Coordinate transformation

  • Thread starter TimeRip496
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  • #1
TimeRip496
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V′μ=((∂yμ)/(∂xν))*Vν

This is a contravariant vector transformation. (Guys I am really sorry for making the formula above looks so incomprehensible as I still new to this.)

For the y in the partial derivative, is y a function in terms of x? In that sense, is it formula that maps x to y? Is it something like how much units of x correspond to each unit of y?
 

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  • #2
Nugatory
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Yes, each ##y^i## is a function of the ##x^i##, and vice versa.

A common and easy example is Cartesian and polar coordinates applied to the two-dimensional Euclidean plane. Using your notation, and with the ##y^i## as the Cartesian coordinates, the ##x^i## as the polar coordinates, and ##i## takes on the values 0 and 1 because this is a two-dimensional space:

##y^0=x^0cos(x^1)##; you're more used to seeing ##x=rcos\theta##
##y^1=x^0sin(x^1)##; you're more used to seeing ##y=rsin\theta##

If you have the components of a vector in polar coordinates ##V^\mu##, the formula you posted will give you the components of the same vector in Cartesian coordinates ##V'^\mu## (using the Einstein summation convention, of course).
 
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  • #3
TimeRip496
254
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Yes, each ##y^i## is a function of the ##x^i##, and vice versa.

A common and easy example is Cartesian and polar coordinates applied to the two-dimensional Euclidean plane. Using your notation, and with the ##y^i## as the Cartesian coordinates, the ##x^i## as the polar coordinates, and ##i## takes on the values 0 and 1 because this is a two-dimensional space:

##y^0=x^0cos(x^1)##; you're more used to seeing ##x=rcos\theta##
##y^1=x^0sin(x^1)##; you're more used to seeing ##y=rsin\theta##

If you have the components of a vector in polar coordinates ##V^\mu##, the formula you posted will give you the components of the same vector in Cartesian coordinates ##V'^\mu## (using the Einstein summation convention, of course).
Thanks! This really help a lot.
 

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