The h Bar

7:55 AM

Trying to think of the Synge trick of defining a family of geodesics between two curves as a foliation, but it is annoying since it depends both on the points of the curves considered and the clocks for each

I guess I could make it depend on a single clock by parallely transporting the original geodesic tangent along one of the curve

but then that would be disregarding the good and valid other clock of the observer

I guess there's a few different ways of foliating such a surface and maybe that yields different useful informations, I don't know

Ooooh apparently the kind of surface used by Synge is a DOUBLY RULED SURFACE

I am wondering if all the terribly vague time derivative rules can be obtained from there

Colourful Spacetime

8:53 AM

I've been trying to evaluate how $\partial_{\mu}\phi$ transforms under *infinitesimal* Lorentz transformations. If $\partial_{\mu}\phi \mapsto (\partial_{\mu}\phi)'=(\Lambda^{-1})^{\nu}_{}_{\mu}\partial_{\nu}\phi((\Lambda^{-1})^{\nu}_{}_{\mu} x^{\mu})$, then for $\Lambda^{\mu}_{}_{\nu}=\delta^{\mu}_{\nu}+\omega^{\mu}_{}_{\nu}$, how does this expression simplify?

One thing that I tried was $$\partial_{\mu}\phi(x^{\mu}) \mapsto \frac{\partial \phi(\bar{x}^{\mu})}{\partial \bar{x}^{\mu}}=\frac{\partial x^{\mu}}{\partial \bar{x}^{\mu}}\frac{\partial \phi(x^{\mu}+\omega^{\mu}_{}_{\nu}x^{\nu})}…

Slereah

none of the latex works!

Colourful Spacetime

Yep, I'll try to fix it

Slereah

I think because you're trying to use superscript twice in a row

that is verbotten

Colourful Spacetime

TEST

$\partial_{\mu}\phi \mapsto (\partial_{\mu}\phi)'=(\Lambda^{-1})^{\nu}_{\ \mu}\partial_{\nu}\phi((\Lambda^{-1})^{\nu}_{\ \mu} x^{\mu})$

Slereah

If you want the secret of doing properly aligned indices

{A_a}^b

${A_a}^b$

Colourful Spacetime

9:09 AM

I've been trying to evaluate how $\partial_{\mu}\phi$ transforms under *infinitesimal* Lorentz transformations. If $\partial_{\mu}\phi \mapsto (\partial_{\mu}\phi)'=(\Lambda^{-1})^{\nu}_{\ \mu}\partial_{\nu}\phi((\Lambda^{-1})^{\nu}_{\ \mu} x^{\mu})$, then for $\Lambda^{\mu}_{\ \nu}=\delta^{\mu}_{\nu}+\omega^{\mu}_{\ \nu}$, how does this expression simplify?

One thing that I tried was $$\partial_{\mu}\phi(x^{\mu}) \mapsto \frac{\partial \phi(\bar{x}^{\mu})}{\partial \bar{x}^{\mu}}=\frac{\partial x^{\mu}}{\partial \bar{x}^{\mu}}\frac{\partial \phi(x^{\mu}+\omega^{\mu}_{\ \nu}x^{\nu})}{\par…

It doesn't recognize $$ \phi $$?

oh it does

ACuriousMind

The fact that there are 5 $\mu$ in your first expression suggests you haven't been careful with your indices :P

Slereah

it is very generous

ACuriousMind

you probably need to look a bit more carefully at how to express the chain rule in index language

in general already $\partial_\mu \phi(x^\mu)$ is questionable notation, since the $\mu$ is not being summed over

Slereah

and it couldn't since the inner one isn't a vector

Colourful Spacetime

Ok let's look at $\partial_{\mu}\phi \mapsto (\partial_{\mu}\phi)'={(\Lambda^{-1})^{\nu}}_{\mu}\partial_{\nu}\phi({(\Lambda^{-1})^{\nu}}_{\ \mu} x^{\mu})$

So

Slereah

9:20 AM

You're still using the same indices for the position and the vector components :p

It may be tricky to do if you do this

Colourful Spacetime

Which one is the "vector component"

Slereah

The derivative of the scalar field is a vector

Colourful Spacetime

ah yes yes ok

let me try again

Slereah

well, a dual vector in this case, but close enough

Colourful Spacetime

Maybe like this?

