# Insights Mathematical Quantum Field Theory - Lagrangians - Comments

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1. Nov 14, 2017

### Urs Schreiber

2. Nov 15, 2017

### strangerep

In Example 5.4,
I would have thought this is because of the 2 derivatives of $\phi$ by the coordinate $x^\mu$. Normally the metric $\eta_{\mu\nu}$ would be dimensionless.

And btw, what are your dimensions of $\phi$? I'm a bit confused about how you're apparently dispensing with physical units here. In ordinary Lagrangian mechanics, we'd have $$S ~=~ \int L \,dt$$where $S$ has dimensions of action ($ML^2/T$), hence $L$ must have dimensions $ML^2/T^2$, i.e., energy. If we express $L$ in terms of a density ${\mathcal L}$ via $$L ~=~ \int {\mathcal L} \; d^3x$$then ${\mathcal L}$ must have dimensions $M/LT^2$.

But perhaps you mean to make $S$ dimensionless by replacing it by $S/\hbar$ ?

3. Nov 15, 2017

### Urs Schreiber

Right, these are two different ways to see the dimension. These two ways lead to equivalent dimensions over Minkowski spacetime, but only the other one, which i use in the series, generalizes well to curved spacetime: On curved spacetime it makes no sense to "rescale coordinates", but it always makes sense to rescale the metric. Even though I don't discuss this generalization in the series, my aim is to present everything in a form such that this generalization will be straightforward.

The alternative perspective that you have in mind I have spelled out on the nLab here

4. Nov 15, 2017

### strangerep

Pardon my ignorance, but,... why not? I would have thought one could at least perform a position-dependent rescaling of the (co-)tangent space at a point, hence rescaling $dx^\mu$ and $\partial/\partial x^\mu$.

5. Nov 15, 2017

### strangerep

Re: "shell": I'm glad you're using this term in a unified way in both the classical and quantum cases. I once used the term "on-shell" in a classical context to mean "in the space of trajectories", but was scolded for doing so.

6. Nov 15, 2017

### strangerep

Prop 5.8, 1st para: Is ${\mathbf {LL}}$ a typo ?

Also, in defn 5.18: typo: mathfb

Last edited: Nov 15, 2017
7. Nov 15, 2017

### strangerep

In the examples from 5.20 onwards: it wouldn't hurt to illustrate the pre-symplectic current explicitly in each case. (I must say I had great difficulty getting my head around $\Theta_{BFV}$ in the abstract.)

8. Nov 16, 2017

### Urs Schreiber

If you carry that thought to the end, you arrive at th rescaling of the metric: Because what does it mean to "rescale the (co)tangent bundle"? This can only mean to 1) have a scale on it and then 2) change that. But a "scale" on the (co)tangent bundle, that's precisely a (pseudo-)Riemannian metric.

9. Nov 16, 2017

### Urs Schreiber

That's spelled out earlier, in examples 5.11 and 5.13!

10. Nov 16, 2017

### Urs Schreiber

Thanks! Fixed now.

11. Nov 16, 2017

### Urs Schreiber

Hm, saying "on-shell" for "when the equations of motion hold" is widely adopted standard. You should feel relaxed about saying it!:-)

12. Nov 16, 2017

### vanhees71

Usually we use natural units in HEP physics. Then $\hbar=c=1$. So actions are dimensionless. In (1+3)-Minkowski space (physical case) thus the Lagrangian (to be precise the Lagrangian density) must be of dimension $\text{Length}^{-4}$. Of course the fundamental form $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ (or the opposite sign, depending on your convention) is dimensionless. The kinetic term for a KG field is
$$\mathcal{L}=(\partial_{\mu} \phi)^*(\partial^{\mu} \phi).$$
Each derivative has dimension $1/\text{Length}$. Consequently the KG field must have also dimension $1/\text{length}$ such that $\mathcal{L}$ has dimension $1/\text{Length}^4$, such that the action is dimensionless.

In natural units energies, momenta, and masses have of course the dimension $1/\text{Length}$ or the other way around length has dimension $1/\text{Energy}$. Usually one counts in terms of energ-momenta-mass dimensions. Then the counting rules for the superficial degree of divergence in renormalization theory becomes more natural, and you can simply state that a Lagrangian leads to a superficially renormalizable theory (in (1+3) spacetime dimensions) if the energy-mopmetnum-mass dimensions of the coefficients in the lagrangian (derivatives, masses, and coupling constants) are not negative.

