Re: Minkowski Metric

John C. Polasek says...

I don't need gij or dij to make a scalar quadratic product.

On the contrary, you certainly do. Let me give you an example:
Suppose that (in 2D Euclidean space) that I tell you that there
are two nearby points A and B. The coordinates of A are (A^1, A^2).
The coordinates of B are (B^1, B^2). What is the distance
between A and B?

That's simple: Let V be the vector with components V^1 = B^1 - A^1
and V^2 = B^2 - A^2. Then the distance is just given by

That shows how fuzzy your thinking is. B1 - A1 is just a number and
you can't make a vector V1 out of it.

D = square-root(V . V)

There are no V's

V is the displacement vector going from point A to point B.



where . indicates the scalar product. So what is D in terms of
V^1 and V^2? Presumably you think it is

D = square-root((V^1 * V^1) + (V^2 * V^2))
No D^2 equals [a1 a2] row vector times [B1 B2] columnvec =
a1b1 + a2b2. Take the square root

You are saying that the distance from A to B is equal to
square-root(a1 b1 + a2 b2)?

That doesn't make any sense. Let's look at a particular
concrete example: In Cartesian coordinates in Euclidean
space, A is the point with x-coordinate 1 and y-coordinate
0. B is the point with x-coordinate 0 and y-coordinate 1.
What is the distance between A and B? Your formula gives

D = square-root(a1 b1 + a2 b2)
= square-root(1 * 0 + 0 * 1)
= 0

So the distance between those points is *zero*? You must
first take the displacement vector, which is the vector
V with x-component V^1 = b1 - a1 = -1, and with y-component
V^2 = b2 - a2 = +1. Then the distance is

D = square-root((V^1)*(V^1) + (V^2)*(V^2))
= square-root((-1)*(-1) + 1 * 1)
= square-root(2)

So did you make a typo, or are you really unable to compute
distances in Cartesian coordinates?

--
Daryl McCullough
Ithaca, NY

.

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