The h Bar - 2020-09-18

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2:28 AM

Can someone just point me in the right direction for my hw q?

"A bottle is dropped from a moving airplane (ignore the effect of air resistance). If the plane from which the bottle was dropped was flying at a height of 500m, and the bottle lands 400m horizontally from the initial dropping point,

a) how fast was the plane flying when the bottle was released?"
So x(0)=0 and x=400 m and y=500 m
I'm not sure what else to do
Bc I don't have time... unless is it just 500*m/9.8(m/s^2)?

Oh wait hold on, I think I got it

3:29 AM

@Færd i like how the picture for gravitational waves is the jamba juice logo

1 hour later…

4:48 AM

@ACuriousMind Hi, are you there? I have a couple of questions...

2 hours later…

7:09 AM

@juliensurel if you had asked your questions directly, I could've answered them now... ;)

5 hours later…

12:21 PM

so my first question is, whether a path integral formulation of classical problems would be a better candidate than the quantization procedure based on the Hamiltonian picture of classical mechanics, in order to avoid the consistency issues with the axioms of the latter.

My second question is regarding Witten's volume formula for moduli space of flat connections. I can follow but not deeply understand his proof. But I recently learned about evaluating stationary phase integrals (I believe this has some similarity with path integral formulation of a classical Lagrangian),

and I think in this context they are related to path integrals in the partition function of the Yang-Mills lagrangian. can his formula be interpreted as asymptotic expansion of the integral in any way?

@juliensurel I have bad news regarding the consistency of path integrals

They're not really any better than the canonical quantization

Is the main issue the so called anomaly in the integral measure?

Well in the case of the path integral you have to build an appropriate measure/integrator

which is about as simple to do as building a theory from canonical quantization

they may not exist or be unique

Both methods have their strengths and weaknesses, but none is so amazing that you could just declare it the winner of quantization

Can you tell me an example where path integral formulation is inconsistent?

12:37 PM

Do you have an example in mind where the canonical quantization is?

Are you talking about operator ordering ambiguity?

Or are you talking about nonlinear theories?

I do not know much about it, for instance, I know the commutator relation on p and x cannot sometimes be extended to even polynomilals.

@ACuriousMind so my first question is, whether a path integral formulation of classical problems would be a better candidate than the quantization procedure based on the Hamiltonian picture of classical mechanics, in order to avoid the consistency issues with the axioms of the latter.
My second question is regarding Witten's volume formula for moduli space of flat connections. I can follow but not deeply understand his proof. But I recently learned about evaluating stationary phase integrals (I believe this has some similarity with path integral formulation of a classical Lagrangian),

1:03 PM

@juliensurel I'm not sure what you mean by a "path integral formulation of classical problems"

the path integral is a quantum mechanical tool - it is an alternative to quantization approaches based on the phase space algebra of observables, but it is not classical

@juliensurel The stationary phase of the path integral forms around the path that is the classical solution of the e.o.m. (since the e.o.m. are the equations for extremizing the action in the first place), so if we can argue that this is a good approximation, indeed the classical solution dominates the value of the integral

Hi guys good evening

danielunderwood

I was reading the Wikipedia entry on $L^p$ spaces and the section on dual spaces mentions

> The dual space (the Banach space of all continuous linear functionals) of $L^p(\mu)$ for $1 < p < \infty$ has a natural isomorphism with $L^q(\mu)$, where $q$ is such that $\frac{1}{p} + \frac{1}{q} = 1$

Since wavefunctions are in $L^2$, is the isomorphism between $L^2$ and $(L^2)^*$ hiding anything (kind of like how a cross product just works in 3D)?

@danielunderwood The $L^2$ space is the only $L^p$ space that is a Hilbert space, and Hilbert spaces are self-dual

What this shows is just that being a Hilbert space is a kind of strong condition for a "random" Banach space to fulfill

1:19 PM

is the one space because are rest are not useful in qm?

@ACuriousMind

Perhaps rephrase that sentence a bit

We use $L^2$ in QM but other spaces have their use, yes

Just not as Hilbert spaces

@JackRod I don't understand the question: Is the one space what? (and at least one of your "are"s should be a "the", I think)

@juliensurel That's a "problem" in the sense of wanting a "nice" and unique quantization map, it doesn't make the resulting quantum theory ill-defined. The problem with path integrals is that we don't know how to rigorously define their measures in arbitrary dimensions for arbitrary fields!

