@Relativisticcucumber sure, you'd be the second h bar user I meet in real life

@ACuriousMind omg. theres someone who met five i think?

thanks XD

no worries :P

@Relativisticcucumber you're thinking of Daniel Sank

@ACuriousMind an animu character

@Obliv This is correct; The Compton wavelength of the electron, which can be found on CODATA, is $\lambda_C=\frac h{mc}$ and so your expression is $2\lambda/\lambda_C=10000/0.0242631023538=412148.4489$ but this value is just uselessly large. Instead, they should have asked for $\frac1{1+\text{that}}=2.426304348\times10^{-6}$ which means that the atom recoils with a KE so much smaller in magnitude than the photon. Clearly reasonable to neglect SR.

Alternatively, you are supposed to work with mass in MeV/c^2 or work with mass times c^2 always, in MeV, and then Planck's constant as $hc$ combination in units of MeV m, so that then things come out nicely.

@imbAF It is that silly thing we call canonical quantisation

@SillyGoose I'm not tooooooo happy with @Mr.Feynman 's answer to you. In the isolated Coulombic H atom, your allowed energies are labelled by $n,\ell$ (or better $n,j$) as the discrete energy eigenvalue indices. In Bloch's theorem, these discrete indices get merged into one, and labels the particular band, from bottom up. i.e. for every $\vec k$ you count upwards from the bottom, including degeneracies. Which means energy bands can overlap, as $\vec k$ varies continuously over 1BZ.

@Amit The greeks would be really confused as to why you are talking about see flee movies

@Amit The word "during" is related to "duration" and normally refers to time not distance.

I suppose you could Lorentz boost it to make it refer to a distance :-)

@JohnRennie @Amit there is actually no physical problem whatsoever. What do you mean, you don't measure your distances with light-seconds and times with light-metres? :P

hi

@naturallyInconsistent can you explain in what point this differs from my answer? Given that I'm studying this now it would be helpful :P

@Mr.Feynman Not very much; I'm pretty much giving context on top of what you stated. For example, you did not mention that there can be degeneracies on the bandstructure diagram. You also did not imply that the bands have anything to do with the free-atomic wavefunctions, nor that there is a merger of $\ell$ degrees of freedom into the band number. Although there is band overlap over the 1BZ, my answer makes it clear that the band index counting upwards is clearly each going to have bounds for

max and min, and that there can never be just one band, since the $n,\ell$ of even just the H atom is an infinite set. Although, of course, the alkali metals, we would just consider the one semi-filled valence band and be done with it.

Thanks =)

huggies

There are too many notations for hom sets

it is about a nonsensical paper

That nonsensical paper had spawned at least 3 different main site questions.

they just write random incorrect equations involving tensors and derivatives

@naturallyInconsistent wow

Does anyone know where I can find an exposition/explanation of Noether's second theorem? I can't find it in Arnold's "Mathematical Methods of Classical Mechanics" for example. I don't think it's in Goldstein or anywhere else I've checked.

section 2: arxiv.org/abs/1510.07038

(as suggested by a comment on a stack question)

thanks!

@qwerty I was looking through community wikis in trying to help you... and it seems like you're on the look for quite a while :)

@Amit hahaha... yes! I was thinking about this sometime late last year, got distracted with some other things and just came back to it.

@qwerty I see :) Well so at least it's a reminder, 'cause Qmechanic in the comments linked to another wiki with lots of possible resources.

Idk if any of them are helpful ofc

@JohnRennie Can you?? :) Time like related events can't be simultaneous in any frame can they? ;)

@Amit none of the ones listed had noether II ;)

@qwerty How odd. Did you happen to take a look at these references in Wikipedia?

thanks I'm looking at the paper silly goose linked.

@qwerty cheers!

if $f$ is a homeomorphism, isn't $\psi f \varphi^{-1}$ guaranteed to be invertible?

each coordinate $\psi, \varphi$ is invertible by definition (they are required to be homemorphisms => bijective). if $f$ itself is also a homeomorphism, then the composition will be a bijection, no?

Sure

it's just the inverse function in the given coordinates

also is there a notion of a morphism in the category of manifolds?

a diffeomorphism seems to be an isomorphism

Smooth maps

Projections would not be diffeomorphisms for instance

oh wait so just drop the bijection condition

but they are morphisms

i was also wondering, when considering if a coordinate is continuous $\varphi: U \to K$ where $U \in (X, \tau)$ and $K \in (\mathbb{R}^m, \tau_d)$, do we consider the subspace topology induced on $U$ and $K$?

what subspace?

