Question from Velocity-Dependent Potential in lagrangian (Goldstein)

[RIgel]

currently working on format.. sor i was not preparedHi
I think this question would be much related to calculus more than physics cause it seems I'd lost my way cause of calculus... but anyway! it says,

[itex]Q=- \frac{\partial{U}}{\partial{q}}+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )[/itex]

but i don't get why ∑[itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex] is a work...

reason why I'm asking this is...

For it says no matter which dimension [itex]Q[/itex] have , ∑[itex]Q⋅δq[/itex] must have a dimension of work,then because [itex]Q=- \frac{\partial{U}}{\partial{q}}+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )[/itex]

so [itex]∑Q⋅δq=∑- \frac{\partial{U}}{\partial{q}}⋅δq+\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex]

should have dimension of work too.

I know [itex]- \frac{\partial{U}}{\partial{q}}⋅δq [/itex] part is work indeed, but i don't get why the latter part [itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex] is having a dimension of work.I kinda have a feeling that i had lost my way because of calculus because i neither can't figure out why [itex]\frac{d}{dt}(\frac{∂r}{∂q})=(\frac{∂\dot{r}}{∂q})[/itex]is right or how it goes like that.

 

[BvU]

Hi RI,

Your post is a bit hard to read; please use ## to enclose inline ##\TeX## and $$ to enclose displayed math.
(my compliments for the typesetting work ...)

It would help if you mentioned the location (in Goldstein ?) of the section you are referring to.
(I found: (1.58) in 3rd edition )

I know ##- \frac{\partial{U}}{\partial{q}}⋅δq## part is work indeed, but i don't get why the latter part ##\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq## is having a dimension of work.

The simplest answer is that the dimensions of ##q_i## and time cancel.
Howerver, I doubt that this is a satisfactory answer for you. If it isn't, can you elaborate a bit on why you struggle with that ?

can't figure out why ##\frac{d}{dt}(\frac{∂r}{∂q})=(\frac{∂\dot{r}}{∂q}) ## is right

This must be from a bit further in the text; where ( I think this needs some context) ?

 

[RIgel]

If it isn't, can you elaborate a bit on why you struggle with that ?

This must be from a bit further in the text; where ( I think this needs some context) ?

Well it seems, because time derived term is on denominator, so if i think 'dt' included in [itex]\partial{\dot{p}}[/itex] as if it is on the numerator of [itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq[/itex] (then it would be like ([itex]\frac{d}{dt}(\partial{U}\frac{1}{∂}(\frac{dt}{q}))⋅δq[/itex]), than it is...or i hope it is right that time eventually cancel out each and 'q's also cancel out each (i think that is pretty much same with your saying that q being canceled with time if i had understood perfectly... is it right?).

But what i expected is more of mathematical procedure or very proof that shows one by one, or pointing out my misunderstanding on partial derivatives. Because it looks so simple if it goes like [itex]\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq=(\frac{\partial{\dot{U}}}{ \partial{ \dot{q}}} )⋅δq=(\frac{\partial{U}}{dt}\frac{dt}{ \partial{q}} )⋅δq=(\frac{\partial{U}}{ \partial{q}} )⋅δq[/itex]...(I pretty much think this equation is very very wrong idea lol apparently here[itex](\frac{\partial{U}}{dt}\frac{dt}{ \partial{q}} )⋅δq)[/itex]

then i think i should say [itex](∂({\frac{dU}{dt})\frac{1}{∂}(\frac{dt}{dq}}))⋅δq[/itex]...and should i understand this as most of dimensions cancel out and only energy remains?...um this doesn't seem to help[itex]\frac{d}{dt}(\frac{∂r}{∂q})=(\frac{∂\dot{r}}{∂q})[/itex] problem is similar too. For it seems it is better to just write the detailed location of actual book, it is on top of page 20 or right above 1.51 on Goldstein classical mechanics third edition

It is the moment of making Lagrangian Equation and there was this equation [itex]\frac{d}{dt}(\frac{∂r}{∂q})=(\frac{∂\dot{r}}{∂q})[/itex] (those 'r's are vectors)

Is it just possible that term [itex]\frac{d}{dt} [/itex] just delve into [itex](\frac{∂r}{∂q})[/itex] and become [itex](\frac{∂\dot{r}}{∂q})[/itex] ??i want to know why it just goes into partial derivationgosh i really need more math here... any suggestion on math book for reading goldstein?(lineear algebra? vector calculus?)

