The discussion focuses on the transformation of basis vectors on manifolds, specifically demonstrating that the transformed basis vector \(\vec{e'}_a\) can be expressed as \(\vec{e'}_a = \frac{\partial x^b}{\partial x'^a} \vec{e}_b\). Participants clarify that the basis vectors \(\vec{e}_b\) correspond to the partial derivative operators \(\frac{\partial}{\partial x^b}\) and \(\vec{e'}_a\) to \(\frac{\partial}{\partial x'^a\). The transformation relies on the invariance of the differential element \(ds\) and the relationship between coordinate systems.
PREREQUISITESMathematicians, physicists, and students studying differential geometry, particularly those interested in the transformation properties of basis vectors on manifolds.
ehrenfest said:Homework Statement
I am trying to show that
[tex]\vec{e'}_a = \frac{\partial x^b}{\partial x'^a} \vec{e}_b[/tex]
where the e's are bases on a manifold and the primes mean a change of coordinates
Yes, I am dealing with coordinate bases.George Jones said:You're dealing with coodinates bases, or else your expression above isn't true. Thus
[tex]\vec{e}_b = \frac{\partial}{\partial x^b}[/tex]
[tex]\vec{e'}_a = \frac{\partial}{\partial x'^a}.[/tex]