If you mean like the projection to a single value then yeah

$U \in \tau$, i.e. $U \subseteq X$ for example

i am thinking: the basic data of a manifold is a topological space $(X, \tau)$ and some $m$-Euclidean space endowed with the Euclidean-norm-induced topology $(\mathbb{R}^m, \tau_d)$. Then a coordinate is definitionally a homeomorphism $\varphi: U \to K$ where $U \in \tau$ and $K \in \tau_d$. But a homeomorphism is a morphism between topological spaces. A priori, I'm not sure what topology to endow $U$ and $K$ with to judge if $\varphi$ is a homemorphism.

It will be the subspace topology yeah

oh okay

> This implies that there are also some impossible judgements, for example, “the apple is straight,” “the apple is not straight,” “the apple is depressed,” “the apple is not depressed.”

@Slereah these are syntactically correct but semantically incorrect en.wikipedia.org/wiki/Colorless_green_ideas_sleep_furiously

> Penrose : it is a good thing to have believed in Many worlds in a certain stage of ur life. The shorter the period, the better

@RyderRude Why is a mouse when it spins?

@Amit it be spinning but not existing

That's the electron

electrons have spirit animals

@RyderRude Wrong notion

ncatlab.org/nlab/show/infinite+judgement

@Slereah they don't give a simple example...

plato.stanford.edu/entries/kant-judgment

> an infinite judgement is of the form Fs are not Gs

philosophers always use incomprehensible words

@RyderRude It's literally the first line

You need subspace topology on both, but to verify continuity you don't need to consider it as a function into $K$

@SillyGoose don't let category theory get into your head too much

@Slereah the example is there but what they mean by that example hasn't been explained properly

@Amit Well you can give it a spatial component :-)

Whales aren't doors, it seems pretty obvious

@JohnRennie That much is true... :)

@Slereah maybe I'm not getting their jargon

why would u name this "infinite judgement"

There's that famous quote by Minkowski.. "...space by itself, and time by itself, are doomed to fade away into mere shadows...", maybe human language is just slowly catching up to that (as @naturallyInconsistent alluded to as well)

analytically continue Lorentz transforms to English sentences @Amit

:P When stuff like the relativity of simultaneity becomes natural to third graders, maybe then...

I'm not sure, but if we ever become an interplanetary civilization this stuff will in some ways have to become a lot more natural to us.

yes. maybe

but QM will never become a reflex

Idk... I mean, this statement is why many people think QM is incomplete. As opposed to relativity that may be incorporated into a larger structure but I think is in less danger of revealing internal inconsistencies. Some insist that the ultimate theories, insofar as they exist, should be elegant and graspable by anyone

If I apply force on an extended object, will it get accelerated in the direction of applied force?

@DebanjanBiswas If its center of mass isn't constrained from moving, in general yes

What if it's not applied on the center of mass?

doesn't matter for a rigid body. the only difference is that you may also create some torque in addition to linear force

@Amit?

@DebanjanBiswas "?" means it wasn't clear?

What will be the magnitude of acceleration?

F/m

Okay, but...

If I apply the same force on it's com, it doesn't rotate

So?

And in other cases it rotes about it's com

Correct, what is the question then

In both cases, they have same linear acceleration

So I apply the force for equal time in both cases

The rotating object will have more energy gained than the previous due to its rotation

It seems kinda weird to me

@DebanjanBiswas energy is about force over distances. The distance of applying away from the centre of mass is slightly longer, and just long enough to give the rotational KE.

There is no weirdness.

Shouldn't both of them have same amount of energy gained if they are equal in mass?

@Amit I just wrote an answer on the main site pointing out how wrong (dangerously seductive!) it is.

