I have seen a vector explanation of time dilation that our net velocity in space-time is the speed of light. Most of that velocity for slow spatial speeds is in the time dimension.

As you go faster in a spatial dimension, your speed in the time dimension reduces, because the net vector magnitude remains constant but the angle changes.

Are we really for practical purposes traveling forward at the speed of light in time?

  • 24
    $\begingroup$ You can say that we’re traveling forward in time at one second per second. Whether it is useful to say that is another question. To me, it seems merely tautological. $\endgroup$
    – G. Smith
    Commented Nov 11, 2019 at 0:25
  • 1
    $\begingroup$ Nothing moves in spacetime. The motion of a point-like particle in 3-D space is represented by a non-moving, infinitessimal thread (a.k.a., the particle's world line) in 4-D spacetime. $\endgroup$ Commented Nov 11, 2019 at 1:35
  • 4
    $\begingroup$ "As you go faster in a spatial dimension, your speed in the time dimension reduces" - note that the time component of the four-velocity is $\gamma c \ge c$. $\endgroup$ Commented Nov 11, 2019 at 3:10
  • 5
    $\begingroup$ Possible duplicate of Why are objects at rest in motion through spacetime at the speed of light? $\endgroup$ Commented Nov 11, 2019 at 4:51
  • 1
    $\begingroup$ related: physics.meta.stackexchange.com/questions/12513/… $\endgroup$
    – user4552
    Commented Nov 11, 2019 at 23:33

6 Answers 6


You, referring to yourself, always have a 4-velocity:

$$ u_{\mu} = (c, \vec 0) $$

which has:

$$ u_0 = c $$

Hence, moving through the time-direction at the speed of light. You can't move through space (otherwise, there would be a preferred frame in which you were not moving through space).

A moving observer that sees you moving through his definition of space at $\vec v $ sees your 4-velocity as:

$$ u'_{\mu} = \gamma(c, \vec v) $$

The magnitude is:

$$ ||u'_{\mu}|| = \frac{\sqrt{c^2-v^2}}{\sqrt{1-v^2/c^2}}=c $$

and that observer sees your clock slowed by $\gamma$.

  • 4
    $\begingroup$ This doesn't seem to address the question in any way, and the question is in any case a duplicate. $\endgroup$
    – user4552
    Commented Nov 11, 2019 at 19:43

For clarity of thought in the physics of distance, time and motion, one should be careful with words such as "moving", especially when people speak of "moving through time". The trouble is that this creates a confusion of two meanings of the word "moving". There is physical motion, when bodies have relative motion and two different worldlines have different slopes, and there is mathematical "motion" as one traces some line using a parameter. "Moving through time" is really the second type of "motion"; it is not physical motion at all, so it can only be assigned a physical speed by some sort of human convention.

If you insist on using the idea of "moving" forward in time, then the only "velocity" of such motion you can expect to find is one second per second. Each non-null worldline shows exactly one second of elapsed proper time per second of elapsed proper time. Of course it does. The practice of multiplying this by $c$ in order to convert it into a speed, so that it has the same physical dimensions of other velocities, is a human convention. Of course it gives the answer $c$. So this is why people say we are moving in time at the speed of light.

Now the velocity four-vector has a scalar invariant associated with it. Of course it does: so do all four-vectors. In the case of four-velocity the value of this scalar invariant is $c$. But does that mean that something is moving at the speed of light? Is that a useful way of expressing the situation? Lewis Carroll Epstein thought that this way of putting it helps us see why you can't have motion faster than light, and he thought it also provides a helpful intuition about time dilation. But I'm not sure if it really helps. The mathematics of four-vectors and invariants helps enormously, but the words we put around this mathematics ought to be well-chosen to express the physics precisely and helpfully. I'm not convinced that the idea of moving through time at light speed is of any help to any physical understanding.

  • 3
    $\begingroup$ This is a fine answer, but it answers a question that duplicates another question. Please consider (a) voting to close this question, (b) deleting this answer, and (c) copying it to make it an answer to the question that this one duplicates. $\endgroup$
    – user4552
    Commented Nov 11, 2019 at 22:44
  • $\begingroup$ @BenCrowell it would be better if a moderator could move the answer directly to the other thread (I think this can be done, but I'm not sure). $\endgroup$ Commented Nov 12, 2019 at 23:19
  • $\begingroup$ @BenCrowell A similar case: physics.stackexchange.com/a/512344/155230 $\endgroup$
    – AlphaLife
    Commented Nov 14, 2019 at 16:24

As Lewis Carroll Epstein explains in Chapter $5$ “The Myth” of his excellent book Relativity Visualized:

There is afoot an errorneous idea. It is that in physics the ultimate reality is a mathematical prescription, an equation. In fact, the ultimate reality is a little story or myth.

Then he divided myths into $2$ categories: good and bad ones. Good myths must be (beside other things) be easy to understand.

And then he introduces the following myth:

To understand the Special Theory of Relativity at the gut level, a good myth must be invented, and here it is.

“Why can't you travel faster than light?

The reason you can't go faster than the speed of light is that you can't go slower.
There is only one speed.
Everything, including you, is always moving at the speed of light.”

How can you be moving if you are at rest in a chair? You are moving through time.

So the movement in time by speed of light is nothing other than a myth. But a good myth — it saves the phenomenon: What is found in nature is explained by it, and what is logically deducted from it is found in nature.

