I Do index gymnastics change a tensor's dimensions?

snoopies622

Suppose I have a velocity field $V^a$ which looks like $< \partial x / \partial t , \partial y / \partial t , \partial z / \partial t >$ and I use the metric tensor to change it into a covector field $$V_a = g_{ab}V^a.$$ Does it then represent the same quantity or $< \partial t / \partial x, \partial t / \partial y , \partial t / \partial z >$ instead? I ask because long ago I read that vectors (like velocity) — when expressed as a set of partial derivatives — have spacial components in their numerators while covectors (like gradient) have them in their denominators. This would suggest - in this case at least - that the covector V would represent time per distance traveled instead of distance traveled per time.

Related Differential Geometry News on Phys.org

andrewkirk

Homework Helper
Gold Member
The thing that goes - in a loose sense - from the numerator to the denominator when we take the dual of a vector is $dx_i$ where $x_i$ is one of the coordinates in the coordinate system. This is apparent by the way the basis vectors and dual vectors are written:

$\frac{\partial}{\partial x_i}$ for basis vector $\vec e_i$ and $dx_i$ for dual basis vector (aka covector or one-form) $\tilde p^i %= g^{ij}\vec e_j = g^{ij}\frac{\partial}{\partial x_j}$.

In the case of the OP the derivatives are with respect to $t$, which is not a coordinate of any coordinate system of the 3D manifold. So this pattern does not apply.

In a 4D spacetime manifold, as used in relativity theory, something analogous to what you write may apply, because in that case $t$ will be one of the coordinates. But in relativity theory velocity vectors ('four-velocities') are unitless.

snoopies622

Thanks, andrewkirk. If you don't mind I'd like to change my example a little:

Suppose the air temperature in a room varies in a smooth, continuous way, and the function T(x,y,z) represents this temperature at a point. Also suppose that a particle in the room moves along a smooth path and at a given moment occupies point (x,y,z) and moves at velocity v(t). If I want to know the rate at which the air temperature in the immediate vicinity of the particle is changing with respect to time, I can

- treat the particle velocity as a vector $v = ( \partial x / \partial t, \partial y / \partial t, \partial z / \partial t )$

- find the gradient of the temperature function to form the covector $G = ( \partial T / \partial x , \partial T / \partial y, \partial T/ \partial z )$

- and then find the inner product in the direct way
$$v \cdot G = \frac {\partial x}{ \partial t} \frac {\partial T }{\partial x } + \frac {\partial y}{ \partial t} \frac {\partial T }{\partial y } + \frac {\partial z}{ \partial t} \frac {\partial T }{\partial z } = \frac {dT}{dt}$$
So I'm wondering what this would look like if I decide to first use index gymnastics to make the velocity into a covector and the temperature gradient into a vector.

If I decide to represent velocity as $v = ( \partial t / \partial x , \partial t / \partial y, \partial t/ \partial z )$
and the temperature spacial derivative as $G = ( \partial x / \partial T, \partial y / \partial T, \partial z / \partial T )$
and find the inner product of these two rank one tensors, I'd get
$$v \cdot G = \frac {\partial t}{ \partial x} \frac {\partial x }{\partial T } + \frac {\partial t}{ \partial y} \frac {\partial y }{\partial T } + \frac {\partial t}{ \partial z} \frac {\partial z }{\partial T } = \frac {dt}{dT}$$, the inverse of what I got the first way.

This all seems logical except that the two tensors above described with partial derivatives make it appear that time is a function of spacial position and that spacial position is a function of temperature, and that gives me a headache.

snoopies622

Re-reading your response, I see that I have repeated my problem in that neither time nor temperature are parts of the spacial coordinate system (x,y,z). Nevertheless, I'm wondering what exactly would happen if I indeed decided to make the velocity a covector and the temperature gradient a vector. Would the quantities that they represent - and the iner product of those quantities - be exactly the same?

andrewkirk

Homework Helper
Gold Member
If the space is ordinary Euclidean space and the coordinate system is the usual Cartesian system then the metric tensor will be the 3D identity matrix, so the numerical components of each vector will be the same as those of its dual (covector). The inner product will be the same.

If on the other hand a different coordinate system such as spherical coordinates were used, the vectors would have different components from their duals.

snoopies622

Afterthought: Since the magnitude ds of a rank one tensor does not depend on whether it is expressed as a vector or a covector (I actually took the trouble to test this myself with a random metric that was not the identity metric), then doing index gymnastics to change a velocity vector into a covector does not change what it represents into its inverse, otherwise this quantity would change as well. That is, if a velocity vector has magnitude 3 and represents 3 meters per second in a particular direction, then the length of its corresponding covector is also 3 and therefore does not represent the quantity 1/3 seconds per meter.

• stevendaryl

"Do index gymnastics change a tensor's dimensions?"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving