I'd have to say that the cup of coffee sitting on the desk next to me has non-zero kinetic energy!

Speaking of my coffee, is it not a universally accepted consequence of SR that it will lose mass as it cools (because its constituents will have less kinetic energy)? The situation with the rotating disk seems similar. It might be useful sometimes to calculate how much more energy a disk has when it's rotating than when it isn't, and you could call that the "relativistic rotational kinetic energy," but either way it contributes to the disk's total energy in its COM frame (aka its mass), doesn't it?

 

[vanhees71]

Of course, it is a universally accepted consequence of SR that your coffee looses mass when it cools.

In general, all quantities referring to intrinsic properties of a material of whatever kind is defined by scalar quantities. Intrinsic properties are all properties needed to characterize this system in its center-momentum frame. E.g., in fluid dynamics, i.e., a liquid, gas, or plasma close to local equilibrium all the intrinsic quantities are defined in the rest frame of each fluid cell, i.e., the usual thermodynamical quantities like internal-energy density, enthalpy density, entropy density, conserved-charge densities (electric charge, baryon number, strangeness, isospin,...) pressure, temperature and chemical potentials associated with the conserved charges.

E.g., an ideal fluid is usually described by the internal energy density (including "rest energy") and pressure via the energy-momentum tensor (west-coast convention)
$$T^{\mu \nu}=(u+p) u^{\mu} u^{\nu}-p \eta^{\mu \nu}.$$
Here $u$ is the internal energy in local rest frame of the fluid cell (LRF), $p$ the pressure (also measured in this LRF) and $u^{\mu}$ the four-velocity field with $u_{\mu} u^{\mu}=1$. The equations of motion (relativistic Fluid equation) is given by energy conservation, i.e.,
$$\partial_{\mu} T^{\mu \nu}=0.$$
There's no additional mass-conservation equation as in non-relativistic fluid dynamics.

In addition you have for any conserved charge $Q$ the corresponding current
$$j_Q^{\mu}=Q n u^{\mu}$$
and an equation for its conservation,
$$\partial_{\mu} j_Q^{\mu}.$$
In addition you need an equation of state to close the system of equations, where $n$ is something like a "net-particle number density" ("particles minus anti-particles" in the fluid cell).

You can of course also split the internal energy in a "invariant-mass density" $\mu$ and the rest $\tilde{u}$. This is advantageous when you want to derive the Newtonian limit. But as stressed already for several times there's no additional conservation law for the total mass!

 

[Sagittarius A-Star]

6) Calling it rest energy were a misnomer for a(n asymptotic) free photon state, because it's never at rest because its invariant mass is 0.

Would it be correct and a good idea, to give it the symbolic expression $\parallel \mathbf {P}\parallel$ and call it "energy-momentum magnitude"?

 

[Dale]

how would you define the kinetic energy of the two-electron system I mentioned? Would you just define it as the sum of the particles' KE?

Yes.

 

[vanhees71]

Would it be correct and a good idea, to give it the symbolic expression $\parallel \mathbf {P}\parallel$ and call it "energy-momentum magnitude"?

No, I'd never ever abuse the mathematically well defined definition of a norm. A norm on a vector space is a map $\|\cdot \|:V \rightarrow \mathbb{R}$ fullfilling the conditions

Positive definiteness: $\|\vec{v} \| \geq 0$ and $\|\vec{v}\|=0 \Leftrightarrow \vec{v}=0$.
Homogeneity: $\|\lambda \vec{v} \|=|\lambda| \|\vec{v} \|$.
Triangle inequality: ##\| \vec{v}_1 + \vec{v}_2 \| \leq \|\vec{v}_1 \| + |\vec{v}_2|.

It is easy to show that for a scalar product (a positive definite bi- (for real vector spaces) or sesqui- (for comoplex vector space) linear form) induces a norm in the usual way
$$\|\vec{v} \|=\sqrt{ (\vec{v},\vec{v})}.$$
This obviously does not work for any more general fundamental form, which is not positive definite. The Minkowski product, which is a fundamental form of signature (1,3) or (3,1) on $\mathbb{R}^4$, cannot induce a norm.

