3:12 AM
@nickbros123 You are considering this wrongly. The averaging is fine, but you are not just averaging over all possible solid angles. You are averaging over the entire phase space. That is a totally different thing to do.
@SillyGoose No, in the ultrarelativistic limit, particles look as if they are massless, but $E^2-p^2=m^2$ is still strictly obeyed. The tiny difference between energy and momentum is still there. When we need that kind of precision, we often work with rapidities, in which case the correct behaviours are easier to tease out, and the correct relations are easier to keep track of, because it is built in.
@DIRAC1930 I have studied some emission of radiation, and I know how the basic stuff go, but since "quasi-stationary states" is not anywhere near the standard terminology, I have no idea what it is you want meow to know about.

@naturallyInconsistent ok so if I hav a function $u(\vec r, \theta, \phi)$ whose probability density function is some $f(vec r, \theta, \phi)$, and if I'm asked to average over all possible theta and phi, I can integrate $u f d\omega$ divided by integral of $f d\omega$ ?
The context is I'm averaging the charge dipole interaction energy over all possible orientation of the dipole
Is this kind of averaging a definition? The usual kind I see integrates over all the dependencies of the integrand, not just a select few

@nickbros123 The function tends to be over the momentum's magnitude and angular components. But yes, that is a valid integration ratio, and it means something very physically pleasing. It is always that standard form. It doesnt even have to be exactly correct, it just needs to be approximately correct for the most important contributors. You do, however, need to know what particular thing you want to average.

3:31 AM
Is there not a problem if we use the density function for the energy itself, like, the definitions of expectations i see are of the kind $\int u f(u) du$, here we use $d\Omega$ the solid angle. Its like, I kinda get it, but also don't

In actuality the expression is this: Firstly, you have that a good probability distribution should be $$\int\mathrm dP=\int_\Omega\frac{\mathrm dP}{\mathrm d\omega}\mathrm d\omega=1$$ and $$\left<h\right>=\int_\Omega h(\omega)\frac{\mathrm dP}{\mathrm d\omega}\mathrm d\omega$$ Now, if you have just the functional form $\frac{\mathrm dP}{\mathrm d\omega}\propto f(\omega)$ rather than strict equality, that means then that $$\int_\Omega f(\omega)\,\mathrm d\omega\neq1$$

Ok I need to view this on a computer, I'll be back in about 30 min

This latter is just the normalisation integral; i.e. Let $N=\int_\Omega f(\omega)\,\mathrm d\omega$, and then $$\left<h\right>=\frac1N\int_\Omega h(\omega)f(\omega)\,\mathrm d\omega$$ and you immediately get the ratio of integrals that you had. i.e. this is a generally correct procedure, suitable whenever you cannot be bothered to compute the normalisation constant to get the probability distribution. Makes for easier reading when $f(\omega)$ nicer than PDF $\frac{\mathrm dP}{\mathrm d\omega}$
You can also think of it as a more generally that it is always the case that $$\left<h\right>=\frac{\int h\,\mathrm dP}{\int\mathrm dP}$$ where both integrals must be over the entire probability space. It is just that the integral at the denominator is, by definition, 1, always.

3:55 AM
well here youre integrating over all the possible values of the random variable in question right? that would be analogous to $\int u f(u) du$, but here we do $\frac{\int u f(u) d\Omega}{\int f(u) d\Omega}$. Im just not able to intuitively understand why we can make use of the PDF=f(u) for the averaging over solid angles
I think perhaps I dont fundamentaly understand the definition of averaging itself.
lets think about, say, volume averaging of, suppose current density: it would be $\vec J=\frac{\int \vec j d\tau}{\int d\tau}$, so its as if we are adding at each small cube the value of $j$ weighted by the probability that is $dP=\frac{d\tau}{\iint d\tau}$
but in my previous scenario, the energy $u$ is a function of $r,\theta,\phi$, and the PDF=$f(u)=g(r,\theta,\phi)$ and I dont know how to intuit this one

