The h Bar

0celo7

15:03

@ACuriousMind Suppose $\alpha$ is a curve w/ $\alpha(0)=p,\alpha'(0)=v$. I have a suspicion that $||v||=\lim_{t\to 0}\frac{1}{t}d(\alpha(t),p)$. ($||\cdot||$ is the Riem. norm, $d(\cdot,\cdot)$ the Riem. distance.)

I have a physicist's proof of this.

Namely one can approximate $\alpha$ with a length-minimizing geodesic to first order in $t$.

And after some magic, you get $||v||$ as the result.

But that needs a Taylor expansion.

@ACuriousMind Or...maybe I can use Taylor's theorem directly?

Emilio Pisanty

@0celo7 Why do you need a full Taylor series? Check your definition of differentiability, which probably reads $\exists \alpha'(0)$ such that $\alpha(t)=\alpha(0)+\alpha'(0)t+$(suitably bounded terms) (possibly only in a chart).

0celo7

@EmilioPisanty Yes, that's what I'm thinking.

Emilio Pisanty

In any case, your proposition looks like it just follows directly from the definition of the Riemannian distance itself.

0celo7

@EmilioPisanty How so?

Emilio Pisanty

@0celo7 Well, how are you defining it?

0celo7

15:08

@EmilioPisanty $d(p,q)=\inf\{L(c)\mid c\in\Omega_{p,q}\}$ (path space).

Alfred Centauri

@JohnRennie we likely have differing views on what is the moral high ground.

Emilio Pisanty

@0celo7 Yeah. So work with that $\mathrm{inf}$.

John Rennie

@AlfredCentauri possibly, though sarcasm is rarely involved in moral elevation :-)

0celo7

@EmilioPisanty But one of the points varies with $t$, this isn't trivial.

And $\alpha$ need not be length minimizing

So $d(\alpha(t),p)$ will be calculated with some other curve

Emilio Pisanty

For instance, the straight path from $p$ to $p+\alpha'(0)t$ (in a chart), and then however you want to make your way to $\alpha(t)$, is one suitable $c$ in $\Omega_{p,\alpha(t)}$.

0celo7

15:11

As $t\to 0$, I can assume there is a geodesic that is length-minimizing

Emilio Pisanty

So that can help you get a bound on $d(p,\alpha(t))$. You can then strengthen this using the boundedness of the residue from the definition of differentiability.

That sort of game.

0celo7

@EmilioPisanty Yeah. But I don't know how the derivative of the "suitably bounded terms" behaves. And the length is calculated from the derivative of $\alpha$.

Emilio Pisanty

@0celo7 But you don't really need that.

Formally you're working on a manifold

but really for all intents and purposes you're working in $\mathbb R^n$

with a slightly wonky metric

0celo7

A position-variable metric, which always makes such computations treacherous.

Emilio Pisanty

So you're free to choose any suitable path between $p+\alpha'(0)t$ and $\alpha(t)$

just choose something that is convenient and which you can still provide appropriate bounds for.

You won't be able to find an exact expression for the length of that residue

but you can still show it's smaller than other stuff

0celo7

15:24

@EmilioPisanty Ok, I'll continue working on it

Emilio Pisanty

@0celo7 👍🏼

ok, yeah, that doesn't work

0celo7

@EmilioPisanty hmm?

Emilio Pisanty

@0celo7 For future reference, copy-pasting emoji into chat doesn't work.

that was a thumbs up

0celo7

I might not need the result -- it's used for a "quick" proof in a geometric analysis book I'm reading. But I already know how to prove it another way.

@EmilioPisanty Looks fine on my screen.

ACuriousMind

@EmilioPisanty I see a thumbs up, followed by a non-displayed unicode

0celo7

15:26

ACuriousMind

Emilio Pisanty

0celo7

Noobs.

Emilio Pisanty

anyways, yeah, whatevs.

Bernard Meurer

@0celo7 You there?

0celo7

15:31

@BernardMeurer Perhaps

Bernard Meurer

$$a = S - (S-a); (S-a)=\epsilon; a=S-\epsilon$$ for $\epsilon >0$

0celo7

Trying to decipher a 1939 paper

it's horrible

Bernard Meurer

Any idea what this means?

