ecce.monkey said:
No I don't agree here. They may be solutions to the same empty mathematical form of an equation, but not to some concrete physical field equation.
I think you misunderstand what it means to be the same form.
Here's an example of a criterion being of the same form.
Consider two tangent vectors
v and
w to some point
P of the Euclidean plane. Let [itex][v]_{x,y} = (v_x, v_y)[/itex] denote the coordinate representation of
v with respect to some orthonormal rectilinear coordinate system (with
P not the origin), and let [itex][v]_{r, \theta} = (v_r, v_\theta)[/itex] be its coordinate representation under the corresponding polar coordinates.
The condition that
v and
w are orthogonal is a condition that has the same form in both coordinate systems. In the (x, y) basis, the criterion is
[tex]v_x w_x + v_y w_y = 0[/tex]
and in the [itex](r, \theta)[/itex] basis, the criterion is
[tex]v_r w_r + v_\theta w_\theta = 0[/tex]
In both cases, the relation has the same form
f(a, b) = 0
where
f is the function defined by
[tex]f(\mathbf{a}, \mathbf{b}) = a_1 b_1 + a_2 b_2[/tex]
For example, in the (x,y) basis, this relation is
[tex]f([v]_{x,y}, [w]_{x,y}) = 0[/tex]
which simplifies to
[tex]v_x w_x + v_y w_y = 0[/tex]
and in the [itex](r, \theta)[/itex] basis, this relation is
[tex]f([v]_{r,\theta}, [w]_{r,\theta}) = 0[/tex]
Note that the only difference between these two relations is the basis with respect to which we represent the vectors as 2-tuples.
Correspondingly, we can observe things like how the coordinate vectors (1,0) and (0,1) represent orthogonal vectors both if we consider them as representing vectors in the (x,y) basis and as representing vectors in the [itex](r, \theta)[/itex] basis.
An example of something not being of the same form is the distance formula of Euclidean geometry. In the (x, y) coordinate system, the distance between two points
P and
Q is:
[tex]d(P, Q) = \sqrt{ \left( x(P) - x(Q) \right)^2 + \left( y(P) - y(Q) \right)^2 }[/tex]
whereas in the [itex](r, \theta)[/itex] coordinate system, the distance is
[tex]d(P, Q) = \sqrt{ r(P)^2 + r(Q)^2 - 2 r(P) r(Q) \cos\left( \theta(P) - \theta(Q) \right) }[/tex]
Again, let [itex][P]_{x,y} = (x(P), y(P))[/itex] be the coordinate representation of the point
P in the (x, y) coordinate system. Now, we can see that does not exist a function
f(a, b) satisfying both:
[itex]f([P]_{x,y}, [Q]_{x,y})[/itex] is the distance between P and Q
[itex]f([P]_{r,\theta}, [Q]_{r,\theta})[/itex] is the distance between P and Q
On the other hand, if we demote the distance function to just another piece of added structure, then we can also ask for [itex][d]_{x,y}[/itex] -- the coordinate representation of the distance function in (x,y) coordinates. (and similarly, in the [itex](r, \theta)[/itex] coordinates) Then, the distance formula is given by the form
f(g, a, b) = g(a, b)
and we see that
[itex]f\left( [d]_{x, y}, [P]_{x,y}, [Q]_{x,y} \right) =[/itex] distance between P and Q
[itex]f\left( [d]_{r, \theta}, [P]_{r,\theta}, [Q]_{r,\theta} \right) =[/itex] distance between P and Q
If it helps to read the above, the three arguments to
f are, in order:
. a function that takes two 2-tuples as input
. a 2-tuple
. a 2-tuple
In case you're curious just what the heck the coordinate representation of
d might be, it's a function that takes two 2-tuples as arguments and is given by:
[tex][d]_{x, y}(\mathbf{a}, \mathbf{b}) = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}[/tex]
[tex][d]_{r, \theta}(\mathbf{a}, \mathbf{b}) = \sqrt{ a_1^2 + b_1^2 - 2 a_1 b_1 \cos\left( a_2 - b_2 \right) }[/tex]
Hurkyl shows you can call x=5 a circle by redefining distance.
I had originally assumed you were taking this latter point of view, because:
(1) You're trying to understand an argument involving GR, where the metric tensor is merely a field which really and truly does have a different coordinate representation in different coordinate charts -- the analogous situation in your counterargument would be a distance function whose coordinate representation is also different in different charts
(2) You invoked a coordinate change that does not leave Euclidean geometry invariant
If you really and truly meant to consider the distance function of Euclidean geometry as having an absolute coordinate form, then you needed to restrict yourself to orthonormal coordinates -- your switch to polar coordinates was illegal, which is why you arrived at a problem in your counter argument.
P.S. I am avoiding tensor notation specifically to oppose the common abuse of thought that confuses a vector with its coordinate representation as a tuple of real numbers. While it is a very practical abuse of thought, it can also be a huge obstacle when you don't understand something. In that vein, I really shouldn't have used the notation [itex]v_x[/itex] and [itex]v_y[/itex] for the components of the coordinate representation of
v w.r.t. (x,y), but alas, no convenient alternate notation sprung to mind.