joigus

Ok. Yes, @RobertSmart is right. There is a little mistake in the constants. Let me display my calculation in detail, because his Latex seems to have been messed up by the compiling engine or whatever and I seem to find a small discrepancy with him.

Your Lagrangian,

\[ \mathscr{\mathcal{L}}=-\frac{1}{2}\phi\Box\phi+\frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4} \]

I prefer to write with an index notation, which is more convenient for variational derivatives:

\[ \mathscr{\mathcal{L}}=-\frac{1}{2}\phi\left.\phi^{,\mu}\right._{,\mu}+\frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4} \]

As we have no dependence on first order derivatives,

\[ \frac{\partial\mathscr{\mathcal{L}}}{\partial\phi_{,\mu}}=0 \]

we get as the only Euler-Lagrange equation,

\[ \frac{\partial\mathscr{\mathcal{L}}}{\partial\phi}=-\frac{1}{2}\left.\phi^{,\mu}\right._{,\mu}+m^{2}\phi-\frac{\lambda}{3!}\phi^{3}=0 \]

Or,

\[ -\frac{1}{2}\Box\phi+m^{2}\phi-\frac{\lambda}{3!}\phi^{3}=0 \]

Or a bit more streamlined,

\[ \Box\phi-2m^{2}\phi+\frac{\lambda}{3}\phi^{3}=0 \]

Sorry I didn't get around to it sooner.

Paraphrasing Sir Humphrey Appleby:

Is that finally final?

I hope so.

PS: BTW, this is a simplified symmetry-breaking Lagrangian. The real thing in the SM is a complex SU(2)-symmetric multiplet \( \left(\phi_{1},\phi_{2},\phi_{3},\phi_{4}\right) \).

8 hours ago, genio said:

Why do you equate religion with god(s)? OP specifically said religion.

OP also specifically formulated it in terms of god/gods and supernatural beings or agencies:

On 10/1/2020 at 10:05 AM, Mnemonic said:

According to the bible Jesus Christ was a supernatural character who could walk on water, occasionally talked to Satan, and could turn water into wine, amongst many other marvels.

Can you be a scientist and still believe in this stuff?

12 hours ago, Genady said:

This is a multi-step exercise. It would be very helpful if somebody could check my step(s) as I go. @joigus, I'm sure it is a child play for you.

I'd like to make sure that I've derived correctly the equation of motion for this Lagrangian:

 

L=−12ϕ□ϕ+12m2ϕ2−λ4!ϕ4

 

The EL equation:

 

∂L∂ϕ+□∂L∂(□ϕ)=0

 

The equation of motion:

 

□ϕ−12m2+λ3!ϕ3=0

 

 

How is it?

P.S. As edit LaTex does not work, I add a typo correction here. The equation of motion is rather

 

□ϕ−m2ϕ+λ3!ϕ3=0

 

Your eq. of motion (once corrected) looks fine. Most people prefer to write (1/2)(grad)phi(grad)phi instead of -(1/2)phi(grad)²phi, but they differ in just a total divergence, so they are equivalent (lead to the same equations of motion).

By grad² I mean the D'Alembert operator. I'll check in more detail later, I would have to do it in my head now and I could miss a sign. This is a famous equation.

1 hour ago, Genady said:

I think, I got it.

The symmetry validates the equation (3), because this equation makes the variation of Lagrangian a total derivative, and this makes the variation of action vanish:

 

IOW, without the symmetry, we can't go from (4) to (5).

Exactly! In fact, that's how you define a symmetry in the action, as a total divergence does not change the "surface" (hypersurface) terms t=t₁ and t=t₂.

You must apply Stoke's theorem first. Perhaps you did, but I didn't have time to check.

It is amazing that you're doing this stuff at this point in your life.

8 hours ago, KJW said:

This is an appeal to your own authority.

No, it was a joke. I thought the emoji had given it away. I'm sorry it didn't, and you took it seriously.

Value judgement is falacious. Like "your analogy is false", "I think you're wrong". That's what I was mocking.

8 hours ago, KJW said:

I think that means you lost the argument.

Funny (and may I say telling), that you consider it a contest. It's not. Nor should it be.

You are just wrong, or you sound to me very much like you are in what seems to be your interpretation of tensors. Take this example: flat \( \mathbb{R}^{2} \) (the plane).