The path integral exists mathematically rigorously only in two dimensions, and sometimes in three. The definitive work on this is still Glimm/Jaffe as far as I know, we haven't made much progress there

Also you still need to renormalize for both path integral and canonical quantization

So all in all neither is quite amazing

Actually, I asked this question to many people why in QM Lebesgue spaces of the second degree is the only one that corresponds to the Hilbert vector space of state functions?

danielunderwood

@ACuriousMind Ah, so being self-dual is why we have a trivial mapping between $| \psi \rangle$ and $\langle \psi |$ just as in the mapping between $v_a$ and $v^a$ in 3d Euclidean space?

1:30 PM

@danielunderwood Yes!

@JackRod The dual space in QM has to be isomorphic to the original space

By Riesz theorem

This is true no matter what representation of the Hilbert space you use

@JackRod We need the physics state space to be a Hilbert space since we need the inner product to compute probability amplitudes, and the $L^2$ space is the only Lebesgue space that's also a Hilbert space. Also, Stone-von Neumann tells us the $L^2(\mathbb{R}^n)$ spaces are the essentially unique representations of the CCR.

Ok when I asked this question to my professors they said I am challenging the fundamentals of physics

lol

for shame

1:35 PM

They misunderstood me

either that or they did not want to admit they didn't know an answer :P

Does it happened with you?

as well like during the college days

no, I never had a prof accuse me of challenging the fundamentals of physics :P

"I actually don't know, let's find out" is an acceptable answer to questions around here, though...

@ACuriousMind good though i accept they were just a high school teachers at that time

so it is not good to say that.

are you a web designer?I read your profile you work for a software firm what kind of work you do?

2:03 PM

Thanks everyone for their answers

2:14 PM

@JackRod As the profile says: "I develop code analysis software at SAP."

What are the requirements of big companies like google? for working with them

Uh...you need to convince them you can do the job they're offering, just like with anyone else?

that's it no other requirements they must be getting thousand of requests every day like me?

well, the specific job description usually tells you what they expect and what's nice to have

like any other company, the requirements are usually some kind of degree + some kind of work experience (or not, if it's an entry-level position) + some kind of experience with the tools you'd be working with

Haven't you ever looked at a job description?

@ACuriousMind Fortunately employers never judge potential recruits by any other metric

2:23 PM

@Slereah I would say they indeed rarely do but many have skewed ideas of what counts as a convincing argument for this :P

@ACuriousMind Well, do you have accurate ideas?

or is it just your gut feeling

my gut is accurate!

What is even the sigma of your gut

@ACuriousMind like a recruiter would be looking for the something.

if there was an objective way to really determine whether someone is a good fit for a position without putting them into it the world would be a different place :P

I've written a grand total of one job application in my life so far and I got the job so I'm really not the person to ask about how to get a job

just kinda stumbled into it :P

2:31 PM

I sent like over a hundred applications to get this one

@Slereah for the same software engineer?

I don't know what that sentence means

2 mins ago, by Slereah

I sent like over a hundred applications to get this one

I still can't parse that sentence

ok sorry for making it hard what i was saying is what kind of job are you doing currently?

@Slereah

have good day all of you ahead as i am leaving

lol

2:50 PM

If you extremist the action, doesn’t that mean that proper time is the minimal time taken? Not the maximum?

assuming we minimise the action and not maximise it

we just search for extrema

Mhm. But wouldn’t the proper time always be the minimum? The maximum could be infinity, no?

I can’t see why proper time maximised is the path taken

depends on whether in your sign convention proper time of timelike paths is positive or negative whether this is technically a minimum or maximum

Time like are negative in tongs notes

so that’s what I’m working with here

user image

Like I don’t see how the action ( which is essentially that integral negated) implies proper time is a maximum

Add a very small deformation to the exact solution, and see what sign this will have

ie pick $x^\mu = \sigma x_0^\mu$

Add some term $+ \varepsilon \delta x^\mu$

and work out the Taylor expansion

2:59 PM

Gotta run but will check it out

quick additional question Whilst I’m away

couldn’t you just go on a slightly larger path? Thus taking more time

i.e. couldn’t you always take a path in Minkowski space that takes longer

Does this being a maximum imply you can take a path shorter than the proper time?

That formula is for the proper time

Are you sure he is talking about the same formula for the action

the action is usually $-m\int d\tau$

3:17 PM

The action is defined similarly

One sec I’ll screenshot

Ugh I can’t send screenshots again

share.icloud.com/photos/0s0qjUtqgd2Zcs1aPe5IQoX8Q

At the top of this. @Slereah

@JakeRose It's written $$S = -mc \int d\tau$$

3:47 PM

Sept 18, 2020: dmckee & David Z are officially no longer Phys.SE moderators.