@nickbros123 The PDF is $\frac{\mathrm dP}{\mathrm du}=\frac1N f(u)$; $$\frac{\frac1N\int uf(u)\,\mathrm d\Omega}{\frac1N\int f(u)\,\mathrm d\Omega}$$ and the $\frac1N$ cancels on both the numerator and denominator.
@nickbros123 correct
@nickbros123 What is there to intuit? You have established that this is a generally correct principle, so now you just apply it everywhere

4:17 AM
@naturallyInconsistent we can just say N=1 here i suppose, that is ok i guess. So here, if I may think of it this way, $$\frac{f(u) d\Omega}{\int f(u)d\Omega}$$ would be the probability "weight" dP multiplied with each value of u and added up? I m not able to see how that would be the dP

@nickbros123 No, we definitely cannot. The reason is that if we had $N=1$, then there is no need for the denominator integral because that would be 1.

i dont see how the denominator integral is 1 though

@nickbros123 IF the N=1 then it is 1

cuz if we integrate over all possible energies it would be 1, but we are only integrating over theta and phi, at fixed r

@nickbros123 You are correct that your ratio there is the probability distribution function. It is $\mathrm dP$

4:21 AM
my understanding of this whole thing is quite shaky i dont like this

@nickbros123 Exactly, this observation is correct. It should not be at fixed $r$; but I think your text is trying to say that $r$ is fixed in the first place and not something you can change, or at least if you change it, that is a tremendously larger energy, and so that part of the phase space can be ignored.
The integral should be over the entire phase space, i.e. over all possible energies.

hmm

bye, good luck

5:06 AM
Schweg
@naturallyInconsistent i see that makes sense
i finally learned what rapidity is like a few weeks ago lol. I remember when i was taking modern it was absolutely not understandable to me. But it makes most sense now as the parameter of the boosts for Lorentz transformations
Everything in physics comes back to representation theory right guys;)

@SillyGoose It clearly is the parameter, but if I am teaching it to beginning students, I would say that it is the parameter, however, for beginning students we can just assert that it is the natural thing to consider, since 1) it brings to closest agreement between rotations and boosts 2) it guarantees the conservation and correctness of some relations like the $E^2-p^2=m^2$ and so it is extremely convenient to use. Not to mention that it means that all the old trigonometric identities
that we have already learnt, can be brought to bear to handle all the plausible numerical headaches. Like, even if the natural statement of something is a cosine, if you want to get high precision, we would change, say, $1-\cos2\varphi=2\sin^2\varphi$ and extract the tiny $\varphi$ that way to extremely high precision if $\cos2\varphi\approx1$. A lot of quantities in SR is wanted with such numerical headaches, i.e. we often want the NR limit or the ultrarelativistic limit, and then these
trickery will pay off beautifully if we know to use them. If you don't learn about hyperbolic parametrisation, then you will not even think of use these beautiful trickery.
In fact, because I will use Minkowski diagrams to teach SR, I would have $\beta=\frac vc=\tan\varphi=\tanh\psi$ where $\varphi$ the angle on the Minkowski diagram is visualised on the Minkowski diagram but not invariant, while the rapidity $\psi$ is invariant but not visualised on the Minkowski diagram. All of these things are just as they are, but I will be clear what is what.
With different symbols so nobody should be confused,
I care about such details when teaching.

2 hours later…
7:14 AM
Absolutely screwed up the spectroscopy exam, well atleast now I can worry about the right things

7:43 AM
hi
this scientist has a good understanding of the problem

8:06 AM
Why do we always call them C* agebra anyway
Aren't we always dealing with von Neumann algebras

1 hour later…
9:30 AM
How to find the percentage for this question 'what is the percentage of 5000 miles being driven per 100 gallon'?
Is it by simply dividing 5000 mile by 100 gallon

10:28 AM
@SnoopyKid show us the full original question. As it is, it makes no sense