All of this should be self-explanatory.

\[ ds^{2}=dx^{i}g_{ij}dx^{j}=\left(\begin{array}{cc} dr & d\theta\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & r^{2} \end{array}\right)\left(\begin{array}{c} dr\\ d\theta \end{array}\right)=dr^{2}+r^{2}d\theta^{2} \]

\[ ds^{2}=dx_{i}g^{ij}dx_{j}=\left(\begin{array}{cc} dr & r^{2}d\theta\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & r^{-2} \end{array}\right)\left(\begin{array}{c} dr\\ r^{2}d\theta \end{array}\right)=dr^{2}+r^{2}d\theta^{2} \]

\[ ds^{2}=dx_{i}\left.\delta^{i}\right._{j}dx^{j}=\left(\begin{array}{cc} dr & r^{2}d\theta\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)\left(\begin{array}{c} dr\\ d\theta \end{array}\right)=dr^{2}+r^{2}d\theta^{2} \]

In fact, \( g_{ij} \) and \( g^{ij} \) will give you more than you bargained for: They will give you a spurious sigularity that isn't there at all. There's nothing wrong at \( r=0 \). The covariant methods tell you that.

But the form of the once-covariant once-contravariant components of the same silly little thing give you a clue, as \( \left.\delta^{i}\right._{j} \) tells you clearly that nothing funny is going on at that point. The Kronecker delta, in this case, is more honest-to-goodness than the other ones.

If you calculate the Riemann, of course, it will tell you that beyond any doubt. It's at that level that you can talk about anoholonomy and curvature.

I'm sure you know all that from what we've talked before. These are cautionary tales that are in the literature.

Another possibility is that we didn't understand each other's point. One can never be sure.

And I apologise to @Genady. For this was his thread on variational derivatives really and wasn't about curvature at all. 🤷‍♂️

7 minutes ago, KJW said:

This is a false analogy.

No, yours is. Wanna know of a better analogy? Mine. Mine is better.

It's been 50 years since I last did this.  

39 minutes ago, KJW said:

Of course.

That's all I wanted to hear, really. If you want to discuss GR, I suggest you open a new thread doing so. Discussing topological aspects of GR on a thread about index gymnastics in flat QFT would be highly misleading.

39 minutes ago, KJW said:

The metric tensor does encode curvature in the sense that one can mathematically obtain the Riemann tensor from it.

One can obtain Shakespeare's Sonnets from the alphabet, but I see no Shakespeare in ABCDEFGHIJKLMNOPQRSTUVWXYZ. Do you?

It's what you do with the alphabet that matters. I can do no nothing like what Shakespeare did. Same happens with the metric.

Yours is actually a common misconception. I'm just saying.

5 hours ago, KJW said:

But they don't encode the same information. The metric tensor encodes the spacetime curvature whereas the Kronecker delta does not. But the inverse of the metric tensor does encode the same information as the metric tensor.

 

What curvature? This is all pseudo-Euclidean metric we're talking about. This is QFT. \( g_{\mu\nu}=\eta_{\mu\nu} \)

I think you mean in GR. But even in GR, the metric tensor does not encode curvature. The Riemann tensor does. And the metric tensor is covariantly constant. Because it is. That should give us a clue. It doesn't really encode much, does it? This is a common misconception, that the metric tensor components "encode" something. In terms of tetrads it's very clear that it's nothing but the identity operator. What's perplexing is that covariantly shifting with the Christoffels obtained from it around an infinitesimally-small closed lood gives you something else, but that's a different story.

No problem.

By the way, I should have written,

\[ \partial^{\mu}\left(\sum_{n}\frac{\partial\mathcal{L}}{\partial\left(\partial^{\mu}\phi_{n}\right)}\partial_{\nu}\phi_{n}-g_{\mu\nu}\mathcal{L}\right)=0 \]

That is correct.

16 minutes ago, Genady said:

Where is my mistake?

Nowhere. You made no mistake. None whatsoever. You are just realising that Schwartz made a mistake, not you. So it was a typo.

He probably meant to write something like,

\[ \partial^{\mu}\left(\sum_{n}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi_{n}\right)}\partial_{\nu}\phi_{n}-g_{\mu\nu}\mathcal{L}\right)=0 \]

which is correct, and consistent with your derivation.

That's the problem with books that don't follow the covariant/contravariant convention. Index gymnastics does that for you automatically.

Sorry, I thought I'd told you:

1 hour ago, joigus said:

Oh. Got you. Yes, you're right. It should be what you say. Classic books in QFT tend to be rather fast-and-loose with the indices.

As matrices, they are. But as tensors, they aren't. They are one and the same basis-independent object, coding the same physical information. This is exactly the same as a vector in one basis looking, eg, like matrix (0 1 0 0) but looking like (0 -1 0 0) in another basis. You are confusing the tensor with its coordinates.

