Re: Length Of A Curve in Spacetime

On Mar 30, 2:09 pm, "Bronwyn" <BB@xxxxxxxx> wrote:
I'm studying relativity at present and have been trying to follow some of
the messages here. I am confused about the way time and clocks are supposed
to vary with speed.
I have read that just as the odometers of two cars will read differently if
they take different routes between A and B, so will two clocks that take
different paths in spacetime. I cannot understand this.

As everyone should know, the length of a curve is given by the equation s =
integral(root[1+dy/dx)^2)dx]
Note that dy/dx is a dimensionless ratio.

For a simple spatial plot like this:

Y
| *
| * * *
|* * *
| *
| *
|
__________________________X

where y is a known function of x, the answer is straightforward. Its units
are those of length.

However if a similar curve in a 2D plane of spacetime is represented by an
X-T graph:

t
| *
| * * *
|* * *
| *
| *
|
__________________________X

we now have the length as: s = integral(root[1+dt/dx)^2)dx]

There is a problem here because the differential term is NOT dimensionless.
It has dimensions T/L and so the equation is dimensionally wrong.
The only way around this is to regard TIME as a fourth spatial dimension but
to do that would render just about all of modern science and technology
useless, which it obviously isn't.
Can somebody please tell me what are the dimensions of a 'length' in
spacetime and what its units might be?

Use RADAR ranging. Send say 10 cycles then
get a reflection, 10 out, 10 back.
Set up a standing wave, the number of cycles
is invariant, so use 10 cycles.

You might use Low Frequency (LF) to establish
that RADAR ranging, but some SOB will fly by
at a gazzillion miles per hour, and can very
well Doppler Shift your LF into gamma rays,
but that SOB cannot change the number 10.

I view the integral "s" in terms of cycles.
Regards
Ken S. Tucker
.

Relevant Pages

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    (sci.physics.relativity)
  • Length Of A Curve in Spacetime
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