Conversion between vector components in different coordinate systems

[Karl Karlsson]

Homework Statement:

I am not completely sure what the formulas $v_j = v^a\frac {\partial x^j} {\partial \chi^a}$ and $v^b = v^a\frac {\partial \chi^b} {\partial x^j}$ mean. Is $v_j$ the j:th cartesian component of the vector $\vec v$ or could it hold for other bases as well? What does the second equation $v^b = v^a\frac {\partial \chi^b} {\partial x^j}$ mean? Is this just the relation between the b:th and j:th component of $\vec v$ being expressed as tangent vectors?

Relevant Equations:

$v_j = v^a\frac {\partial x^j} {\partial \chi^a}$ and $v^b = v^a\frac {\partial \chi^b} {\partial x^j}$

I am not completely sure what the formulas $v_j = v^a\frac {\partial x^j} {\partial \chi^a}$ and $v^b = v^a\frac {\partial \chi^b} {\partial x^j}$ mean. Is $v_j$ the j:th cartesian component of the vector $\vec v$ or could it hold for other bases as well? What does the second equation $v^b = v^a\frac {\partial \chi^b} {\partial x^j}$ mean? Is this just the relation between the b:th and j:th component of $\vec v$ being expressed as tangent vectors?

 

[stevendaryl]

Hmm. The "relevant equations" seem slightly wrong, to me. I think it should be:

$v^j = v^a \frac{\partial x^j}{\partial \chi^a}$ and $v^a = v^j \frac{\partial \chi^a}{\partial x^j}$

Having said that, the equations are good for translating between any two coordinate systems, whether Cartesian, or not.

 

[stevendaryl]

I'm surprised that this question is considered "precalculus". It seems like calculus, to me.

 

[Mark44]

I'm surprised that this question is considered "precalculus". It seems like calculus, to me.

To me, also. I have moved this thread, although the OP might not still be interested in it.