13. Nov 16, 2017

### Urs Schreiber

A detailed study of QFT with respect to "physical scaling" of the metric tensor is in https://arxiv.org/abs/1710.01937, following https://arxiv.org/abs/1411.1302. This is well adapted to QFT on curved spacetimes. Maybe I find time to come back to this later.

14. Nov 17, 2017

### strangerep

I did read those examples, of course, but failed to get a grip on what the presymplectic current means physically. I've been able to (more-or-less) follow your other stuff, since I could relate it to ordinary classical mechanics and pedestrian Lagrangian field theory. But the presymplectic current didn't ring a bell with anything I'd learned before. Probably I'm just missing something important.

Last edited: Nov 17, 2017
15. Nov 17, 2017

### strangerep

I don't think "dimensionless" is the right word here. Just because you give something a value of 1 doesn't cause it to lose its physical type. E.g., if both $\hbar$ and $c$ were dimensionless, I should be able to add them together. But that could never make sense because they're different types. You can't sensibly add a velocity to an action.

16. Nov 17, 2017

### strangerep

Actually, when I carry that thought further, I think of $$ds^2 ~=~ g_{\mu\nu} dx^\mu dx^\nu ~,$$and I want $ds$ to have the same dimensions as $dx$. Also, if I re-scale $dx^\mu \to k dx^\mu$ I want that to induce $ds \to kds$. That seems to force $g$ to be dimensionless.

OTOH,... I suppose that if one wanted $ds$ to be invariant under $dx^\mu \to k dx^\mu$, then $g$ must scale (inversely) also.

17. Nov 17, 2017

### vanhees71

I don't know, how you call a quantity which is just a number like, e.g., the Sommerfeld fine-structure constant $\alpha=\frac{e^2}{4 \pi} \simeq 1/137$ (where $e$ is again a pure number in the Heaviside-Lorentz units with $\hbar=c=1$), which conventionally is defined as ($\alpha=\frac{e^2}{4 \pi \epsilon_0 \hbar c}$ in SI units) or radians for angles. Of course, it doesn't make physical sense to add $\hbar$ and $c$. Also in conventional units you have examples for this: E.g., it doesn't make sense to add an energy (dimension Force times Distance) to a torque (also dimension Force times Distance).

The usual and most convenient convention in HEP is to set $\hbar=c=1$. These are just conversion constants depending on the used system of units. It is expected that next year the entire SI will be redefined by giving these constants a fixed value (as is already the case for the speed of light, coupling the definition of the metre to that of the second, which is defined by a hyperfine transition of Cs).

The only thing you need to know to convert from these natural units of the theoreticians to SI units is that $\hbar c \simeq 0.197 \text{GeV}\,\text{fm}$. In the natural system you have the choice of only one unit left. Usually one works with GeV for masses, energies, momenta (in the conventional system multiplied with the appropriate powers of $c$) and fm for times (conventionally $\text{fm}/c$) and distances. This, together with the value of $\hbar c$, given in these units, is all you need in HEP, and the natural units make dimensional analysis, as exemplified in the previous posting, easier.

18. Nov 17, 2017

### Urs Schreiber

Oh, I see. i should maybe add some outlook explanation in that section. The role of the presymplectic current will be elucidated in the chapter "Phase Space": it is the jet bundle avatar of the symplectic form on the phase space. In other words, it is the current whose conserved charge with respect to some spacelike Cauchy surface is the symplectic form for the phase space corresponding to that Cauchy surface.

You can see this well in the example of the scalar field: if you write "q" for the field "phi" and "p" for the field derivative, then the presymplectic current has the form $\delta p \wedge \delta q \wedge volume$. After transgression this becomes the familiar $d P \wedge d Q$ on the phase space of the scalar field.

I don't have time today to add an outlook remark on this in the present chapter. But please remind me to do so.

19. Nov 17, 2017

### strangerep

I call $\alpha$ dimensionless because, in full gory detail, it is: $$\alpha ~:=~ \frac{1}{4\pi\varepsilon_0} \, \frac{e^2}{\hbar c} ~.$$

20. Nov 17, 2017

### vanhees71

Yes, so what's wrong with what I wrote before? In natural units velocities are measured with dimensionless numbers, because you set $c=1$. In the SI it's just $\beta=v/c$ (with $v$ and $c$ measured in metre over second, m/s).