Scleronomic system when the work function depends on generalised velocities.

user116211

Sep 22, 2016 20:08

$$V= \sum_i\frac{\partial U}{\partial \dot q_i}~ q_i -U$$ 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

@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

Sep 22, 2016 20:10

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

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

Sep 22, 2016 20:12

@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

Screw rigor, lemme compute these derivatives.

user116211

Could I make it clear @ACuriousMind?

0celo7

Sep 22, 2016 20:14

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

@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

Sep 22, 2016 20:19

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

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

Sep 22, 2016 20:27

@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

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

ACuriousMind

So what is the definition of $V$?

user116211

Sep 22, 2016 20:29

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

user116211

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

ACuriousMind

Sep 22, 2016 20:39

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

@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

Sep 22, 2016 20:49

@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

@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

Sep 22, 2016 20:56

@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

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

user116211

Sep 22, 2016 21:10

Oh, so, can I say $U= U(q_1,\ldots,q_n ; \dot q_1,\ldots,\dot q_n)$ does not explicitly depend on time $t\,?$

dmckee

::facepalm::

Comment not posted:

user116211

@dmckee :(

ACuriousMind

@MAFIA36790 Pretty sure he's not talking about you.

dmckee

> Oh, you've done the [first ultrasimplified model system] problem. Well, that must make you an expert, so I'll take my experience doing [graduate level textbook problem on the same subject], [almost but not quite cutting edge problem], and working on extensions to [best of breed cutting edge simulation of similar problem] and defer to you.

user116211

@ACuriousMind yes, i know...?

dmckee

Sep 22, 2016 21:17

::pounds head on desk::

Sometimes it like pounding sand.

ACuriousMind

@MAFIA36790 Then I have no idea what your ":(" was supposed to mean, but whatever

user116211

Anyways, @ACuriousMind, I think I got what he wanted to mean.

dmckee

And there were textual clues that it was going to be that way, but no. I just had to stick my two cents in.

user116211

He meant that since $U$ doesn't depend explicitly on $t$, the total energy remains constant provided $V$ is written like $(18.5).$

user116211

And, yeah, @ACuriousMind, he is talking about scleronomic systems.

user116211

Sep 22, 2016 21:20

But @ACuriousMind, where did he get the motivation to write the first term as$\displaystyle\sum_i\dfrac{\partial U}{\partial \dot q_i}~ q_i\,?$

ACuriousMind

@MAFIA36790 Presumably by computing the total conserved energy for a scler. system?

user116211

@ACuriousMind okay.

user116211

@ACuriousMind Does it involve calculus of variation?

user116211

I would start that chapter tomorrow.

ACuriousMind

Since the standard method would be to apply Noether's theorem, probably yes.

user116211

Sep 22, 2016 21:25

@ACuriousMind okayish!

Scleronomic system when the work function depends on generalised velocities.

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