I understand that everything in the universe travels at the speed of light through spacetime. As a layperson I'm going to refer to this velocity as the spacetime velocity.

Some examples..

  1. Objects with mass have a spacetime velocity that point mostly in the time direction, but also in the 3 space directions.

  2. Light has a spacetime velocity that points the maximum amount in the 3 space directions and minimum in the time direction.

Now, why I think the answer to my first question may be yes..

If an object had its spacetime velocity pointed the maximum in the time direction, wouldn't this object be traveling at the maximum speed something could travel through time?

If I'm thinking about it correct and there is in fact a speed limit in the time direction, is this speed limit of time also out of reach like the speed limit of space?

Or asked in another way, could something ever travel the maximum speed through time?

  • 2
    $\begingroup$ Why do you think light only moves only in spatial direction and not in time direction? This would mean light could be everywhere at an instant, implying infinite speed. But the speed is c, meaning it is not everywhere at the same time, hence the velocity vector also points in time direction. $\endgroup$
    – Koschi
    May 19, 2021 at 16:48
  • $\begingroup$ @Koschi I will update my question with this fact $\endgroup$ May 19, 2021 at 19:11

2 Answers 2


So the spacetime velocity refers to the 4-velocity. Most of our favorite 3D vectors have a 4-vector counter-part, and it is this 4-vector counter part Lorentz transforms between various interial frames.

The most basic one is spatial separation:

$$ \vec r = (\Delta x,\Delta y, \Delta z)$$

Different coordinate systems have different values of the $\Delta$'s, but they all agree on the length:

$$||\vec r||^2 = \Delta x^2+\Delta y^2+ \Delta z^2$$

in 3D space. The 4-vector version is:

$$ r^{\mu} = (c\Delta t, \vec r)$$

All inertial frames agree on the so-called invariant interval:

$$\Delta s^2 = r^{\mu}r_{\mu} \equiv c^2\Delta t^2 - ||\vec r||^2 $$

If you differentiate a world-line defined by $r^{\mu}(\tau)$ with respect to $\tau$, you get the 4-velocity:

$$ \frac{dr^{\mu}}{d\tau} = v^{\mu} = (\gamma c, \vec v) $$

where $\gamma=1/\sqrt{1-(/c)^2}$ is the usual Lrentz factor. It's magnitude is:

$$v^{\mu}v_{\mu} = \gamma^2c^2 - \gamma||\vec v||^2 = c^2$$

which is constant for all $\vec v$. This is the origin of "we all move through spacetime at the speed of light".

In the rest frame, it is:

$$ v^{\mu} = (c, 0, 0,0) $$

That is, "we move through time at $c$".

Now there is no frame that moves at $c$, and we can't write down the 4-velocity of light because $\gamma \rightarrow \infty$, but as you approach $c$, you don't move through space only (your point 2).

We can rewrite it (in the $z$-direction) as

$$ v^{\mu} = \gamma c(1, 0, 0, \sqrt{1-1/{\gamma^2}}) \rightarrow \gamma c (1, 0, 0, 1) $$

so light moves through space and time equally. The scale factor $c\gamma$ diverges, which is why people say light doesn't move through time, and sees the Universe Lorentz contracted into a 2D plane, but that view point gives no insight into relativity. It is a viewpoint from an inertial frame that does not exist.

As seen in the twin paradox, the path between two time-like ($\Delta s^2 > 0$) events is the inertial path in-which both events occur at the same location. That way $\Delta r=0$, so that:

$$ \Delta s^2 = \Delta t^2 $$

and $\delta t^2$ is maximized. (Remember, all inertial frames see the same $ \Delta s^2$).


The maximum speed is $c$. An object that is at rest relative to you will have a spacetime velocity $c$ that only points in the time direction. Here moving through time at $c$ means that time flows at the same rate for both you and that object , i.e., there is no time dilation. To be exact though, you're not really moving through $t$. Your motion is in the direction of $ct$ in spacetime since that is the proper fourth component in spacetime.


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