- #1
spaghetti3451
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Homework Statement
If ##\bf{v}## is a vector and ##\alpha## is a covector, compute directly in coordinates that ##\sum a_{i}^{V}v^{i}_{V}=\sum a_{i}^{U}v^{j}_{U}##.
What happens if ##\bf{w}## is another vector and one considers ##\sum v^{i}w^{i}##?
Homework Equations
The Attempt at a Solution
##\alpha(\bf{v})=\alpha(\bf{v})##
##\implies (a_{i}^{V}\ \sigma^{i}_{V})(v^{k}_{V}\ \vec{e}_{k}^{V})=(a_{j}^{U}\ \sigma^{j}_{U})( v^{l}_{U}\ \vec{e}_{l}^{U})##,
where the Einstein summation convention has used to sum over the ##i##, ##j##, ##k## and ##l## indices and the left-hand side uses the ##U## coordinate system and the right-hand side uses the ##V## coordinate system, so that
##\implies a_{i}^{V}\ v^{k}_{V}\ \sigma^{i}_{V}(\vec{e}_{k}^{V}) = a_{j}^{U}\ v^{l}_{U}\ \sigma^{j}_{U}(\vec{e}_{l}^{U})##
##\implies a_{i}^{V}\ v^{k}_{V}\ \delta^{i}_{k} = a_{j}^{U}\ v^{l}_{U}\ \delta^{j}_{l}##
##\implies a_{i}^{V}\ v^{i}_{V}\ = a_{j}^{U}\ v^{j}_{U}##For the second part, again using the Einstein summation convention,
##v^{i}w^{i}=v^{i}w_{j}\delta^{ji}=\langle \vec{v},\vec{w}\rangle## for ##g^{ij}=\delta^{ij}##.
Now, ##g^{ij}=\delta^{ij}## is true only for Cartesian coordinates in flat space, so that the metric changes under a coordinate transformation. Therefore, the transformation of ##v^{i}w^{i}## depends upon the specific change of coordinates.
What do you think?