@DebanjanBiswas I specifically explained why it is not. What are you even asking

Actually mine and your messages were posted in same time

Sorry, for the confusion

oh ok

@naturallyInconsistent The EPR related one?

correct

@Amit yes, QM doesn't give us an ontology

@DebanjanBiswas I thought this had been asked before:

it is an incomplete theory. But the non intuitiveness of it has more to do with the fact that it is about the microscopic world

and we are evolved in macroscopic circumstances

but u could make the same argument for relativity being non intuitive

maybe it's just that QM is more complicated mathematically

SR is somewhat less complicated. GR is on par with QM in complications

Nice answer @naturallyInconsistent, one day I'll understand what the Bell tests are all about and I'll be able to appreciate it even better. But I really liked the historical commentary

@Amit yay thanks. But you should watch some nice youtube videos on the Bell's theorem. IIRC, Veritasium's version is tolerable.

@naturallyInconsistent ty, I'll do that

@naturallyInconsistent ur answer is good, but also, the compromise can't fall entirely on the SR side as QM is an incomplete theory

@RyderRude anybody who still falls for QM being an incomplete theory is totally unqualified to discuss the subject.

I have been deliberately ignoring your utterances on the topic but you had to come poke meow

@naturallyInconsistent u should say that to Penrose

@RyderRude he believes in gravitational wavefunction collapse. That is already falsified by experiments. There is no talking to him about this.

@naturallyInconsistent he believes in more than one thing, just like any normal person. One of those things being that QM is incomplete

anyone speak fortran here?

@SirCumference miao miao does

@RyderRude of course, just like the red queen...

@naturallyInconsistent just want to know, if I see something like 5d5, is that the equivalent of python's 5e5?

i.e. scientific notation

and gravitational collapse hasn't been falsified, as it is not a fixed theory. It is a category of theories

he would know better than u if his theory is falsified

@SirCumference this has nothing to do with fortran. d just implies that it is double precision, not just single precision floating point. You can see the same thing with C#

@SirCumference but yes, that is true.

@RyderRude Again, you can very much continue to showcase how much you wish to be the same as the red queen, but you cannot force other people to respect you when you do so. You are perfectly free to do as you wish.

@naturallyInconsistent the notation is pretty much peculiar to Fortran in that there the D takes the place of the e; in the other languages like C# it's just a suffix - i.e. 2D is a valid C# double literal, but my understanding is that in Fortran you'd have to write 2D0, right?

@naturallyInconsistent idk what Red Queen is. But QM is incomplete.

u r maybe too unqualified to understand Penrose

@ACuriousMind Yes and no. It is not the case, because the newer versions prefer if you do stuff like 8_dp or with KIND or REAL(n) or whatnot. I actually checked Google before I typed that, precisely because I know some other languages also allow the d notation.

@naturallyInconsistent my point is that I only know other languages where it's a suffix, not an infix exponential operator like apparently in Fortran, and I think it's the infix nature that @SirCumference was asking about more than that it's a double

@ACuriousMind Ah, I see what you mean now. I am not sure if any other languages use it as an infix; I'd simply check via Google whenever I encounter this issue. But yes, I have also answered him that this infix notation is a double precision version of the infix scientific notation that is usually denoted as e.

@ACuriousMind Yeah, that's what I was confused on

@naturallyInconsistent Yep, thanks

Ugh my messages just get sent out of order in this internet

@SirCumference it seems to be quite okay from here

I like seeing questions about Fortran because it reminds me there are many others having to work with ancient programming languages :P

Yeah I was kind of surprised seeing how much it's used in astro

Well, modern Fortran is not ancient at all.

It looks and behaves like Python

dinosaurs that someone has bolted an entire "modern" syntax extension onto are even better!

lol

@naturallyInconsistent I can see similarities. Does feel kind of different from other languages I know though

@SirCumference of course! It has to be Fortran!

Every language has gotta have its weird stuff

But at the very least this one is fairly readable for someone new to it like me

@ACuriousMind I am not even sure if this is true any more. I mean, If you feed F77 and older stuff into a F90 and newer compiler, insisting that the compiler treat it as F90 and newer, then it will crash and burn. Needless to say, F90 and newer files sent to F77 and earlier compilers would also crash and burn.

@naturallyInconsistent oh, if it doesn't have to be backwards-compatible it's another thing

It is extremely not backwards compatible

totally different beast

I'm more used to cases like C++ where the committee keeps adding new stuff but they can't introduce syntax errors for any obsolete constructs because they are committed to not splitting the ecosystem

@SirCumference if you are not faint of heart, glance at F77 code from behind a truck windshield.

