Hello World...

Hello @JohnRennie Sir...

@Charlie The classic on that is the '73 Epstein-Glaser paper leading to what's today called "causal perturbation theory". We've known the nature of renormalization for almost half a century, but strangely many people still wave their hands and consider it mysterious/a questionable trick.

(this is not a recommendation for you to read that paper right now; it's rather technical and will do nothing to help you pass a standard QFT course :P)

Renormalization in my mind is still "subtract infinities from infinities", I'm not very far into it yet though :P. It will be interesting to read some more of the historic stuff when I'm done with the mandatory things I need for exams

How do I look for research internships/summer schools more effectively? I'm in my 2nd year and I think I've missed the deadline for most programmes. Does anyone know of any summer programmes that are still open to applications?

Not to kick you while you're down but the irony of your name made me laugh

But if you want actual help you'd need to at least say which country you live in and what kinds of things you're interested in

@Charlie Haha, my timeline jumping days are behind me. After my antics lead to a global pandemic, I've given up the job.

@Charlie I am in India studying Engineering Physics. I haven't fully zeroed in on my field of interest yet, but I've done a few projects on astronomy over the year. I've developed a sudden interest in condensed matter physics, but my knowledge is surface level at best.

Hello. I need some advice regarding choosing books. I started learning QED with many books and papers and now I realize that I cannot finish everything. So I felt that I should choose a single book and complete it first. Is this fine? Among the books I'm reading I like most of them. Then how should I choose the book ? (Preference to modern books?) Thank you.

Also, Engineering Physics at my place is a cocktail of physics and electrical coursework, if that's important.

@blahblah Peskin and Schroeder or David Tong's QFT notes are generally a good place to start. Peskin and Schroeder is considerably heavier, Tong's notes are a much gentler introduction.

Oh but if you specifically want QED maybe not Tong, Peskin and Schroeder does cover QED though.

@Charlie Would you recommend reading old books? (Like Fermi's paper/ Feynman's book/Landau)

I haven't read any of them so I don't know, it probably wouldn't hurt though, presumably they are all well written

@ACuriousMind this link here ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/… shows that t=const surface are timelike and r=const surfaces spacelike. It says that if the normal to a surface is timelike then the surface is timelike ....isn’t this wrong and opposite

@ACuriousMind is it correct

@Shashaank there's no need to ping me with such general questions that anyone knowing a bit of relativity can answer - just ask the question here, someone who knows the answer and wants to answer it will answer

for the question itself, if that's really the definition given there, it's non-standard- the usual terminology is the other way around: Timelike surfaces have spacelike normals and vice versa (but null surfaces do have null normals)

Sure I Will keep that mind.

I wanted to ask, how do I figure out if a coordinate is timelike, spacelike or null

it's just the sign of the corresponding $(\mathrm{d}x^\mu)^2$ term in the metric

Well that’s what I was doing but in ds^2=x^2du^2 - 2dudr+ 4r/x dudx -r^2dx^2 -x^2dy^2, the sign of du^2 is positive so I thought it’s a timelike coordinate but the answer given is null

@timetraveller69 are you studying at an IIT ?

@Shashaank At $x=0$ it's null

The question just asks what are the nature of the coordinates in the above metric. The answer given is u is null and all others are spacelike. Any ways I don’t see how either the answer or even at x=0 follows from just seeing the sign of the coordinate.

@satan29 Yep

Shouldn’t we find the norm of normal vectors to hypersurfaces of constant coordinates rather than checking the sign too the terms in the metric

@Shashaank ah well, if the metric is not diagonal it's not so simple

The normal vector to a hypersurface of constant $u$ is $n_\mu = (1,0,0,0)$, so you need to compute $g^{\mu\nu}n_\mu n_\nu$

note that that $g^{\mu\nu}$ is the inverse of the $g$ in $\mathrm{d}^2 = g_{\mu\nu}\mathrm{d}x^\mu \mathrm{d}x^\nu$, so the relevant sign is the sign of the $uu$-component of the inverse metric

when the metric is diagonal, both signs are the same

Yeh and I don’t know how to calculate the inverse very quickly because the metric is not block diagonal

well you can just flip the indices and compute $g_{\mu\nu}n^\mu n^\nu$

then you don't need to invert the metric, you just need to contract $g^{\mu\nu}$ with $n_\mu$, which is straightforward