 

[BvU]

Well it seems, because time derived term is on denominator, so if i think 'dt' included in ##\partial{\dot{q}}## as if it is on the numerator of ##\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq ##

Correct for the dimensions. Differentials have a dimension, just like normal physical quantities, yes. Manipulating the dimensions is fairly trivial, but with the differentials themselves you have to be a lot more careful.

(then it would be like (##\frac{d}{dt}(\partial{U}\frac{1}{∂}(\frac{dt}{q}))⋅δq##),

You can not write ##{d\over dt}{\partial U\over \partial \dot q} ## like that. You really have to take the partial derivative of ##U## wrt ##\dot q## and then the time derivative of that (*).
Dimension time cancels. Multiplication with ##\delta q## cancels the dimension of ##q## and you are left with the dimension of ##u##, i.e. work.

(*) Interchanging the order of differentiation may be possible -- see below.​

Because it looks so simple if it goes like $$\frac{d}{dt}(\frac{\partial{U}}{ \partial{ \dot{q}}} )⋅δq=(\frac{\partial{\dot{U}}}{ \partial{ \dot{q}}} )⋅δq=(\frac{\partial{U}}{dt}\frac{dt}{ \partial{q}} )⋅δq=(\frac{\partial{U}}{ \partial{q}} )⋅δq$$ ...(I pretty much think this equation is very very wrong idea lol apparently here$$

You bet it's wrong ! It only goes like that for the dimensions.
If U depends on ##\dot q## only and not at all on ##q## itself you 've lost the whole thing !
And: How does time depend on ##q## ?

Re

##\frac{d}{dt}(\frac{∂r}{∂q})=(\frac{∂\dot{r}}{∂q})## problem

##{\bf r}_i## is a function of the ##q_j## and time (1.38 and 1.45').
Taking the partial derivative (1.46) of ##{\bf r}_i## wrt ##q_k## means that you leave all the other ##q## and also t constant and take the derivative.
Write it out and take the time derivative to check that ##\frac{d}{dt}(\frac{∂r_i}{∂q_j})=(\frac{∂\dot{r_i}}{∂q_j})##

I realize I'm not helping you very much here. Check the math books forum for a decent calculus book.​

Note that ##\bf r## is a vector, but ##\bf q## is also a vector ! Writing ##\partial r\over \partial q## is error-prone (to put it mildly (**) ).(**) I remember convincing a math tutor that physicists write ##dr\over dq## when they mean the Jacobian ##d\vec r\over d\vec q##, just to get higher marks in a test. I'm still somewhat ashamed of that: it was a genuine fault/oversight of mine...

 

[RIgel]

Yep thanks thanks ( ´∀｀)

I think i got it

[itex]\frac{d}{dt}(\frac{∂r_i}{∂q_j})=(\frac{∂\dot{r_i}}{∂q_j})[/itex] problem is perfectly settled

Gosh i can't believe why I was wandering, you really woke me up

And it indeed is a mistake to write those vectors as normal font, but making math symbols neat as this level was my limit it seems... I was too exhausted to find the itex code(?) for bold.

It felt bad when i couldn't put all of sub scripts and details...but i had too much of me in figuring out itex already

Anyway i really appreciate you pointing out that [q][/qi] is vector!For first question...to be honest, I think i 'can' get used to fact that dimensions works like that, and utilize it, but still i think i will think about that again if I have some moments.lol [itex](\frac{\partial{U}}{dt}\frac{dt}{ \partial{q}} )⋅δq[/itex] indeed seems awkward. i was half laughing writing that

I had been using this website just reading, but it is my first time writing here and what i found out is that you were amazing on writing all those math symbols...i never knew it before... it really made me respect for all those things

Again thank you so much!

 

[BvU]

You're very welcome. I appreciated the opportunity to freshen up what I thought I knew a bit but didn't really master (still don't ).

My advice: take simple examples to work out the theory text and do lots of exercises.