@ACuriousMind That's actually a thing that also tripped meow up. Miao miao learnt C++ before there was the ISO standardisations. i.e. back in the C with classes era. Learnt a lot of crappy old C hacks that really hindered understanding of how to write good code these days. And then came back to it after 2014 for the first time. Idiomatic C++ is soooooo goooddd

if you think that, have you heard the word of our lord and saviour Rust? :P

Yes; sadly, my office did not want to adopt my Rust code and instead went back to Python.

>:(

I was also playing with Kotlin and some others. I might try out Julia soon.

Never did get to play with Erlang, though.

Of course, Mathematica is the most beautiful of them all.

ew

However, the one and only widely used language that I really really really cannot tolerate, is Matlab. Absolutely disgusting.

okay, sure, compared to Matlab, Mathematica is beautiful

Maybe COBOL would go with that.

@ACuriousMind It isn't just that. I mean, when I'm coding in Mathematica, I'd use the escapes, so then my code would render as traditional, looking closer to LaTeX renders than code. And then I get to prettify the codebase, sprinkle lots of commentary, etc

hey, my daily work is in what is essentially a strange German-built COBOL dialect that's been frankensteined into a general purpose programming language over 40 years!

The printouts are nice. Not as good as LaTeX, but nice enough.

@ACuriousMind is it the horror known as MUMPS?

no, it's ABAP

MUMPS will literally give you mumps

You will need a baptism, then

well, good thing I'm vaccinated against that :P

measles and rubella too?

sure, and a bunch of other stuff that's obligatory for children to be vaccinated against

Sadly, the MMR vaccine was the centrepoint of Wakefield. Absolute scum.

honestly when you asked that question I briefly feared I'd get a bunch of MMR scaremongering if I said yes :P

lol

@naturallyInconsistent am I right that what Veritasium uses here is actually not the original Bell version, but some later simplified version? Because I remember trying to read the Bell version once and getting lost. It seemed more complicated

the answer to "is this pop-sci presentation simplified" is always 'yes' :P

Oh, one thing: Mathematica somehow has amazing defaults for plotting. If I have a horrible function to plot, there is some chance I'd just give up, send the data to Mathematica to plot, and then copy how it picks the viewport, and then replot the same data in TikZ using the Mathematica viewport

haha ACM right. but in this case I think it's simplified in a way that also got simplified in the official publications, not a dumbing down

I am just curious what this specific version is called, that's the simplified Bell

@Amit Bell's original had no choice but to be air-tight, and so of course that presentation had to be very technically heavy. This simplified version is the essence of it, and it is true enough that you can understand the whole scheme and understand why it is true as it is.

And then there are a lot of modern explorations around the topic, like CHSH

And the conversion of Bell's stuff into a box-colouring problem, which then makes it a logic problem that we can brute-force

Anyway, as long as the simplification does not mislead you into thinking that you understood something that isn't the case, I'm perfectly fine with the simplification.

Okay. Well I think Veritasium used a version very similar to the one explained here, apparently due to Mermin

That's some expert sleuthing

lol

take the star, you rascal

:)

For the record ACM you were right to the extent that after reading this, Veritasium did dumb down even this version but really just a little bit. He essentially took into account only a single "row" from the table relating spin state & detector setting, in order to avoid a more messy probability computation ;)

@Amit but you definitely can see that the relevant computation is only just that one predicts $p>0.5$ without allowing for equality, and the other predicts $p=0.5$, and this disagreement is sufficient to understand the whole thing~

Yes, no doubt. For a 10 minutes video, he did an outstanding job

exactly.

the basic data of a manifold is euclidean space $\mathbb{R}^m$, a topological space $(X, \tau)$, and some atlas structure $\{(U_i, \varphi_i)\}$

the tangent space can be defined just from this

the cotangent space can then be defined as the space of linear functionals over the tangent space (at each point)

I miss the good olde days when I could count on Veritasium videos to be amazing. These days, the visual quality has increased, but I don't know when he is just shilling, or the treatment is pop-sci.

@SillyGoose wait, wait, why do you need the Euclidean space again?

@naturallyInconsistent to be able to talk about homemorphisms into $\mathbb{R}^m$

Ok, then where is your metric?

i think i am confused about that precisely: what structure does the metric induce on the manifold

you need no metric for (co)tangent space

@ACuriousMind surely there should be something to find the length of vectors?

the metric of a (pseudo-)Riemannian manifold is an additional structure on a smooth (or at least $C^2$) manifold

@naturallyInconsistent no, it's not necessary; smooth manifolds do not come with any such notion by default

Ok, I think I see where you are going with that. It is not yet a system we would be doing physics on.

