Re: Why TWLS= 0 <> OWLS in any ONE Frame.

On Mon, 30 Jan 2006 15:34:08 -0600, Tom Roberts wrote:

> sal wrote:
>> "scalar" is often -- perhaps usually -- taken to mean "scalar field
>> on the manifold" and hence something which is also "invariant", or
>> independent of choice of coordinates.
>
> Yes. A real-valued field on the manifold (occasionally a complex
> value is used).
>
>
>> Time's not, so time's not a scalar, either.
>
> Well, be careful. The time coordinate of a specified observer _is_ a
> scalar field on the manifold.

Yes, of course, you're right, at least as long as we stick with SR.
In GR, where particular coordinate systems may cover a subset of the
manifold, the statement is a little less clear-cut.

And of course we shouldn't forget that other neglected invariant,
either: the relativistic mass (OH NO NOT THAT!): the inner product of
the 4-momentum of an object with the 4-velocity of a particular
observer is the mass/energy of the object, aka relativistic mass, as
measured in that observer's rest frame.... so, since the inner product
of two vectors is an invariant, so's the (frame-dependent)
relativistic mass.


> Specifically: for observer A, observer A's coordinates apply a real
> value t_A(P) to every point P of a given region of the manifold, and
> is therefore a scalar field on that region; all observers will agree
> that A's time coordinate has the value t_A(P) at each point P in
> that region (of course different observers will naturally use
> different time coordinates; that's irrelevant).
>
> This has been called an "observer-dependent invariant", which
> about sums it up.
>
> In the abstract, "time" without specifying an observer has no
> meaning in relativity (or at least no _observable_ meaning).

If a tree falls in the forest and nobody's there to observe it ....
what makes you think it really fell?


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