10:44 AM
@SnoopyKid u shud calculate the number of miles covered in 100 gallons and express that as a percentage of 5000

11:04 AM
@naturallyInconsistent to add to this in 2d integrable quantum field theories the rapidity is a very natural quantity to use as you can parametrize both mass and momentum in terms of a single rapidity

Hello All...

hello

11:45 AM
Elastic collision between very very big mass like earth and very very small mass like ball compared to earth.
Suppose $m_1$ = mass of earth and $m_2$ = mass of ball
if found Earth initial velocity equal Earths final velocity. Which seems natural.
But for ball i got strange results which is not the case. I found
Ball final velocity = 2$\times$Earth initial velocity $+$ Ball initial velocity
It means ball final velocity increases by twice of earths velocity.

3 hours later…
2:27 PM
Apparently the thing corresponding to the difference between classical and quantum physics as far as logical theories go is the weakening rule
Where additional context will change the provability of a statement
So if some assumption implies another, $A \vdash p$, then additional assumptions will keep this true, $A, B \vdash p$
From the weird correspondence between category theory and logic, this corresponds to the existence of a projection for the composition of two systems
This is true for classical mechanics apparently, but not quantum mechanics
Presumably this relates to the whole change of the state during measurement
Tough finding papers on the topic

2:56 PM
Another one of these 'quantum GR is classical' papers out today

@bolbteppa you really dont like those huh?

Not just me

3:15 PM
Ugh
That internal logic stuff seems interesting but the people writing it do not have outsiders in mind

@bolbteppa i think it's fine to explore all possibilities. people who find this line interesting will keep building up on it
one doesnt hav to pay attention if one doesnt find the base interesting

Does anyone know what Gell-Mann meant by the Decoherent histories viewpoint? youtube.com/watch?v=R-7VOAVJjt0

> the modalities contain the three pairs of opposites

possibility - impossibility

being - nothing

necessity - Zufälligkeit
Ah yes
The Zufälligkeit

If QCD was able to be written using the operator formalism, I think I would be way more interested in it

@RyderRude this is going to turn into a massive thing that will be compared with Einstein and Professor Lisi it already has started before this one, but this one is 'renormalizing postquantum gravity' aka the post-holy-grail

3:23 PM
And also if it didn't have a silly name e.g. 'quarks'

11

Dear sb1, the top experts behind the Consistent Histories often credit Hugh Everett - the father of the many-worlds interpretation - with making the foundational quantum physics community focus on the "histories" as the basic entity whose probability quantum mechanics predicts. However, this is w...

When people do QFT, do they automatically "think" using path integral methods or is the operator formalism more ingrained in their minds?

Without an external classical measuring device, there is no theory of QM, there is nothing, these guys are just pretending that isn't a fatal flaw to going beyond Copenhagen QM and trying to axiomatize a classical measuring device so we can apply wave functions to the universe, it really doesn't make sense to me but I could be missing something

@DIRAC1930 It is more archetypical to see operators for QFT
Path integrals are more of a specialty thing

I thought you can't do QCD using operator methods

3:28 PM
Because they basically say it's Copenhagen with some frills I don't waste time worrying about it but I really doubt what they are saying is right

@DIRAC1930 Well they don't typically use QCD as the archetypical QFT theory

note that in the classical quantum divide physics, the mathematics still needs to be discovered
it's not clear yet what mathematics a classical-quantum divide universe follows

That Garrett Lisi guy is so annoying
What are peoples thoughts on the proposed Future Circular Collider?

