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12:10 AM
yes! googled it and maybe it's from here: "Jean Le Rond d’Alembert" which has developed and used partial differential equations.
not sure
 
Just call it d
 
@0celo7 I always have to look up the meaning of $Q$ and think it through, so I can't answer that without turning on my brain after working hours.
 
@dmckee Ok. I did it, it seems correct.
$Q$ is the mass deficit
 
I do. but I used to call it "rond". you know it's kinda weird when you study sth in english and read a half of its symbols in another language(what you have learned at school to call them) because you are used to it!
 
12:25 AM
@dmckee Is 331MeV a lot of binding energy?
I don't have a feel for these numbers
 
@0celo7 Compare it to the mass of the nucleus. It's about a third of a nucleon, but for a high $Z$ nucleus it's a relatively small perturbation.
 
@dmckee For potassium 40?
 
12:55 AM
It's less than one percent. I don't know if that's right, but I don't think it's so big you can dismiss it as unreasonable.
 
@dmckee I'm sure it's right, the computation is pretty trivial
But I just don't know how much energy that actually is
 
anyone happen to know how long a room can sit around before it's frozen?
 
@DanielSank Two weeks I think.
Mods can unfreeze them if there is a good reason. Just let us know.
 
1:12 AM
@DanielSank what?
@dmckee oh
I thought he meant a room room.
 
user218912
1:35 AM
lol
 
user228700
2:26 AM
@JohnRennie: Morning sir :-) Only 30 more points to go, wohoo! (:P)
 
user228700
I had a very small follow-up question; let's consider a ball kept at rest on the ground. I apply some force, $F$, to take it from $H=0$ to $H=h$, where $H$ is its height from the ground. If the magnitude of this force $F$ is equal to the weight of the body(Hence, its not accelerating), then our equation ∆$U=$∆$KE+W_{NC}$ would give me ∆$U=mgh=W_{NC}$, no? Is this actually correct? The "nonconservative" forces are responsible for the change in potential energy of the body..?
 
@KaumudiHarikumar If your extra force $F$ is conservative, then it would have a height dependent potential energy of its own, which would cancel out that due to gravity.
 
user228700
@Anthony But um, external forces applied by machines/ourselves tend to be nonconservative, no?
 
@KaumudiHarikumar ;)
 
user228700
@0celo7 I'm trying :P
 
2:39 AM
Yes! But if it were, say, an electric field, it would be conservative. In the case where it's a nonconservative force, then work done by it shows up as $W_\mathrm{NC}$.
 
user218912
@0celo7 does it make sense to write $\eta^{\epsilon\delta}\delta_{\mu\alpha}\delta_{\alpha\beta}\partial_\epsilon A_\delta$?
 
user228700
@Anthony I'm afraid I don't fully understand that...
 
no
maybe
 
user218912
no because the indices aren't paired right?
 
it's bad notation
 
user218912
2:40 AM
how should I write it?
 
user218912
and what does it equal?
 
@KaumudiHarikumar Sorry. You're right about externally applied forces being nonconservative, and the work done by them enters your equation as $W_\mathrm{NC}$.
 
@IceLord In geometric analysis we write stuff like $e_ie_i$ (sum implied), but don't do that in relativity
@IceLord I don't know what it's supposed to be
 
@KaumudiHarikumar Don't worry about electric fields if that's a bad example for you.
 
user228700
@Anthony Um, no, that's OK. So, if I were to apply an external force $F$, being nonconservative, it would be responsible for the change in potential energy? Bearing in mind that I have considered the work done on the body by the net force.
 
user228700
2:45 AM
The work done by the net force on the body would just be the net work done on the body, correct..?
 
@KaumudiHarikumar Yes, if you apply some external nonconservative force $F$, it can cause the potential energy to change.
@KaumudiHarikumar The net work done on the body will then be the same as the work done by the net force.
 
user228700
Oh crap, I think I made the mistake of including the nonconservative force $F$ in the term $W$ on the left. I shouldn't have done that.
 
user228700
@Anthony So I can't just "add all the forces". I need to check which one's conservative and which one is not, correct? (@JohnRennie)
 
@KaumudiHarikumar If the potential energy due to some force (like gravity) enters your equation, then the work done by that force should not.
 
user228700
3:02 AM
@Anthony OK. Thanks :-)
 
@KaumudiHarikumar Sure!
 
user218912
@0celo7 what about $\eta^{\alpha\beta}\eta^{\epsilon\delta}\delta_{\mu\alpha}\delta_{\alpha\beta} \partial_\epsilon A_\delta$
 
user228700
@Anthony Actually, one more, sorry (:P) Am I correct in my understanding that conservative forces can't change the total mechanical energy of the system?
 
