Re: A Geometrical Proof of the Non-invariance of the Spacetime Interval

[David Rutherford <drutherford@SOFTCOM.NET>]

John Mallinckrodt wrote:
> David Rutherford writes:
>
> > I'm not claiming that the magnitude of the spacetime interval is not
> > invariant.
>
> It seems to me that that was almost exactly what you were claiming.
> In your first message you wrote
>
> > I think I can show that the spacetime interval ... is not invariant ...
>
> Except for the addition of the phrase "magnitude of the" in your
> latest message, these two statements would seem to be explicitly
> contradictory.  If you really *are* trying to draw a distinction
> between the spacetime interval and its magnitude, then I would remind
> you that the spacetime interval *is* a scalar quantity--it is the
> magnitude of a spacetime four vector that "locates" one event
> relative to another.

Maybe I'm using the wrong terminology. I'm referring to lines drawn
between E1 and E2 in the two frames. I hesitate to call them vectors,
since in order to be called vectors, the components in each frame would
have to describe the same geometrical entity. In this case, in my
opinion, they don't.

> > I'm claiming that, since the assigned positions of E1 in F
> > and F' don't coincide in space (meaning they are not independent of any
> > reference frame), at times later than t = 0, lines drawn between E1 and
> > E2 representing the spacetime interval between E1 and E2 in F and F'
> > don't coincide in spacetime (meaning they are not independent of any
> > reference frame). In other words, they aren't the same line.
>
> I can't make sense of this.  I think you may be suffering from some
> fundamental confusion about the geometry of spacetime, the nature of
> spacetime four-vectors, and, particularly the distinctions between
> *vectors*, *components* of vectors in particular coordinate systems,
> and *magnitudes* of vectors.
>
> The spacetime interval is the magnitude of a four-vector that
> "locates" event E2 relative to event E1 in spacetime.  That four
> vector exists independently and apart from coordinate systems; it can
> not (and does not) "depend" on the choice of a coordinate system.

The definition of a vector (or first rank tensor) is a quantity
whose components transform between frames in the same way as the
coordinates. But how do you know that the coordinates (which, in this
case, are the components of the spacetime 'four-vector'), themselves,
transform correctly. You can't say that the coordinates transform
between frames in the same way as the coordinates. That's circular.
You have to use a different argument in the case of the spacetime
four-vector to show that it is (or is not) independent of coordinate
systems.

I'm trying to show that the tails of the representations (?) of the
spacetime 'four-vector', in each frame, don't coincide in spacetime,
independently and apart from either coordinate system.

> Of course the *description* of that four vector (in the form of its
> components) will be different in different coordinate systems in
> analogy to the more familiar case of vectors in three dimensional
> Euclidean space.

I think you have to do things differently in the case of moving
reference frames in spacetime. Observers in F and F' agreed on the
location of E1 at t = 0, since their measuring devices were
'synchronized' at that time, so I say that they must _always_ agree on
its location, in order to describe the same event. According to the
Lorentz transformations, they definitely don't agree on the location of
E1 at t = t. It's okay if they don't agree on the coordinates of E2, in
order to describe the same event, because their measuring devices are no
longer necessarily synchronized at E2.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

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