- #1
gordon831
- 9
- 0
Hey all,
I've never taken a formal class on tensor analysis, but I've been trying to learn a few things about it. I was looking at the metric tensor in curvilinear coordinates. This Wikipedia article claims that you can formulate a dot product in curvilinear coordinates through the following:
[itex]\textbf{u} \cdot \textbf{v} = g_{ij} u^i v^j [/itex] where [itex]g_{ij}[/itex] is the covariant metric tensor for the coordinate system.
In particular, I'm interested in 2-D polar coordinates, therefore, [itex]g_{ij} = \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)[/itex] which makes sense: the polar coordinates are orthogonal so [itex]g_{ij}[/itex] is diagonal, and an infinitesimal change [itex]ds[/itex] occurs with a proportional change in [itex]dr[/itex] and varies with [itex]r[/itex] in [itex]d \theta[/itex]. Now I expand the expression for the dot product and using Einstein summation:
[itex]\textbf{u} \cdot \textbf{v} = u^1 v^1 + (r)^2 u^2 v^2 [/itex]
Now this really doesn't make much sense to me. To me, this implies that the length of a vector in polar coordinates depends on [itex]\theta[/itex], which isn't true, the length is contained entirely within the [itex]r[/itex] component. If we transform a vector in polar to Cartesian and then write our dot product, we find [itex]\textbf{u} \cdot \textbf{v} = u^1 v^1 \cos(u^2 - v^2) [/itex], but I don't see this happening any time soon with this metric and the given dot product formulation. Is there any way to get the same dot product using the metric of the coordinate space? If not, why isn't this working? Thanks!
I've never taken a formal class on tensor analysis, but I've been trying to learn a few things about it. I was looking at the metric tensor in curvilinear coordinates. This Wikipedia article claims that you can formulate a dot product in curvilinear coordinates through the following:
[itex]\textbf{u} \cdot \textbf{v} = g_{ij} u^i v^j [/itex] where [itex]g_{ij}[/itex] is the covariant metric tensor for the coordinate system.
In particular, I'm interested in 2-D polar coordinates, therefore, [itex]g_{ij} = \left( \begin{array}{cc} 1 & 0 \\ 0 & r^2 \end{array} \right)[/itex] which makes sense: the polar coordinates are orthogonal so [itex]g_{ij}[/itex] is diagonal, and an infinitesimal change [itex]ds[/itex] occurs with a proportional change in [itex]dr[/itex] and varies with [itex]r[/itex] in [itex]d \theta[/itex]. Now I expand the expression for the dot product and using Einstein summation:
[itex]\textbf{u} \cdot \textbf{v} = u^1 v^1 + (r)^2 u^2 v^2 [/itex]
Now this really doesn't make much sense to me. To me, this implies that the length of a vector in polar coordinates depends on [itex]\theta[/itex], which isn't true, the length is contained entirely within the [itex]r[/itex] component. If we transform a vector in polar to Cartesian and then write our dot product, we find [itex]\textbf{u} \cdot \textbf{v} = u^1 v^1 \cos(u^2 - v^2) [/itex], but I don't see this happening any time soon with this metric and the given dot product formulation. Is there any way to get the same dot product using the metric of the coordinate space? If not, why isn't this working? Thanks!