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

Space time 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.

Qmechanic
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james2018
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  • Follows https://en.wikipedia.org/wiki/Minkowski_diagram, https://arxiv.org/pdf/physics/0703002.pdf – caverac Dec 07 '18 at 14:41
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    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 –  Dec 07 '18 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 https://en.wikipedia.org/wiki/Spacetime – BobtheMagicMoose Dec 07 '18 at 14:49

1 Answers1

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[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
orthogonal to the radius vector.

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!

robphy spacetime diagrammer

robphy
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  • 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. – Alexander Cska Aug 18 '19 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.) – robphy Aug 18 '19 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. – Alexander Cska Aug 18 '19 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). – robphy Aug 18 '19 at 19:00