0celo7

sure

not sure why you need it

Bernard Meurer

I don't get it

0celo7

15:32

what part

who writes like that anyway

is that how they're teaching you to write in your analysis class

ACuriousMind

@BernardMeurer What do you mean, "what this means"?

It's a series of equations, the third follows from the first two, but as written it's just three equations.

0celo7

that's my confusion too

Bernard Meurer

@ACuriousMind We were defining the maximum of a given set, and we said that there is: $$S\text{ s.t. }\begin{cases} \forall x\in K: x \leq S \\ \forall\epsilon>0 \exists x\in K: x>S-\epsilon\end{cases}$$

For $K\subset \mathbb{R}$

0celo7

The maximum?

Or the supremum

Christ

Bernard Meurer

Yeah, so we defined the supremum

brb, test time

0celo7

15:37

@ACuriousMind Is it normal to get drowned in logical quantifiers in European analysis?

ACuriousMind

@0celo7 Yes

0celo7

@ACuriousMind It looks terrible

ACuriousMind

It's mainly due to the profs being too lazy to write full sentences on the blackboard, I think

0celo7

@ACuriousMind My topology prof writes "exists" and "for all" out.

Elven Lord style

Why not just say $s=\sup K$ is the least upper bound? That is, $s\ge k$ for any $k\in K$, and $s\le b$, for $b$ any upper bound of $K$?

ACuriousMind

Well, some define "upper bound" first (with quantifiers) and then do that, yes

0celo7

15:40

@ACuriousMind I did. An upper bound is a number s.t. $b\ge k$ for any $k\in K$.

ACuriousMind

However, learning to read and deal with quantifiers is a valuable skill

0celo7

No need for quantifiers.

@ACuriousMind I can read them just fine.

But no one actually writes mathematics like that.

ACuriousMind

Okay, I don't have a really good reason why this is done

0celo7

Good. Admit defeat.

@ACuriousMind My algebra prof took off points for too many quantifiers. Said it's "informal."

Some people were writing stuff like $\forall r\in R\exists k\in K\Leftrightarrow g\in G:\epsilon>0\implies a\land b\in \rho$ and stupid stuff

@ACuriousMind If we extend a map of vector spaces "by proportion," does that mean demanding $F(cv)=cF(v)$?

ACuriousMind

I have never heard that term

0celo7

15:49

Me neither

what the heck is $\overline{\lim_{s\to 0}}$

and $\underline{\lim}_{s\to 0}$

ACuriousMind

Never seen that either

0celo7

PhD level analysis...

I think it means either limsup/liminf or one-sided limits.

user116211

@0celo7 Really weird if they indeed denote that.

0celo7

Yes. Limsup.

user218912

16:25

@0celo7 I figured out everything. I am now level 100 maturity.

0celo7

Doubtful.

Explain everything the.

*then

user218912

not much to explain, the prof taught us how to do derivatives of the lagrangian density today. so all my issues are cleared up.

user218912

and no, it's not that obvious and simple because

user218912

when he asked the class what ___ term becomes

user218912

nobody raised their hand

user218912

16:36

lel

Xasel

dead?

user218912

dead?

ACuriousMind

dead?

user116211

dead?

John Rennie

That explains the smell

Xasel

16:46

zombies out there?

physicist zombie designinga vaccum tube missile

is that a secret spell to raise the dead from their grave ? :P

ACuriousMind

Are you trying to communicate with us or are you just saying random words? :P

Xasel

depends on your perspective :P

John Rennie

Does radiation kill zombies? If not you could use zombie students and save a fortune on shielding.

ACuriousMind

@Xasel No, whether you are trying is certainly not dependent on my perspective. What's dependent is whether you're succeeding.

Xasel

@John Rennie Well I googled a lot about local velocity but couldn't find any info on it and the answer you linked seemed to quite above my mathematical background

Well what's the probability of both Curious Mind

John Rennie

16:50

@Xasel What about local velocity? Is this something you asked yesterday?

Xasel

yeah

John Rennie

Does this mean anything? I've read it several times and I'm still not sure.