Considering what you said, should it be written as:

$(\partial_{\mu}\phi)' {(\Lambda^{-1})^{\nu}}_{\mu}\partial_{\nu}\phi({(\Lambda^{-1})^{\kappa}}_{\ \sigma} x^{\sigma})$

oops

Slereah

9:24 AM

Yes

ACuriousMind

$\kappa$ is an unusual choice, but sure

Colourful Spacetime

$(\partial_{\mu}\phi)' ={(\Lambda^{-1})^{\nu}}_{\mu}\partial_{\nu}\phi({(\Lambda^{-1})^{\nu}}_{\ \sigma} x^{\sigma})$

NO

ACuriousMind

no, with the $\kappa$ it was better, now you have three $\nu$s again :P

Colourful Spacetime

$(\partial_{\mu}\phi)'={(\Lambda^{-1})^{\nu}}_{\mu}\partial_{\nu}\phi({(\Lambda^{-1})^{\rho}}_{\ \sigma} x^{\sigma})$

Slereah

also a bit of advice, if you're doing a lot of tensor notation on the chat, I find that it's easier to use $abc...$

Less to write

Colourful Spacetime

9:26 AM

Ok, I will come back in due time when I've corrected my calculations

Slereah

Remember that $$\frac{\partial x^a}{\partial x^b} = \delta^a_b$$

Its likely to come up a lot

Colourful Spacetime

ye ye

Colourful Spacetime

10:01 AM

Hopefully latex will work this time

Ok here is my first problem. We have

$$\partial_a \phi(x^b) \mapsto \frac{\phi(\bar{x}^b)}{\partial \bar{x}^a}=\frac{\partial x^k}{\partial \bar{x}^a} \frac{\partial \phi(\bar{x}^b)}{\partial x^k}.$$

There are two ways to evaluate $\frac{\partial x^k}{\partial \bar{x}^a}$, using $\bar{x}^k=x^k+{\omega^k}_{c}x^c$, we have \\

$$1. \quad x^k=\bar{x}^k-{\omega^k}_cx^c \implies \frac{\partial x^k}{\partial \bar{x}^a}= \delta^k_a$$
$$2. \quad \bar{x}^k-x^k={\omega^k}_c x^c \implies x^c={\omega_k}^c(\bar{x}^k-x^k) \implies x^k= {\omega_c}^k(\bar{x}^c-x^c) \implies \frac{\partial x^k}{\partial …

ACuriousMind

@ColourfulSpacetime You can't invert the infinitesimal $\bar{x} = x + \omega x$ by solving for $x$ (or, well, you can, but it leads to nonsense as you've just demonstrated).

You get the form of $x(\bar x)$ by taking the finite inverted transformation $x = \Lambda^{-1}\bar{x}$ and then passing to its infinitesimal version $x = \bar{x} - \omega \bar{x}$

Colourful Spacetime

10:20 AM

If ${\Lambda^a}_b = \delta^a_b +{\omega^a}_b$ then ${(\Lambda^{-1})^b}_a = \delta^b_a - {(\omega^{-1})^b}_a=\delta^b_a - {\omega_a}^b$?

ACuriousMind

well, not equals, but infinitesimally that's true

no wait

you don't need to invert the $\omega$

Colourful Spacetime

So instead of ${(\omega^{-1})^b}_a$ I should write ${\omega^b}_a$?

