Coordinate transformation derivatives

[topsquark]

I've had to hit my books to help someone else. Ugh.

Say we have the coordinate transformation $$\bf{x}' = \bf{x} + \epsilon \bf{q}$$, where $$\epsilon$$ is constant. (And small if you like.) Then obviously
$$d \bf{x}' = d \bf{x} + \epsilon d \bf{q}$$.

How do we find $$\frac{d}{d \bf{x}'}$$?

I'm missing something simple here, I'm sure of it.

-Dan

 

[Fantini]

Perhaps I'm getting lost in notation, topsquark. I'm not sure I understand your question.

Let us assume that $\mathbf{x}', \mathbf{x}$ and $\mathbf{q}$ are in $\mathbb{R}^3$. Therefore we have that $d \mathbf{x}' = \text{rot } \mathbf{x}' = \nabla \times \mathbf{x}'$.

What are you trying to compute?

 

[topsquark]

Perhaps I'm getting lost in notation, topsquark. I'm not sure I understand your question.

Let us assume that $\mathbf{x}', \mathbf{x}$ and $\mathbf{q}$ are in $\mathbb{R}^3$. Therefore we have that $d \mathbf{x}' = \text{rot } \mathbf{x}' = \nabla \times \mathbf{x}'$.

What are you trying to compute?

Simple example then. Say we have
$$x = e^t$$

Then
$$dx = e^t dt$$

Formally we have
$$\frac{d}{dx} = e^{-t} \frac{d}{dt}$$

I'm looking for something along those lines. Basically I am trying to simplify the operator
$$\frac{\partial }{ \partial ( \bf{x} + \epsilon \bf{q} )}$$

-Dan

 

[Fantini]

Say we have
$$x = e^t$$

Then
$$dx = e^t dt$$

Formally we have
$$\frac{d}{dx} = e^{-t} \frac{d}{dt}$$

This is something I don't agree with. You seem to be saying that $dx = e^t dt$ implies $\frac{1}{dx} = e^{-t} \frac{1}{dt}$.

 

[alyafey22]

$$dx = e^t dt$$

Formally we have
$$\frac{d}{dx} = e^{-t} \frac{d}{dt}$$

\(\displaystyle \frac{d}{dx}\) what do you mean by that ?

we know that \(\displaystyle \frac{d}{dx} ( f(x) ) = f'(x) \)

 

[Mathwebster]

Are you looking for how derivatives transform under infinitesimal transformations?

 

[topsquark]

This is something I don't agree with. You seem to be saying that $dx = e^t dt$ implies $\frac{1}{dx} = e^{-t} \frac{1}{dt}$.

Not quite.
$$dx = e^t dt \implies \frac{d}{dx} = \frac{d}{e^t dt} = e^{-t} \frac{d}{dt}$$

And I did say "formally." (Wasntme) I understand there are technical issues with this "derivation." It's a substitution technique I was taught when solving Euler differential equations.

-Dan

- - - Updated - - -

Are you looking for how derivatives transform under infinitesimal transformations?

Originally yes. But certainly there is an approach to simplify the operator in the "macroscopic" case?

-Dan

 

[Mathwebster]

So if you're transforming from $(t,x,u) \rightarrow (t'x',u')$ where $t$ and $x$ are the independent variables and $u$ the dependent variable, you could use Jacobians.

 

[I like Serena]

Sounds indeed as if you refer to the Jacobian matrix, which in this case is the identity matrix.

Assuming you mean that \(\displaystyle \frac d {d\mathbf x'} = \left(\frac \partial{\partial x_1'}, ..., \frac \partial{\partial x_n'}\right)\), that would be the same as \(\displaystyle \frac d {d\mathbf x}\)

 

[Mathwebster]

Not what I meant. Suppose that

$\bar{t}=t+T(t,x,u)\varepsilon + O(\varepsilon^{2}),$
$\bar{x}=x+X(t,x,u)\varepsilon + O(\varepsilon^{2}),\;\;\;(1)$
$\bar{u}=u+U(t,x,u)\varepsilon + O(\varepsilon^{2}),$

and I wish to calculate $\displaystyle \dfrac{\partial \bar{u}}{\partial \bar{t}}$ then an easy way is to use Jacobians. I.e.

$\displaystyle \dfrac{\partial \bar{u}}{\partial \bar{t}} = \dfrac{\partial(\bar{u},\bar{x})}{\partial(\bar{t},\bar{x})} = \dfrac{\partial(\bar{u},\bar{x})}{\partial(t,x)} / \dfrac{\partial(\bar{t},\bar{x})}{\partial(t,x)}\;\;\;(2)$.

Now insert the transformations (1) into (2) and expand. The nice thing about (2) is that it is easy to calculate. Furthermore, a Taylor expansion for small $\varepsilon$ is also fairly straight forward (if this is what topsquark was thinking)

 

[I like Serena]

Ah. I sort of assumed $\mathbf q$ was a constant.
I guess I shouldn't have, since you did refer to $d\mathbf q$, suggesting $\mathbf q$ depends on $\mathbf x$.

 

[topsquark]

To all that mentioned Jacobians: (sighs) Yup. That's what I was looking for. I thought it would be something simple!

Thanks to all!

-Dan