3:50 PM
Personally I think they should just wait until they can access even higher energies

Perhaps the best way to access even higher energy is to learn from building an accelerator

They might as well just go slightly bigger since 1) If nothing is found, they will say, we just need to go slightly bigger or II) if something is found, they will want to immediately go bigger to find more details which will just take more time and money than if they had just built the bigger one to begin with

4:05 PM
Using law of conservation of momentum and conservation of kinetic energies for perfect elastic collision the relation of change in momentum.
$\Delta{P_{12}} = 2\Big(\frac{m_1m_2}{m_1 + m_2}\Big) (U_1 - U_2)$
If we divide the above relation with time it becomes force which can be used for N3L for $F_{12} = - F_{21}$. Also i found above relation as $\Delta{P_{12}} = - \Delta{P_{21}}$

@123 What if you go into the frame that the Earth is stationary?
See if you get that the magnitide of the velocity of the ball is the same before and after

@RyderRude From above relation i can say change in momentum depend on the momentum which created by difference of initial velocities of both object times product of masses divided by the total mass and twice the result.
@DIRAC1930 Yes i thought about it from this relation. But this equation comes naturally by solving law of momentum and conservation of KEs.

In this way the transformed velocity $\tilde{v}_B$ of the incoming ball is $\tilde{v}_B=v_B - v_E$, $\tilde{v}_E =0$ where non-tilded expressions mean in the frame you were using before

$U_1 + V_1 = U_2 + V_2$ You can use $U_1 = 0$ when Earth is stationary

4:21 PM
So that is saying that the kinetic energy of the ball before and and after the collision is the same
Ignore the above comment
Do you have a kinetic and potential energy?

@DIRAC1930 Ookay.... But i am confused about what happened during collision. Different momentum means both objects have different force. So how much force both objects deliver with different forces and momentum exchange?
@DIRAC1930 No potential energy. Both objects are isolated and no any other interaction beside collision in one dimension.

If there is no potential energy, you will have that $E_E^i + E_B^i =E_E^f + E_B^f$ where $E_B, E_E$ are the kinetic energies of the earth and ball respectively and $i,f$ refers to initial and final energies

In the above equation i can clearly see both objects have different momentum and change in depend on both objects momentum by above relation , because during collision both spent same amount of "$time$" so by dividing time it becomes force.
@DIRAC1930 Let me share you the calculation by pictures.

What are you trying to find out as in what is the homework question?

@DIRAC1930 im excited for them to not find supersymmetry again so maybe we finally drop using supersymmetry in phenomenology and understand it is just a nice maths thing
i say this while studying supersymmetric gauge theory myself

4:33 PM
What is the general consensus regarding the likelihood of supersymmetry being true or no?
It seemed like before the LHC, it was pretty much expected that it will eventually be found but nowadays the consensus seems to have changed

to me it seems lower in general, but there are still many people thinking its right beheind the corner

@DIRAC1930 Pls see the above calculation

@DIRAC1930 already many physicists are not very approving of the FCC
at least ive heard many being agains it in my uni
all the money it is going to cost could be used into financing loads of other reasearch

@123 you are not being careful about speeds v.s. velocities. Your relation here is roughly correct, but it is about speeds, not velocities.

4:45 PM
@naturallyInconsistent Hello.. Yes but we can choose sign (+ or -) to make these velocities and set direction. I did this in 2nd picture.

@123 No, you are not allowed to do that. Velocities carry direction information. The velocity of the ball swaps signs.

From the above relation can i conclude if person-1 apply 10N and person-2 apply 15N force on each other it is possible. Because $\Delta{P}$ results shows both objects momentum contributing

@123 No, it is just wrong.

I am equating different momentum with force because during collision both objects will spent same amount of time on each other. Then change in momentum divide by time become force.
@naturallyInconsistent Oooh Okay. Pls clear me. Where i am wrong. I also have knk example and your yesterday discussion.

You obtained that $\Delta p_{12}=-\Delta p_{21}$. Whatever force acting on those objects are that exchanged that momentum, it is going to be acting for the same amount of time. Same magnitude, divided by same time, how can it be 10N and 15N? You are so wrong.

4:50 PM
But equations telling me both objects momentum contributing in change in momentum.

It is necessary that the momentum of both objects come into play. But you observed that it is the same momentum change, just opposite signs, and same time. It cannot be different numbers.
It is either both 10N, different signs, or both 15N, different signs. It can never be 10N and 15N
and your working is so long. It is a lot simpler than you think.

