Why is change of variables in the proof of Noether's Theorem legit ?

[Leb]

I have looked up a few derivations of Noether's Theorem and it seems that chain rule is applied (to get a total derivative w.r.t. q_{s} ( = q + s ) is often used. What I do not understand is why this is legitimate ? If we start with L=L(q,q^{.},t) how can we change to L=L(q_{s}, q_{s}^{.},t} ? Is it because we can say that dq_{s} = dq + ds (and since s is just a number if I understand correctly) = dq ?

I attach a derivation of Noether's theorem from my lectures.
View attachment noether.bmp

 

[Leb]

Maybe I should rephrase the question.

Let's say we have s function F=F(x,y,t) and we change the variables just slightly to $x_{s}=x+\epsilon s and y_{s} = y + \epsilon s$. wouldn't the total derivative of F(x_{s},y_{s},t) w.r.t. s

$\frac{dF}{ds} = \frac{\partial F}{\partial x_{s}}\frac{dx_{s}}{ds} + \frac{\partial F}{\partial y_{s}}\frac{dy_{s}}{ds}$ I do not see how this is equivalent to
$\frac{dF}{ds} = \frac{\partial F}{\partial x}}\frac{dx_{s}}{ds} + \frac{\partial F}{\partial y}\frac{dy_{s}}{ds}$

 

[Dick]

Maybe I should rephrase the question.

Let's say we have s function F=F(x,y,t) and we change the variables just slightly to $x_{s}=x+\epsilon s and y_{s} = y + \epsilon s$. wouldn't the total derivative of F(x_{s},y_{s},t) w.r.t. s

$\frac{dF}{ds} = \frac{\partial F}{\partial x_{s}}\frac{dx_{s}}{ds} + \frac{\partial F}{\partial y_{s}}\frac{dy_{s}}{ds}$ I do not see how this is equivalent to
$\frac{dF}{ds} = \frac{\partial F}{\partial x}}\frac{dx_{s}}{ds} + \frac{\partial F}{\partial y}\frac{dy_{s}}{ds}$

It's just notation. You are given F is a function of x, y and z. $\frac{\partial F}{\partial x}$ is the partial derivative of F wrt to its first variable. The thing you mean by your expression is exactly the same the meaning of the lecture expression.

 

[Leb]

Thank you for the reply.

Sorry, but I am too daft to understand. How can we jump from x, y variables to x_{s}=x+εs,y_{s}=y+εs, differentiate w.r.t. x, y and get an expression as if we were differentiating w.r.t. x_{s}

Or do you mean that whether I write $\frac{d}{dx_{s}} \text{or} \frac{d}{dx}$ does not matter ? But it does matter what is on top, isn't it ?

Could you maybe talk more about this "just the notation" matter ?

 

[Dick]

Thank you for the reply.

Sorry, but I am too daft to understand. How can we jump from x, y variables to x_{s}=x+εs,y_{s}=y+εs, differentiate w.r.t. x, y and get an expression as if we were differentiating w.r.t. x_{s}

Or do you mean that whether I write $\frac{d}{dx_{s}} \text{or} \frac{d}{dx}$ does not matter ? But it does matter what is on top, isn't it ?

Could you maybe talk more about this "just the notation" matter ?

If your question is how can you get away with just putting x into the partial derivative instead of x_{s}, remember you are eventually going to let s->0. Write down a concrete example like F(x,y)=xy and work through everything.

 

[Leb]

I mean, yeah, if we let s go to zero, then at the limit I would not argue with such an approximation, and I guess in this case it is correct, since if I understand correctly, we are talking about infinitesimal changes.

But if s was not that small ? Would this notation still hold ? For some reason I would have been much happier if they change the bottom partial x to partial x_{s}.

Thanks again !

 

[Dick]

I mean, yeah, if we let s go to zero, then at the limit I would not argue with such an approximation, and I guess in this case it is correct, since if I understand correctly, we are talking about infinitesimal changes.

But if s was not that small ? Would this notation still hold ? For some reason I would have been much happier if they change the bottom partial x to partial x_{s}.

Thanks again !

Sure, if s isn't infinitesimal you can't make that approximation. $\frac{\partial F}{\partial x} |_{x=x_s,y=y_s}$ would be an explicit way to write it. But that's actually what they mean when they write the chain rule anyway and usually don't say bother specify what to substitute for x and y.

 

[BucketOfFish]

I was thinking this same thing today too. I think it just comes down to the fact that $\frac{\partial x_s}{\partial x}=1$, so by the chain rule: $$\frac{\partial f}{\partial x_s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x_s}=\frac{\partial f}{\partial x}$$.