  • 2
    $\begingroup$ Is it wrong if I imagine a rabbit in a waistcoat explaining this? $\endgroup$
    – user126527
    Commented Nov 11, 2019 at 14:59
  • 5
    $\begingroup$ @JollyJoker yes - you should be imagining a caterpillar taking a moment from his hookah to turn and explain this to you. $\endgroup$
    – MD-Tech
    Commented Nov 11, 2019 at 15:26
  • 3
    $\begingroup$ Is this more of a myth than, say, $E=mc^2$ or $F=ma$? If so, why? $\endgroup$
    – usul
    Commented Nov 11, 2019 at 17:55
  • $\begingroup$ If you think this answers the question, then you might want to delete it and cut and paste as an answer to the question that this one duplicates. But this author's explanation seems truly awful to me -- "not even wrong." $\endgroup$
    – user4552
    Commented Nov 11, 2019 at 19:45
  • $\begingroup$ @usul: $F = ma\ $ is not a myth — a corresponding myth is that force causes acceleration. Very understandable, but force and acceleration are not separated in time. So this myth is not perfect. $\endgroup$
    – MarianD
    Commented Nov 12, 2019 at 3:37

You are correct, the four vector (velocity) is defined as a four vector in four dimensional spacetime, that represents the relativistic counterpart of velocity (3D).

Physical events correspond to mathematical points in time and space, the set of all of them together forming a mathematical model of physical four-dimensional spacetime. The history of an object traces a curve in spacetime, called its world line. If the object has mass, so that its speed is less than the speed of light, the world line may be parametrized by the proper time of the object.

The four velocity is the rate of change of four position with respect to proper time along the curve.

The value of the magnitude of the four velocity (quantity obtained by applying the metric tensor g to the four velocity U) is always c2.

In SR, the path of an object moving relative to a certain reference frame is defined by four coordinate functions, where the timelike component is the time coordinate multiplied by the speed of light.


Thus, you are correct, you could say that us, who have rest mass we do experience time, and we are moving at the speed of light in the temporal dimension.

We just have to accept that the universe is built up so, and the four vector is built up so, that its magnitude is c always.

  1. spatial speed affects temporal speed

In your case, as you start moving faster in the spatial dimensions, you have to slow down in the temporal dimension, because the magnitude of the four vector has to be c always. This is the very reason that we talk about spacetime, and not space and time. This is the very reason that spacetime consists of causally linked space and time.

As you realize that your relative speed in the spatial dimensions affects your relative speed in the temporal dimension, you realize that space and time are not independent anymore.

Your example is about spatial speed affecting your temporal speed. But the two are causally linked the other way around too.

  1. temporal speed affects spatial speed

This is GR time dilation and when you realize that if you place an object into a gravitational field (stationary relative to the source of the field), then the object is relatively at rest in space, still it is moving through time at speed c. The four vector's magnitude is c always, thus, if the gravitational zone (gravitational potential) affects the temporal speed of the object, thus slows it down in the temporal dimension (relatively), then its spatial speed needs to compensate. What will the object do? It will start moving in space (relative to the source of the field). This is because the universe is built up so and the four vector is built up so, that its magnitude needs to be c always. If the temporal speed of the object reduces because of the gravitational field (relatively) then its spatial speed needs to compensate. The object will start moving along a geodesic towards the center of gravity. This is when you realize that time can affect space too (space curvature affects time and that affects the objects spatial dimension), thus we need to talk about spacetime as a causally linked phenomenon made up of space and time.


One way to describe Special Relativity in terms of first principles is

  1. There are two different types of dimension.

  2. All directions of the same type are equivalent.

From this you find that the distance formula is defined by the equation $$D^2=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2-(w_2-w_1)^2}$$

with $D$ being the spacetime distance.

So the distance formula is different from the Euclidean Distance Formula as one plus sign is replaced by a minus sign. So as an object moves faster through space it moves faster through time as well, so that your total speed through space time remains the same.

Thinking of relativity in terms of Geometry, and thinking in terms of 4 velocity instead of velocity through space has some advantages, as I find it easier to understand the first principles if I define first principles in terms of Geometry, and think of the physics principles as emergent from the Geometry Principles, than the other way around.

Thinking in terms of Geometry, and in terms of 4 velocity, and in terms of 4 acceleration makes it easier for me to understand how to derive equations, as once I know an equation in Euclidean Geometry I can adjust it to get a similar equation in Lorentzian Geometry. It also makes it easier to come up with ways to test equations I derive to make sure it is consistent with the principles of relativity.

Some secondary principles you can figure out from thinking about relativity in this sense is that if an object accelerates at a constant rate, then there is a point in space time that all points on the objects world line are the same spacetime distance from, the 4 velocity of an object is perpendicular to the 4 acceleration of that object, and to accelerate at a constant rate and direction through space means to turn at a constant rate through spacetime. It can also help with understanding relativistic Electromagnetism if you think about a group of electrically charged particles in Euclidean Space that move at a constant speed, and with each electrically charged object being everywhere along its path at once.

  • $\begingroup$ This doesn't seem to have much to do with the question, which in any case was a duplicate of another question. $\endgroup$
    – user4552
    Commented Nov 11, 2019 at 21:51

The answer to your question seems to be agreed as yes with one small proviso.

Observers in different inertial frames, who are also moving forward (for them) at speed c are moving in a different direction from yours in four dimensions. Equivalently, your moving forward in your inertial frame is at an angle to theirs.

The small proviso is your use of the word really. That depends on whether the mathematical model is really right or whether it merely gives the right answers.

  • $\begingroup$ This doesn't seem to address the question. The question involves the misconception that it's meaningful to talk about "moving" through spacetime, as opposed to space. $\endgroup$
    – user4552
    Commented Nov 11, 2019 at 22:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.