@naturallyInconsistent Aaaah okay...
It means my equation just showing how momentum exchange. Not different 10N and 15N forces exchange between objects.
I am assuming exchange of change in momentum equal force because both object spent same amount of time.
@naturallyInconsistent It means because we don't know the information of duration of time both objects spent on each other. Then we can't say how much force objects applied on each other. My equation just tell about exchange of momentum not force. This is what i concluded with your discussion.

please stop associating individual objects with a force. forces are only defined between a pair of objects, and they are equal and opposite
u can talk about the momenta associated to the colliding objects, P1 and P2. there is no such thing as F1 and F2. there is only F12 and F21

5:10 PM
Oooookay...

@123 yes. the magnitude of the average F12 and F21 depends on the duration

@123 I have not made that assertion.
It is however, a good idea to concentrate on momentum exchange.

@naturallyInconsistent Oookay...
Is there anything important to know about "$time$" information . Because if we don't know time how can i calculate force by just looking change in momentum. If i can calculate the force then i can use this force in N2L.

@123 as for you Earth and ball problem, you should consider the problem in two steps. First, let us consider the centre-of-momentum frame of reference. Let the incoming ball have speed $V$; the Earth will then have speed $\varepsilon=\frac mMV\approx0$. Let the left side of the zero be initial, and right side is final. Then the Conservation of Linear Momentum (CoLM) reads as $(+M\varepsilon)+(-mV)=0=(-M\varepsilon)+(+mV)$
The Conservation of Energy reads $\frac12M(+\varepsilon)^2+\frac12m(-V)^2=\frac12M(-\varepsilon)^2+\frac12m(+V)^2$

right..

5:22 PM
Now, we don't just want the centre-of-momentum frame. We want the laboratory frame. We do that by boosting the whole system to a certain velocity $v$. In that case, we will have $$\text{CoLM}:\qquad M(v+\varepsilon)+m(v-V)=M(v-\varepsilon)+m(v+V)$$ which is obviously true and conserved, provided the centre-of-momentum frame equation is true. We just have to look at the energy and check that it is also conserved.

Okay

$$\text{CoE:}\qquad\frac12M(v+\varepsilon)^2+\frac12m(v-V)^2=\frac12M(v-\varepsilon)^2+\frac12m(v+V)^2$$

okaaay

\begin{align}\text{LHS}&=\frac12Mv^2+Mv\varepsilon+\frac12M\varepsilon^2+\frac12mv^2-mvV+\frac12mV^2\\&=\left(\frac12Mv^2+\frac12mv^2\right)+\left(\frac12M\varepsilon^2+\frac12mV^2\right)+v(M\varepsilon-mV)\\\text{RHS}&=\left(\frac12Mv^2+\frac12mv^2\right)+\left(\frac12M\varepsilon^2+\frac12mV^2\right)-v(M\varepsilon-mV)\end {align}

okay

5:30 PM
It is clear that the LHS and the RHS only differ in the sign of the last bracket. However, by conservation of momentum in the centre-of-momentum frame of reference, we worked out that $\varepsilon=\frac mMV$ so that this last bracket is zero anyway, so they energy is conserved.

Yes
I can see

just did this on tikz

So, this solution is a valid solution to the entire problem. Now, if you look at the velocity of the Earth, it just went from $v+\varepsilon$ to $v-\varepsilon$, which is a tiny change from $v$, so, let us ignore the $\varepsilon$ and consider them as all the same. In the case of the ball, however, it changed from $v-V$ to $v+V$

i'd like to be fluent in tikZ:P

okay...

5:33 PM
Note that $v-V$ has to be negative, i.e. $0<v<V$, so this change is rather big. But fret not, $v+V=2v+(V-v)$ i.e. 2 Earth's speed plus the initial speed of the ball. As expected.