ACuriousMind

you have $\Lambda = \mathrm{exp}(\epsilon\omega)$ and so $\Lambda^{-1} = \mathrm{exp}(-\epsilon\omega)$, so to first order you get $\delta + \epsilon\omega$ and $\delta - \epsilon\omega$

@ColourfulSpacetime yes

Colourful Spacetime

11:06 AM

So the correct answer would be $\frac{\partial x^k}{\partial \bar{x}^a}=\frac{\partial(\bar{x}^k-{\omega^k}_c \bar{x}^c)}{\partial \bar{x}^a}=\delta^k_a-{\omega^k}_a$

Colourful Spacetime

11:20 AM

Then, the other problem I have is how to evaluate $\frac{\partial \phi(\bar{x}^b)}{\partial x^k}.$ I thought about doing a taylor expanion on $\phi$: $$\frac{\partial}{\partial x^k}\phi(x^b-{\omega^b}_cx^c)= \frac{\partial}{\partial x^k} (\phi(x)-{\omega^b}_cx^c\partial_b \phi(x))=\partial_k \phi -{\omega^b}_k \partial_b \phi(x) - {\omega^b}_cx^c\partial_k \partial_b \phi(x)$$

ACuriousMind

why do you need to "evaluate" it?

what are you trying to do here, exactly?

Colourful Spacetime

I want to find out how the Lagrangian transforms. My text says that $\delta \mathcal{L}= -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu} \mathcal{L})$, I want to prove this

I started by varying the Lagrangian $\mathcal{L}(\phi,\partial_{\mu} \phi)$, so I want to find out what $\delta(\partial_{\mu} \phi)$ is

Basically I've been trying to answer this question here

0

Q: Problem 6 of Sheet 1 - Quantum field theory David Tong - Variation of Lagrangian density

The Problem reads: Consider the infinitesimal form of the Lorentz transformation derived in the previous question: $x^\mu \rightarrow x^\mu +\omega^{\mu}_\nu x^\nu$. Show that the scalar field transforms as $$\phi(x)\rightarrow \phi'(x)=\phi(x)-\omega^{\mu}_\nu x^\nu \partial_\mu \phi(x)$$ and h...

ACuriousMind

11:38 AM

ah, sure

so what's wrong with your Taylor expansion?

Colourful Spacetime

Is it wrong and you are asking me to find out why it's wrong, or is it right and you are wondering why I would think that it's wrong?

ACuriousMind

the latter - I'm confused what you think the problem is

(but also, note the arguments in that question - neither the asker nor the answerer need to explicitly evaluate $\delta(\partial_\mu \phi)$, so you're making this more difficult for yourself than it has to be :P)

Colourful Spacetime

the two derivatives $\partial_k \partial_b$ seem very awkward to me, because I have never seen them before. Whenever we have two derivatives it's usually of the form $\partial^k \partial_k$

Hmmm... ok I will keep that in mind

ACuriousMind

@ColourfulSpacetime you'll have to be more specific than "awkward" - I don't see any problem with that

Colourful Spacetime

Ok that's fine, I mean logically I don't find a problem, it's just that I haven't seen it before.

Slereah

11:51 AM

By the way I think one thing you may recognize here is that ${\omega^b}_c x^c \partial_b$ is an infinitesimal Lorentz generator, here

you may recognize it from en.wikipedia.org/wiki/…

So it sounds plausible that it would be here

You're doing an infinitesimal Lorentz transform on the vector $d\phi$

You have your original vector at a given point, you move that point from its original place with that rotation, and then you rotate the original vector

One thing you can do in those situations btw is simply to test it on a simple function and see what happens

An interesting thing to note here is what happens at the rotation point, $x = 0$

The rotation leaves that point fixed, but the derivative of the new field still changes because the values around it are moving

and then you are just left with $\partial_k \phi(0) -{\omega^b}_k \partial_b \phi(0)$

So the second term only deals with how the vector is moved, while the third term will deal with how the base point is moved

Colourful Spacetime

12:16 PM

@Slereah Thanks, although I find it a bit difficult to follow. I'll read more about the lorentz generators etc

For the variation of the Lagrangian I have $\delta \mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi}+\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)} \delta(\partial_{\mu} \phi) = \partial_{\mu}(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\delta\phi)$, where the equation of motion has been imposed. But I have no idea how to continue from now on.