IOW: All metrics, no matter the dimension and signature, look exactly like the identity matrix (the Kronecker delta) when the scalar product is expressed under the convention that the first factor is written in covariant components, and the second one in contravariant ones.

A tensor is a physical object. A matrix is just a collection of numbers used to represent that object.

Let me put it this way:

\[ U_{\mu}V^{\mu}=U_{\mu}\left.\delta^{\mu}\right._{\nu}V^{\nu}=U^{\mu}\left.\delta_{\mu}\right.^{\nu}V_{\nu}=U^{\mu}g_{\mu\nu}V^{\nu}=U_{\mu}g^{\mu\nu}V_{\nu} \]

Oh. Got you. Yes, you're right. It should be what you say. Classic books in QFT tend to be rather fast-and-loose with the indices. \( \partial_{\nu}\mathcal{L}=\partial_{\mu}\left(g_{\mu\nu}\mathcal{L}\right) \) is not a tensor equation. \( \partial_{\nu}\mathcal{L}=\partial_{\mu}\left(\left.\delta^{\mu}\right._{\nu}\mathcal{L}\right) \) is.

Although I should say there is no fundamental difference between \( g \) and \( \delta \) really. \( \delta \) is just \( g \) (viewed as just another garden-variety tensor) with an index raised (by using itself).

Bogoliubov is similarly cavalier with the indices if I remember correctly. Is it from the 50's?

30 minutes ago, Genady said:

My question is about the following step in a derivation of energy-momentum tensor:

When the ∂νL in (3.33) moves under the ∂μ in (3.34) and gets contracted, I'd expect it to become δμνL . Why is it rather gμνL ? Typo?

(In this text, gμν=ημν )

No, no typo. It is actually a theorem (or lemma, etc) of tensor calculus that the gradient wrt contravariant coordinates is itself covariant. It is just a fortunate notational coincidence that the "sub" position in the derivative symbol seems to suggest that.

Proof:

10 hours ago, Genady said:

Just to answer the OP question,

It would not.

Without the step function it would be

∫dk0δ(k2−m2)=1ωk

rather than 12ωk .

 

It would.

!!!

The 1/2 factor doesn't change Lorentz invariance of the metric measure, but it's quite essential to the formalism that comes later. One would think we're done with negative energies/frequencies, and such. But no. They keep biting our buttocks later with the Fourier transform. That's where the Stueckelberg-Feynman prescription for antiparticles comes in.

You want energies to be positive. As k⁰ (the zeroth component) of the 4-momentum is the energy component, all states must be decreed to have zero amplitude for that choice. That's achieved by the step function trick.

You missed a well-known trick for delta "functions"...

The delta function satisfies,

\[ \delta\left(f\left(x\right)\right)=\sum_{x_{k}\in\textrm{zeroes of }f}\frac{f\left(x-x_{k}\right)}{\left|f'\left(x_{k}\right)\right|} \]

for any continuous variable \( x \) and "any" well-behaved function \( f \) of such variable. Taking as your corresponding function and variable both \( k^{0} \) and,

\[ f\left(k^{0}\right)=\left(k^{0}\right)^{2}-\left(\omega_{\mathbf{k}}\right)^{2} \]

you get,

And that's why you need the step function: to kill the un-physical \( k^0 \)'s. Negative energies do appear again in the expansion of the space of states, but they're dealt with in a different manner. This is just to define the measure for the integrals. All kinds of bad things would happen if we let those frequencies stay.

I'm sure there are better explanations out there. But the delta identity is crucial to see the point.

6 minutes ago, joigus said:

All kinds of bad things would happen if we let those frequencies components stay.

1 hour ago, Genady said:

Checking with the physicists here:

On the p.17 it says,

Shouldn't it say force rather than potential? Isn't any potential rather quadratic close to equilibrium?

Absolutely. It's an erratum. The potential is quadratic, so it's the force that's linear.

Close to equilibrium the Taylor expansion of the potential must be quadratic, as at the equilibrium position, the gradient (the force) must be zero. So the next-to-zero^th-order term for the force is proportional to V''(x₀).

V(x)=V(x₀)+(1/2)V''(x₀)(x-x₀)²+...

Reminds me of this priceless piece of comedy:

Quote

 

“Boy these Conservatives are really something aren’t they? They are all in favor of the unborn! They’ll do anything for the unborn but once you’re born, you’re on your own!

Pro- life conservatives are obsessed with the fetus from conception to nine months. After that they don’t want to know about you! They don’t want to hear from you! No nothing!

No neo-natal care, no day care, no head start, no school lunch, no food stamps, no welfare, no nothing!

If you’re pre-birth you’re fine! if you’re pre-school, you’re fucked!”