@SillyGoose idk i really just hardcoded loads of numbers to get that

Now, it needed to be speeds. If you used velocities, then you will have the wrong sign.

it also took me thirty minutes, so i wouldnt say im so good

@naturallyInconsistent Oooh okay , it means my equation is not valid for velocities.

@lucabtz If your hardcoded numbers are just zeros and ones, and you used overall scaling of the entire picture to get the picture bigger than 1cm by 1cm, then you would have already been good at it.

5:35 PM
@naturallyInconsistent its actually done like that

@123 It is not "not valid for velocities". You have misinterpreted what your equation is trying to tell you because you were not careful about the difference between velocities and speeds. That is why I emphasised to you from the beginning that this is the case.

except for the curves

@lucabtz so you did well! Yay!

three of the curves use ellipse (one is rotated with rotate)
the weird one is using plot[smooth cycles] and a hell of harcoded numbers
the pictures are three independent scopes, shifted next to each other

@lucabtz I would have also thought that that one is horrible anyway and deserved to be in hardcoded numbers hell.

5:37 PM
i wish there was just an interface between ipad drawing and latex images
i mean there is screenshotting and sending to laptop and uploading but it is a bit of a cumbersome process

@naturallyInconsistent oooh okay. You gave same results at which i was confused in my equation when i considered ball and earth example. I found ball final speed $V_B = 2U_E + U_B$ where $U$ represent initial speed

It would be useful if AI can convert input prompts into half-functioning TikZ code that then generates an approximately useful picture. However, it will be difficult to get them to output pristine code suitable for humans to edit.
@123 And I have also obtained that $V_B=2U_E+U_B$ with far less work.

@naturallyInconsistent Yes. But i am confused with the results. Because the result shows ball final velocity has twice the earth velocity with ball initial velocity. But in reality we see only Earth velocity + ball initial velocity. why 2 here

good god dewitt's volumes on qft look gnarly

@123 No, you are literally wrong. The maths is correct. This experiment is done, and the results agree with the maths.

5:45 PM
@naturallyInconsistent Can you pls explain the results in your wordings. So i can correct myself.

I already did. I know you are also struggling with English at the same time, but it is more helpful if you can express where your confusion is.
In reality, we see 2x Earth speed + ball initial speed. The 2 is correct. You say you see 1x, and that is wrong.

@naturallyInconsistent agreed

You think you are correct. You are just wrong.

@naturallyInconsistent Yes i mostly struggling in reading english if wordings are not explained in easy way.
@naturallyInconsistent How we see this

@123 There is a toy that has the same behaviour, and when we drop it on the floor, it does this same thing. The result of the experiment of dropping it on the floor is that the maths prediction is perfectly correct, we can draw the expected heights on the wall and the balls will fly to the correct place.

5:50 PM
It means if we drop the ball on earth from height $h$ in perfectly elastic collision then ball will go above the height where we drop?

Yes, and very much higher. It is very dramatic.

@naturallyInconsistent Oooooh... I see ..... Can you share the toy name. So i can search it on google or youtube.

even though the $\mathcal{L}$ symbol is used (which usually refers to Lagrangian density these days), is $\mathcal{L}$ meant to be an action-like quantity in the above?

Finally pls clear me ... I have learnt about exchange of momentum from you. But we used force in N2L how we calculate force. because N3L doesn't allow me to calculate force rather change in momentum.
By the way thanks for your precious time and sharing your knowledge. It helped me a lot. Your spent of few words helped me.

5:59 PM
@123 No, N3L will be applicable to forces too. You just havent considered situations where forces are appropriate.
However, it is better to consider everything in terms of momentum. Forces are the first to die in the quantum revolution. Momentum is what is actually going to be the way forward.
@SillyGoose No, the text is just omitting some stuff. The $\mathscr L$ is just the Lagrangian density, which I remind you, is the integrand of the action integral / principle.

@naturallyInconsistent I didn't see any example where force is used in N3L in any textbook. But the force in N3L used in N2L for single object. How can we calculate the force?

