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



Robert Cohen wrote:

I hesitate to get into this fray because I know very little about
relativity but here goes.

David Rutherford wrote:

I'm just saying that they lead to a paradox when you compare the
assigned locations of breakfast, in the two frames, at the time of
lunch, which is when you "draw the lines" between breakfast and lunch,
in the two frames.

[snip]

Therefore, both lines (from breakfast) start at time zero, but at
different spatial locations. That means that they start at different
points in spacetime, thus, they don't coincide in spacetime and
cannot describe a geometically invariant quantity (the spacetime
four-vector between the events).

to which Hugh Haskell wrote:

No.

[snip]

That is, assuming that the x-axis for both frames lies along the
LA-San Diego vector, that both will get the same value for x^2 -
c^2*t^2, as measured by instruments in their own reference frames.
It doesn't mean that either x or t will be seen to be the same in
both frames, because they won't.

Help me out here.

One person is going at v=0. The other is going at v<<c.

Each measures event A (breakfast) occurring at a particular point in
space-time (x1,t1)

No, F measures (x1,t1) and F' measures (x1',t1'), however, since the
origins of the two frames coincide at A, x1 = x1' = 0 and t1 = t1' = 0.

and event B (lunch) occurring at a particular point in space-time
(x2,t2).

No, F measures (x2,t2) and F' measures (x2',t2'), but this time x2 is
not equal to x2' and t2 is not equal to t2', since F and F' are in
motion relative to each other.

Are x1 and x2 measured differently by each observer, assuming v<<c?

x1 and x2 can only be measured by observers in F. Those are the spatial
coordinates of A and B in F. x1' and x2' can only be measured by
observers in F'. Those are the spatial coordinates of A and B in F'.

Are t1 and t2 measured differently by each observer, assuming v<<c?

t1 and t2 can only be measured by observers in F. Those are the time
coordinates of A and B in F. t1' and t2' can only be measured by
observers in F'. Those are the time coordinates of A and B in F'.

DR says yes to the first and no to the second, which implies that
the four-vector is not invariant.

I say that the questions are meaningless for the reasons I gave. I think
what you meant to ask was:

Are the (spatial) distances between A and B measured differently by each
set of observers, assuming v<<c?

and

Are the time intervals between A and B measured differently by each set
of observers, assuming v<<c?

My answers, assuming that you are referring to the SR description of the
situation, are yes to the first and yes to the second.

HH says yes to both, but that they change in just the way to keep
the four-vector invariant.

The way that HH measures the distance between A and B, in the two
frames, definitely, assures that the magnitude of the SR spacetime
interval will _not_ be invariant (I'm talking about the magnitude of the
interval, here).

It seems to me that at v<<c, the answer is no to both and so the
four-vector is invariant.

The answer (in SR) to the first question is that it depends on the time
interval between A and B. You can have a slow speed, but if the time
between A and B is great enough in F, there can still be a big
difference between the distances measured between A and B in F and F'.

The answer (in SR) to the second question is almost yes. As v/c
approaches 0, (t2 - t1) approaches (t2' - t1').

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

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf
"Energy Density Correction"
http://www.softcom.net/users/der555/enerdens.pdf
"Proposed Quantum Mechanical Connection"
http://www.softcom.net/users/der555/quantum.pdf