On May 15, 7:04*pm, JanPB wrote:
On May 14, 10:08*pm, Koobee Wublee wrote:

On May 14, 7:10 am, Mike wrote:

On May 13, 5:46 pm, Koobee Wublee wrote:
The spacetime described by the Schwarzschild metric only applied to
the spherically symmetric polar coordinate. *You are welcome to
transform to another coordinate system with a different metric.

No, the Schwarzschild metric IS the metric. There is no such thing as
Schwarzschild metric in a coordinate system with a metric.

Mathematically, you are just wrong. *For example, describing flat
spacetime using the linear rectangular (Euclidean) and using the
spherically symmetric polar coordinate systems require you to supply
different metric for each choice of coordinate system. *shrug

No, it's the same metric in both cases, e.g. in the plane the
following are equal:

* * dx^2 + dy^2

and:

* * dr^2 + r^2 dtheta^2

...where x = r cos(theta) and y = r sin(theta) (polar coordinates).

The above are two different coordinate decompositions of the same
metric. To see that they are equal, just evaluate them on an arbitrary
vector and see you get the same value in both cases.

This is where KW gets the small chicken bone stuck in his throat,
completely unable to swallow the distinction between a vector or
tensor and its coordinate decompositions. To him, if you have a
different component representation, you're talking about a physically
different thing, and he pretends to sniff that it is "mathematically
so".


For example, consider vector v defined in Cartesian coordinates like
this:

* * v is situated at point (1,1),
* * v = (1,2)

Then, applying ds^2 = dx^2 + dy^2:

* * ds^2(v) = v.v = dx(v) * dx(v) + dy(v) * dy(v) = 1^2 + 2^2 = 5

Now calculate using the polar representation of ds^2:

* * v is situated at (sqrt(2), pi/4),
* * v = (3/sqrt(2), -1/2)

Then, applying ds^2 = dr^2 + r^2 dtheta^2:

* * ds^2(v) = v.v = dr(v) * dr(v) + (sqrt(2))^2 * dtheta(v) *
dtheta(v) =

* * * * * * = (3/sqrt(2))^2 + 2 * (-1/2)^2 =

* * * * * * = 9/2 + 2 * 1/4 =

* * * * * * = 9/2 + 1/2 = 10/2 = 5

Same result! Yout hink that's an accident? :-) You'd obtain the
corresponding same results with any vector. It must be so because both
expressions for ds^2 are equal.

Same thing with Schwarzschild, etc.

The term "metric" as used by everyone refers to that mapping that
takes tangent vectors (like v) to their squared lengths (like v.v).
That mapping is coordinate-INdependent. OTOH its _coordinate
representation_ of course depends on coordinates (by definition). But
dx^2+dy^2 and dr^2 + r^2 dtheta^2 are two different coordinate
representations of the same metric (the same mapping taking v to v.v,
or equivalently, a mapping taking vector pairs v,w to v.w - that's
what metric _is_).

--
Jan Bielawski