I would simply stick to the modern conventions and call it invariant mass defined by
$$M^2=P_{\mu} P^{\mu}/c^2=s/c^2,$$
where $P^{\mu}$ is the total four-momentum (of a closed system).

 

[Sagittarius A-Star]

I would simply stick to the modern conventions and call it invariant mass defined by
$$M^2=P_{\mu} P^{\mu}/c^2=s/c^2,$$
where $P^{\mu}$ is the total four-momentum (of a closed system).

That's a possibility. But I want to find out, if a simple symbol exists, that indicates intuitively, that it stands for the (context-dependent Euklidian or pseudo-Euklidian) magnitude of the vector, which the symbol $m$, without an additional explanation, doesn't. And of course, I want to avoid "ict" as a possible workaround.

A good example from the 4-position is $\Delta s$. It is intuitively regarded as some kind of "distance".

 

[vanhees71]

[EDIT: corrected confusing typos below: the Minkowski fundamental form is not a metric but a pseudo-metric; and it's $\mathrm{d}s^2$ rather than $\mathrm{d}s$.]

The Minkowski pseudo-metric does not induce a metric, because it's not positive definite. Why do you want to introduce totally useless and confusing ideas?

I guess you mean
$$\mathrm{d}s^2 = g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}?$$
That's not a distance (squared), because it's not positive definite.

 

[Sagittarius A-Star]

I might call Definition 1 "the kinetic energy in a system" for short in some contexts, but never "the kinetic energy of a system"

That disagrees to the formulation of Wikipedia, but I think, your formulation is the systematic one.

 

[Sagittarius A-Star]

The Minkowski metric does not induce a metric, because it's not positive definite. Why do you want to introduce totally useless and confusing ideas?

I don't want to introduce something new, I investigate, if such a thing already exists.

I guess you mean
$$\mathrm{d}s = g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}?$$
That's not a distance (squared), because it's not positive definite.

And what about calling it spacetime interval ?

 

[Dale]

Would it be correct and a good idea, to give it the symbolic expression $\parallel \mathbf {P}\parallel$ and call it "energy-momentum magnitude"?

I'd never ever abuse the mathematically well defined definition of a norm

Many others would. It is accepted terminology, and why not? It is a perfectly reasonable generalization of a norm to manifolds with mixed signature.

We generalize many things in pseudo Riemannian geometry by relaxing some requirement of other geometries. So this is commonly done. That you personally prefer not to generalize the norm does not make it invalid.

Instead of explaining that the Minkowski norm violates the standard definition of a norm (since that is obvious for any generalization of any concept) why don’t you argue where such a definition can cause confusion or something?

The Minkowski metric does not induce a metric

Hmm

 

[vanhees71]

You got me. It's of course the Minkowski pseudo-metric.

I've never seen any scientific paper or textbook introducing a "norm" in this sense. The first time I've seen it is from you pointing to the Wikipedia article. Is this "norm" then imaginary for space-like vectors (in the west-coast convention) or for time-like vectors (in the east-coat convention). Why do you think it's useful to introduce an unneeded concept against well-established mathematical definitions in use for decades/centuries?

 

[Dale]

Why do you think it's useful to introduce an unneeded concept against well-established mathematical definitions in use for decades/centuries?

I am glad you asked. The reason why is because it highlights the geometry. IMO, the key concept of relativity is the generalization of geometry to spacetime.

All generalizations inevitably go against some established definitions, so that in itself is not particularly problematic. The question is if the insight gained by the generalization is valuable. My experience is that it is valuable.

For example, people who understand the geometry never have trouble with the twin paradox. People who understand the geometry can move to GR more easily. People who understand the geometry can see the link between charge density and current or between energy and momentum easier. Etc.

 

[vanhees71]

I agree, since the most fundamental concept of contemporary physics are symmetry principles and thus in a wide sense geometry.

