Genady

\[\phi(x) \rightarrow \phi(x+\xi)=\phi(x)+\xi^{\nu} \partial_{\nu} \phi(x) + ...\]

\[\frac {\delta \phi} {\delta \xi^{nu}} = \partial_{\nu} \phi\]

\[\frac {\delta \mathcal L} {\delta \xi^{nu}} = \partial_{\nu} \mathcal L\]

\[\frac {\delta \mathcal L[\phi, \partial_{mu} \phi]} {\partial \xi^{\nu}}=\partial_{mu} (\frac {\partial \mathcal L}{\partial (\partial_{mu} \phi)} \frac {\delta \phi} {\delta \xi^{nu}})\]

Thank you. All's well. 

14 minutes ago, joigus said:

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

Yes, I like the book otherwise, but it would be so much easier to follow if the indices were where they should be.

31 minutes ago, joigus said:

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:

 

UμVμ=UμδμνVν=UμδμνVν=UμgμνVν=UμgμνVν

 

Please, I really, really know this. I know this index gymnastics, lowering and raising indices, tensors vs. basis representations, etc. I appreciate your time, but there is no need to teach basics here. Let's focus.

Back to my question. 

Let's take \(\nu=1\).

If \(\partial_{\nu} \mathcal L = \partial_{\mu} (g_{\mu \nu} \mathcal L)\),

then \(\partial_1 \mathcal L = \partial_{\mu} (g_{\mu 1} \mathcal L) = -\partial_1 \mathcal L \).

Where is my mistake?

Just now, 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. ∂νL=∂μ(gμνL) is not a tensor equation. ∂νL=∂μ(δμνL) is.

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

 

They are different: \[\delta=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\] \[g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}\]

1 hour ago, joigus said:

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:

  Reveal hidden contents

 

xμ=(Λ−1)μβx^β

 

 

∂∂x^μ=∂xα∂x^μ∂∂xα=(Λ−1)αμ∂∂xα

 

 

I understand this but I don't think it answers my question. This is what I mean:

I can rewrite (3.34) so: \[ \partial_{\mu} (\sum_n \frac {\partial \mathcal L} {\partial (\partial_{\mu} \phi_n)} \partial_{\nu} \phi_n) = \partial_{\mu} (g_{\mu \nu} \mathcal L)\]

Then, from this and (3.33), we get \[\partial_{\nu} \mathcal L = \partial_{\mu} (g_{\mu \nu} \mathcal L)\]

I think, it is incorrect.

It rather should be \[\partial_{\nu} \mathcal L = \partial_{\mu} (\delta^{\mu}_{\nu} \mathcal L)\]

P.S. Ignore positions of indices; Schwartz does not follow upper/lower standard. The difference is between \(g\) and \(\delta\).

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 \(\delta^{\mu}_{\nu} \mathcal L\). Why is it rather gμνL ? Typo?

(In this text, gμν=ημν )

3 hours ago, Sensei said:

I wonder, homework at your age and skills.. A new reincarnation has taken over your account..

 

It is not technically homework, but it could've been if I were technically student. Just a new textbook to work on.

I don't anymore read books that don't have equations. 🙃

Just to answer the OP question,

4 hours ago, Genady said:

Wouldn't the result be the same without it?

It would not.

Without the step function it would be \[\int dk^0 \delta (k^2-m^2) =\frac 1 {\omega_k} \] rather than \(\frac 1 {2 \omega_k}\).

36 minutes ago, KJW said:

One thing that is not generally recognised is when you see an image representing a black hole, the outer edge of the black disc is not the event horizon. In fact, it is the photon sphere at r=3rs2 , where rs is the Schwarzschild radius (the radius of the event horizon). Note that the photon sphere is the minimum radius for light to escape transversely away from a black hole.

 

 

A light that is produced by hot infalling matter between the photon sphere and the event horizon can still escape radially, right?

33 minutes ago, joigus said:

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,

 

δ(f(x))=∑xk∈zeroes of ff(x−xk)|f′(xk)|

 

for any continuous variable x and "any" well-behaved function f of such variable. Taking as your corresponding function and variable both k0 and,

 

f(k0)=(k0)2−(ωk)2

 

you get,

  Hide contents

 

δ(f(k0))=f(k0−ωk)2ωk+f(k0+ωk)2ωk

 

And that's why you need the step function: to kill the un-physical k0 '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.

 

Thank you. I got it. My mistake was that when I replaced \(k^0\) with \(\omega_k\) I've missed that it can be + or - \(\omega_k\). The step function is needed to kill one of them.