@IceLord Three alphas, nope.
 
user218912
oh i made a typo
 
user218912
3:07 AM
sorry
 
user218912
let me rewrite it with different indices
 
user218912
$\eta^{\sigma\tau}\eta^{\epsilon\delta}\delta_{\mu\sigma}\delta_{\alpha\tau} \partial_\epsilon A_\delta$
 
@KaumudiHarikumar If you include potential energy due to those forces in your total mechanical energy, then yes.
 
user218912
is that equal to $\partial_\mu A_\alpha$?
 
Christ. let's see...
probably not, no
 
user218912
3:11 AM
what is it then?
 
$\delta$ shouldn't have two lower indices in SR
 
user218912
okay so
 
user218912
if you do
 
well...maybe
where did you get them from?
 
user218912
lagrangian density example in class.
 
3:12 AM
you need to be careful with indices
 
user218912
i'm filling in the missing steps.
 
@IceLord I'll blame physicist rigor ofc
Deltas should not have two lower indices
that's all I can say...
 
user218912
@0celo7 alright so
 
user218912
if I write
 
user218912
$\delta^\mu_\sigma$
 
user218912
3:13 AM
and similarly for the other delta
 
user218912
then I will get
 
user218912
$\partial^\mu A^\alpha$
 
user218912
is that right?
 
user218912
because it works.
 
user218912
i think.
 
user218912
3:15 AM
or maybe not...
 
user218912
just not sure what it is...
 
user228700
@Anthony@JohnRennie: If my understanding is correct, then I will be correct in writing the following equation: $PE_f+KE_f=W_{ext}+PE_i+KE_i$. By rearraranging this, I get ∆$KE=W_{ext}+(PE_i-PE_f)=W_{ext}+W_C}$ (Where W_C=Work done by conservative forces=-∆$U$). Is this equation correct?
 
@KaumudiHarikumar That all looks good!
 
user228700
@Anthony Phew. Thank you! Now I just need @JohnRennie to comment on this.
 
user218912
see you always doubt people.
 
3:26 AM
why do you trust JR?
@IceLord bruh I need Chavel Riem. Geo
introduction my ass this thing has lots of good geometric analysis
 
user218912
tell me where I went wrong and help me 15 more times i'll buy you it.
 
user218912
depends on how much it costs though
 
like $40
 
user218912
okay sure
 
user218912
deal?
 
3:27 AM
$47
@IceLord Hmm, I thought you said something about me helping you before and you owing me...
 
user218912
@0celo7 well add 15 more helps to it.
 
hmm
 
user218912
every question answered is a "help" btw.
 
user218912
so 15 is not a lot
 
ok
 
user218912
3:37 AM
a first question would be do you know what the answer is and where I went wrong?
 
user218912
do you want the thing from before as well?
 
user218912
(before I got the expression)
 
uh, I'm reading a book right now
 
user218912
fine I need to know by next class monday. since if I don't i'll get lost in the other field theory stuff and it will be a slippery slope to game over.
 
user218912
so please tell me by then thanks :)
 
3:39 AM
ugh
what is it
 
user218912
the prof did an example in class for the same lagrangian on the problem set.
 
user218912
but didn't show the steps
 
user218912
so, i already showed my steps for the $\frac{\partial \mathcal{L}}{\partial A_\alpha}$
 
user218912
and got it right
 
user218912
but for the $\frac{\partial \mathcal{L}}{\partial(\partial_\mu A_\alpha)}$
 
user218912
3:41 AM
I don't get it right (i think)
 
user218912
since I get a factor of 2 in my final answer but he doesn't.
 
user218912
and i'm not even sure if my indices are right since i kind of hand waved it.
 
user218912
so the term is
 
user218912
$\mathcal{L} = -1/2 \partial_\alpha A_\beta \partial^\alpha A^\beta$
 
user218912
I did
 
user218912
3:44 AM
$\mathcal{L} = -1/2\eta^{\epsilon\delta}\eta^{\sigma\tau} \partial_\epsilon A_\delta \partial_\sigma A_\tau$
 
user218912
is it right for far?
 
user218912
then I took the derivative and got
 
@IceLord no
 
user218912
oh
 
you need to contract $A_\tau$ with $A_\delta$
 
user218912
3:45 AM
okay so i'll stop there
 
user218912
where did i go wrong
 
user218912
oh
 
user218912
@0celo7 why?
 
because that's what you wrote
4 mins ago, by IceLord
$\mathcal{L} = -1/2 \partial_\alpha A_\beta \partial^\alpha A^\beta$
 
user218912
so how should it look?
 
user218912
3:48 AM
oh i see
 
user218912
so it should be
 
user218912
wait no
 
user218912
:(
 
user218912
can you tell me how to do it please?
 
user218912
then i'll do the next term myself.
 
