Re: New version of a relativity FAQ

Pmb says...

I don't believe that Einstein was aware of the coordinate-free
way to define vectors and tensors.

The very fact that they are tensors means that they are coordinate
independent, i.e. their definition does not depend on a particular
coordinate system.

In Einstein's time, tensors were defined by how they *transform*
under changes of coordinates, so the definition was not coordinate-free.
It required coordinates to even state what a tensor was. That's
very different from the modern coordinate-free approach, which
defines vectors and tensors without reference to a coordinate system.
I don't believe that Einstein knew about this approach.

For reference, here's my take on how to define tensors in a
coordinate-free way:

1. Start with a manifold M.

2. Define a "scalar field" on M to be a smooth function from
M to R (the reals). For example, if M is the surface of the Earth at a
given moment, then the function giving the temperature at
each point on the surface is a scalar field.

3. Define a "parameterized path" on M to be a smooth function
from R to M. The obvious example is the function giving an
object's position as a function of time, but the parameter
doesn't need to be time, it could be any quantity that varies
along the path. For example, if you have a long highway, and
pick a starting point and a starting direction, then your
position on the Earth as a function of the distance down
the highway is a parameterized path.

4. Define a "tangent vector" to a parameterized path as follows:
If P(s) is a parameterized path, then the tangent vector to P
at s0 is that operator v which acts on any scalar field Phi(X)
as follows:

v(Phi) = d/ds (Phi(P(s)) evaluated at s0

This approach identifies a vector with its corresponding
directional derivative. So a tangent vector to a path
uniquely determines how (to first order) any scalar field
varies along that path.

5. Define a "one-form" or "covector" as a kind of gradient
of a scalar field. If Phi is a scalar field, and X0 is some
point on the manifold, then the gradient to Phi at X0 is
the operator w which acts on any tangent vector v as follows:

w(v) = v(Phi)

Since a tangent vector is a kind of local approximation
to a parameterized path, a gradient of a scalar field
uniquely determines how (to first order) that field varies
along any path.

6. Define a n-m tensor T to be a multilinear function that takes
n vectors and m one-forms and returns a real. Since a one-form
is just a function from vectors to reals, we can see that a
1-0 tensor is the same as a one-form. Similarly, a 0-1 tensor
uniquely determines a vector (and is uniquely determined by one)
so it is common to identify 0-1 tensors with vectors.

All of these definitions require no coordinates at all. Vectors
and one-forms and tensors are not defined in terms of their
components. But the usual component description is easily
recovered: A coordinate system on a region of a manifold
specifies an indexed collection x^u of scalar fields. So
if v is some tangent vector, then we can define the component
v^u to be the result of v operating on the field x^u. For each
index u, we define the corresponding basis vector e_u
to be that vector whose components are zero except in
the direction u, and whose component in that case is 1.
Then if Phi is any scalar field, we can define
@/@x^u Phi = e_u(Phi).

Then, if w is any one-form we can define its components
as follows: w_u = w(e_u). The significance of this is
that if Phi is any scalar field, and w is its gradient
one-form, then w_u = @/@x^u Phi.

Because one-forms are linear
operators, we can write, for any vector v:

w(v) = sum over indices u of w_u v^u

--
Daryl McCullough
Ithaca, NY

.

Relevant Pages

  • Re: New version of a relativity FAQ
    ... defines vectors and tensors without reference to a coordinate system. ... Define a "scalar field" on M to be a smooth function from ... Define a "parameterized path" on M to be a smooth function ... n vectors and m one-forms and returns a real. ...
    (sci.physics.relativity)
  • Re: Eric Gisse insists Temperature is Not a Tensor!
    ... quantities are tensors of rank 0;" ... Wikipedia is wrong and you are stupid for treating Wikipedia like a ... measures the rate and direction of change in a scalar field; ... the gradient of a scalar field is a vector field. ...
    (sci.physics.relativity)