4:17 PM

evening

So necessarily a stationary point of the action is a stationary point of the proper time. The minus sign changing it from a minimum to a maximum. Why can’t you always choose to just go a bit further though?

Is maximising proper time another way of saying you can’t break light speed?

It's a property of globally hyperbolic manifolds that for any two timelike separated points, there exists a curve that extremize their proper time

the same way that a geodesic in a Riemannian manifold is a minima of curve length

There's a proof in uuuuh O'neill I think?

also probably Hawking Ellis

What kind of phones y'all guys use?

oh also you're dealing with special relativity

Basically you just have to prove that a non-straight line will have a larger proper time

@Slereah action integral

4:26 PM

By which I mean, a smaller in value but negative proper time

That's basically the point of the twin paradox

Any accelerated observer will experience a smaller proper time

0

Q: Problem with deletion of comments

Someone decided to delete all comments posted under my question here: Do "almost black holes" exist? This is insane, there were valuable comments. They should have been moved to chat if they were too numerous as this is usually done, not deleted! Edit: I have seen rob's comment at the location of...

support comments

5:20 PM

When we deal with non-linear coordinate systems, we still need a notion of covariance and contravariance (for example cartesian to polar coordinates), but we're no longer working with vector spaces, what are we formally doing here? Is this "covariance" under the diffeomorphism group?

in order to be able to tell what you're doing formally, we would first need to know what you're doing non-formally ;)

"deal with non-linear coordinate systems" is a bit vague

I'm really just thinking any coordinate system that isn't well defined on a vector space, eg. polar

What's bugging me is the treatment of polar coordinates as "the basis vectors change at each point", the slightly troubling image of $\hat\theta$ and $\hat r$ "basis vectors" defined at each point and rotating as we move along $\hat\theta$ comes to mind

No coordinate system is "as on a vector space" on manifolds

If I had to be more specific I'd say coordinate systems in which we have to treat $\Bbb R^n$ as a manifold rather than a vector space

@Charlie it's not a troubling image - you're treating $\mathbb{R}^n$ as a manifold that has a tangent space at each point and you're choosing polar coordinates $(r,\phi,\theta)$ for it so at each point you have a basis $(\partial_r, \partial_\phi,\partial_\theta)$ for the tangent space at that point

5:28 PM

It's not particularly bothersome, and it's what you do in classical mechanics, too!

Did you never do orbital mechanics?

With basis vectors in spherical coordinates?

@ACuriousMind ah it's only troubling if we're still thinking of $R^n$ as a vector space, this is likely just that the distinction isn't explicitely made that we're working on a manifold, especially in introductory physics linear algebra courses

when people aren't doing differential geometry they tend to sweep this under the rug and just talk about "varying" basis vectors $\hat{r}, \hat{\phi},\hat{\theta}$

@Slereah I haven't taken a lot of standard physics undergrad courses but I have briefly looked at classical orbits when trying to understand them in gr

I'm not a particular fan of this pedagogic approach but that's how it is

@Charlie Bit of a tall order to do GR now then!

5:32 PM

Tall order or not it was sufficient to pretty convincingly pass the gr exam :P then again orbits were in the later part of the course where things become more complicated and less detailed in lectures

I'm pretty happy to drop the idea of vector spaces altogether, but unfortunately my understanding of what constitutes covariance was pretty carefully resting on the idea of representations on vector spaces

The substitution I was expecting to make when working on manifolds instead was that covariance/contravariance was related to the diffeomorphism group

Since "covariant/contravariant [...]" is almost always followed by the word vector, this becomes a problem if we're on a manifold

Well you still have vector spaces

just at every point of the manifold

hmm that's true

5:47 PM

The tangent space has many many definitions

if you so wish, you can check that it is a vector space for every one

and for every definition you can also check that the change of coordinates induce a change of coordinate basis

6:26 PM

Ok I went for a walk and thought about it a bit and I actually think I get it

We define the coordinate functions on the manifold, these naturally induce a basis $\partial_i,\mathrm dx^i$ of the tangent spaces at each point. Then when we ask if our vector field transforms covariant/contravariantly we are asking how the components of the vectors in the field change under the coordinate transformation, since the coordinate transformation, as you say, induces a change of coordinate basis.

And I haven't given it a huge amount of thought, but I assume the Jacobian matrix for the coordinate transformation acts on the tangent spaces, and whether the components transform as the Jacobian (vectors) or it's inverse (covectors) determines whether we have a covariant or contravariant vector field

that sounds just about right

that has been bugging me for literally months

ty

Basically at every point of the manifold, the change of coordinates correspond to some invertible matrix

it just find the terminology a bit weird - a "contravariant vector field" is what we usually call a 1-form :P

which encapsulates the change of basis

@ACuriousMind or dual vector

or covector

or cococovector

the dual co-1-form

6:36 PM

the co/contravariant talk is usually used in some setting that doesn't use diff.geo. concepts so that the transformation behaviour is the defining property

@ACuriousMind You could say that "covariant" applies more to the components than the 1-form itself

it's not wrong, but expect mathematicians to stare blankly at you if you try to talk to them like this

oh wait I thought covectors were covariant

@ACuriousMind I bet "contravariant vector" was created before 1-forms!

oh wait it doesn't matter reallyt does it

6:38 PM

the $\mathrm{d}x^i$ stuff is the covectors/1-forms. I honestly use co/contravariant so rarely that I might've mixed them up

If they're not covectors, why are they part of the cotangent space

@Slereah sure, doesn't mean it's the standard way to still talk about it

Answer me that

@ACuriousMind what would you instead call a "covariant vector field"?

@Charlie just a vector field

6:39 PM

oh, fair enough

there's vectors and covectors, their fields are "vector fields" and "1-forms"

As I mentionned, "covariant" mostly refers to the components

Not the vectors themselves

not the most symmetrical of naming conventions, but that's how it is :P

yeah fair enough

1-forms are covariant because they vary with the basis

that's what co-variant means

for your basis $\partial_\mu$, take your Jacobian $J^\mu_\nu$

Such that the new basis is $\partial_\nu = J^\mu_\nu \partial_\mu$

The components of a dual vector vary the same way, ie $\omega_\nu = J^\mu_\nu \omega_\mu$

On the other hand, vectors are contravariant

They vary against

6:42 PM

see, this variance stuff is confusing, why don't we call them covectors and contravectors :P

ie for a vector $V^\mu$, your new components are $V^\nu = (J^{-1})^\nu_\mu V^\mu$

You're using the inverse matrix

that's what the co/contra business means

where the reference you're using is the tangent space coordinate basis

That way, the vectors themselves are invariant

ie $$V = V^\mu \partial_\mu = V^\nu (J^{-1})^\sigma_\nu J^\mu_\sigma \partial_\mu = V'$$

@ACuriousMind Don't confuse co- as in "dual" and co- as the latin root for "with"!

Covectors are duals to vector, but it's unrelated to co/contravariance

6:57 PM

Is there anything we can say about the Jacobian of the coordinate transformation acting on a single tangent space as being a representation of a particular group?

haven't we been here before? :P

Sep 7 at 12:48, by ACuriousMind

I mean, you can certainly "freeze" the map at one point - the map that maps a diffeomorphism to its Jacobian at a particular point would form a representation on the tangent space at that point, but that's not a very natural viewpoint to take

mmmyes probably, those were less enlightened times for me :P

ah I remember this conversation I'll go and reread itty

The Jacobian at a point is just in $GL(\mathbb{R}^n)$

But then again, that's the largest possible group on $\mathbb{R}^n$

so that's not saying much

but really this isn't interesting from a representation-theoreticl viewpoint - the possible Jacobians are all invertible matrices $\mathrm{GL}(\mathbb{R}^n)$, whose unique irreducible representation is precisely $\mathbb{R}^n$

ok it makes more sense this time

7:04 PM

Don't worry too much about groups

Unless you're doing some fairly complicated things, GR doesn't use a lot of groups

You don't really need groups for doing basic GR

@ACuriousMind, so about Witten's formula for partition function int e^{-L(A)}dA, where L(A) is the classical yang mills action, would it be possible to expand the integral as an asymptotic series? I think the expansion is only possible in the semi-classical limit when you have int e^{iL(A)/h} dA so that the exponent is very large...

Remember to use dollar signs for the mathjax

yes

you can get mathjax for the chat here

@juliensurel the trick with topological theories is that if they compute a topological invariant like the volume here, you can take any limit you want and it can't change what it computes because the limits are smooth but the invariants can only change discontinuously

at least that's the physicist's handwaving for why this works

it's never convinced me on a deep level but Witten's success proves him right, I guess :P

7:11 PM

@Charlie, unfortunately the computer I'm using is not mine and I don't have admin credentials to dl anything.

You don't need to download anything for mathjax

tbf if you put the dollars in it will at least appear in mathjax to acm or whoever you're trying to talk to

might as well minimize the suffering :P

also if you put in a big latex formula, nobody is gonna read it without mathjax :p

7:24 PM

do I just need to click on the those tabs to be able to run latex?

