On May 15, 8:04*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).



What KW is saying, which is obvious to a 10 years old buy NOT TO YOU,
it's that the metric, without the corresponding set, does not specify
a geometry.

YOU ASSUME a priori --- somethign that he does not --- that if someone
tells you the metric is:

dr^2 + r^2 dtheta^2

then you know that:

x = r cos(theta) and y = r sin(theta)

Although this is a trivial example you presented and in most cases
someone will assume that, he is saying that in general, without the
coordinate transformation to (x,y), given the above metric, you cannot
deduce it is equivalent to dx^2 + dy^2.


I think it is obvious that you manipulate words in a way to suit you.
You distort other peoples arguments and you present trivial examples
that do not correspond to the original problem posed.

You, PD, Roberts, Carlipp and other not worth mentioning all exhibit
the same exact pattern of behavior.

I wonder if behind all these nicknames there is just one person, or
several persons instructred by the same person on how to act.

Remember, a 10 year old understands you are wrong. How long are you
going to play this game?

Mike






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.

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