The Energy - Momentum Equation vs the Energy - Mass Equation

[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.

 

[vanhees71]

I'm still puzzled about the question, what all this effort should be good for.

Of course, some of the theorems (or rather lemmas) about the Minkowski geometry are useful, e.g., the "causality property" that the temporal order of events is unchanged under proper orthochronous Lorentz transformations if and only if the events are spacelike or lightlike separated or that a vector Minkowski-orthogonal to a time-like vector is spacelike and a Minkowski orthogonal vector to a light-like vector is parallel to this light-like vector (maybe there are more), but all these are more easily formulated with Minkowski products of the various vectors rather than with any square roots of them.

 

[Sagittarius A-Star]

I'm still puzzled about the question, what all this effort should be good for.
...
but all these are more easily formulated with Minkowski products of the various vectors rather than with any square roots of them.

The norm of a 4-vector is the square-root. It is, except for photons, a directly measureable quantity in the object's rest frame. That could be for example the proper time of a time-like twin, which is measured by that twin's wristwatch or my proper energy, also called my "invariant mass", measured by the bathroom scale, if I am standing still on it.

The instrument for measuring this norm directly must have the same worldline as the object, to that the related 4-vector applies. For example is the wristwatch of a tachyon also a tachyon (from the standpoint of us, the tardyons).

 

[vanhees71]

I stick to the mathematically sound and solid usual definitions. What you declare as norm is no norm and it is overly complicated and pretty useless.

 

[Sagittarius A-Star]

I stick to the mathematically sound and solid usual definitions.

That's O.K. Only, in SR it is not "unusual" to use the term "norm" also for 4-vectors, according to some examples, which were linked or mentioned here.

What you declare as norm is no norm

I meant "pseudo-norm". I omitted the "pseudo", because I assumed, that it's clear from the context.

it is overly complicated

I find it easy.

and pretty useless.

I find, in the context of a geometry-based theory like SR, a concept of 4-vectors without a (pseudo-)norm is incomplete.

 

[Dale]

The instrument for measuring this norm directly must have the same worldline as the object

I wouldn’t read too much into directness. Most measurements are rather indirect. For example, the current mise en pratique for measuring mass is very indirect, involving measurements of voltage, current, speed, and frequency, along with previous measurements of acceleration. Even the more common, most direct technique, is actually measuring torque.

 

[Dale]

I stick to the mathematically sound and solid usual definitions. What you declare as norm is no norm and it is overly complicated and pretty useless.

Nonetheless this particular generalization of terminology is well established and is not just personal “What you declare” terminology, but standard and well accepted.

If you wish merely to grumble then that is fine. But I f you wish to persuade others to adopt your viewpoint then you will need to come up with persuasive arguments. That it doesn’t meet the formal definition of a norm is not persuasive, because people using this terminology already know that and are content with the generalization.

So far your most persuasive argument was about confusion engendered in students. I would develop or expand that type of argument. Unless you are interested only in grumbling.

 

[vanhees71]

I find, in the context of a geometry-based theory like SR, a concept of 4-vectors without a (pseudo-)norm is incomplete.

To the contrary, it is utmost important for the consistency of the theory that the geometry of relativistic spacetimes is pseudo-Riemannian and not Riemannian!

 

[vanhees71]

Nonetheless this particular generalization of terminology is well established and is not just personal “What you declare” terminology, but standard and well accepted.

If you wish merely to grumble then that is fine. But I f you wish to persuade others to adopt your viewpoint then you will need to come up with persuasive arguments. That it doesn’t meet the formal definition of a norm is not persuasive, because people using this terminology already know that and are content with the generalization.

So far your most persuasive argument was about confusion engendered in students. I would develop or expand that type of argument. Unless you are interested only in grumbling.

It's the first time in this thread that I've seen that somebody defines $\| \cdot \|$ as a "norm" based on an indefinite fundamental form. It's not needed and even more confusing than the standard use of the word "metric" for the bilinear form. The standard terminology is already confusing enough for students, as you can see on the example of this useless discussion.

