Is it meaningful to draw $ct$ and $x$ axes by two lines intersecting at right angles?

Question

In the Cartesian coordinate system, the x-axis is really perpendicular to the y-axis, by construction. Also, under a rotation of the coordinate system, the transformed coordinate axes $x',y'$ remain orthogonal. Let us now talk about spacetime diagrams as drawn in textbooks. The spacetime diagram i.e. $ct$ versus $x$ diagram is often drawn as two perpendicular axes i.e. as two lines interacting at right angles. Is there any sense in which $ct$ and $x$ axes are perpendicular? Moreover, under Lorentz transformations, the transformed axes, $ct', x'$, are shown to intersect not at right angles but at an acute angle.

So my question is, does it make sense to draw with the $ct$ and $x$-axis as two lines interacting at right angles? I am not even sure what "$ct$ perpendicular to $x$" is supposed to mean. I am sorry if this question sounds dumb.

Answer 1

The $x$ and $t$ axes are perpendicular, in the sense that the scalar product of a vector pointing along the $t$ axis, $(1,0,0,0)$, and a vector pointing along the $x$ axis, $(0,1,0,0)$, is zero:

$$ \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}=0. $$

(The matrix in the middle is the metric tensor.)

Transformed $x^\prime$ and $t^\prime$ axes are also perpendicular:

$$ \begin{pmatrix} \gamma & \gamma v & 0 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \gamma v \\ \gamma \\ 0 \\ 0 \end{pmatrix}=0. $$

Indeed Lorentz boosts, like spatial rotations, manifestly preserve scalar products.

Answer 2

You can draw any picture you want :) The question is whether the picture gives rise to useful visualization strategies.

One important rule about spacetime diagrams is that orthogonality (defined via the Lorentzian inner product) is not the same as perpendicularity (defined by interpreting the spacetime diagram as a 2D Euclidean space). Therefore, you have to be careful not to assume your intuition from Euclidean geometry directly applies in a spacetime diagram. Two vectors can be orthogonal (have a zero inner product), even if the angle between them on the diagram is not 90 degrees. Similarly, a Lorentz transformation that mixes $t$ and $x$ (a boost) will not act on a spacetime diagram in the same way that a rotation would act on the $xy$ plane.

I should note that there is a sense in which the $t$ and $x$ axes being perpendicular (in the 2d Euclidean sense) does make sense. With this choice, we can view the $t$ axis as the time for events that occur at the same spatial location, $x=0$, in the frame chosen to draw the diagram. Furthermore, we can draw grid lines to connect events that occur at the same time, or at the same location, in the frame of the diagram.

Furthermore, spacetime diagrams are still very useful.

• First, we make spacetime diagrams even in non-relativistic physics -- any time you have plotted the position of a particle versus time, you have made a non-relativistic spacetime diagram. (Perhaps just with the $t$ and $x$ axes flipped relative to what is commonly done in relativity).
• Second, a spacetime diagram makes the causal structure manifest.
  • Any two events separated by a line with slope $1$ are null separated, meaning that a light ray can travel from one to the other.
  • Events which are separated by a line with a slope larger than 1 $1$ are timelike separated, meaning you can find an inertial coordinate system where both events happen at the same spatial location and different times. And, in any frame, one event will always happen before the other.
  • Events which are separated by a line with a slope smaller than $1$ are spacelike separated, meaning that you can find an inertial coordinate system where both events happen at the same time and different spatial locations. Also, in some frames one happen will occur first, and in other frames the other event will occur first.
• Finally, spacetime diagrams can be used to solve problems in special relativity very elegantly. So long as you understand the rules of the diagram, it is possible to understand the solution of problems like the twin paradox and the pole-in-a-barn paradox purely by drawing spacetime diagrams. It is well worth working through the solutions to some of these paradoxes and trying to ask and answer your own questions to get an intuition for how spacetime diagrams can help.

Answer 3

Yes, time is perpendicular to space because they are two independent parameters, or shall we say two separate directions objects can be separated. You can travel through time without traveling through space, simply by standing still, and two objects can be separated by a time interval $t_2-t_1$. Likewise two objects at a given instant on time can be separated by a distance $x_2-x_1$.

Now, Minkowskian spacetime (which is described by Special Relativity) is a hyperbolic spacetime, so the usual rules of Euclidian geometry do not apply. Hyperbolic geometry applies. So while a moving observer will appear (to a stationary observer) to have tilted $x't'$ axes that are not visually 90° apart, they are still perpendicular in a Hyperbolic geometry because they are symmetrical about the 45° line (asymptote of the hyperbola).

Answer 4

So my question is, does it make sense to draw with the and -axis as two lines interacting at right angles? I am not even sure what "$ct$ perpendicular to $x$" is supposed to mean.

Here is a geometrical interpretation of "orthogonality", which is related to the inner product (mentioned in the other answers).

From my answer to Can we show that time is orthogonal to space? ,

• 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 vector with the unit-hyperbola (the Minkowski circle) centered at the tail of the observer's 4-velocity vector.
  
  The tangent line to that hyperbola is Minkowski-orthogonal 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. On the diagram below [with time running upwards], I drew the coordinate-grid of the blue inertial observer. The lines parallel to the blue tangent line are "simultaneous" ("have constant $t'$ ") for the blue inertial observer.
  
  
  

• The "intuition" to have is that the tangent to the "circle" in that geometry
  is
  orthogonal to the radius vector.
  Visit my visualization: https://www.desmos.com/calculator/kv8szi3ic8 and tune the $E$-parameter from the Minkowski case ($E=+1$) to the Euclidean case ($E=-1$),
  with the Galilean case (a PHY 101 position-vs-time diagram) as the intermediate case $E=0$.
  
  
  
  

In the $E\neq -1$ cases, the visual representation of E-orthogonality coincides with the familiar Euclidean orthogonality only in the case of "zero"-relative velocity for the blue inertial observer.

See also Meaning of simultaneity in special relativity

Answer 5

All that really matters is the set of events that is spacetime. This set is a four-dimensional manifold, which means that you can locally form a one-to-one correspondence of each event with a point in the $R^4$ set.

So it makes sense that we visualize spacetime events as a $R^4$ space. The exact way you visualize the events does not matter. The co-ordinates are meaningless labels. One can do an aribtrary co-ordinate transformation without changing the physics.

In introductory relativity, we draw the diagrams such that the co-ordinates themselves correspond to space and time measurements. This is not necesasary and it is not even generally possible to do in General Relativity. Generally, the co-ordinate labels are meaningless labels and the information about space and time measurements is encoded not in the co-ordinates but in the metric tensor. For example, you can calculate the proper time experienced by an observer as an integral of the metric tensor along the worldline of the observer.

To answer your question, there is nothing deep about us drawing space and time measurements in a perpendicular way. It's merely a nice choice of a co-ordonate system that makes the co-ordinates co-incide with space and time measurements. In general, we use meaningless co-ordinates and we draw them orthogonally because that's just how we visualize $R^4$