Even if I input $\delta \phi$, there is still the $\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi)}$ which doesn't go away

Slereah

Well, if the variation is a total derivative, then the Lagrangian varies, but not the action

So it is indeed a symmetry

The remaining $\delta L$ leads to conserved currents, which you can find using Noether's theorem

Colourful Spacetime

If it's a symmetry should I use Noether's theorem? then $\partial_{\mu} j^{\mu}=0$ might give the answer

Ahh

Does that mean that any Lagrangian $\mathcal{L}=\mathcal{L}(\phi,\partial_{\mu}\phi)$ has Lorentz symmetry?

Slereah

If you're lazy you can check physics.stackexchange.com/questions/12559/…

@ColourfulSpacetime Symmetries are usually defined up to a total derivative for the Lagrangian, so yes

Colourful Spacetime

I don't see why $\partial_{\mu} (\frac{\partial L}{\partial (\partial_{\mu} \phi)}\delta \phi) = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu} L)$

ACuriousMind

@ColourfulSpacetime it's not

that r.h.s. comes from when the Lagrangian is explicitly coordinate-dependent, but you didn't include a $\partial_\mu L \delta x^\mu$ in your derivation

Colourful Spacetime

12:33 PM

So $\partial_{\mu} (\frac{\partial L}{\partial (\partial_{\mu} \phi)}\delta \phi) +\frac{\partial L}{\partial x^{\mu}} \delta x^{\mu} = -\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu} L)$

Slereah

check out exercize 6 here btw if you need help : sns.ias.edu/~mschmittfull/teaching/qft/QFTSolutions1.pdf

ACuriousMind

@ColourfulSpacetime well, only if your $L$ is a Lorentz scalar so the first term is zero :P

it's just the second term that's equal to that total derivative, and the answer to the question you already linked already explains how

(that's why I said that computing $\delta \partial_\mu \phi$ is unnecessary work)

Colourful Spacetime

Now things make more sense, I just found unconvincing the claim "L transforms as a scalar field" without explicitly showing that it does. That's what I've been trying to do all along

@Slereah I didn't know there were solutions! Thanks!

ACuriousMind

@ColourfulSpacetime if it wasn't a scalar it wouldn't make a lot of sense to integrate over it

i.e. "$L$ is a scalar" is an assumption/axiom you must put at the very start of field theory, it's not something you can "show"

(of course, you can show it for any particular choice of $L$)

Colourful Spacetime

I keep wondering why in the lecture notes (page 16) it does not explicitly state that the Lagrangian is coordinate depended, while usually in the Lagrangian of a field there is usually no $x^{\mu}$ explicitly.

Slereah

12:43 PM

Well Noether's theorem is usually written for very general Lagrangians

It doesn't hurt to include such terms

Colourful Spacetime

@ACuriousMind Ah I see!

Ratman

Does anyone knows how can I have in c++ something that works like a pointer to function, without having to write explicitely the arguments of the function? since those could change with different applications?

ACuriousMind

you can cast any pointer to/from void* :P

Colourful Spacetime

I have no idea my friend

ACuriousMind

I'm not sure what you mean by "working like a pointer to a function" - the main point of a function is that you can call it, but if you don't know the arguments how could you call it?

see stackoverflow.com/q/16508749/3929857, stackoverflow.com/q/16508749/3929857 for more qualified responses on why what you want probably isn't possible/doesn't make sense within the C++ framework

Slereah

12:50 PM

@ACuriousMind A real language would allow currying

Ratman

@ACuriousMind the main point is that I am writing a Metropolis class to sample pdf's, so there is a function pointer as data member to save the function to sample. But the arguments of the pdf may vary for different applications, so I was trying to being more general. Thanks a lot for the links, i'll give it a try

ACuriousMind

@Slereah What do you mean? You can "curry" in C++, since you can implement currying via lambdas and returning function (pointers)

Slereah

You can curry in all languages, but it is generally not very natural

ACuriousMind

it's not a first-class concept in the language, but it's possible (and I'm sure you can write some horrifying template that curries arbitrary functions :P)

Slereah

Well, not all languages

Probably not languages that aren't functional at all

ACuriousMind

12:55 PM

Yeah, there are languages where you can't return something that represents a function at all

Slereah

One time I tried my best to implement type-free combinatorial calculus in C#

Boy what a nightmare

ACuriousMind

@Ratman I think in that case you probably need template programming to express that the class varies over the function

Slereah

Apparently C# really doesn't like it if you don't have a type to return eventually

Colourful Spacetime

1:53 PM

@ACuriousMind I don't see why this total derivative is zero, can you elaborate? In general it shouldn't be zero, right?

