Chronology | Current Month | Current Thread | Current Date |
[Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |
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.
http://www.esu.edu/~bbq____________________________________________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;