It to find n^mu I need to raise an index on n_MU for which I need the inverse metric

ah, right

well, computations can be annoying :P

But the point is g^{uu} isn’t zero so the coordinate isn’t null in any case

Dou you think the answer is wrong. Because I think r is a null coordinate

Rather than u

that's also true

yes, $r$ is certainly null because there's no $\mathrm{d}r^2$

whatever texts you're working with don't seem to be very careful about what they're doing :P

Result- I think u is timelike ( I don’t know at x=0 because I don’t have the exact value of g^{uu} so it can blow up at x=0) , r is null because there is no dr^2 and x and y are spacelike...let me know if I am wrong

no need to invert the metric or anything

So you are saying that u isn’t null it could be timelike only since x^2 is always positive , right

Btw, the book is Ray d inverno

could be, if we're not covering the position $x=0$

if $x=0$ is part of our chart, then $u$ is null at that point, and overall neither timelike nor spacelike nor null

Sorry, why is u not timelike at points other than x=0. If g^{uu} isn’t zero then the norm of the normal is positive suggesting it’s a timelike coordinate

it is timelike at those points

I didn't say anything about it being not, I just said it is overall neither of the three types because it changes its type at $x=0$

Ok yeah I get it

also, please have a look at the mathjax link in the room description - reading non-rendered super- and subscripts is a bit annoying, and you can get mathjax typed between $ to render in here easily

But one of the answers here on the site said that to check whether a coordinate is timelike or space like or null is to check the the norm of the corresponding normal vector to constant coordinate hypersurface. But another answer said that to figure out the nature of the coordinate just vary that coordinate and put all other other coordinates constant and then see the sign. Are these both methods equivalent...

Sure, I was of the impression that mathjax doesn’t work on the chat...is there a way to render the expressions in the chat as well

After working out a bit they give the correct answer for the above metric ( the latest correct version above). So I guess they are equivalent. But I can’t see how the two methods are equivalent...

I must be missing something here, in $\phi^4$ path integral formalism you calculate the $n$-point Green's functions by just taking functional derivatives of the generating functional with the source, something like (not being particularly careful about how I write it): $$G^{(n)}(x_1,x_2,...,x_n)=(-i^n)\frac{1}{Z[0]}(\delta/\delta J_1) ...(\delta/\delta J_n) Z[J]|_{J=0}$$ and you find through basically a Gaussian integral that $Z[J]=Z[0]\cdot \exp(\int d^4xd^4x'J(x)\Delta_F(x-x')J(x'))$

This is fine, my problem is that it seems like the Z[0]'s just cancel, which is odd because I thought you needed the Z[0] on the denominator to basically cancel the bubble diagrams in the numerator. Which it seems like you're not going to be able to do if it just cancels with the Z[0] in the second formula

Because going through and doing the functional derivatives on the exponential, you still get the vacuum diagrams

Uh I may be missing something in there for the functional derivatives in the interacting theory, I hope that doesn't make it completely nonsensical but they should be in there

Hello all! I have a "yes/no" question that would probably be off-topic on the site, so I figured I'd just ask it here (and maybe post it if the discussion is more complex of what I'm thinking)

@Charlie I don't know what you mean by "you still get the vacuum diagrams"

It's regarding this question of mine physics.stackexchange.com/questions/633144/… but from a more general perspective

The action for a point particle is integrated in one dimension (proper time), while the action for, say, the EM field is integrated in 4 dimensions. Is this because a particle propagates on its own worldline while a field propagates in spacetime?

