# A Scale in mathematics vs in quantum physics

1. Oct 27, 2016

### RockyMarciano

What is the difference between the concept of scale in the SM QFTs and the scale concept in geometry?
In other words, QFT is a scale-dependent theory that is living on a scale-invariant space(either R4 or M4, related by Wick rotations). Mathematically scale-invariance and scale-dependence appear as incompatible, are they congenial in QFT? and if so how is that achieved?

2. Oct 27, 2016

### vanhees71

I don't know how to answer you concrete question, because I've no clue what you mean by "difference of concept" here. A scale is a scale in geometry, and spacetime is a special kind of (in this case pseudo-Euclidean) geometry.

The reason why QFTs are very unlikely to be scale invariant is that scale invariance usually is anomalously broken, i.e., while the classical field theory may be scale invariant (e.g., pure Yang-Mills theory or QCD in the massless-quark limit), the quantized theory isn't.

3. Oct 27, 2016

### Demystifier

Realistic QFT's are not scale invariant. Some toy QFT's are scale invariant, and they are used as an approximation to real physics because this symmetry simplifies calculations.

4. Oct 27, 2016

### RockyMarciano

Right, so my question is how does the scale dependence of the QFTs that sustain the Standard model fit mathematically in a scale invariant spacetime?

5. Oct 27, 2016

### vanhees71

Spacetime is not scale invariant to begin with, because in order to measure space and time intervals you need a unit of time (or space). In the SI units time is defined by a hyperfine transition in Cs, and that sets the scale "second" and defines also a length scale "light second" (in natural units you measure space and time intervals in seconds).

6. Oct 27, 2016

### RockyMarciano

That's what I wrote, and it follows naturally to wonder about the insertion of such theories in a scale-invariant Minkowski specetime. I'm aware that there are important difficulties in making certain physics theories mathematically sound, I'm just trying to corner some apparently evident from the mathematical point of view possible sources of those difficulties.

7. Oct 27, 2016

### RockyMarciano

When I say spacetime I'm referring to Minkowski spacetime or R4, I'm only referring to the mathematical definition of the mathematical spaces used in definitions of spacetime in QFT, sorry if that was a confusing shorthand.

8. Oct 27, 2016

### Demystifier

Usually the scale dependence in QFT is due to a mass term in the action. The mass does not depend on the spacetime metric, so the insertion is kind of trivial.

9. Oct 27, 2016

### RockyMarciano

I can see that its appearance as a term in the action doesn't depend on the spacetime but it seems reasonable that it must be compatible with it if we are going to claim that the local gauge and the global symmetries that belong to that metric are perfectly congenial.

10. Oct 27, 2016

### Demystifier

Well, the mass term is compatible in the sense that it transforms as a spacetime scalar.

11. Oct 27, 2016

### RockyMarciano

Yes, but you surely are aware that scale in QFT involves also the energy scale and also vector bosons.

12. Oct 27, 2016

### Demystifier

How are vector bosons relevant to the scale?

13. Oct 27, 2016

### RockyMarciano

It is the interaction that introduces the scale in the 4-dimensional QFTs, no?.

14. Oct 27, 2016

### vanhees71

The point is that even in theories without any dimensionful quantity as pure Yang Mills or massless QED/QCD, the scale invariance of the classical theory is broken when quantizing the theory. You can understand this formally in two ways:

(a) perturbatively: Evaluating quantum corrections, i.e., loop diagrams, you encounter divergences and thus have to renormalize the theory. E.g., in massless QED you have to renormalize the fermion self energy, which is linearly divergent. Now you cannot subtract the divergency at zero four-momentum, because there is an essential singularity in the self-energy function, because the threshold is at 0, because both the photon and the fermion are massless. So you have to subtract at a space-like four-momentum which introduces the renormalization scale $\Lambda$, and your running coupling and wave-function normalization becomes scale dependent, and scale-invariance is broken. Also the fermion doesn't stay massless, but you generate a mass dynamically, which also solves the problem with IR and collinear divergences (there's a famous paper by Eric Weinberg and Coleman on this, getting famous under the buzzword "mass without mass"). It's clear that through the introduction of a scale the scale invariance is broken.

(c) In the path integral the scale invariance transformation leads to a non-trivial change in the path-integral measure and thus the quantum action doesn't fulfill the naive Ward-Takahashi identity of the classical theory (which is the tracelessness of the energy-momentum tensor). Thus the symmetry is broken also in this "non-perturbative" sense. That's known as "trace anomaly".