I didn't seem to work

3 hours later…

10:20 PM

I've run into a problem, the tangent space to spacetime is $\Bbb R^{1,3}$, but the canonical coordinate basis is $\partial_i,\mathrm dx^i$ (which uses the coordinate of the charts, which are in $\Bbb R^4$), what am I missing?

I have assumed, maybe wrongly, that coordinates on $\Bbb R^4$ can't be treated identically to coordinates on $\Bbb R^{1,3}$

what do you think the difference is between coordinate on one space compared to the other?

the only difference between them is the metric/inner product on them, but the definition of coordinates or the tangent space never refers to a metric - it just needs the properties of $\mathbb{R}^n$ as a smooth space

do we not run into issues when computing the norm of the vectors? zero norm vectors in R^1,3 are different to the zero vector in R^4 no?

hmm

saying "the tangent space is $\mathbb{R}^n$" is a statement about vector spaces, not about inner product spaces

a priori there is no notion of inner product or norm on the tangent space

oh

but we have to introduce one at some point no?

that's what a metric is, in its proper mathematical definition - a notion of inner product for each tangent space

10:24 PM

otherwise the tangent space isn't much use

the $\mathrm{d}s^2$ stuff is just physicist mumbo-jumbo :P

@Charlie that's not true - you need the tangent space to define a vector field, and why would the notion of vector field only be useful if you have a metric?

oh yeah that's a good point

you can study e.g. the flows generated by vector fields, or their zeros (cf. hairy ball theorem)

none of that needs a metric - differential geometry/topology is much larger than (pseudo-)Riemannian geometry where you have a metric

once we do have a metric, then do the coordinates become a problem?

depends on what you think the coordinates do :P

10:28 PM

at the moment to me they just provide a canonical basis on $T_pM$ and $T_p^*M$

unless it's useful to define the tangent space in the target space of the charts?

but then they aren't even 4-vectors, which might defeat the point

then I don't see where you think you need the notion of an inner product on the coordinates, i.e. how you could distinguish "coordinates from $\mathbb{R}^n$" and "coordinates from $\mathbb{R}^{1,n-1}$"

yeah I guess I can't think of a way to distinguish them, the only problem I can imagine is in the norms

which, even if we're not equipping the tangent spaces with an inner product at first, still has to come into play at some point if we want to know whether vectors are lightlike etc.

the definition of coordinate charts $\mathbb{R}^n \to U\subset M$ makes no mention of norms

the map is a homeomorphism, or a diffeomorphism, it is not a morphism of things that know about norms or inner products

wait is that back to front?

hm?

oh, you want to write the chart as $U\to \mathbb{R}^n$?

10:32 PM

why is the coordinate chart going from $\Bbb R^n$ to $U\subset M$?

doesn't matter, it's an invertible map :P

oh ok :p

I mean, if you don't see why it doesn't matter you should probably think about it, but the "spirit" of the definition really doesn't care about which way you write the map

so if I define a tangent vector in the tangent space at a point $p$ and I want its norm, I have to define an inner product on $T_pM$ right

yes

that's what a metric is - an assignment of an inner product to each tangent space

10:36 PM

And if we say it's a lightlike vector, I expect it to have vanishing norm

oh wait hmm

ok now I'm confused about how the tangent space can even know about the coordinates in the chart in the first place

unless we define the tangent space on $\Bbb R^n$

which would be news to me

I'm confused what you're confused about :P

what do you think we mean when we say things like "$\partial_r$ is a tangent vector"?

oh no

to my understanding there is a 1-to-1 correspondance between directional derivative operators at a point and the vectors that can be defined extrinsically in the tangent space

yes

and do not obviously the directional derivatives w.r.t. to the coordinates form a basis for the directional derivatives?

yes

so that's how the tangent space "knows" about the coordinates

if that doesn't answer your question you'll have to ask it differently :P

10:47 PM

oh ok so to define the tangent space at a point $p$, we need a chart containing $p$, then the tangent space to the euclidean $n-$space that is the target space of the chart is "the tangent space at p"

there is no "tangent space to the euclidean n-space"

oh

but it's a manifold right?

the $\mathbb{R}^n$ in the definition of a chart is not a manifold

oh

well that is a horrifying revelation

it would be a recursive (in a bad way) definition if we defined a manifold in terms of local isomorphisms to $\mathbb{R}^n$ as a manifold

the charts are just homeomorphisms of topological spaces

10:50 PM

oh yeah I guess that is circular

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