Ah I mean you still get the disconnected diagrams in the numerator, from which (as I've seen by hand) you can factor out the disconnected vacuum bubble parts and cancel them with the denominator, but it seems like the denominator is already cancelled

I've been using the world "action" so much in the last few years that now I'm not sure of what it even means.

@Charlie writing $Z[J] = Z[0] \cdot \dots$ is precisely the analog of that factoring out here

@MauroGiliberti Is this not just the distinction between the action for a point particle and the action for a field?

@ACuriousMind Oh

@Charlie Well I think it is, but I was asking for confirmation because I'm not sure.

@MauroGiliberti The solutions to the point particle's equation of motion are functions of (proper) time $q(\tau)$, the solutions to the field equations of motion are functions of spacetime $\phi(x)$. The action integrates over the "spacetime" that is the argument to the solutions of the equations of motion

@ACuriousMind Ok that makes total sense. It's what I thought but I wanted to be sure I wasn't ignoring something important

I'm almost certain I'm looking at conflicting information in these notes

Because I'm (as close as possible to) 100% sure that going through with these functional derivatives at first order for the two point function, you're going to get something like: $$\delta/\delta J(x_1)\delta/\delta J(x_2)\left(\delta/\delta J(y)\right)^4\exp(\int d^4xd^4x' J(x)\Delta_F(x-x')J(x'))$$ which is still going to give you disconnected diagrams because I've done this calculation by hand several times now

wait...I'm not sure why there's a term quadratic in $J$ there

and even that would be fine, because we expect it to cancel with the denominator, but the denominator is no longer there because we've just cancelled it with the $Z[0]$ from the formula above

in the exponential?

ah, nevermind, I remember now

sorry there should be a factor of one half in the exponent

@Charlie so why are you doing the $\delta_{J(y)}^4$ there?

They replace the fields $\phi(y)^4$ in the expansion of the exponential containing the interacting Lagrangian

the n-point function is $G(x_1,\dots, x_n) = \delta_{J(x_1)}\dots J_{J(x_n)} \frac{Z[J]}{Z[0]}$

(perhaps some factors of $\mathrm{i}$ missing)

there are no internal functional derivatives when you compute the full n-point function

+and you get the order-by-order expansion simply by expanding the interaction term $\exp(\int L_\text{int})$

And in the interacting theory that generating functional is $Z[J]=\int [D\phi]\exp(S_{int}[\phi,\partial\phi]+i\int d^4x J(x)\phi(x))$ no?

yes

sure

but throw in a $\sim\epsilon\int\phi^2$ for convergence pls

So then you separate the free and interacting lagrangian and expand the exponential now containing the interacting lagrangian and replace the fields $\phi^4$ with functional derivatives no?

My integrals come with a 0% guarantee of convergence :P

@Charlie yes

Ok if I'm correct up until there, how come "there are no internal functional derivatives" in that case? Because I would have thought the internal derivatives were the ones coming from the interacting lagrangian expansion

which is the $\delta/\delta J(y)^4$ in what I wrote way above

ahh I get what you mean now

we have to look closer at how you get the $Z[0]$ prefactor for that

@Charlie I don't know what you're reading but I can't make it any clearer than pp. 188 of Weigand's script - note that there's only a $Z[J] = Z_0[J]\cdot \dots$ factorization there, and $Z_0[0]$ contains not the vacuum bubbles of the interacting theory

the argument that $Z[J]/Z[0]$ doesn't contain vacuum bubbles is not that there are is a simple factor you could pull out and still have neat function expressions - as you correctly say you still get bubbles from the perturbative expansion at each order, that cancels out order by order against the perturbative expansion of the $Z[0]$ denominator

Oh yeah wait a second the denominator still contains the interacting lagrangian

So that $Z[0]$ factor that I've got above is the $Z[0]$ of the free Lagrangian so they definitely can't cancel

@ACuriousMind so if I have an action integrated in p+1 dimensions it can either be a p-dimensional "particle" propagating on its "time", or an action for fields propagating in (p+1) spacetime, right?

