Why definition of V is not required.

user116211

Sep 23, 2016 11:00

@ACuriousMind, You remember the work function from yesterday's discussion?

user116211

As you know, $V,$ the potential energy was defined to be the negative of $U$ i.e., $V= -~U\,.$

user116211

But it is the case when $U$ doesn't depend explicitly on time $t\,.$

user116211

The relation doesn't hold for time-dependent $U\,.$

user116211

So, how should I define $V$ for $U:= U(q_1, q_2, \ldots, q_n; ~t )\,?$

Secret

ok I am running out of time, thus cross SE analysis stops here for today. But it seems that contrary to what I thought, while the phenomenon is very common across the internet as in there is always at least one example in any media, it seems that for SE, it is only concentrated enough on PSE and MSE to be able to allow me to pick them up via random sampling

user116211

Sep 23, 2016 11:04

I dug the book but didn't find any relation between the two in the time-dependent case.

ACuriousMind

@DanielSank cc @Secret I think you massively overanalyzed what happened here. I think the question was just not particularly clearly written. I didn't have to crawl through the chat at all to understand what it really meant to ask, the comment from the OP was more than enough.

user116211

It is only written that $V=- ~U$ is true only for time-independent $U\,.$

ACuriousMind

@MAFIA36790 Who's saying that there is a $V$ for time-dependent $U$?

yuggib

@MAFIA36790 \o

user116211

@ACuriousMind ah!

user116211

Sep 23, 2016 11:10

But @ACuriousMind, can $V$ be defined for rheonomic cases?

user116211

He writes:

user116211

> $\delta V = \mathrm dV$ would no longer be true if $V$ were rheonomic depending through $t$ explicitly.

Secret

@ACuriousMind This is one facet of what the meta post is stressing, but there is also another important thing to brought focus on by the meta post is that often (at least in MSE, PSE), when the OP tries to clarify the question, eventually the comment flow stops as the answerers stops following up. The question is then effectively ignored no matter how many edits the OP followed. In a few lucky cases, ~2 months later, there will be a new answerer to answer the question, but often when the question is

user116211

where $\delta$ denotes virtual variation.

Secret

being erroresnously treated, people stopped trying to check how the question is fixed and thus the question is effectively being abandoned

user116211

Sep 23, 2016 11:12

I mean he introduced $V$ first through this expression $V= -~U$ without giving separate definition of $V\,.$

user116211

Now, that $U$ is time-dependent and the relation doesn't hold true, how can I then define $V$ for rheonomic/time-dependent cases?

ACuriousMind

@Secret That only happens when the follow-up comment by OP don't actually make it clearer. In this case, OP made a comment to me explaining I misunderstood, but I was asleep at the time and you were acting in the whole chat discussion as if I had ignored that comment when I hadn't even seen it.

Slow down the horses here

@MAFIA36790 Why do you want to define $V$?

user116211

@ACuriousMind Because, as he said in the above quote 'if $V$ were rheonomic', there must be a general definition of $V$ which works for rheonomic cases also.

user116211

@ACuriousMind In fact, can it be defined for rheonomic cases?

Secret

@ACuriousMind No, I never said you ignored the comment, I am saying that the situation of that questions reminds of the more widespread situation I mentioned that I saw across MSE and PSE, even when the OP does end up clarifying the question. As you have said, you closed the question not because of duplicate, but because of being unclear. Daniel later realised that you are not the one who cast the duplicate, but the general point about the phenomenon remains

So our purpose of the metapost is to raise awareness that often questions get erroreously judged because answereers and closers sometimes overlooked the crucial part of the question

and the phenomenon of abandoned questions

ACuriousMind

Sep 23, 2016 11:18

@MAFIA36790 I don't know, I don't have these parts of mechanics memorized. But from what you quoted, the sentence "$\delta V = \mathrm{d}V$ would no longer be true if $V$ were rheonomic" could equally well be read as an argument why rheonomic $V$ doesn't make sense. That's why I asked why you want to define a $V$ - what do you need it for?

