# Space time diagrams: Length contraction

I am trying to get a better understanding for length contraction in a geometric sense and came across this diagram:

but what I am having trouble with is why in the ref frame of the blue vector are the $$ct'$$ and the $$x'$$ not perpendicular to eachother?

I mean are they, and it just what the axis looks like in the diagram. I of the understanding this diagrams are not a simple rotation, but more of a sheer, stretch, rotation complex. But I can seem to wrap my head around that concept enitrley and was wondering if anyone could expand on how the diagram are actually drawn.

• Commented Dec 7, 2018 at 14:41
• remember that the lorentz transformation is not a rotation in euclidean space, but in Minkowski space, thus it cannot be represented in the way you are used for spatial rotations
– user65081
Commented Dec 7, 2018 at 14:44
• Essentially you have two different planes - the x-ct plane and the x'-ct' plane. The x'-ct' plane is a 'pinched' (my word) version of the x-ct plane. The graphs in the wiki should provide some great intuition en.wikipedia.org/wiki/Spacetime Commented Dec 7, 2018 at 14:49

[While this may be unfamiliar...] the $$ct'$$ and $$x'$$ are Minkowski-perpendicular to each other in all frames.
In the reference frame of the $$ct'$$-axis, the axes will also appear to be [ordinary-Euclidean]-perpendicular to each other.

Here's how Minkowski describes this...

From Minkowski's "Space and Time"...

We decompose any vector, such as that from O to x, y, z, t into four components x, y, z, t. If the directions of two vectors are, respectively, that of a radius vector OR from O to one of the surfaces ∓F = 1, and that of a tangent RS at the point R on the same surface, the vectors are called normal to each other. Accordingly, $$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$$ is the condition for the vectors with components x, y, z, t and $$x_1$$, $$y_1$$, $$z_1$$, $$t_1$$ to be normal to each other.

In other words,
locate the intersection of an observer's 4-velocity with the unit-hyperbola (the Minkowski circle) centered at the tail of the observer's 4-velocity.

The tangent line to that hyperbola is Minkowski-perpendicular to that observer's 4-velocity. That observer's x-axis is drawn through the tail of her 4-velocity, parallel to that tangent line.

The "intuition" to have is that
the tangent to the "circle" in that geometry
is

You can play around with this idea in my visualization [screencaptured below].

• The unit-hyperbola (the "Minkowski circle") is in blue.
This figure is unchanged by a Lorentz boost.
• The red dotted line is the observer-worldline.
• The red tangent line is the prototype for "simultaneous events according to the red-observer".
• The red-observer's x-axis is drawn parallel to that tangent line.

https://www.desmos.com/calculator/r4eij6f9vw
Play around with the E-slider to see the Galilean and Euclidean analogues!

• the hyperbola has the equation $(ct)^2-x^2=1$. But why exactly do we chose this curve from a a family of many curves. This hyperbola is just the interval $ds=1$. We actually have two hyperbolas with reciprocal tangent lines $\frac{v}{c^2}$ and $\frac{c^2}{v}$. Each tangent line is the transformed coordinate axis. The one which intersects the $x$ axis infers that the proper length $ds=1$. Therefore, i assume that in your picture you , a meter stick with proper length = 1 is represented. Commented Aug 18, 2019 at 12:07
• The hyperbola $(ct)^2-x^2=1$ describes the tips of the 4-velocities of all inertial observers that met at a common event O and traveling in the x-direction. The hyperbola marks all of their individual events “1 second since their common meeting event”. (This hyperbola has spacelike tangent lines, which determine the x-axis for the observer from O meeting a particular event on that hyperbola.) Commented Aug 18, 2019 at 12:10
• I don't get the point with the 4 velocity tip. The four velocity is given by $\frac{dx^{\mu}}{ds}$. But the point with the tip etc. i don't understand. Commented Aug 18, 2019 at 13:22
• The 4-velocity is a unit vector in Spacetime. It is akin to a unit vector from the origin... the tips of all such unit vectors trace out a circle. Similarly, the tips of all 4-velocities with tail at a common event trace out one branch of a hyperboloid (whose tx-slice is the future timelike hyperbola with spacelike tangents). Commented Aug 18, 2019 at 19:00