The question:

Show that \[\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0)=\frac 1 {2 \omega_k}\] where \(\theta(x)\) is the unit step function and \(\omega_k \equiv \sqrt {\vec k^2 +m^2}\).

My solution:

\(k^2={k^0}^2 - \vec k ^2\)

\(\omega _k ^2 = \vec k^2 +m^2\)

\(k^2 - m^2 = {k^0}^2 - \omega_k^2\)

\(dk^0= \frac {d{k^0}^2} {2k^0}\)

\(\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0) = \int_{-\infty}^{\infty} \frac {d{k^0}^2} {2k^0} \delta ({k^0}^2 - \omega_k^2) \theta (k^0) = \frac 1 {2 \omega_k} \theta (\omega_k) = \frac 1 {2 \omega_k}\)

However, the point of the unit step function there is unclear to me. Wouldn't the result be the same without it?

5 hours ago, swansont said:

Part of the issue here is the general lack of math you have in pop-sci descriptions. (IIRC Hawking noted the adage that your audience drops in half for each equation in a book; he only had E=mc^2) which forces one into imprecise language.

 

Take “Hawking particles are not created in the near vicinity of the horizon, but instead come from a region surrounding the black hole with a few times the black hole’s radius”

What does “near the vicinity” actually mean? It’s not quantified. A 10 solar mass BH has a Schwarzschild radius of under 30 km. Are Hawking particles created 50 or 100 km away in the vicinity, or not? 100 km is quite a small distance in astronomical/cosmological terms, and these numbers would be small even for a 10x or 100x bigger BH. We get reassured that some rock passing within 50,000 km or so of earth will exhibit a near-miss. That’s almost 10x the radius, and it’s considered nearby.

 

I agree, the vagueness of such statements is an issue. Next time somebody says, "Hawking radiation is generated just outside the event horizon", I will ask first, how far is "just".

Yes, for an audience that thinks that Event Horizon is an actual physical structure, there is no difference between an approximation and a myth.

I've arrived to an expected answer, but I am not sure at all that the process was what the problem statement wants.

First, I considered \(0=(t+\delta t)^2-(x+vt)^2-(t^2-x^2) \approx 2t \delta t - 2xvt - v^2t^2\). Ignoring \(O(v^2)\) gives \(\delta t=vx\), i.e., \(t \rightarrow t+vx\).

Keeping \(O(v^2)\) gives \(t \rightarrow t+vx+\frac 1 2 v^2t\), which is the correct expansion of the full transformation to the second order.

Now, taking \(x \rightarrow x+ \delta x, t \rightarrow t+vx\) gives by the similar calculation \(x \rightarrow x+vt+\frac 1 2 v^2x\).

Is it what the exercise means?

No, I don't call approximations, myths. They are different things. Is it an approximation to say that all life forms were created at once from scratch?

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?

24 minutes ago, TheVat said:

We just scratched the surface in my biology courses.

We too, just touched it.

58 minutes ago, TheVat said:

If you are male, you also got a little more nuclear DNA from mom anyway, since the Y chromosome is much shorter than the X.

OTOH, if you are male, you have X and Y chromosomes, while if you are female, you have only one X, since the other X gets inactivated (see Barr body).

The other myth is, for example, that the Hawking radiation is result of virtual pair production when one of the virtual particles becomes real.

I've found this article that tries to straighten some misconceptions:

Sabine Hossenfelder: Backreaction: Hawking radiation is not produced at the black hole horizon.

Quote

All this supports the conclusion that Hawking particles are not created in the near vicinity of the horizon, but instead come from a region surrounding the black hole with a few times the black hole’s radius.

On 2/19/2024 at 10:32 PM, Genady said:

Well, the instantaneous change of acceleration, in the case of circular motion, happens to be also "along the direction of motion, perpendicular to the instantaneous acceleration". However, I don't know how they (i.e., the time derivative of acceleration and the radiation) are related and don't claim anything in this regard. My point in this post was that AFAIK, a constant acceleration of charged particle can be insufficient to cause radiation.

After a further contemplation. It appears that I was wrong anyway.

16 minutes ago, joigus said:

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,

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.

Yes, I certainly agree with this.

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. 

4 minutes ago, joigus said:

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.

I'd refer to the following analogy.

QFT is like GR when SM is like LCDM.

GR can fit many different cosmological models, and the open questions are about actual contents and history of the universe, which are the aspects I refer to as being "in addition" to the GR framework. 

1 hour ago, joigus said:

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. 

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.

On 2/19/2024 at 6:43 AM, joigus said:

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.

Are these QFT or rather SM issues?