0

Q: Question related to average velocity

While i was solving a problem in physics [![Snapshot of the question][1]][1] [1]: https://i.sstatic.net/543UG.jpg from P.A Tipler sir's book , i found in one problem ( which i included here as snapshot).Here(in problem) in underline it is written that "this is not the average of running and j...

kinematics

Xasel

oops goota go....discuss about it later

user218912

all of a sudden my shoes are really hurting the sides of my feet and I'm in my dorm and my class is a 30 min walk away.

user218912

idk what to do

ACuriousMind

16:56

@JohnRennie See my edit & comment

user218912

if I go to class I'll probably end up having blisters but if I don't go to class then I'll lose marks since it's a lab.

user218912

i'll just go, bye

0celo7

@IceLord I only answer the questions if I want to impress the prof

Which happens in three classes right now

Xasel

Can somebody explain me how black holes can be created in violation of pauli exclusion principle?

John Rennie

You have first to explain how they violate the exclusion principle

ACuriousMind

17:08

^

0celo7

Lol

Xasel

becuse as far as I know pauli exclusion principle states that no two electron can occupy the same state

0celo7

@ACuriousMind pls remind me later to rant

Like around 5 pm. Thanks.

ACuriousMind

@0celo7 About what?

I will not remind you about a rant I don't want to hear :P

Xasel

and in the creation of blackholes mass gets concentrates at certain point due to high gravitational force

ACuriousMind

17:09

Also, 5pm was four hours ago :P

0celo7

Graduate students trying too hard to impress the prof/find mistakes.

dmckee

@Xasel And presumably you think that they must being doing so somewhere inside the horizon?

0celo7

Ok, like 11PM your time.

ACuriousMind

@Xasel Ah, see, at the point where the black hole has formed, no one really knows what's going on inside the horizon

dmckee

Two issues with that: (a) you haven't shown that this is the case and (b) nothing that happens on the other side of a horizon is even communicated to our region.

ACuriousMind

17:10

In particular, the validity of naive quantum theory in such a highly general relativistic regime is...questionable.

Xasel

so say if n elecron get trapped there might be possibility of atleast 2 having same estates thus inturn violation ?.......

confused?

naive quantum theory?

dmckee

If you ignore the "can't know what happens on the other side" thing and take the naive balckhole solutions to GR serious the space-time on the other side of the horizon is rather different in character than our own.

Xasel

on what basis do quantum theory can be questioned? @ACuriousMind

@dmckee:how it can be so that a cocentrated mass alters the property of space-time?

dmckee

@Xasel It has a built in assumption about space-time that is violated (or at least not obviously true) in a strong gravitational field region. The two theories are not compatible in a straight forward interpretation.

ACuriousMind

@Xasel There's a lot to consider here: 1. The Pauli exclusion principle is derived from standard QFT, which does not incorporate general relativity, so it's not clear whether or not it applies here. 2. Even before collapsing to a black hole, we believe the matter goes through phases such as neutron stars and quark gluon plasma, which exist precisely because the gravitational attraction is stronger than the exclusion principle.

3. The interior of a black hole is a rather weird spacetime where the former radius has become the time coordinate. I'd be careful to make any statement about it.

dmckee

17:15

@Xasel If you accept the existence of black-hole you have already conceded that space-time is vastly altered in such a region. As far as we know the effect of mass/energy/momentum on space-time just is. That's the bottom of the current understanding.

ACuriousMind

The exclusion principle only says that no two fermions can be in the same state, it does not forbid that the fermions occupy arbitrarily highly energetic states, which is precisely what happens in the degenerate matter that forms neutron stars

dmckee

@0celo7 The one person I know personally who has been the victim of a random mugging in an alley is a big muscular guy. It doesn't help much if you get coshed from behind.

Situational awareness is key.

Basically, you should flee any situation involving a squirting lapel flower.

Xasel

hmm...interesting:gravitational attraction is stronger than the exclusion principle

user116211

What to do with this:

user116211

0

Q: Can you build a room at the bottom of the ocean?

This is the scenario: You travel to the bottom-most depths of the sea. Naturally this environment is not viable for human life outside of some kind of submarine. So you then punch a hole in the sea floor. You hollow out the area under the sea floor to create a room. You block off the room from th...

pressure

user116211

17:26

?

Xasel

Then why EM force can be used to bring two fermion at same state @ACuriousMind

ACuriousMind

@Xasel I don't understand the question.

