Orthogonal diretion on the Minkowski diagram

[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!?

 

[DrGreg]

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!?

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.)

 

[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!