- #1

snoopies622

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- TL;DR Summary
- How to calculate the size of a vector, confusion with basis vectors.

I'm stumbling on something rather basic here, will explain with an example. (Pardon the LaTeX problems, trying to fix..)

Suppose I have a plane, and in the plane I put the familiar (x,y) Cartesian coordinate system, and the metric is the usual Euclidean metric with [itex] ds^2 = dx ^2 + dy^2 [/itex].

Now suppose I add into this another coordinate system defined by

u=x+2y

v=x-y

and so it follows that

x=(1/3)(u + 2v)

y=(1/3)(u-v).

The basis vectors for u and v are

[itex] \vec{e}_u = < \frac {\partial x}{\partial u},\frac {\partial y}{\partial u} > ,= < 1/3 , 1/3 > [/itex]

[itex] \vec{e}_v = < \frac {\partial x}{\partial v},\frac {\partial y}{\partial v} > ,= < 2/3 , -1/3 > [/itex]

and the corresponding covectors are

[itex] \bar{e}^u = < \frac {\partial u}{\partial x},\frac {\partial u}{\partial y} > ,= < 1 , 2 > [/itex]

[itex] \bar{e}^v = < \frac {\partial v}{\partial x},\frac {\partial v}{\partial y} > ,= < 1 , -1 > [/itex].

The inner products of the basis vectors and covectors are

[itex] \vec{e}_u \cdot \bar{e}^u = < 1/3 , 1/3 > \cdot < 1 , 2 > =1 [/itex]

[itex] \vec{e}_v \cdot \bar{e}^v = < 2/3 , -1/3 > \cdot < 1 , -1 > =1 [/itex]

. . as one would expect.

My confusion is this: Isn't the magnitude of a vector equal to the square root of its inner product with itself?

[itex]

\vec {v} \cdot \vec{v} = v^a v^b g_{ab} =

v^a (v^b g_{ab}) = v^a v_a

[/itex]

This would imply that [itex] \vec{e}_u [/itex], for example, has a magnitude of 1, when clearly it's the square root of [itex] (1/3)^2 + (1/3)^2 = 2/9 [/itex].

Suppose I have a plane, and in the plane I put the familiar (x,y) Cartesian coordinate system, and the metric is the usual Euclidean metric with [itex] ds^2 = dx ^2 + dy^2 [/itex].

Now suppose I add into this another coordinate system defined by

u=x+2y

v=x-y

and so it follows that

x=(1/3)(u + 2v)

y=(1/3)(u-v).

The basis vectors for u and v are

[itex] \vec{e}_u = < \frac {\partial x}{\partial u},\frac {\partial y}{\partial u} > ,= < 1/3 , 1/3 > [/itex]

[itex] \vec{e}_v = < \frac {\partial x}{\partial v},\frac {\partial y}{\partial v} > ,= < 2/3 , -1/3 > [/itex]

and the corresponding covectors are

[itex] \bar{e}^u = < \frac {\partial u}{\partial x},\frac {\partial u}{\partial y} > ,= < 1 , 2 > [/itex]

[itex] \bar{e}^v = < \frac {\partial v}{\partial x},\frac {\partial v}{\partial y} > ,= < 1 , -1 > [/itex].

The inner products of the basis vectors and covectors are

[itex] \vec{e}_u \cdot \bar{e}^u = < 1/3 , 1/3 > \cdot < 1 , 2 > =1 [/itex]

[itex] \vec{e}_v \cdot \bar{e}^v = < 2/3 , -1/3 > \cdot < 1 , -1 > =1 [/itex]

. . as one would expect.

My confusion is this: Isn't the magnitude of a vector equal to the square root of its inner product with itself?

[itex]

\vec {v} \cdot \vec{v} = v^a v^b g_{ab} =

v^a (v^b g_{ab}) = v^a v_a

[/itex]

This would imply that [itex] \vec{e}_u [/itex], for example, has a magnitude of 1, when clearly it's the square root of [itex] (1/3)^2 + (1/3)^2 = 2/9 [/itex].

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