Is the "spacetime" the same thing as the mathematical 4th dimension?

We often say that time is the fourth dimension, but I am wondering if it's means that time is like the fourth geometrical axis, or it's something different than a geometrical axis and something that's used to represent it graphically even though time has no geometrical feature.

If they differ, can you tell me in what way it differs so that a layman can understand?

  • 9
    $\begingroup$ Spacetime has 4 dimensions, 3 spatial and 1 temporal. The temporal (time) dimension is treated slightly different from the spatial dimensions. $\endgroup$ Commented Sep 24, 2019 at 16:24
  • 1
    $\begingroup$ chapter 1 of Wheeler and Taylor's Spacetime Physics is your best friend. $\endgroup$
    – Shing
    Commented Sep 25, 2019 at 10:49
  • 5
    $\begingroup$ Incidentally, although sci-fi usually calls time the fourth dimension, it's arguably more common for physicists to call it the zeroth dimension (but I've seen $4$ used). $\endgroup$
    – J.G.
    Commented Sep 25, 2019 at 13:22
  • 2
    $\begingroup$ @Kevin Actually, if you know the value of a wavefunction over a plane over all time, that determines its value across all space. $\endgroup$ Commented Sep 25, 2019 at 17:10
  • 2
    $\begingroup$ @Kevin If I draw a line segment and have no other information you can't tell what 2d shape it is part of, but given appropriate extra information you certainly can. (for example a differential equation describing the edge of the shape would be a very natural form for the extra information to take). In the case of space time we do have such information (we call it the Laws of Physics) which allow us to work out what the universe will look like some distance into the time direction. $\endgroup$ Commented Sep 25, 2019 at 17:22

11 Answers 11


Yes, time can be treated as a fourth axis- that idea was developed by a German mathematician called Hermann Minkowski not long after Einstein published his theory of special relativity (Minkowski was Einstein's supervisor for a time).

Representing time as a fourth axis- along with the usual three spatial axes- is now standard in text books and scientific papers. I saw a quote from Einstein implying that he didn't like it at first, something along the lines of 'Now that mathematicians have got hold of relativity I'm not sure I understand it myself any more.'

The Minkowski institute has a website where you can read English translations of his papers.

Minkowski's spacetime is in some ways analogous to 3D space. For example, in 3D space there is no predefined value of 'up', so you can pick any direction you like to orient your Z axis, for example. Likewise in Minkowski's space there's no predefined direction for the T axis- if two observers are moving relative to each other then their respective T axes diverge, with the divergence increasing with their relative speed.

You can use the concept of 4 D spacetime to get a feel for things like time-dilation and length contraction in a way that's analogous to measurements in ordinary space. For example if you use the normal 'Z equals up' orientation you might tell me that a certain flagpole is a hundred feet high and a foot wide. If I have my Z axis tilted away from yours I will say that the height of the flagpole is less than a hundred feet, but it's a lot wider than a foot. Similar things happen in spacetime, where a diverging direction for the T axis would mean that observers measure different elapsed times.

However, you can't take the analogy too far, as the geometry of Minkovski space (ie the rule for calculating distances etc) isn't the same as the geometry of Euclidian space, which is what we were all used to before we were introduced to relativity. In that respect you can't really think of time as something you can treat exactly like the three spatial dimensions.

That said, it turns out that mathematically you can represent 'flat' spacetime as Euclidian if you make your fourth dimension iT (ie T multiplied by the square root of -1). I read somewhere that representing spacetime that way used to be more popular.

  • 2
    $\begingroup$ Since the 3 spatial axes use the same "units" of measurement, is there some kind of mapping from distance units to time units? How many meters are there in an hour? How many seconds are there in a meter? $\endgroup$ Commented Sep 24, 2019 at 22:34
  • 12
    $\begingroup$ @KyleDelaney The space and time dimensions are joined by c, the speed of light. $\endgroup$ Commented Sep 24, 2019 at 23:55
  • 3
    $\begingroup$ @KyleDelaney There are many possible unit systems, but several of them are called "natural units" and set c=1. Planck units are such a system, but the size and time scales are inconvenient outside of theoretical physics: en.wikipedia.org/wiki/Natural_units#Planck_units $\endgroup$ Commented Sep 25, 2019 at 0:39
  • 21
    $\begingroup$ @Kyle Yes, 1 second of time duration is the same size as 1 light-second of space distance. However, although relativity combines space & time, it still maintains an important distinction between them. Spacetime intervals that span a greater space distance than time distance are called spacelike, those that span a greater time distance than space distance are called timelike, and those spanning equal space & time are called lightlike. All observers agree on whether a given interval is spacelike, timelike, or lightlike. $\endgroup$
    – PM 2Ring
    Commented Sep 25, 2019 at 4:10
  • 2
    $\begingroup$ @ChatterOne Keep in mind that reference frames are just tools to simplify problems. Reality doesn't have any reference frames. It's just that if you can pick a useful reference frame, you save yourself a lot of calculations and headaches. If you want to preserve a "time reference frame", you can - time dilation and length contraction are just two ways of looking at the same thing. $\endgroup$
    – Luaan
    Commented Sep 25, 2019 at 12:10

So the direct answer is a great NO, because spacetime is something physical which has a deep mathematical meaning.