It's really difficult to maintain that balance between pop-sci and sci-integrity. I assume, more so when the millions of likes and subscribers start coming (though I don't really know, to what extent that's also an actual living for anyone, fame and popularity are hard to let go of)

@SillyGoose if the metric is Riemannian, it straightforwardly gives you a metric in the sense of metric topological spaces: The distance between two points is the length of the curve of minimal length between them.

if the metric is Lorentzian (or otherwise not positive-definite), there isn't really such a pithy characterization except the physics idea of it defining what we consider "spacetime", i.e. a notion of causality with lightcones and spacelike and timelike separations etc.

@ACuriousMind I am reading that in the internal logic, negation of a proposition is equivalent to the internal hom to the initial object, $[-, \varnothing]$

many are saying this

But in say Set, wouldn't that be empty itself for every object?

I'm not seeing how you'd have $[[X, \varnothing], \varnothing] = X$

can we have a $V$ and dual space $V^*$ that are not "canonically" isomorphic?

I am thinking of what the metric brings to the (co)tangent structure

@SillyGoose they are never canonically isomorphic unless you have a structure like a metric

so a metric $g$ let's you write $T_pM \ni v \mapsto \langle v, - \rangle \in T_p^*M = g(v, -)$ or something like this right

it brings isomorphism if i am not mistaken. musical one...

since a metric will induce an inner product on any tangent space which then induces a natural way of defining the dual space (as written above)

@Slereah I'm pretty sure that the internal logic of Set is rather weird; it's not a topos, for one

Isn't it just classical logic?

no

i thought Set was like the basic category for internal logic

fuuuck

Why is everything hard

also what do you mean it's not a topos?

isn't set the basic topos

no, sheaves of sets are the basic topos

Isn't Set = Sh(*)

oh, true

it's a topos

but this explains why its internal logic is weird: It's logic about the space with one point :P

Ah I see

So the logic of a topos depends heavily on the site?

I...think so?

I'm not very good at this "logic" part of category theory

ahhh, wait

the logic of Set is classical logic, i.e. the logic of true and false

is there a slick way to deduce the coordinate basis for cotangent space $T_p^*M$?

given that $T_pM$ is spanned by $\partial_\mu$

@SillyGoose It's the basis given by $dx$, ie the exterior derivative of the functions $x$ which are the value of a coordinate at that point

@Slereah nLab actually does this example explicitly

@SillyGoose it's the dual basis

i.e. $\mathrm{d}x^\mu(\partial_\nu) = \delta^\mu_\nu$

also that

ah right okay

The coordinate function $x : M \to \mathbb{R}$ is dual to the coordinate line $\gamma : \mathbb{R} \to M$

It's the line of constant coordinate $x$

and that duality lifts to one on the derivative

@ACuriousMind Ah I think I get it

It only works on the subobject classifier?

And while almost every object has an empty hom to $\varnothing$ (including the terminal object), that's not true for the empty object which has the identity morphism

Although I'm guessing there are Things where other objects pass through the subobject classifier to determine something or other idk

need to look into it

@ACuriousMind hm wait actually how do i parse this. $dx^\mu$ should be a linear functional over $T_pM$ for each $\mu$. so i mean formally the relation you wrote makes sense to define the dual basis, but i don't get how $dx^\mu$ (the differential of $x^\mu$) actually acts as a ma

@SillyGoose Not only does the relation I wrote down make formal sense, it suffices as the definition of $\mathrm{d}x^\mu$.