ACuriousMind

@ColourfulSpacetime When your $L$ is a Lorentz scalar, then it will be zero, because a scalar transforms only like the second part - i.e. if it is non-zero, you don't have a scalar

I'm not sure what sort of elaboration you're looking for there

Colourful Spacetime

@ACuriousMind So according to that, L is always a lorentz scalar and therefore this term always vanishes?

ACuriousMind

well...not "always" always - e.g. if you have a non-relativistic Lagrangian that will of course rarely be a Lorentz scalar

but the defining feature of a relativistic field theory is pretty much that the Lagrangian is Lorentz-invariant, i.e. a scalar

Colourful Spacetime

I think I found a better way to understand this, would it be correct to say that the term $\partial_{\mu} (\frac{\partial L}{\partial (\partial_{\mu} \phi)}\delta \phi)$ vanishes because of Noether's theorem? Because the "inside" is Noether's current

Ahh i think no

Basically I want from $\partial_{\mu} (\frac{\partial L}{\partial (\partial_{\mu} \phi)}\delta \phi) +\frac{\partial L}{\partial x^{\mu}} \delta x^{\mu}$ to show that it is equal to $-\partial_{\mu}({\omega^{\mu}}_{\nu}x^{\nu} L)$. The argument is that it vanishes because the outcome should be a lorentz scalar?

I can see why the second terms leads to the answer

Just not that the first term vanishes

ACuriousMind

2:13 PM

As I said, that $L$ is a scalar is an assumption for relativistic field theory

you can't "show" it

If you have a specific $L$ (e.g. the Lagrangian of the free scalar field) you can plug it in and show that the term vanishes, but for generic $L$ there's nothing you can do - we just demand that $L$ is a scalar, and that's it

Colourful Spacetime

Does $L$ being a scalar imply that the first term vanishes?

ACuriousMind

yes, because that's what being a scalar means

Colourful Spacetime

Ok perfect

I understood

Finally 😂

ACuriousMind

a scalar has the transformation behaviour $\delta f = \partial_\mu f \delta x^\mu$, and so $L$ is a scalar if and only if that term vanishes

Colourful Spacetime

Even better explanation, thank you very much!

Here is a small gift, a song that I love:

https://www.youtube.com/watch?v=JfRo9G4rPHA

EnthusiastiC

2:57 PM

Does two body nuclear interaction include pairing force?

EnthusiastiC

3:27 PM

How can we interpret the asymmetry energy part of the liquid drop model? I couldn't understand it's importance of compensation for Coulomb repulsion?

Especially for neutron excess nuclei

Colourful Spacetime

I have no idea unfortunately

EnthusiastiC

If what I understand is logical, nuclei favor neutrons over protons because of the latter yields extra energy which is repulsion. Hence, if every proton in nucleus decays into neutron, the latter requires to occupy above Fermi level which is unfavorable.

@ColourfulSpacetime OK np then

Waiting for someone who can help

Some people consider it as a balance energy which again I couldn't figure out. If we have proton excess then we have extra energy in addition to the repulsion force

Monty

3:57 PM

can anyone discuss the basics idea non-equilibrium statistical mechanics with a maths guy

basic*

PM 2Ring

6:04 PM

@EnthusiastiC That doesn't quite work. Eg, tritium has 1 proton + 2 neutrons, but it's unstable, decaying to He-3, with 2 protons + 1 neutron, which is stable.

rob

11:08 PM

@PM2Ring Sure, but tritium is really too small for the liquid-drop model to apply. The statement by @EnthusiastiC is a good explanation for why the "valley of stability" bends away from the $N=Z$ diagonal.

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