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[SOLVED] Coordinate transformation derivatives

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
I've had to hit my books to help someone else. Ugh.

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

How do we find [tex]\frac{d}{d \bf{x}'}[/tex]?

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

-Dan
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Perhaps I'm getting lost in notation, topsquark. I'm not sure I understand your question. :confused:

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? :confused:
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Perhaps I'm getting lost in notation, topsquark. I'm not sure I understand your question. :confused:

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? :confused:
Simple example then. Say we have
[tex]x = e^t[/tex]

Then
[tex]dx = e^t dt[/tex]

Formally we have
[tex]\frac{d}{dx} = e^{-t} \frac{d}{dt}[/tex]

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

-Dan
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Say we have
[tex]x = e^t[/tex]

Then
[tex]dx = e^t dt[/tex]

Formally we have
[tex]\frac{d}{dx} = e^{-t} \frac{d}{dt}[/tex]
This is something I don't agree with. You seem to be saying that $dx = e^t dt$ implies $\frac{1}{dx} = e^{-t} \frac{1}{dt}$.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
[tex]dx = e^t dt[/tex]

Formally we have
[tex]\frac{d}{dx} = e^{-t} \frac{d}{dt}[/tex]
\(\displaystyle \frac{d}{dx}\) what do you mean by that ?

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

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Are you looking for how derivatives transform under infinitesimal transformations?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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.
[tex]dx = e^t dt \implies \frac{d}{dx} = \frac{d}{e^t dt} = e^{-t} \frac{d}{dt}[/tex]

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
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
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 \(\displaystyle \frac d {d\mathbf x}\)
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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