-George Carlin

 

Sigh

1 minute ago, KJW said:

that radiation from a charge is a non-local phenomenon

Ahem

On 2/21/2024 at 4:33 PM, Moontanman said:

Why would any of this matter? The number of people who accept something as true has no bearing on the actual truth of it. 

Fifty billion flies...

9 minutes ago, Genady said:

To clarify the analogy in my previous post. I don't mean that QFT is like GR, nor that SM is like LCDM. I mean that SM relates to QFT like LCDM relates to GR. 

Yes, very much so. I'm re-reading what I said as well as your comment that motivated it. I was kinda losing track of what I was trying to say, and thinking 'why the hell did I mention QFT?' And (after re-reading) I see it's because you said,

On 2/17/2024 at 3:12 PM, Genady said:

[...] mass of an affected body was introduced into the equation to make it fit into the force-based model. Then, it disappears when the model is not force-based anymore.

The reason I mentioned QFT (or the SM as a particular case) is because I wanted to point out that sometimes, even though force and mass are not pillars of the theory, you still have to do a lot of work with this mass, so it's very far from disappearing from most considerations. But it's not like the theory is telling you what this parameter actually is or does.

4 hours ago, Genady said:

QFT is a framework that fits models with massive as well as massless neutrinos, with one as well as five generations of particles, with photons as well as phonons, with vacuum as well as solid state, etc. I mean that all the missing questions are specific to the model, i.e., SM, and are in addition to QFT.

SM is but one particular QFT. Are you sure you're not splitting hairs here?

SM cannot be in addition to QFT the same way the statistical mechanics of an Ising magnet wouldn't be in addition to statistical mechanics. It's given by a particular choice of Hamiltonian within the general procedures of quantum statistical mechanics. SM is QFT under a particular choice of Lagrangian, including gauge groups, global gauge groups, and Higgs multiplets. 

Unless I overlooked an essential point you made, which is certainly possible, especially of late.

27 minutes ago, Genady said:

Are these QFT or rather SM issues?

Ultimately it's a major SM issue, I think. But there are very general arguments in QFT in which Yang-Mills pretty much appears as the only interesting generalisation for gauge invariance. So what I mean I suppose is that from QFT to SM there's "just" (ahem) a choice of symmetry groups, generations, and mixing parameters. A very wise expert in QFT nothing fundamentally different from the general principles of QFT conveniently generalised. 

On 2/17/2024 at 3:12 PM, Genady said:

Another way to look at it is that mass of an affected body was introduced into the equation to make it fit into the force-based model. Then, it disappears when the model is not force-based anymore.

That's certainly what happens in GR. In QFT I think the process is much more painful. The theory is not force-based either, but we must start with mass being a parameter that discriminates between different types of fields (massless vs massive). But the physical mass (inertia) becomes more of a dynamical attribute that depends on the state and has to be calculated perturbatively. And there is no explanation for the spectrum of masses.

23 hours ago, joigus said:

have given to Newton's third law as an equation

I should have said Newton's 2nd law, obviously. I have this kind of dyslexic-like glitch that makes me do that very much like @studiot's problem with the typing.

23 hours ago, MigL said:

This is actually a deep, and surprising, result, as it shows inertial mass is equivalent to gravitational mass.
Even A Einstein pondered why this should be so, and may be what Joigus was referring to here

Yes, that's exactly what I meant. Now, if all forces of Nature were like that, I wouldn't find it surprising at all. After all, the word "surprise" has to do with contrast in comparison to previous experience, or inference from that. Electricity is not like that, nor any other interaction. Thereby the word "amazed".

23 hours ago, Genady said:

A gravitational acceleration being independent of a body's mass also follows from the Kepler's third law.

True. In fact Newton used Kepler's third law to guess his inverse-square law. Textbooks generally point out that the power law is implied. But the equivalence principle is too. The thing is, because the mass on the receiving end of the gravitational interaction (not the mass as a source) disappears from all the physics, it is almost inescapable that the distorsion that a source introduces around it can be described in some geometric way as a distorsion of space-time itself.

I think this is amazing even after one learns about GR.

17 hours ago, MigL said:

F=ma simply defines the relationship between two ( or possibly 3 ) variables.

Exactly (my highlight in bold red).

This is at no detriment to the use that @Janus and @Genady here in particular (and most physicists elsewhere as well) have given to Newton's third law as an equation. Any definition, identity, or formula can be postulated as an equation the moment one gives numerical values to any of the terms involved, or values in terms of further parameters. So, for example: sin²a+cos²a=1 is an identity. It says something obvious. A further substitution, eg sin²a=1/2 makes it an equation.