I'm trying to understand the Minkowski spacetime better.

If an object is not undergoing acceleration, and is far from any large mass, does it travel maximally "fast" through time? Can we calculate a 'speed' of some sort? Put another way, is there some 'Minkowski unit' that is conserved as one accelerates?

Does the question as currently formulated even make sense?

EDIT: Just adding a few clarifications.
1. The object can be considered 'at rest' (or at least moving at a very slow non-relativistic constant speed) compared to (a set of) very distant galaxies.
2. Imagine the spacetime hyperplane intersecting Earth's current spacetime. I know it's outside our lightcone and all, but imagine you had big-bang-triggered clocks every here and there. Would there be a zone of spacetime where the clocks had reached a maximum time? Would that be (significantly: minutes, days, year) different from Earth's estimation of time since the Big Bang?

  • $\begingroup$ Define 'stationary'. There is no absolute reference frame. $\endgroup$ – Time4Tea Jan 12 '15 at 3:59
  • $\begingroup$ Non-accelerating. If we need a reference frame, a distant set of galaxies. $\endgroup$ – Serban Tanasa Jan 12 '15 at 4:01
  • $\begingroup$ Non-accelerating can still be moving. It will be moving maximally fast through time from its own point-of-view. Although, it will seem to be progressing more slowly through time from the point-of-view of an observer moving relative to it. $\endgroup$ – Time4Tea Jan 12 '15 at 4:15
  • $\begingroup$ The shirt answer is imho yes, although it seems maybe conceptually unprecise. $\endgroup$ – peterh - Reinstate Monica Jan 12 '15 at 4:37
  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Serban Tanasa Jan 12 '15 at 17:08

The speed that we use in relativity is the four-velocity. Just as ordinary velocity is a 3D vector that describes the rate of change of your position in space with time, the four-velocity is a 4D vector that describes the rate of change of your position in spacetime with proper time. This four-velocity is an invarient and all observers in all inertial frames will agree on its value.

Proper time is a concept that doesn't exist in non-relativistic physics. You'll be familiar with the fact that clocks on fast moving spaceships run slow due to time dilation, and this makes defining time a bit tricky. Your proper time is the time shown on a clock you are carrying i.e. the time shown by a clock that is stationary relative to you. In special relativity the proper time is an invariant and all observers in all frames will agree on its value. This makes it a very useful concept, and if you study relativity you'll run into proper time all over the place.

Anyhow, in 3D space where your position is given by a vector $(x, y, z)$, the velocity is given by:

$$ \vec{v} = \left(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right) $$

In 4D spacetime we choose an inertial frame using coordinates $x$, $y$, $z$ and $t$, and the position some object is then given by the four-vector $(t, x, y, z)$. The four-velocity is then given by:

$$ \vec{U} = \left(\frac{cdt}{d\tau}, \frac{dx}{d\tau}, \frac{dy}{d\tau}, \frac{dz}{d\tau}\right) $$

There are a few interesting things to note about the four-velocity. Suppose you are measuring your four-velocity in your own inertial frame. In this frame you are stationary so $dx = dy = dz = 0$, and the time $t$ is the same as the proper time $\tau$ (because that's how proper time is defined). So your four-velocity is:

$$ \vec{U} = \left(\frac{cd\tau}{d\tau}, 0, 0, 0\right) = \left(c, 0, 0, 0\right) $$

So even when you're standing still your four-velocity is $c$. That's because even you you're not moving in space you are moving in time, and in fact you're moving though time at the speed of light.

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  • $\begingroup$ Thanks! So my intuition was essentially correct, and $\vec{U}$ is the preserved unit? $\endgroup$ – Serban Tanasa Jan 12 '15 at 16:22
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    $\begingroup$ @SerbanTanasa: the magnitude of $\vec{U}$ is conserved because it's always $c$. The direction of $\vec{U}$ will change as you accelerate. $\endgroup$ – John Rennie Jan 12 '15 at 16:25
  • $\begingroup$ So a gravity well acts by essentially bending your future towards itself? Fascinating. Thanks for the answer! $\endgroup$ – Serban Tanasa Jan 12 '15 at 16:29
  • $\begingroup$ what book would you suggest for a more in-depth study of this? Assume basic calculus and rusty linear algebra. $\endgroup$ – Serban Tanasa Jan 12 '15 at 16:47
  • $\begingroup$ @SerbanTanasa: I learned all this stuff decades ago (the ink was still wet on Albert's original paper) so I have no idea what would be a good modern introductory book on special relativity. I would just Google for suggestions. $\endgroup$ – John Rennie Jan 12 '15 at 16:57

Using natural units, an object moves through time at a speed of $\frac{\tau _{proper time}}{t_{observer time}}$. For an object that is stationary as you describe, it is moving forward through time at a speed of $1\frac{second}{second}$. This has an upper limit of 1, and an asymptotic lower limit of 0, just as speed in natural units has a lower limit of 0 and an asymptotic upper limit of 1. Also note that in natural units, velocity is unitless.

$1= (\tau / t)^2 + \beta ^2$

Using conventional units:

$c^2 = (c\tau / t)^2 + v^2$

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