3:55 AM
what am I doing
 
user218912
taking the derivative of it
 
user218912
wrt. to $\partial_\mu A_\alpha$
 
user218912
I know there are 2 kronecker deltas
 
user218912
but the metrics idk
 
user218912
are there 4 metric tensors? @0celo7
 
4:01 AM
Use $$\frac{\partial A_\mu}{\partial A_\nu}=\delta^\nu_\mu.$$
That's the correct index placement.
 
user218912
yes I know that.
 
user218912
I said it ^
 
user218912
Idk about the metric tensors though.
 
user218912
what is the full expression with the metric tensors?
 
of what?
dude there's too many equations
 
user218912
4:04 AM
of $\mathcal{L} = -1/2 \partial_\alpha A_\beta \partial^\alpha A^\beta$
 
user218912
we want to lower all the indices right?
 
user218912
what is the correct expression?
 
...isn't that basic SR?
 
user218912
yes but you said i'm wrong
 
$\mathcal L=-\frac{1}{2}\eta^{\mu\alpha}\eta^{\nu\beta}\partial_\alpha A_\beta \partial_\mu A_\nu$
 
user218912
4:09 AM
oh
 
user218912
i see so the indices are not matched at the top
 
user218912
thanks @0celo7
 
user218912
how many help points is that?
 
how many questions was that
 
user218912
idk
 
user218912
4:10 AM
I repeated myself a lot of times so
 
user218912
about 3?
 
@IceLord you're lucky I didn't make that bet with you.
 
user218912
@0celo7 you sure?
 
user218912
I think i'm doing quite well tbh
 
user218912
some kids in my class didn't even know how to do #1
 
4:20 AM
I wouldn't be helping you if I did
 
user218912
oh I see
 
user218912
Sep 19 at 0:55, by 0celo7
If I make this bet, then I'm betting you fail. I don't want you to fail.
 
user218912
<3
 
user218912
@0celo7 always looking out for me.
 
user228700
5:13 AM
@JohnRennie: Congratulations!!! :-D
 
Thanks, I just got the last upvote about ten minutes ago :-)
I'm just working through the chat log. Give me five minutes and I'll have a look at your question ...
@KaumudiHarikumar: I can't help feeling you're worrying a bit too much about non-conservative forces.
In your question you talk about a ball rising a distance $h$ so it's PE increases by $mgh$.
If you or I lifted the ball up then you could argue non-conservative forces are involved because we are complicated systems and tend to convert all forms of energy into heat. So we are non-conservative in the sense that energy leaves the system as heat.
But the ball could have been lifted by a spring or lever that is a conservative system.
 
user228700
@JohnRennie Oh, I see :-) BTW, I was one among ur last 5 upvotes to glory (:P).
 
Thanks :-)
 
user228700
5:29 AM
@JohnRennie Yeah, I kind of was. I think I've got it now after having brainstormed for the past hour.
 
In practice we don't worry that much about non-conservative forces in physics. I think you put your finger on it when you said:
> Am I correct in my understanding that conservative forces can't change the total mechanical energy of the system?
 
user228700
OK. Phew. Thanks :-)
 
If we have a non-conservative force acting it means we aren't keeping track of where all the energy is going so it looks as if total energy is increasing or decreasing.
 
user228700
Yes, OK.
 
Outside of unusually sadistic problem sheets we are usually dealing with situations where it's normally obvious that energy is conserved.
 
user116211
5:46 AM
Hey @JohnRennie, did you cross 200k?
 
user116211
yes \o/
 
user116211
Okay, I would be making the post this afternoon....
 