@Secret I'm not sure how people making mistakes is news, or something that needs particular "awareness".

Users are human, and they are fallible. I think everyone is aware of that. Communicating such that everyone involved understands the same thing is difficult, and often fails. What's the point?

Secret

Well, for MSE and PSE, most questions either get closed or abandoned can be treated back to this, especially when a stream of OP comments followed that apparently get no responses for days or months. (and perhaps we are wrong) but we do noticed that when people answer things in general, they often missed the crucial point in the question that is crucial to kill the question, which is why our idea to raise the awareness of this specific mistake into view.

user116211

@ACuriousMind As why I need for, see, this whole thing started why the law of conservation of energy doesn't hold true for time-dependent $U$ and whether $V=-~U$ holds true for the time-dependent cases. Then digging up the unread chapters of the book, I came across that quote above. the next line confirmed that $V=-~U$ is true only for time-independent $U\,.$

Secret

Perhaps for you it is clear, as evidenced by most of your answers (provided ithe question is clear to you) you often managed to address the crucial point (and hence answer accepted and upvoted as expected), but it seems in general often a couple of answerers failed to do this

ACuriousMind

@Secret I think what I don't get is why you think there's a "specific" mistake here. The mistake is people missing the point of the question, which doesn't seem all that specific to me.

I agree most misunderstandings that happen happen due to this, but that's because it's a really rather broad class of mistakes

user116211

As that of conservation law, I saw that the total energy doesn't remain constant when the Lagrangian $L$ does depend on time. So, that query seems to get answered as well as the second one that $V=-~U$ doesn't hold true for rheonomic cases.

ACuriousMind

Sep 23, 2016 11:26

@MAFIA36790 When you say "doesn't hold true", you must have a definition of $V$ other than that. What is it?

Secret

I think the specific mistake is that people often tries to address only the statement that forms the question, but actually to answer the question one need to address the relevant chain of logic that lead to the question, not just the question itself, which is something that most people overlooked

ACuriousMind

I'm not convinced that's something most people overlook.

user116211

@ACuriousMind Yes, that's my question! What is the other definition?

ACuriousMind

@MAFIA36790 Why do you think there is one?

(We're going in circles here!)

user116211

@ACuriousMind yeh; I'm also confused with that ;/

ACuriousMind

Sep 23, 2016 11:29

What reason do you have to believe that there must be a meaningful general definition of $V$?

user116211

@ACuriousMind There must be some reasonable point why he said "if $V$ were rheonomic"... What does that mean? What is the definition of $V$ when it is rheonomic? He didn't touch those queries :(

user116211

In fact the whole point of potential energy is for the conservative cases....

ACuriousMind

@MAFIA36790 Without more context I can't really tell, but I already said that that sentence sounds to me like an argument why $V$ doesn't have a good definition. In my interpretation, he's saying that $\delta V = \mathrm{d}V$ is something we want from $V$, but if $V$ (as a function, however it may be defined) would be rheonomic, it can't fulfill that, so it's pointless to try and define $V$.

user116211

@ACuriousMind "... why $V$ doesn't have a good definition."- This answers my query.

ACuriousMind

I already said the same thing here:

15 mins ago, by ACuriousMind

@MAFIA36790 I don't know, I don't have these parts of mechanics memorized. But from what you quoted, the sentence "$\delta V = \mathrm{d}V$ would no longer be true if $V$ were rheonomic" could equally well be read as an argument why rheonomic $V$ doesn't make sense. That's why I asked why you want to define a $V$ - what do you need it for?

user116211

Sep 23, 2016 11:34

@ACuriousMind ;P

ACuriousMind

...why are you starring those?

user116211

But he could have written that straight ;(

user116211

@ACuriousMind Sorry, haven't meant the second one.

user116211

Anyways, I would conclude that $V$ is defined only for the conservative time-independent/scleronomic cases and not for the other one as it doesn't make sense to define it for the rheonomic cases.

The h Bar

Participants