Xasel

You said the gravitation attraction is stronger than the ecxlusion principle

and we know that EM force is stronger than Gravitational

ACuriousMind

That sentence doesn't mean what you think it means.

Xasel

then why can't we violate the exclusion principle with the help of EM force....hmm then enlightne me on what it means?

^sorry if i sound annoying

ACuriousMind

17:29

Try to describe a scenario in which the EM force gets so strong. In the case of gravitation it's easy: Put a lot of mass close together. But how would you do it for the electric force?

0celo7

@dmckee There's no alley. Just a big parking garage

Xasel

hmm..in Halliday Resnick there is a numerical compared the gravitation and EM force...uhuh I get it how can we get a concentrated charge?(this is the problem ain't ti)

0celo7

And Knoxville has a lot of methheads in the surrounding countryside...

@dmckee A girl was attacked by a clown last night about 50 feet from my place

It sounds silly but it's real...

Obliv

18:03

Can anyone explain what specific heat is a function of here -> $C_v = (\frac{\partial U}{\partial T})_v$ Like, what is $v$ and why isn't it just $\frac{dU}{dT}$?

I read on the wiki page that it's a function of the structure of the substance itself

ACuriousMind

It's specific heat at constant volume

Obliv

oh okay

what about the partials

ACuriousMind

What about them?

Obliv

Oh I think I get it. It's the change in $U$ with respect to $T$ only. $U$ might be changing with respect to other variables right

so $\frac{dU}{dT}$ would be incorrect notation

Xasel

hey i just learnt about charged BH..Can they generate a large EM force on aaccount of large charged mass concentrationd

Obliv

18:08

ah and $\frac{\partial U}{\partial T}$ would depend on the degrees of freedom that the particles of the substance has to store thermal energy in.

Xasel

hmm....dead again?:P

ACuriousMind

@Obliv Yes.

@Xasel Define "large".

Obliv

@acuriousmind what are you up to

ACuriousMind

huh?

Obliv

I meant that as in what's up

ACuriousMind

18:14

not much, it's a lazy Thursday evening

Obliv

I wish I could be lazy : ( @acuriousmind Did you start your 3rd year of grad school? I forget

ACuriousMind

Officially it's my fourth semester in the master's course. I've done everything except writing a thesis, which after months of inaction will hopefully happen soon (at least I met with my prospective supervisor, which is progress)

Obliv

Oh right you'd need to do a thesis to get a master's degree, correct? What's the difference between that and doing a PhD?

ACuriousMind

@Obliv The difference is that a master's degree is seen as the prerequisite for a PhD here.

Obliv

@acuriousmind so you'd have to do another thesis after the master's? that doesn't make much sense to me

ACuriousMind

18:25

@Obliv Yes. A typical PhD takes three to five years of mainly research, at the end of which a PhD thesis is supposed to come out.

Obliv

Whoah that's not what happens here lol. My friend is in his 4th year of grad school and he's supposed to have a thesis ready by the end of it. The first 2 years was spent taking courses.

ACuriousMind

I know, the American system is completely different and very strange to me.

Obliv

I like your system better. Though, I disagree with needing to write a thesis for a master's degree. @acuriousmind I would prefer if you get all of the coursework done first, then taking 3-5 years for your thesis.

ACuriousMind

You have to take into account that the bachelor/master nomenclature is rather new here

Formerly, there was only a single five-year course at the end of which stood a "diploma", for which you had to write a diploma thesis.

There already was much concern about the loss of quality when transitioning to the bachelor/master system, and dropping the thesis requirement for the equivalent degree would have been an obvious downgrade.

Obliv

Aw why would you guys choose that nomenclature too? If it's new then you guys could have made it sound way cooler. Bachelor, master, doctorate doesn't make any sense. It should be like an MMO and go novice, adept, master or something haha

ACuriousMind

18:34

@Obliv To make transitioning between different European universities easier. The introduction of a more-or-less uniform bachelor/master/doctorate progression was a Europe-wide project called the Bologna process.

Obliv

that's a rather pork-uliar name for a process.. ahaha...

I should probably get back to doing this report.

ACuriousMind

@Obliv It's named after the place, not the sausage! :P

Obliv

is that where the sausage came from?