Hi, everybody.
 
user116211
@DanielSank o/
 
@MAFIA36790 about 6 a.m. this morning :-)
 
5:48 AM
@MAFIA36790 \o
@MAFIA36790, this high-five ritual with you... it's strangely comforting.
 
user116211
@DanielSank ;D
 
@DanielSank though I note mafia seems to be left handed ...
 
user116211
@JohnRennie yes!!
 
user116211
You are a wizard-nerd @JohnRennie ;)
 
@KaumudiHarikumar the issue of how to model non-conservative systems can be very interesting. I won'g go into details for fear of distracting you from your goal of acing the entrance exams, but I'll say that the challenge of applying quantum mechanics to non-conservative systems is still an area of research.
 
user116211
5:52 AM
Anyways, @JohnRennie, ACM answerd my query on why Lanczos told to keep $t$ constant when the work-function $U$ is time-dependent:
 
user116211
3
A: Why is $t$ considered constant when forming the differential $\mathrm dU(q_l; t)$ for $\mathrm dU= \overline{\mathrm dw}$ to be true?

ACuriousMindThe differential form $\delta w$ for infinitesimal work (I switched notation because I don't want to draw that silly bar each time) is defined on the space of Lagrangian mechanics, i.e. the space spanned by the $(q,\dot{q})$. When you form $\mathrm{d}U$ as $$ \mathrm{d}U = \partial_{q^i} U \mathr...

 
user116211
@DanielSank Lanczos wrote at some point non-conservative forces can be expressed as the function of time-dependent work function $U\,.$
 
@MAFIA36790 Not sure what you mean.
 
user116211
@DanielSank hmmm. Which part?
 
I don't know "work function".
 
5:55 AM
@MAFIA36790 Cool, though I'm not sure i understand the answer. I'd have to sit down and read Lanczos to have any hope of understanding what's going on.
 
user116211
@DanielSank ah!
 
Are you saying that the non-conservative force is the gradient of the time dependent work function, similar to how conservative forces are the gradient of a potential energy?
 
user116211
Wait, @Daniel, it bothered ACM also while I was discussing with him; apparently this term has only been used by Lanczos in his treatise.... lemme give you the permalink....
 
By the way, I added my reading list to my public journal.
I will continue to update it as I think of more good books.
 
user116211
@DanielSank good!
 
user116211
5:57 AM
@DanielSank kinda.
 
user116211
10 hours ago, by MAFIA36790
@ACuriousMind Well, Lanczos introduces $U$ as the function whose differential is the infinitesimal work $\overline{dw}.$
 
user116211
10 hours ago, by MAFIA36790
@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.
 
user116211
The general case being this:
 
user116211
$$V= \sum_i\frac{\partial U}{\partial \dot q_i}~ q_i -U$$
 
user116211
Here $U$, though, doesn't explicitly depend on time.
 
6:01 AM
Ok
The dimensions look messed up.
$$\frac{\partial U}{\partial \dot{q}}q$$
has dimensions of $U$ times time.
 
user116211
@DanielSank :(
 
user116211
@DanielSank This is the bothering part; but ACM told it was added so as to keep the sum of potential energy plus kinetic energy constant.
 
So certainly you can't subtract $U$ from it.
@MAFIA36790 Something here is screwed up.
 
user116211
@DanielSank Would I upload the snapshot of the excerpt?
 
user116211
Wait.... I did it yesterday...
 
6:04 AM
Is this a mechanics book?
 
user116211
@DanielSank yep.
 
I have never seen a mechanics book that used good notation. I wonder if that's the issue here.
 
user116211
Opera is not responding
 
user116211
;((
 
awwwww
 
user116211
6:10 AM
 
user116211
@DanielSank: Could you read it? Or I would upload a more zoomed version....
 
The size is fine.
Yes, well, the notation is terrible.
Somehow time is dimensionless here.
 
user116211
@DanielSank ohh ;/
 
user116211
But why would the author take $t$ to be dimensionless?
 
@MAFIA36790 Я не знаю.
Authors do such things often.
 
user116211
6:14 AM
@DanielSank using google translate...
 
In quantum mechanics, authors will often simply omit $\hbar$ by claiming that it is magically equal to one.
 
user116211
@DanielSank Okay, I'm very much accustomed to that.
 
user116211
In relativity, they take $c= 1\,.$
 
user116211
But taking time dimensionless is, frankly speaking, totally weird and new to me.
 
@MAFIA36790 Yes, they do. However, I would point out that constants of Nature cannot possibly be said to be equal to a dimensionless number.
This is utter nonsense. We can talk about why I say that if you want, but I guess it's not the most interesting topic right now.
@MAFIA36790 Well, actually, I often set up my equations to use a dimensionless time. it is convenient
However, to do this we need a time scale of some sort. We can't just arbitrarily declare $t$ to be dimensionless.
 
user116211
Lanczos was actually a mathematician.
 
user116211
@DanielSank okay.
 