Surely if you're doing the equivalent of a point particle action functional in $>1$ dimensions you're basically doing the worldsheet action in string theory

(for example, the p+1 dimensional brane action can be seen as the action of the p-brane propagating on its own time, or the action of the string fields propagating on the p-brane (that is the spacetime where they live))

@MauroGiliberti yes, although a "p-dimensional paritcle" is usually called a p-brane and the two viewpoints are kind of dual to each other - I talk a bit in this answer about how the action for the string can be viewed as either the string propagating through spacetime or just some action of an sCFT that happens to have a certain number of fields. There is nothing intrinsic to the action that would tell you which interpretation is the "physical" one

fields in an action aren't "tagged" with whether they are "fields" or "coordinates", this is semantics the pure formalism neither knows nor cares about

@Charlie Yes I am :P

@ACuriousMind Thank you for the clarification and the link!

aha

Hello Guys...

1drv.ms/u/s!AozWlUoG8z4tngG9jFZwPdeveyfX?e=0A1YxH

Pls see link in simple equation of motion. Why i don't get the exact distance. What is the mistake here.

@ACuriousMind Pls look at this link.

Distance should be 44m is correct as per diagram. But equation gives me 33m. What is mistake here?

Are neural networks hierarchical? Let me clarify, I'm reading a book that explains that the biological brain is hierarchical in the sense that the first layer finds the edges and lines, the second layer uses those edges and lines to construct the nose, mouth, eyes, and the third hierarchical level uses nose, mouth, eyes to construct the face. My question is, if you have a neural network trained to classify faces, if you get one of the layers and plot the values, will the plotted layer represent

the features (let's say eyes, mouth, ears...) corresponding to the hierarchy?

Nevermind, I googled :P

Thanks for giving us the molecular rotations question @ACuriousMind !

If Vi = 0m/s, a = 6 m/s^2, t=1s.

Why in just distance of s = 3m, object achieve final velocity Vf = 6m/s.

And after that in every second it added 6m distance in previous distance as per acceleration. Which is not following in 1st second.

1drv.ms/u/s!AozWlUoG8z4tngIOQgOLtV99dWI_?e=9kntim

Pls see the link for description.

@123 Use the SUVAT equation $s=ut+\frac{1}{2}at^2=0\cdot 1+\frac{1}{2}\cdot 6\cdot 1^2=3m$

@Charlie How it is possible only 3m distance object achieve higher velocity Vf = 6m/s compared to distance covered.

You're probably confused because the final velocity that your object reaches after one second is $6m/s$, but it doesn't have that velocity for the entire second that it is accelerating for.

@Charlie No i am not confused with time 1s. I am confused with distance. 3m distance and object achieve 6m/s.

What $V_f=6m/s$ means is that after that first second the object's velocity is $6m/s$, but if it is still accelerating it's final velocity after a further one second will be $12m/s$.

velocity is the measured according to distance. How distance is shorter and velocity is higher.

The 6 and the 3 aren't really related there (in the way you seem to think they are)

If the objects initial velocity were $6m/s$ and it weren't accelerating it would travel $6m$ in one second, but the velocity is increasing from $0m/s$ to $6m/s$

@Charlie Yes. Pls see my link diagram. It is best explanation of my question.

1drv.ms/u/s!AozWlUoG8z4tngIOQgOLtV99dWI_?e=9kntim

Just because the final velocity is $6m/s$ doesn't mean the object has to have travelled $6m$ during it's acceleration to that velocity

@Charlie You can see $\Delta{s}$ in my diagram. it is just 6m addition in previous distance every time. Except for 1st second.

@Charlie Oookay.. But after 1st second relation exactly follow 6m distance addition every time in previous distance. As in $\Delta{s}$ in my diagram. Which is same as i have acceleration of $a = 6 \frac{m}{s^2}$

It's just how the function $y=3x^2$ scales between $0\leq x\leq 1$

1drv.ms/u/s!AozWlUoG8z4tngPNEDx5A35EANck?e=C16lGI

You can also see in this picture when i start with Velocity Vi = 2m/s , Just after 1st second object follow same relation of $\Delta{s}$ added 6m distance in previous one. As per acceleration is 6m/s^2

If i change the acceleration a = 4m/s^2, Just after 1st second every time 4m distance is added in previous distance.