@naturallyInconsistent isn't $\delta \mathcal{L} = 0$ a little strong a requirement for minimizing the action? Or, I guess it is omitting the boundary conditions in which case you don't know what the precise condition should be

@123 Again, there will be scenarios where that is applicable. I don't care. As long as you get the momentum correct, you will always be correct.
@SillyGoose Again, shorthand for "up to boundary terms and gauge freedom"

I see

@naturallyInconsistent Okay.. In one of the book. I have read finding the forces on object is always a challenge.

6:14 PM
How is this derivation done? i dont understand how to derive s in regards to the function p(v)

@Mad This is standard functional derivative. I know you have maths background, and I also know that maths treatment of functional derivatives is a tremendously long treatise. Have you learnt that yet?

@SillyGoose What book is that
"I can tell you've read a book!"

@naturallyInconsistent Oh jeeze, i gotta google up that again, term is not ringing any bills.
Yup no idea lol never had that! (Based on the wikipedia page)(My mathbackground consists of some undergraduate math courses, not all the syllabus)

6:32 PM
I just know that you are happier if you do things the maths way, so go knock yourself out on that.

@naturallyInconsistent is there a quick way on doing that derivation ? (the physicist way?)
I would be fairly surprised if the prof really wanted us to know about this part of maths.

Yes. Just ignore the integral signs. The derivative of $-p\ln p$ is $-\ln p-\frac pp$, the derivative of $-\alpha p$ is $-\alpha$ and the derivative of $-\beta|v|p$ is $-\beta|v|$

@DIRAC1930 this is this paper

Is this on the Periels bracket?

yes
supposedly the og paper
it sounded interesting

6:37 PM
If it's from the 50s ACuriousMind will get angry

so here $\delta \mathcal{L} = 0$ is your linear system of diffeqs and the inhomogenous equation is $\delta \mathcal{L} = -\lambda \delta A$, right?
@DIRAC1930 hehe how come

@Mad it is physics. The prof prefers if we think as little of the maths background as much as possible.

@naturallyInconsistent oh okay, so i just remove integral signs and derive. When can i do this generally?

He gets angry when I try to learn field theory properly from 1950s texts
lol

@Mad Actually, I dont know. I only know the physicist's arguments, which make sense. At any point of the integration region, we can just add a tiny bump function that is zero everywhere else but for a tiny positive blip somewhere. That then destroys the integral and forces the result to be correct as we want it. It is quite general, and it helps that physics only cares about functional of the integral form.

6:41 PM
oh i didn't realize weinberg's volume were published in 1990s

@naturallyInconsistent Alright. thanks!

@SillyGoose it has nothing to do with ACM. It is just DIRAC and blobteppa doing their stupid shit. ACM is perfectly happy with good 1950s texts.

Tbf Weinberg's first book is similar to the approach from L&L 4
However I way prefer L&L 4
And also Weinberg focusses on the connection to QM in the same way L&L does
@naturallyInconsistent It seems stupid until you want to actually understand what's going on

@DIRAC1930 hahahahahahaha

what other 1950s qft textbooks ar ethere
the modern qft textbooks seem to be written for people who already know qft
XD

6:50 PM
Well 1950s was a bit of an exageration lol

in the picture i posted above, are we just trying to define a first derivative of the field $\phi_a$?

@SillyGoose nah, that is not what is happening. There are many authors who are writing to introduce to beginners. However, they might not really know how to teach the subject. They think that students need hand-holding where they really do not, and cannot handle rigour where they can. It is the exposition that is requiring work, not the students.

The biggest issue for me was not reading the section on 2nd quantization in L&L 3 (QM) first
Basically, the issue is that everyone who claims to understand QFT doesn't, unless they can talk about it in the language of QM
Unfortunately, L&L 3 and L&L 4 are the only books that explain this all starting from scratch

3 hours later…
9:48 PM
@DIRAC1930 i think you're having the wrong impression that people don't know about second quantization. It's just that discussion high energy physics in those terms doesn't work as well as in condensed matter