Like I said, you need a time scale.
Here's an example:
$$\ddot{x}(t) = -\omega_0^2 x(t)$$
Suppose we want to make time dimensionless.
Define $\xi \equiv \omega_0 t$.
 
user116211
okay...
 
6:21 AM
Then $d/dt = \omega_0 d/d\xi$
So:
$$\ddot{x}(\xi) = -x(\xi)$$
where now the dot means $d/d\xi$.
This is nice. See, the $\omega_0^2$ went away.
 
user116211
@DanielSank yes!
 
But to do this, I had to explicitly make up a new dimensionless quantity.
 
user116211
@DanielSank yeh, got that.
 
@MAFIA36790 Ok, well, the problem (often) is that physicists are too lazy to be explicit. See, with relativity, we have the Lorentz factor:
$$\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}$$
Often, this is just written
$$\gamma = \frac{1}{\sqrt{1 - v^2}}$$
with the ridiculous claim that $c=1$.
 
user116211
@DanielSank yeh; a very common practice.
 
6:24 AM
It is simply impossible for $c$ to be equal to 1.
 
user116211
@DanielSank agrees...
 
The quantity $c$ doesn't even have a numerical value unless we choose a system of units.
The Lortentz factor equation is valid without any reference to a unit system.
 
user116211
They must choose a unit such that $c= 1\,.$
 
Obviously, all the author has really done is implicitly decided that what he writes as $v$ is really $v/c$.
 
user116211
@DanielSank yes.
 
6:25 AM
We could say $\xi \equiv v/c$ and then
$$\gamma = \frac{1}{\sqrt{1 - \xi^2}}$$
 
user116211
@DanielSank sounds legit and more reasonable.
 
It would even be ok to write at the beginning of the book "In this book, the symbol $v$ means velocity divided by speed of light". That would be fine.
But they don't do this. Instead they make the nonsense statement that $c=1$.
I wish they would not do that.
The language should be clear. If the language is unclear I have no chance to understand anything.
 
user116211
@DanielSank They never do this.
 
@MAFIA36790 I know. It's incredibly stupid.
We have a choice between two sentences:
1) In my book, v means velocity divided by speed of light.
2) In my book, we work in units such that c=1.
Option 1 is sensible, option 2 is complete nonsense.
But they pick option 2.
¯\_(ツ)_/¯
 
user116211
@DanielSank ;P
 
user116211
6:34 AM
@DanielSank Classical books like Feynman Lectures gloriously use option 2.
 
What's worse, when I point this out, most of the time the person I'm talking to says I'm wrong.
 
user116211
@DanielSank O.o
 
user116211
You must be talking with physicists ;P
 
@MAFIA36790 What can I say? People get used to something and then don't want to think it's wrong.
@MAFIA36790 I don't actually much like the Feynman lectures as books for learning.
 
user116211
@DanielSank hmmm.
 
6:35 AM
They're good if you already know the subject.
 
user116211
@DanielSank Well, I haven't read thoroughly Vol 1; but Vol 2 and Vol 3 are actually quite good; but of course, not at all rigorous.
 
user116211
@DanielSank Hmm, I knew nothing about QM when I started to read Vol 3; the journey was amazing but rough and enlightening for me, I would say.
 
user116211
He starts to develop right from the base; so anyone can read them.
 
user116211
But yeh, it is good to have a pre-knowledge on the concerned topic; but I won't say it is mandatory to read his lectures.
 
@MAFIA36790 Oh, ok good!
I always wanted to be able to calculate.
 
user116211
6:39 AM
The maths is not too hard; the language is lucid; and covers many topics. But as I said, they are not rigorous.
 
user116211
@DanielSank ;)
 
Even though I'm an experimentalist now, I always wanted the math and the physics together as one.
 
user116211
@DanielSank agree!! agree!!
 
I like books that do not shy away from math. If I need a math concept, the book should either give it to me first, or tell me to go pick it up myself.
 
user116211
@DanielSank sure.
 
user116211
6:41 AM
One quick query, @DanielSank....
 
Ahhhh, finished my daily Russian lesson.
@MAFIA36790 Yes?
 
user116211
@DanielSank Are you learning Russian?
 
@MAFIA36790 Re-learning.
 
user116211
@DanielSank Ah!!
 
user116211
@DanielSank Well, Lanczos wrote that if $U$ is time-dependent, then the law of conservation of energy can't be applied....
 
user116211
6:43 AM
Is there a formal proof of it?
 