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.44.1733
http://journals.aps.org/prd/abstract/10.1103/PhysRevD.23.2262

15. Oct 27, 2016

### RockyMarciano

Thanks for the informative post with wich I obviously agree. It helps understanding how the scale dependence arises in renormalized QFTs.
My question was seeking a different thing though. As I said I'm interested in spelling out what's mainly behind the difficulties to make the renormalizable QFTs mathematically sound.
And one candidate that is obvious to me (I am trying to find out if for anyone else here) is the fact that when trying to reconcile the global symmetries of the mathematical space our theory lives in with the structure of the local gauge renormalizable QFTs that produce accurate predicitions, one finds this contrast between the scale invariance of the former and the scale dependence of the latter, irrespective of how this dependency comes about.

16. Oct 27, 2016

### Staff: Mentor

Under what definition of "scale invariant" are R4 and M4 (by which I assume you mean $R^4$ with a Euclidean and Minkowski metric, respectively) scale invariant? Any metric space has a length scale and so is not "scale invariant" under the obvious interpretation of that term.

17. Oct 28, 2016

### RockyMarciano

One precision first, R4 and M4 were meant to refer to affine Euclidean and Minkowskian spaces in 4 dimensions respectivley. But anyway the Euclidean geometry as a metric space is too scale-invariant. Also Minkowski space is not a metric space according to the mathematical definition of metric space, wich is related but independent of whether it has a metric tensor or not, it is better not to conflate the two,
The concept of scaling transformation or dila(ta)tion and its correspondent symmetry leading to the concept of similarity in euclidean geometry is such a basic one that I didn't think it needed further explanation in an thread labeled advanced. Look up https://en.wikipedia.org/wiki/Similarity_(geometry) or https://en.wikipedia.org/wiki/Scaling_(geometry) or https://en.wikipedia.org/wiki/Dilation_(metric_space).
It is directly related to the notion of unitarity in the case we are discussing here, QFT, in the sense that without it you don't have the possibility of normalizing vectors in the presence of a gauge connection.
By the way, what specific length scale were you referring to? AFAIK neither Euclidean nor Minkowski geometries have an absolute scale length.

Last edited: Oct 28, 2016
18. Oct 28, 2016

### Staff: Mentor

Again, under what definition of "scale invariant" is this true? Under the definition that's most obvious to me, that distances should be invariant under scale transformations (dilations), it is obviously false. The Wikipedia link you gave to the article on Dilations gives the definition of a dilation transformation which supports what I just said.

Yes, that's correct if we insist that a metric be positive definite; under this terminology Minkowski space is only a pseudometric space. But I don't see what this has to do with scale invariance; you can still define dilation transformations in a pseudometric space.

The length scale defined by the metric; given a metric, there is an invariant length between any two points in the space. A dilation transformation changes this length.

19. Oct 28, 2016

### RockyMarciano

What is false? You just wrote the definition everyone uses:"that distances should be invariant under scale transformations", this is scale-invariance and of course the definition I'm using. I honestly don't know what you are up to. But I would respectfully suggest for you to discuss these very basic geometric questions in a different thread as to not get very offtopic here and not distract from my OP question. Perhaps this migh help you too: http://math.stackexchange.com/questions/23129/why-is-euclidean-geometry-scale-invariant

And here you seem to be mixing up the physical notion of dimensionful length referred to some physical standard and the mathematical concept of distance that is a dimensionless quantity in the physical sense.

20. Oct 28, 2016

### Staff: Mentor

I assume that by "scale transformations" you mean dilations? If so, then Euclidean N-space is obviously not scale invariant. A dilation transformation is a map from Euclidean N-space to itself such that the distance $d(x, y)$ between two points gets mapped to $r d(x, y)$, where $r$ is some nonzero real number. Any such transformation where $r \neq 1$ obviously changes distances, and therefore Euclidean N-space is not scale invariant by your own definition.

It seems to me that the definition of "scale invariant" that you are using, which is what I am trying to clarify, is crucial for understanding your question. If you mean "invariant under dilations", then that's fine, we can discuss what quantities in a QFT are invariant under dilations. But if that's the definition you are using, then your claim that Euclidean space is scale invariant is incorrect; so I'm trying to figure out whether you are just mistaken about that, or whether I'm not correctly understanding the definition of "scale invariant" that you are using.

This only shows that some geometric quantities in Euclidean geometry are scale invariant. It does not show that all geometric quantities in Euclidean geometry are scale invariant, which is what you are claiming.

I haven't talked about "physical lengths" at all. The metric is a mathematical function; it takes two points and gives you a number. "Scale invariant" means that that number should not change under scale transformations. But dilation transformations with $r \neq 1$, as above, change the number that you get when you plug in the same two points.