One crucial point of relativistic spacetime models is that they are not Euclidean/Riemannian affine spaces/manifolds but pseudo-Euclidean/pseudo Riemannian affine space (SR/GR). Uncounted textbooks and paper indeed call corresponding the fundamental forms a "scalar product" or $\mathrm{d}s^2$ "a metric", but I've really never seen that they introduce $\|\mathrm{d} x\| =\sqrt{\mathrm{d} s^2}$. It's already confusing to call the indefinite fundamental forms scalar product, but to introduce a symbol $\| \cdot \|$ for something that's definitely not a norm, is too much.

At least, I'd not accept such a thing for any textbook I'd recommend to my students. It's hard enough for them to learn to read a Minkowski diagram correctly, i.e., not mistaken it as if it could be read as having "distances" and/or even "angles" as in the Euclidean plane with a Cartesian basis.

 

[Dale]

One crucial point of relativistic spacetime models is that they are not Euclidean/Riemannian affine spaces/manifolds but pseudo-Euclidean/pseudo Riemannian affine space (SR/GR).

Agreed. And many basic proofs that hold in Riemannian geometry with its positive definite norm do not hold in pseudo Riemannian geometry precisely because of the missing positive definiteness.

When working with a generalization it is indeed essential to keep track of what theorems were lost in the generalization. That is true of any generalization.

It's hard enough for them to learn to read a Minkowski diagram correctly, i.e., not mistaken it as if it could be read as having "distances" and/or even "angles" as in the Euclidean plane with a Cartesian basis.

And that is a good reason. Thank you.

 

[Sagittarius A-Star]

I've really never seen that they introduce $\|\mathrm{d} x\| =\sqrt{\mathrm{d} s^2}$.

Wolfgang Rindler wrote generally for 4-vectors:

and define the magnitude as
$$A = | \mathbf {A}^2|^\frac{1}{2} \geq 0.$$

Source (see equation 21):
http://www.scholarpedia.org/article/Special_relativity:_mechanics#The_4-vector_calculusSo, can I write $P$ for the magnitude of $\mathbf {P}$?

 

[vanhees71]

Ok, I still don't think that such a definition is desirable, no matter how famous the author might be.

 

[PAllen]

Well, I just checked. MTW, Synge, Bergmann, and James L. Anderson (4 out of 4 of the GR textbooks closest to me) all use norm for the contraction of vector with itself and the metric for spacetime metrics (pseudo-Riemannian). Thus, as @Dale said - the ship has sailed on this. It is viewed as one of the many Riemannian terms extended to pseudo-Riemannian, with context avoiding any confusion.

[edit: for the sake of completeness, Pauli's famous 1921 GR article agrees with @vanhees71, and reserves the term norm of the case of positive definite metric. Also, Synge pulls a trick and defines the norm to be the contraction multiplied by an indicator to make it positive for both timelike and spacelike vectors, but still 0 for null.]

 

[vanhees71]

Well, to call $\eta_{\mu \nu}$ a "scalar product" and $\eta_{\mu \nu} \mathrm{d}x^{\mu} \mathrm{d} x^{\nu}$ a "metric", you can't get out of the common use of physicists' sloppy language, but to take the square root and calling this a norm then is beyond the line I want to accept. Since when can a norm be imaginary? It doesn't fulfill all the properties that makes a norm a useful concept. It's not needed!

I always call the product a pseudo-scalar product or the Minkowski product. This doesn't hurt in any way and makes the concept much clearer.

 

[Sagittarius A-Star]

to take the square root and calling this a norm then is beyond the line I want to accept. Since when can a norm be imaginary?

In the definition of Rindler, that I cited in above posting #108, and, as I understand PAllen # 110, also in that from Synge, the square-root cannot become imaginary.

But I don't like this part of the definition of Rinder and Synge. I prefer, if the norm of the 4-momentum always equals to the "invariant mass" $m$, for tardyons, photons and for (hypothetical) tachyons. Tachyons have an imaginary $m$ and a real energy $\gamma * m$.

[Edit: Deleted last (wrong) sentence, that PAllen cited.]

 

[PAllen]

...To avoid, that the norm becomes imaginary for tardyons, the norm's definition should be made dependent on the sign convention (west-/east coast).

I like this typo. Ordinary particles are tardyons, and FTL are tachyons!

 

[Sagittarius A-Star]

I like this typo. Ordinary particles are tardyons, and FTL are tachyons!