Я люблю русский язык
@MAFIA36790 Yes.
 
user116211
Well, I dig a bit on that and found this post:
 
user116211
4
A: Can a force in an explicitly time dependent classical system be conservative?

ShaktyaiIf the force is time dependent, you do not have energy conservation: $$m \frac{\mathrm dV}{\mathrm dt}= F(t)$$ $$\mathrm d\left(\frac{1}{2}mV^2\right)=F V~\mathrm dt=F~\mathrm dx$$ For a time dependent potential: $$\mathrm dU=\mathbb{grad}(U(t))~\mathrm dx + \frac{\partial U}{\partial t}~\mat...

 
If $U$ is time dependent then we can magically add potential energy to the system whenever we want.
 
user116211
WoW:
 
user116211
6:46 AM
0
Q: Can Black Holes process quantum information?

user98848Seems that spinning black holes are capable of complex quantum information processes https://arxiv.org/abs/1608.06822 - Photonic Bell states creation around rotating black holes It might be possible that rotating black hole to "write" quantum algorithms?

 
user116211
Anyways, in that post, it is written:
 
user116211
$$\mathrm dU=\mathbb{grad}(U(t))~\mathrm dx + \frac{\partial U}{\partial t}~\mathrm dt $$
 
Oh no.
 
user116211
in Cartesian coordinates.
 
user116211
@DanielSank What?
 
6:48 AM
This $dU$ notation. What does it mean?
Ugh, fine.
Go on.
 
user116211
@DanielSank The differential of work-function.
 
@MAFIA36790 Yeah, ok fine.
 
user116211
And then he concluded $$\mathrm d\left(\frac{1}{2}mV^2+U(t)\right)=\frac{\partial U}{\partial t}~\mathrm dt\,.$$
 
user116211
What stings me is that Lanczos told to keep $t$ constant while evaluating $\mathrm dU\,.$
 
user116211
Although, $U$ in my context is the work function while in that post it is the potential energy, I guess this doesn't make the difference.
 
6:51 AM
Is $V$ velocity?
 
user116211
@DanielSank yes. I don't know why the poster used $V$ to denote velocity.
 
@MAFIA36790 It's fine.
 
user116211
@DanielSank Generally capital v denotes volume; but well, nevermind.
 
user116211
Anyways, the whole point was that $\partial U/\partial t = 0$ for $\overline{\mathrm dw} = \mathrm dU$ to be true.
 
user116211
Here, $\overline{\mathrm dw}$ is the infinitesimal work.
 
6:54 AM
I'm having trouble with the notation.
 
user116211
Here is ACM's answer on this:
 
user116211
3
A: Why is $t$ considered constant when forming the differential $\mathrm dU(q_l; t)$ for $\mathrm dU= \overline{\mathrm dw}$ to be true?

ACuriousMindThe differential form $\delta w$ for infinitesimal work (I switched notation because I don't want to draw that silly bar each time) is defined on the space of Lagrangian mechanics, i.e. the space spanned by the $(q,\dot{q})$. When you form $\mathrm{d}U$ as $$ \mathrm{d}U = \partial_{q^i} U \mathr...

 
Answer on what?
What's the question?
 
user116211
2
Q: Why is $t$ considered constant when forming the differential $\mathrm dU(q_l; t)$ for $\mathrm dU= \overline{\mathrm dw}$ to be true?

MAFIA36790 The definition of work function on the basis of $$U:= U(q_1,q_2,\ldots,q_n)\tag{17.6}$$ is too restricted. We have forces in nature which are derivable from a time-dependent work function $U(q_1,q_2,\ldots,q_n; \,t)\,.$ the equation $$\overline{\mathrm dw} = \mathrm dU\tag{17.5}$$ still holds,...

 
user116211
@DanielSank the bar, right?
 
6:56 AM
@MAFIA36790 Among other things, yes.
But ACM's answer was accepted, by you.
So what is left to ask?
 
user116211
@DanielSank Well, $\mathrm dw$ is not always integrable; that's what's the bar meant for.
 
user116211
@DanielSank yes; and I got that. But from that point, how to derive for the time-dependent $U,$ the conservation law doesn't apply?
 
@MAFIA36790 Oh, yes I remember that my thermodynamics book used a similarly stupid notation. Ok.
@MAFIA36790 Well, first, you see that it is actually true that if $U$ depends on $t$ then of course energy is not conserved, right?
 
user116211
@DanielSank I believe that; but I want to have a formal derivation from that point.
 

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