# How is time illustrated in a Cartesian coordinate system?

## Main Question or Discussion Point

I am teaching myself math and have a question about cartesian coordinate systems. How is time illustrated in such a graph?

[Moderator's note: Moved from a math forum after post #13.]

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fresh_42
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Usually as the ##x-##axis. However, specific cases might require different solutions.

Orodruin
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It is not. Cartesian coordinates are coordinates on Euclidean space and in (special) relativity spacetime is not Euclidean but Minkowski.

That does not mean you cannot draw it in a graph on a piece of paper. A search for ”spacetime diagram” should give a number of examples.

fresh_42
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It is not. Cartesian coordinates are coordinates on Euclidean space and in (special) relativity spacetime is not Euclidean but Minkowski.

That does not mean you cannot draw it in a graph on a piece of paper. A search for ”spacetime diagram” should give a number of examples.
Yes, but in a math forum it is probably not a light cone. And Minkowski is only the metric, not the space.

Can it be illustrated in the coordinates? (x,y,z, time)?

Orodruin
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Yes, but in a math forum it is probably not a light cone.
I did not see the forum. I assumed we were talking relativity. The OPs posts seem to indicate to me that this question would be better off in the relativity forum.

And Minkowski is only the metric, not the space.
I disagree. Minkowski space is the affine space equipped with a pseudo metric of signature (1,3). Cartesian coordinates indicate orthogonal directions in Euclidean space, something that also requires a metric.

fresh_42
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Can it be illustrated in the coordinates? (x,y,z, time)?
In case you want to draw a spacetime diagram, then follow @Orodruin 's hint and search for it. We cannot illustrate four dimensions, and as time has a special meaning in special relativity - see Minkowski remark above - things are more difficult than just a Euclidean plane is. The ##(x,y,z)## components are often represented by one direction, and time by another. But in a spacetime diagram we speak about events, so there are areas outside a person's light cone (past and future), and events within. In a Euclidean space everything is within, but special relativity has the restriction of a finite light speed.

Orodruin
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Can it be illustrated in the coordinates? (x,y,z, time)?

fresh_42
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I just wanted to point to the diagram. It illustrates that things in a spacetime diagram are different from the usual ##(x,y)## diagrams. I already said to follow your advice and google for spacetime and Minkowski.

I first thought we were talking about Newton and things like velocity. Or earth science and things like ice ages, or nuclear physics and radioactive decay with a logarithmic scale. That's why I just said: ##x-## axis and different areas require different solutions.

At least this discussion is again a good example how simple questions can evolve to complicated answers. And that the simplicity of a question is often a lack of specification.

Orodruin
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And that the simplicity of a question is often a lack of specification.
Which is why I wrote #8. Let us wait until OP gets back regarding that.

I am reading a book called "Special Relativity and Classical Field Theory" and it is talking about cartesian coordinates. It says we need a "t axis"

A quote from the book:

"In order to specify WHEN something happens, we need a time coordinate. A reference frame is a coordinate system for both space and time. It consists of an x-, y-, z-, and t axis. We can extend our notion of concreteness by imagining that theres a clock at every point in space. We also imagine that we have made sure that all the clocks are synchronized, meaning that they all read t=0 at the same instant, and that the clocks all run at the same rate. Thus a reference frame is a real or imagined lattice of metersticks together with a synchronized set of clocks at every point."

Ibix
In that case, you do need a Minkowski diagram if you want to draw it (as @Orodruin suggested). Note that the time direction is certainly not what one might naively regard as a Cartesian coordinate, because you aren't looking at a Euclidean plane.

We would regard the rods and the clocks as a coordinate system on spacetime, regardless of whether or not we actually draw a diagram on paper. They do collectively define four orthogonal directions (orthogonal in the Minkowski sense, not in the Euclidean sense).

vanhees71
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I am teaching myself math and have a question about cartesian coordinate systems. How is time illustrated in such a graph?

[Moderator's note: Moved from a math forum after post #13.]
The answer depends on within which theory you look.

In non-relativistic physics we use the Galilei-Newton model of spacetime. Mathematically you have time as a oriented one-dimensional real parameter, which you can depict in the usual way as a line. At each point of this line you must think to have put a 3D Euclidean affine space, which describes the physical space at each instance of time. There's no closer relation between time and space in Newtonian physics, which is known since Newton as the idea of absolute time and absolute space, which both are just there unaffected by anything going on in the physical world (mathematically it's a fiber bundle). A non-relativistic space-time diagram is thus just two axis, one labeled with time ##t## and one with a component of Euclidean spatial vectors, describing the value of this component at each instant of time, such as you draw a function graph for a function ##x=x(t)##. There's no geometric meaning of this plane,

In special-relativistic physics things become already more beautiful. There space and time are described by a real space-time continuum as a pseudo-Euclidean affine continuum. It's pretty much like a 4D Euclidean affine continuum with the very important qualification that the "scalar product" is no longer positive definite but a bilinear form of signature (1,3) (or equivalently (3,1) depending on which convention you use; as a relativistic nuclear-particle physicist I prefer the "west-coast convention", which is (1,3)). I.e. the "distance" of spacetime intervals is given by the product of the corresponding four-dimensional vectors with components ##(x^{\mu})=(x^0,\vec{x})## with ##x \cdot y=x^0 y^0 - \vec{x} \cdot \vec{y}##, where ##\vec{x} \cdot \vec{y} = x^1 y^1 + x^2 y^2+x^3 y^3## is the usual Euclidean scalar product for the components with respect to a Cartesian basis system.

Here a Minkowski diagram, describing again the one-dimensional motion of a particle by drawing one axis for time (or more conveniently ##x^0=c t##) and one spatial component, has a geometrical meaning, but not in our usual terms of a Euclidean plane but of a "Minkowski plane", where the metrical properties are given by the pseudo-scalar product rather than the proper scalar product of Euclidean vectors. The most important result to keep in mind is that instead of circles, defining positions of equal distance from a given point, you get hyperbolae, ##(c t)^2-x^2=A=\text{const}##. You have to distinguish time-like "distances" (##A>0##), lightlike ones (##A=0##, for which the hyperbolas degenerate into straight lines, the "light cone, and spacelike (##A<0##).

Changing from one pseudocartesian basis to another and thus using new components ##(c t',x')## in this Minkowski plane describes the transformation of the spacetime vector components of one inertial reference frame to those of the other inertial frame moving with constant velocity in ##x##-direction with respect to the former system. It's a pseudocartesian basis change (i.e., the analogies of rotations in Euclidean space) if and only if the pseudoscalar product looks the same in both sets of coordinates, i.e., the spacetime "distances" don't change. This leads to the Lorentz transformation and the fact that the relative velocity between inertial frames cannot exceed the speed of light. Particularly in the Minkowski diagram the light cone of the origin is unchanged, and it's always given by the bisecting lines between the space-time axes (one cone describing light signals travelling in positive the one those travelling in negative ##x##-direction), which means that for any observer the light signal travels according to ##x=\pm c t##, i.e., for any observer light travels with the same speed ##c##, no matter how the light source might move relative to him or her.

For more details, see my SRT writeup:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

SiennaTheGr8
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