# Thread: Hello, is distance invariant in Euclidean space?

1. ## Hello, is distance invariant in Euclidean space?

Hello!
I need a reliable link to proof someone that distance doesn't depend on a reference frame in Euclidean space.

For example: if there are two points on the Earth the distance is the same no matter if we use parallels and meridians as coordinates or Cartesian coordinates, associated with the centre of the Earth.

2. ## Re: Hello, is distance invariant in Euclidean space?

Surely this works straight out from the galilean transformations.

x' = x-vt
y' = y
z'= z

distance [between P1, (x1,y1,z1) and (x2,y2,z2)
is defined as

$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$

and in the primed system, substituting the transformations

$d' = \sqrt {{{\left( {{x_2} - vt - \left[ {{x_1} - vt} \right]} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$

$d' = \sqrt {{{\left( {{x_2} - vt - {x_1} + vt} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$

which equals d

3. ## Re: Hello, is distance invariant in Euclidean space?

Euclidean space doesn't depend on coordinate systems. The distance is a property of the space.

You can make transformations that don't preserve distance though (think about measuring in feet or metres), but that's a property of the transformation.