While learning about Special Relativity I learnt that we use the Transformation matrix to alter the space .This matrix differs for $Contravariant$ and $Covariant$ vectors.Why does it happen?,Why one kind of matrix ($Jacobian$) for $basis$ $vectors$ and other kind($Inverse$ $Jacobian$) for $gradient$ ,$divergence$ etc.Why does the matrix change ,why does changing quantities require a new matrix and why does $Inverse$ $Jacobian$ satisfy this?

$My$ $understanding$ $of$ $how$ $Jacobian$ $achieves$ $this$:


Here Grant says that by moving a small dist $\partial x$ ,we produce a change in $\partial f_1$ direction and $\partial f_2$ direction , $\partial f_1$ in being $x+\sin(y)$ direction & $\partial f_2$ being in $y+\sin(x)$ ,(ie) $\hat{x}$ transforms to $x+\sin(y)$ and $\hat{y}$ axis transforms to $y+\sin(x)$, by moving a small distance $\partial x$ in $x$ axis in input we produce a effect on both the axes in output space, similar thing happens in y direction .

$$(x,y) \rightarrow (x+\sin(y),y+\sin(x)) $$ Moving a small distance DX in x direction

Moving a small distance <span class=$\partial x$ in x direction in input space (input direction) causes a change in both the axes in output space" />

$$ J = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\ \end{bmatrix} $$

Is my interpretation a right one?

And why does the $Jacobian $ Matrix change for covariant and contravariant vectors?


1 Answer 1


For a small distance, the coordinates $$ \left(\begin{matrix}\hat x\\ \hat{y}\end{matrix}\right)=\left(\begin{matrix}f_1(x,y)\\ f_2(x,y)\end{matrix}\right)=:f(x,y) $$ change by $$ \left(\begin{matrix}\Delta \hat x\\ \Delta \hat y\end{matrix}\right)=J\cdot \left(\begin{matrix}\Delta x\\ \Delta y\end{matrix}\right)\,. $$ That's the transformation law for contravariant vectors. Conversely, by the chain rule, \begin{align} \partial_{x}(F\circ f)&= ((\partial_{\hat x}F)\circ f)\,\partial_x f_1+((\partial_{\hat y}F)\circ f)\,\partial_x f_2\\[3mm] \partial_{y}(F\circ f)&= ((\partial_{\hat x}F)\circ f)\,\partial_y f_1+((\partial_{\hat y}F)\circ f)\,\partial_y f_2\,. \end{align} This can be written as $$ \nabla (F\circ f)=J^\top\cdot ((\hat\nabla F)\circ f) $$ or as $$ \nabla =J^\top\cdot\hat\nabla\,. $$ This is the transformation law for covariant vectors (gradients).


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