The four axes of spacetime are $x, y, z$ and $ct$, where $c$ is the speed of light, and $t$ is time. Why is the speed of light (not any other speed) used to define the fourth axis of spacetime? If someone answers that the speed of light is the maximum speed among all speeds, such answer does not satisfy me. I would further ask: why is the maximum speed used to define the fourth axis of spacetime?
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4 Answers
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Suppose in three space you want to know the distance, $s$, to some point in space $(x,y,z)$. This is simply given by Pythagoras' theorem:
$$ s^2 = x^2 + y^2 + z^2 \tag{1} $$
Now consider 4D spacetime, and ask the same question: what is the distance to the spacetime point $(t,x,y,z)$. We call this distance the proper distance (or the norm of the four-vector) and in special relativity it is given by the Minkowski metric:
$$ s^2 = -(ct)^2 + x^2 + y^2 + z^2 \tag{2} $$
Ignoring the minus sign for the moment, this looks just like equation (1) except that in addition to the three components $x$, $y$ and $z$ we have the extra time component $ct$. That's why when we are graphing points in spacetime we conventionally use the four axes $x$, $y$, $z$ and $ct$.
To a degree this is simply converting your question to a different one to because you'll now be asking: yes, but why the Minkowski metric? And I don't think there is a good answer except to say that experiment confirms that the Minkowski metric does indeed describe the geometry of flat spacetime. That's just the way the universe is constructed.
Incidentally starting from the Minkowski metric we can easily see that $c$ is the maximum speed. See for example my answer to What is so special about speed of light in vacuum?
answered Oct 23, 2018 at 7:31
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$\begingroup$ I think that one element of justification to answer "why the Minkowski metrics ?" could be that it provides a Lorentz-invariant norm for the four-vectors. $\endgroup$ Commented Oct 23, 2018 at 7:52
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$\begingroup$ @BarbaudJulien yes, but I guess we'd then ask why is the geometry Lorentzian? At the end of the day you just have to shrug and say well that's what we find from experiment. $\endgroup$ Commented Oct 23, 2018 at 7:54
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1$\begingroup$ Sure, but if i'm correct, the Lorentz transformations can be derived only from the simplified principle that "light should go at the same speed in every inertial reference frame'. So i think this makes quite a lot of physical sense to go from relativity principle -> Lorentz transformations -> Minkowski metrics (not sure that the last step is logically valid though) $\endgroup$ Commented Oct 23, 2018 at 7:59
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3$\begingroup$ @BarbaudJulien The Minkowski metric is just one of many possible metrics that are compatible with (general) relativity. It just happens to be the one in which space is everywhere flat, but there is in principle no reason why some other metric wouldn't be the one that approximates the geometry of the universe. It just so happens that we live in a universe whose background curvature in the absence of significant mass-energy density is very close to (if not exactly) flat. If this were not true, then "Special Relativity" (i.e. relativity on the background spacetime) would use a different metric. $\endgroup$ Commented Oct 23, 2018 at 9:28
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1$\begingroup$ @BarbaudJulien And, just to make sure we're on the same page here, the norm of a vector, $v_\mu v^\mu$, is "Lorentz-invariant" (well, really we mean diffeomorphism-invariant) regardless of the metric, just because covariant and contravariant vectors transform in complementary ways. $\endgroup$ Commented Oct 23, 2018 at 9:41
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If someone answers that the speed of light is the maximum speed among all speeds, such answer does not satisfy me.
But $c$ is an invariant speed. Let me be clear about this, if an object is observed to have speed $c$ in an inertial reference frame then, according to the Lorentz transformation, it has speed $c$ in all inertial reference frames.
Thus, $c$ is a universal constant with dimensions of $[LT^{-1}]$. As such, it seems natural for $c$ to be a conversion factor for measuring time in units of length (or length in units of time).
Indeed, in geometrized natural units, $c$ is the dimensionless number $1$ and time has dimension of $[L]$.
Note that this does not mean that $c$ is used to "define the fourth axis of spacetime" in any way that I can think of.
answered Oct 23, 2018 at 11:44
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$\begingroup$ The statement "the speed of light is used to define the fourth axis of spacetime" is just a rough and not very strict statement. A more strict statement is: the speed of light along with time are used to define the fourth axis of spacetime. $\endgroup$– ShenCommented Oct 24, 2018 at 12:59
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The speed of light $c$ defines the scaling factor for all spacetime coordinates. We can see this by using a system of units where $c$ is a pure unitless number, usually by setting $c=1$. By imposing that $c$ is a pure number, we also establish that distance and time are exactly equivalent (time is converted into distance when you multiply by 1). So, in these natural units, space and time have exactly the same units (you can choose whether you call them units of distance or time, since those two quantities are equivalent). In SI units, this is equivalent to measuring distance in light-seconds and time in seconds, or measuring distance in meters and time in meters that light travels in a time interval.
The fundamental idea of relativity is that space and time are not distinct. Rather, they mix together as components of an overall spacetime. The fact that we have chosen different units to measure space and time is a historical artifact of our earlier belief that space and time were distinct, and therefore carries no fundamental relevance.
answered Oct 23, 2018 at 12:11
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I believe that you should turn your question around.
It's an accident of history that we first discovered $c$ as the speed of light, but its true significance (according to SR) is that it's the space / time conversion factor.
Once that's established, it can be shown that $c$ is invariant, and that massless particles must travel at $c$.
BTW, in relativity, time is conventionally the zeroth component, not the fourth.