That's not a typo.
[Edit: Sorry you are right. I made an error.]

 

[Dale]

Also, Synge pulls a trick and defines the norm to be the contraction multiplied by an indicator to make it positive for both timelike and spacelike vectors, but still 0 for null.]

Notice that this trick does not fix the norm in the sense that would be required for most proofs based on the norm to hold. Specifically, it is still degenerate, meaning that two points having a difference with norm zero are not necessarily the same point. So even with this approach you still must be aware of which theorems have been lost in the move from Riemannian to pseudo-Riemannian geometry.

 

[Sagittarius A-Star]

Thus, as @Dale said - the ship has sailed on this.

Notice that this trick does not fix the norm

So, I prefer the following formulation and definition(s):

"Energy-momentum magnitude" $\left\|\mathbf {P}\right\| = \sqrt{\mathbf {P} \cdot \mathbf {P}} = \sqrt{(\frac {E}{c})^2 - p_x^2 - p_y^2 - p_z^2}$

Then, for all kinds of objects (tardyons, photons, tachyons) is valid:
$$\left\|\mathbf {P}\right\| * c = E * \sqrt{1 - \frac {v^2}{c^2}}$$

 

[vanhees71]

Very misleading notation, as I stated often enough now . For tachyons it's an imaginary number.

 

[Sagittarius A-Star]

Very misleading notation, as I stated often enough now . For tachyons it's an imaginary number.

I think, if it's used in the context of SR and 4-vectors, then it's already by this implicitely clear, that not the Euklidean "norm" is meant, but a pseudo-Euklidea "pseudo-norm".

 

[Sagittarius A-Star]

No, I'd never ever abuse the mathematically well defined definition of a norm.
...
Triangle inequality: $\| \vec{v}_1 + \vec{v}_2 \| \leq \|\vec{v}_1 \| + \|\vec{v}_2\|$.

There can be found an analogous statement in Minkowski geometry:

Proposition 8 (Minkowski triangle inequality). If $U$ and $V$ are future-pointing, timelike four-vectors, then $U + V$ is also future-pointing timelike and satisfies
$$\sqrt{g(U+V, U+V)} \geq \sqrt{g(U , U)} + \sqrt{g(V , V)}\ . \ \ \ \ \ \ \ \ \ \ (6.17)$$
Proof ...
...
Thus the four-momentum can be a future-pointing timelike vector, with the pseudo-norm dictated by the rest mass of the particle

Source, see page 34 and page 43 (Warning! Please ignore "Definition 18" on page 44 ):
https://courses.maths.ox.ac.uk/node/view_material/44180

BTW, the Minkowski triangle inequality is also a direct proof of the "twin paradox".

If the plane defined by x and y is spacelike (and therefore a Euclidean subspace) then the usual triangle inequality holds.

Source:
https://en.wikipedia.org/wiki/Triangle_inequality#Reversal_in_Minkowski_space

 

[Sagittarius A-Star]

For tachyons it's an imaginary number.

I am currently looking, how to to get rid of imaginary numbers in the Minkowski triangle inequality in the most elegant way. The appoach for this by W. Rindler (see my related postings #112 and #108) would make it necessary, the replace for tachyons the symbol "$\geq$" by "$\leq$", to address the Euklidean property of the spacelike 4D-subspace of the 4D-Minkowski space, as mathematicians call the spacetime (see also Wikipedia link in #119). I would find it ugly, to have different triangle inequalities for the time-like- and the space-like subspaces.

Wouldn't it be a more elegant appoach, to replace the (eventually imaginary) numbers in the Minkowski triangle inequality generally by their squares, as written in Wikipeda?

$\|u+w\|^2 ... \geq ... (\|u\| + \|w\|)^2$
The result now follows by taking the square root on both sides.

Source:
https://en.wikipedia.org/wiki/Minkowski_space#The_reversed_triangle_inequalit

Then, the Minkowski triangle inequality would compare for tachyons negative real numbers, which would be a substitute for exchanging - only for tachyons - the symbol "$\geq$" by "$\leq$", as described above for the Rindler appoach.