It's the map that sends the vector $\partial_\mu$ to $1$

Oh apparently to actually get logical propositions, you map your sets to subsingletons

And your proposition $f : A \hookrightarrow B$ gets translated to a proposition via the function that maps... $B \to \{\bullet\}$ I think?

and from there you can do valuations and operations on your propositions

The valuation is something like $[f] = u(A)$ with $u : B \to \{\bullet\}$

Or something like that

I think so, it's a bit hard because i'm not 100% sure that all the different sources do it the same way

@ACuriousMind i get that it provides a definition for dx but i am having trouble seeing how this is consistent with viewing dx as the exterior derivative of the coordinate function x

@SillyGoose The exterior derivative of a function produces a 1-form; a 1-form is, at every point, a cotangent vector, that is a linear map from tangent space to the underlying field

it's also the "standard" derivative of the coordinate function: For the coordinate function $x^\mu : M \to \mathbb{R}$, its derivative ("Jacobian") is a linear map of vector bundles $Dx^\mu : TM \to T\mathbb{R}\cong \mathbb{R}$, i.e. a linear map of the tangent space at each point to $\mathbb{R}$, i.e. a 1-form.

hm well I get that domains and codomains match

I guess I don't get concretely what is $dx^\mu$

the linear map defined by any of these definitions

I'm not sure what you're looking for

i would think if $x = x(y_1, y_2, ...)$ then $dx = \sum_i \frac{\partial}{\partial y_i} x dy_i$

using the usual notion (i think) of the exterior derivative

Yes, for any function $f$ you have that $\mathrm{d}f = \partial_i f \mathrm{d}x^i$ in terms of the coordinates $x^i$

right so then how would $dx \frac{\partial}{\partial x} = 1$

I don't understand the question

these two statements do not contradict in any way

well i guess i am trying to concretely compute that

concretely compute $dx \partial_x$ using the definition of the exterior derivative

@SillyGoose and what definition of the exterior derivative are you using?

the "problem" with such elementary definitions is that there is usually not a single one, but that there are several equivalent characterizations of them, and each of them has some properties that are obvious (or part of the definition) and some you have to do a bit of work to derive, but which of them are easy and which are difficult depend on the exact definition from which you start

i guess i am very confused because it seems we have some natural structures on our manifold which coincide with other mathematical structures. and i'm not sure what is literally coming from the manifold structure and what is just an identification with other math

I'm not sure what that means, so yes, I agree you're confused ;)

heh

well the $dx^\mu$ for instance I can just define as the dual basis of $\partial_\mu$

it has nothing a priori to do with a notion of an exterior derivative or whatever

well...I mean the notation is like that because it turns out to be, if you take that as the definition

conversely, if you define it via the exterior derivative, it turns out it's the dual basis

what is the definition of exterior derivative used for that?

@SillyGoose that's also not unique, of course there are several different possible definitions of the exterior derivative :P

in this case, you have to choose one that does not pre-suppose the existence of the $\mathrm{d}x^\mu$, but already Wiki has at least two ("in terms of axioms" and "in terms of invariant formula")

so is the exterior derivative i wrote down a naive version that is just a special case of the definitions you are referring to

hm

wait nvm

I mean what you wrote down is the "definition in local coordinates", specifically for a the derivative of a 0-form

you can either take that as the definition (and linearly extend it to n-forms), or you have to derive it from any other definition you make

this is why, when a proper math course makes such a definition, you usually have a lemma shortly after the definition where you prove the equivalence between the definition the course has chosen and common other definitions

and once you've proven this lemma, you can always use the definition that's most convenient for the proof you're trying to do

I understand that if you come at this in an unorganized fashion it can feel like you have to make sure this stuff is not circular, but that's why people write textbooks that start from one definition and carefully derive the other; it's again a matter of getting used to it - once you've seen this kind of "definition switching" for some things, you don't really think about it anymore once you've proven the equivalence of all the different definitions

i guess i am still confused because in the (simplest?) case we might want to show that $dx(y) (\partial / \partial x) = 1$

if i expand this, i get $(\partial x/\partial y) dy (\partial/\partial x)$

and then im not sure how to further simplify this expression into a $1$

again, what definition of the exterior derivative are you using?

the "expand" part suggest you're using the definition in terms of local coordinates

yesh

but then, of course, this is trivial: By definition $\mathrm{d}x^\mu(\frac{\partial}{\partial x^\nu}) = \delta ^\mu_\nu$ for any choice of coordinates $x$

there is literally nothing to show

but i obtain that relation from linear algebra

I don't know what you mean

if we work in coordinates, that's the definition

we don't "obtain" it, it's the definition

er hm

you mean that is part of hte definition of exterior derivative?

if you want to show that this is consistent, i.e. that you can express coordinates $x$ as a function of $y$ and still get the delta by just applying the definition in terms of $y$, it's just a bunch of applications of the chain rule