@Charlie I understand the mathematical relation. But in term of physics what happened in 1st second. Why object not covered the same distance as per acceleration in 1st second. After 1st second it added same distance as acceleration i have.

I don't think there is a physical reason for it, it's just the fact that $y=ax^2$ for $0\leq x\leq 1$ is always such that $y\leq a$

physics is mathematics applied to the real world

@Charlie Thank you very much. Hope you understand my point what i am trying to ask. My English is not very good.

if you think something different should happen here, then you don't understand what acceleration is - it's by definition the second time derivative of position, and if you compute what that means, you get that this means the particle has covered a certain distance

@ACuriousMind You are right. Just i wanted to try to understand. What special happened in 1st second.

@123 there is nothing special happening, we're literally just applying the definition of constant acceleration

the mathematical relation between acceleration and position is all there is; I don't know what you mean when you ask for a "physical" reason

@ACuriousMind Oookay... :-)

@ACuriousMind After 1st second i can always tell the distance covered by adding the same distance value (as acceleration value) to the previous distance. There is relation but in 1st second i don't know the distance until i compute.

Once i know the distance for 1st second. Then i add distance value (same as acceleration value) to previous distance and i can tell the distance for next second.

I don't know what you mean. For constant acceleration $a$, the distance covered is $x(t) = \frac{1}{2}at^2$. So $x(2t) = 4x(t)$, i.e. if you have the distance after 1 second then the distance after 2 second will be quadruple that distance

there is also no such thing as "distance value (same as acceleration value)" - distance and acceleration are different physical dimensions and have different units, you cannot meaningfully compare their values

1drv.ms/u/s!AozWlUoG8z4tngIOQgOLtV99dWI_?e=9kntim Pls see the link and check $\Delta{s}$.

I have no idea what "check $\Delta s$" is supposed to mean

$\Delta{s}$ is distance covered only in that second.

If i set acceleration a = 4m/s^2, i found it exactly follow the same relation. After 1st second 4m distance is added to the previous distance every time i found the distance covered in next second.

What you're trying to say is that the additional distance is linear in time. This is simple to see: You're looking at $x(t) - x(t - 1\mathrm{s}) = at\cdot 1\mathrm{s} - \frac{1}{2}a\cdot (1\mathrm{s})^2$, which is linear in $t$ with slope $a$

@ACuriousMind I don't understand how you come up with this relation. But is it not quadratic due to $t^2$

@123 I just plugged in $x(t) = \frac{1}{2}at^2$ and used the binomial formula for the $(t-1)^2$ term

@ACuriousMind Ookay...

I don't know whether that means you understand or you have no idea what I'm saying :P

@ACuriousMind I don't get the relation meaning.

Your $\Delta s$ is the difference between the distance covered at time $t$ and at time $t-1$. I simply computed it by computing $x(t) - x(t-1)$. What exactly is unclear?

@ACuriousMind Aaah i see. Yes

My $\Delta{s}$ is the distance covered only in that 1second.

Let me correct by putting subscript in that and description in picture.

@ACuriousMind 1drv.ms/u/s!AozWlUoG8z4tngToY9FQ3FWYwi9d?e=j64RJS Pls see the link how i compute $\Delta{s}$

I'm not sure why you're using subscripts instead of function notation, but I understood that. In your notation, $\Delta s = s_t - s_{t-1} = \frac{1}{2}at^2 - \frac{1}{2}a(t-1)^2 = at - \frac{1}{2}a$.

@ACuriousMind Ookay now i understand. I means this relation is linear in time.

In your 1st term you write (1/2)at, Is not it (1/2)at^2

that's a typo, fixed :P

@ACuriousMind Thank you for your time. It means my $\Delta{s}$ is linear in time.

It means. t = 1s it is just half of acceleration.