# Minkowski normal v. Euclidean normal

1. Mar 8, 2014

### flyinjoe

What is the difference between the normal in Minkowski spacetime and the normal in Euclidean space?

2. Mar 8, 2014

3. Mar 8, 2014

### flyinjoe

Yes, sorry, I am sort of using normal as a synonym for orthogonal. Essentially, how would vector that is normal to a surface in Minkowski spacetime differ from a vector normal to the same surface embedded in Euclidean space?

I was thinking it had something to do with the magnitude function in the determination of the normal vector being different in each space (i.e., the denominator of the equation for a normal vector).

I'm sorry if this isn't making much sense.

4. Mar 8, 2014

### Fredrik

Staff Emeritus
In a typical 1+1-dimensional spacetime diagram, a line through the origin with slope v is "orthogonal" (with respect to the Minkowski metric) to a line with slope 1/v. In particular, the x' axis in the image below is orthogonal to the t' axis.

In other words, as you tilt a line through the origin, the line through the origin that's orthogonal to it is tilted by the same amount as in the Euclidean case, but in the opposite direction.

5. Mar 8, 2014

### George Jones

Staff Emeritus
In Minkowski space, some non-zero vectors are orthogonal to themselves, i.e., the hypersurface normal to a vector can include the vector!

When does this happen?

6. Mar 8, 2014

### robphy

The best definition for "normal" is the geometric one, as Minkowski defines in
en.wikisource.org/wiki/Translation:Space_and_Time
http://en.wikisource.org/wiki/Page:De_Raum_Zeit_Minkowski_018.jpg

In Euclidean geometry,
the tangent line to a circle at a point P is perpendicular to the radius to that point
(http://en.wikipedia.org/wiki/Tangent_lines_to_circles)

By analogy, in Minkowski spacetime,
the tangent line to a Minkowski-circle (a hyperbola) at a point P is Minkowski-perpendicular to the radius to that point.
This has an immediate physical intrepretation: an inertial observer's sense of "space [at an instant]" (what events are simultaneous for that observer) is perpendicular to that observer's sense of "time [along his worldline]".

7. Mar 8, 2014

### bcrowell

Staff Emeritus
One good way of thinking about it is that when a timelike vector o is orthogonal to a spacelike vector s, an observer whose velocity four-vector is o considers s to be a vector of simultaneity. In other words, o points along that observer's time axis, and s could point along one of her spatial axes. (Of course this doesn't cover lightlike vectors or orthogonality of spacelike vectors to other spacelike vectors.) I have a treatment in this style in my SR book, http://www.lightandmatter.com/sr/ , at the beginning of the first chapter.

8. Mar 8, 2014

### Naty1

flyinjoe...I'm not sure I really understand the answers so far....

maybe this will help a bit...

http://en.wikipedia.org/wiki/Minkowski_diagram

The first three sections, at a minium, provide insights.

You know that Minkowski is four dimensional and Eucledean is just the three of space, right?

So as I see it [simple minded perhaps] is that just as a normal to a plane has one less degree of freedom than a normal to a line embedded in that plane, so too is a normal restricted by the addition of a time component.

If you read between the lines in Wiki where it says

and you can probably visualize different paths leads to different tangents and normals....