ACuriousMind

Yes

Bernard Meurer

I'm back

ACuriousMind

18:41

~~Oh no~~ Hey, Bernhard

Loong

@dmckee in case you need a precedent: meta.chemistry.stackexchange.com/a/94/7951

ACuriousMind

@Loong Tempted to flag that as "not an answer"

Loong

:-D

user116211

-3

Q: (b) Normalize the wave function e-lxl sin ax

Estimate the size of the' hydrogen atom and the ground-state energy from the uncertainty principle?

quantum-mechanics homework-and-exercises atomic-physics uncertainty-principle estimation

user116211

OP should have given the page number of the book also along with no. (b) in the title.

user218912

18:58

@ACuriousMind hello

ACuriousMind

heyhey

Obliv

19:30

Hey I think there's a mistake here: imgur.com/a/y9b7r @acuriousmind can you confirm that the word in the red underline should be small not large

for if the einstein temperature ${\theta}_E$ were to be large, the ratio $\frac{{\theta}_E}{T}$ would be smaller

ACuriousMind

@Obliv no

Obliv

but how

$$C_v = 3Nk\left(\frac{\theta_E}{T}\right)^2 \frac{e^{\theta_E/T}}{(e^{\theta_E /T}-1)}$$

user116211

@Obliv \left( \right)

Obliv

thanks @mafia

ACuriousMind

@Obliv I don't know how to tell you - if $\theta_E$ is large, then $\theta_E/T$ is also large (unless $T$, for some reason, is even larger, but then against what is $\theta_E$ large?). That's such an obvious fact that I don't know what there is to explain

Obliv

19:35

oh my god..

Sorry @acuriousmind my brain shut off

@acuriousmind Oh okay my confusion was the statement before. I thought it was going to explain the lower specific heats at higher temperatures for certain solids. But, it explains that for larger specific heats a greater oscillator frequency is needed (for a larger $\theta_E$)

That's why my brain was saying $C_v$ must be smaller so $\frac{\theta_E}{T}$ must get smaller

0celo7

19:51

@ACuriousMind Please give me a nice $2$D curve that passes through 0

smooth

ACuriousMind

y(x) = x

0celo7

nontrivial

it should not be a geodesic

ACuriousMind

y(x) = x^2

0celo7

what's that in parametric form?

ACuriousMind

x(t) = t; y(t) = t^2?

Not quite sure what you mean

0celo7

19:53

good.

I will be back after some computations.

performs geodesic ball surgery

Obliv

@0celo7 how long was ur report from yesterday

0celo7

@Obliv 1600 words

Obliv

mines due in 2 hours and I'm still writing the theory section :'D

0celo7

wtf are we going to the same school?

wait

Obliv

you didn't know?

0celo7

19:56

no labs on Thursday at my school

Obliv

I'm at the school next door. The University of Tennessee 2

0celo7

@ACuriousMind Ok, I need help with a limit :(

I'm dumb

can u pls help

Bernard Meurer

We Didn't Start The Cancer (Rare Meme Folksong)

►

ACuriousMind

@0celo7 Which one?

0celo7

@ACuriousMind It's related to the problem from earlier. I want to understand the theorem in Euclidean space first. So the claim is that if $\alpha:\Bbb R\to\Bbb R^n$ is a smooth curve w/ $\alpha(0)=0$, and $\alpha'(0)=v$, then $||v||=\lim_{t\to0}\frac{1}{t}||\alpha(t)||$

For the life of me I cannot compute that limit.

ACuriousMind

20:00

You're trying to compute that for the parabola right now, right?

0celo7

No...that limit I can do (somewhat)

But there is an issue with the parabola limit: it does not exist

Specifically, the one-sided limits disagree.

To see this, note that $||\alpha(t)||\ge 0$ but not so for $t$

So I'm inclined to believe the theorem/result needs to be modified a bit.

ACuriousMind

@0celo7 Ah, yes, that's right.

0celo7

Perhaps $||v||=\lim_{t\to 0}\frac{1}{|t|}||\alpha(t)||$.

ACuriousMind

Sounds reasonable

0celo7

I can live with that, the thing I'm trying to ultimately prove (isometry group is a Lie group) will still work with this modification.