 

[Dale]

The standard terminology is already confusing enough for students, as you can see on the example of this useless discussion

Have you ever seen actual students actually struggle with the concept of the Minkowski norm? What indication do you have that it is truly confusing to students? If you have not seen that then exactly what struggles would you anticipate they would have and why? Perhaps the reason that the standard terminology seems confusing to students is insufficient emphasis of the similarities and so adopting this terminology would help. What makes you think it wouldn’t?

Personally, for me it is still about geometry. Anything that we can do to emphasize the geometry is a good thing. Yes, it is important to explain that Minkowski geometry is not the same as Euclidean geometry, but most Euclidean geometric concepts have a close analogue in Minkowski geometry. The similarities are every bit as important as the differences because the similarities allow them to reason geometrically.

How many students who understand the Minkowski norm geometrically will be fooled by the twin paradox?

 

[vanhees71]

My experience is that it is hard for students (including myself when I was a student) to forget about the Euclidean heuristics built from long training in Euclidean (plane) geometry when it comes to read a Minkowski diagram. My students told me that it was important for them that I emphasized the important difference between Euclidean and Minkowski and that instead of circles you need time- and space-like hyperbolae to determine the "unit tick marks" on the axes of different inertial reference frames depicted by them. The said, it's much easier to understand in this way than it was presented to them at high school. After some thinking I came to the conclusion that the way I treat the kinematics in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

is the shortest and simplest way without too high-level math.

In Minkowski geometry there's no norm induced by the fundamental form (Minkowski product), and that makes all the difference. There are also no angles in the Eulidean sense (though there are rapidities which are similar but not the same).

I think, the right didactics is a good mixture of algebraic/analytical and geometric concepts. The geometry of spacetime, as used in physics, is analytic geometry anyway!

 

[Dale]

My experience is that it is hard for students (including myself when I was a student) to forget about the Euclidean heuristics built from long training in Euclidean (plane) geometry when it comes to read a Minkowski diagram. My students told me that it was important for them that I emphasized the important difference between Euclidean and Minkowski and that instead of circles you need time- and space-like hyperbolae to determine the "unit tick marks" on the axes of different inertial reference frames depicted by them. The said, it's much easier to understand in this way than it was presented to them at high school. After some thinking I came to the conclusion that the way I treat the kinematics in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

is the shortest and simplest way without too high-level math.

In Minkowski geometry there's no norm induced by the fundamental form (Minkowski product), and that makes all the difference. There are also no angles in the Eulidean sense (though there are rapidities which are similar but not the same).

I think, the right didactics is a good mixture of algebraic/analytical and geometric concepts. The geometry of spacetime, as used in physics, is analytic geometry anyway!

This is a much more persuasive argument. Instead of just complaining about not meeting a particular definition (which we already know and don’t find objectionable) you have provided personal experience and expertise (which we didn’t know and which explains your position).

When discussing semantics, simply quoting a definition is not persuasive precisely because the best definition is in dispute. Your explanation here is far better because it explains why you prefer the non-generalized definition.

 

[Sagittarius A-Star]

It's the first time in this thread that I've seen that somebody defines $\| \cdot \|$ as a "norm" based on an indefinite fundamental form.

Do you mean the https://courses.maths.ox.ac.uk/node/view_material/44180 #119? If, yes, then I don't understand, how you come to "indefinite fundamental form".

In the paper, they define a pseudo-norm only for the limited scope of positive $A \cdot A$.

Proposition 8 (Minkowski triangle inequality). If $U$ and $V$ are future-pointing, timelike four-vectors ...
...
The pseudo-norm of a future-pointing timelike vector is ...

It was only me, who tried in #120 to extend the scope, based on Wikipedia, which describes an extended scope.

I just found another paper. There, the "norm" is not defined as the square-root. I think, "norm squared" would be a more logical name for that:

The norm of a vector is defined to be inner product of the vector with itself; unlike in Euclidean space, this number is not positive definite
...
(A vector can have zero norm without being the zero vector.)

Source (at end of page 15):
https://preposterousuniverse.com/wp-content/uploads/grnotes-one.pdf