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Re: Breakfast in LA - lunch in San Diego ( was Re: A Geometrical Pro of of the Non-invariance of the Spacetime Interval)



So we agree that |t2'-t1'| ~= |t2-t1|, for very, very small v.

Does |x2'-x1'| ~= |x2-x1|, for very, very small v?

Note: "~=" means approximately equal.

Note: The distance from LA to San Diego is *not* approximately
equal to zero.

Okay. So, |x2'-x1'| != |x2-x1|, even for very, very small v.

Does |t2'-t1'| ~= |t2-t1|, for very, very small v?


If I read "!=" as "does not equal"

and "~=" as "approximately equals" then I agree for a itsy
bitsy v. But
its important in SR that they aren't exactly equal, they have
a fractional
difference of order v^2/c^2.

What is the magnitude of the four-vector? Is it (x2'-x1')^2 -
c^2(t2'-t1')^2 for F' and (x2-x1)^2 - c^2(t2-t1)^2 for F?

Yes.


Is the magnitude invariant?


Yes.
____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301