@ACuriousMind But this is a much harder problem now

Because when it was $\frac{1}{t}||\alpha(t)||$ it was basically a derivative

the abs val makes the limit pretty hard to compute

let me check this with some more curves.

let's go for the cubic.

oh...not good

the square root goes imaginary.

Seems wrong.

Ah, $2^1=2\ne 3$.

ACuriousMind

20:05

@0celo7 I think you can just "drop" it in the sense that you can take the absolute value of the limit instead of the limit of the absolute value, and then you have your derivative again

0celo7

@ACuriousMind But I've shown that limit does not exist (at least in some cases)

ACuriousMind

@0celo7 Take it only from above

0celo7

@ACuriousMind Sure, so what is $$\frac{d}{dt}||\alpha(t)||?$$

Obliv

WOW I just noticed I learned what a space curve was in calc yesterday @0celo7 it's the set of all points that a continuous vector function traces out right?

0celo7

It seems to be $$\frac{\alpha'(t)\cdot\alpha(t)}{||\alpha(t)||}$$

@ACuriousMind So what's the limit of that as $t\searrow 0$?

ACuriousMind

20:07

Eugh

0celo7

@Obliv Never heard of "space curve" before.

My gf is taking calc 3 and whenever she asks me stuff I have to look up the definitions. No serious mathematician uses calc 3 terminology.

user116211

$$V= \sum_i\frac{\partial U}{\partial \dot q_i}~ q_i -U$$ where $V$ is the potential energy and $U$ is the work function.

0celo7

@ACuriousMind My thoughts exactly. It's a hard limit!

user116211

I'm not getting how they got the first term ;/

0celo7

@ACuriousMind I'm assuming a direct $\epsilon-\delta$ assault would only lead to tears and anguish?

Is there a Hospital rule for vector-valued limits of this form?

ACuriousMind

20:09

@0celo7 Not sure, I have terrible intuition for limits

0celo7

It's an indeterminate form.

Wait

It's not even vector-valued

HOSPITAL RULE

@Obliv how does the Hospital rule work?

user116211

Is this due to calculus of variation? .

ACuriousMind

Does that yield an easier expression, though?

@MAFIA36790 What is a "work function"?

0celo7

@ACuriousMind Only one way to find out. The derivative of $||\alpha(t)||$ in the denominator is nasty.

So is it $f/g$ has the same limit as $f'/g'$?

ACuriousMind

if the latter exists, then it's the same as the former, yes

0celo7

20:11

ok, let's check

ACuriousMind

Note that non-existence of the limit of the derivatives doesn't imply non-existence of the original limit

0celo7

@ACuriousMind Wait

user116211

@ACuriousMind Well, Lanczos introduces $U$ as the function whose differential is the infinitesimal work $\overline{dw}.$

0celo7

can we only use the Hospital rule for genuine limits

Or does it work for one-sided limits

ACuriousMind

Good question

I think it needs a proper limit, but I'm not sure

0celo7

20:13

Screw rigor, lemme compute these derivatives.

user116211

Could I make it clear @ACuriousMind?

0celo7

Lol, this is not helpful

$$\frac{\alpha''(t)\cdot\alpha(t)+||\alpha'(t)||^2}{\alpha'(t)\cdot\alpha(t)/|| \alpha(t)||}$$

ACuriousMind

@MAFIA36790 No, because I have no idea what the setting is. What are we doing here? Thermodynamics? Classical mechanics? If the work along paths has an anitderivative $U$, then that's what I would usually call a potential energy, so what is the definition of potential energy here?

user116211

@ACuriousMind Ah! Classical Mechanics.

user116211

@ACuriousMind He then defines potential energy $(V)$ as the negative of the work function i.e., $V= -~U\,.$ This is the special case of the above general case.

ACuriousMind

20:18

@MAFIA36790 If $V=-U$, then your equation there is just a silly way of writing $\sum_i \frac{\partial U}{\partial \dot{q}_i}q_i = 0$, no?

user116211

@ACuriousMind sure.

0celo7

So I think when I write out the fractions, the first term dies.

Don't know how to prove it, but seems reasonable.

So now

ACuriousMind

@MAFIA36790 Since you haven't at all told us from what this equation is supposed to be derived, what's your question, exactly?

