Orthogonal diretion on the Minkowski diagram

1. Sep 1, 2010

mersecske

I have a trivial question:

Let assume a world sheet of a time-like spherical shell in Minkowski space-time.
On the 2D-Minkowski diagram (R,T), where R is the radius and T is the time,
the world line is represented by a time-like curve.
Let assume that the shell collapse and its 4-velocity is

$$u^\alpha=(\dot{T},\dot{R},0,0)$$

where $$\dot{T}=\mathrm{d}T/\mathrm{d}\tau >0$$ and $$\dot{R}=\mathrm{d}R/\mathrm{d}\tau <0$$.

Let $$n^\alpha$$ be an orthogonal vector to the 4-velocity and the surface.
With the metric diag(-1,1,1,1) we get:

$$n_\alpha=s (-\dot{R},\dot{T},0,0)$$
$$n^\alpha=s (\dot{R},\dot{T},0,0)$$

where s is the sign depending on the orientation.

I expect that $$n^\alpha$$ is orthogonal to $$u^\alpha$$ on the Minkowski diagram, and its space-like. For example, I expect that the outward oriented normal vector has positive coordinates. But since $$\dot{T}>0$$ and $$\dot{R}<0$$ this is not true, and for example for s=+1, $$n^\alpha$$ is a past directed vector on the diagram!?

2. Sep 4, 2010

DrGreg

Orthogonality in Minkowski space does not "look like" Euclidean orthogonality on a diagram. Two vectors drawn on a diagram which are Minkowski-orthogonal won't (usually) have an angle of 90° on your Euclidean piece of paper.

In fact, because of the way that lowering of indexes works, you will find that the covector $n_\alpha$ is "Euclidean-orthogonal" to the vector $u^\alpha$, if you draw both on the same diagram, since

$$n_\alpha u^\alpha = 0$$​

(Of course, strictly speaking vectors and covectors live in different spaces, so you shouldn't really draw them on the same diagram, but you can get away with it in Minkowski coordinates.)

3. Sep 4, 2010

mersecske

If somehow you draw them on the same diagram, and they are Euclidean-orthogonal, this means that you suppose the basis unit vectors are the same (representations are the same) in the two kind of vector space, which is not a good choice, I think!