0celo7

$$\lim_{t\to 0}\frac{||\alpha(t)||\cdot||\alpha'(t)||^2}{\alpha'(t)\cdot\alpha(t)}$$

@ACuriousMind I mean, if we cancel terms stupidly that looks like $\lim_{t\to 0}||\alpha'(t)||$ to me.

user116211

@ACuriousMind He didn't write any equation prior to this; he was discussing rheonomic constraints and then he told that if $U$ is time-dependent, then the potential energy $V$ can be written as $V= \displaystyle\sum_i\dfrac{\partial U}{\partial \dot q_i}~ q_i -U\,.$

user116211

20:22

My question is how he got the first term. This is what I'm not getting.

0celo7

Or $$\lim_{t\to 0}\frac{||\alpha'(t)||}{\cos\theta_t}$$

where $\theta_t$ is the angle between $\alpha(t)$ and its derivative.

But I'm thinking $\theta_0=0$

@ACuriousMind So we get $||v||=\lim_{t\to 0}\frac{1}{|t|}||\alpha(t)||$.

Maybe lol

I just gave a good physicist proof.

@ACuriousMind Is the curve $(\cos t,t^3)$ smooth?

I'm asking for a friend who wants to know if he's insane.

ACuriousMind

@MAFIA36790 That doesn't make any sense. If indeed $V=-U$ by definition, then there's no way that term could suddenly appear there. Either that's a typo or you have gotten something rather wrong here.

0celo7

Oh I'm stupid.

ACuriousMind

@0celo7 Certainly

0celo7

It doesn't go through 0

Ok, the result holds for $(\sin t,t^3)$.

Good

I think I'd better give an $\epsilon$-$\delta$ proof.

user116211

20:29

@ACuriousMind He said $V= -~U$ is a special case of that general case.

ACuriousMind

So what is the definition of $V$?

user116211

Wait, @ACuriousMind, lemme give you the link of the book...

ACuriousMind

If it's that equation, as I am beginning to suspect, then there's nothing to explain.

No, I'm not skimming through a book to find the definition you're talking about

user116211

@ACuriousMind No, you don't have to; I'm uploading a snapshot....

user116211

20:38

user116211

Done @ACuriousMind; are they looking good to you? Or I'll take a zoomed snapshot....

ACuriousMind

Yeah, so that equation is the definition of potential energy in the general, rheonomic case. What's the question?

user116211

@ACuriousMind The question is in $(18.5)$ what is the first term? I mean how he got that? He didn't just write it by its own, did he?

user116211

@ACuriousMind Yeah. But after writing $(18.5)$ he wrote that when $U$ is independent of velocities, the first term disappears and $V$ again becomes the negative of $U\,.$

ACuriousMind

@MAFIA36790 Well, in this text, he did. But as the paragraph preceding it tells you, this is done because it will turn out that the sum of this and the kinetic energy will be conserved during the motion as "total energy".

Obliv

20:43

@0celo7 yeah well space curve makes more sense intuitively than smooth curve anyhow. dood it's crunch time i have an hour 20 minutes left. also apparently i wasn't supposed to wear nice clothes to this lab since we're working with some kind of oils that will ruin clothing

:[

user116211

@ACuriousMind reading again between the lines...

user116211

@ACuriousMind okay! Got the point. This was all done so that the total energy remains the same even when the constraints are rheonomic and $U$ depends on time and velocities.

user116211

But @ACuriousMind, if the total energy remains the same, can't I say then that the conservation law of energy satisfies the rheonomic system?

ACuriousMind

@MAFIA36790 I don't know what it means for a law to satisfy a system

user116211

But this shouldn't be true though....

user116211

20:53

@ACuriousMind As you can see in the second snapshot, he writes that for rheonomic systems, law of conservation of energy doesn't satisfy.

ACuriousMind

@MAFIA36790 Your grammar is still broken, but I see now that I misspoke: In a rheonomic system, this isn't conserved, but in a scleronomic system it will be.

user116211

@ACuriousMind surely true.

user116211

But didn't he write $V$ in $(18.5)$ such that the "total energy" remains constant?

user116211

Was he talking about scleronomic system?

ACuriousMind

@MAFIA36790 Yes, read the sentence again

user116211

20:59

Well, I can see he didn't mention time, but rather mentions velocity.

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