I) Intuitive Idea

Intuitively and roughly speaking, spacetime is the "place" of all events, or the set of all events. An event is something that "happens in a time $\tau$ and takes place somewhere". You can grasp the main concept with a simple example: You have a physics test, Friday 11:00 AM, at the Physics Departament building, on floor 5. Well, if you go to the right place but wrong time you will miss the test. To access the event "test" you have to be in the right place at right time. So, you necessarily must to deal with four numbers: one for time and three for space.

Because of relativity, the time are not just a parameter, but a coordinate! In Lorentz transformations you transform time as a usual coordinate. You must consider time as just another coordinate as the usual spatial ones.

II) Dimension

The most elementary definition of dimension comes from a mathematical subject called linear algebra, which is one of the "mathematical tools" used to properly describe general relativity (GR) in mathematical terms. In GR, we basically deal with finite-dimensional vector spaces, so the concept of dimension is the most elementary one:

A dimension is the number of the basis vectors of a given vector space.

So a "mathematical 4th dimension" is just a 4-dimensional vector space.

III) Spacetime: A brief explanation

Well, here is where we use math to describe physics. The physics of spacetime was introduced in 1905 with Einstein's paper. But spacetime was born in 1906 with Minkowski's paper. Now, there are some facts that we will use to construct the proper idea of spacetime:

1) In physics we can measure lengths and time and a mathematical object which have this property of "measure" is the norm given by a inner product. In Newtonian mechanics, the norm is the Euclidian one:

$$ \|v\|^{2} := \langle v,v\rangle = \sum^{3}_{i=1}\sum^{3}_{j=1} \delta_{ij}v^{i}v^{j} \tag{1}$$

where $\delta_{ij}$ is the matrix:

$$ \delta_{ij} = \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix} $$

In a sense, this norm together with a vector space gives the geometrical structure of Newtonian mechanics because we can calculate lengths, define vectors, calculate velocities and accelerations, and so on....

2) This norm sets what we call "Euclidean space" or "Euclidean geometry". Note that if you define another dimension, "the 4th dimension", you will just construct a 4-dimensional Euclidean space.

Now, the physical fact is: the geometry of spacetime isn't Euclidean, because we use a particular norm called the "Minkowski norm" or "Lorentz norm" on a 4-dimensional vector space. Because of this fact all the "conventional linear algebra" must to be adaptated to the Lorenztian geometry, given by the norm:

$$ \|v\|^{2} := \langle v,v\rangle = \sum^{3}_{\mu=0}\sum^{3}_{\nu=0} \eta_{\mu\nu}v^{\mu}v^{\nu} \tag{2}$$

where $\eta_{\mu\nu}$ is the matrix:

$$ \eta_{\mu\nu} = \begin{bmatrix} -1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} $$

III) Spacetime: The general picture

The matrix of inner product $(2)$ (and in general) is called the components of the metric tensor $g$. The metric tensor is (roughly speaking) a bilinear map which produces a particular scalar called a line element, which is simply the value of the norm of differential line element vectors, i.e.


$$ ds^{2}\equiv g\Bigg(dx^{\mu}\frac{\partial \vec{r}}{\partial x^{\mu}},dx^{\nu}\frac{\partial \vec{r}}{\partial x^{\nu}}\Bigg) := \|d\vec{r}\|^{2} =: \langle d\vec{r},d\vec{r}\rangle = \sum^{3}_{\mu=0}\sum^{3}_{\nu=0} g_{\mu\nu}dx^{\mu}dx^{\nu} \tag{3}$$

Now, in general metric tensors aren't easy matrices like $\delta_{ij}$ and $\eta_{\mu\nu}$. In fact the metric tensor can become a tensor field which varies through space (and then the geometry varies pointwise).

To describe this general behaviour of "a tensor field which varies through space (and then the geometry varies pointwise)" we need the manifold mathematical framework (which is beyond the scope of this answer).

$$ g_{\mu\nu} = \begin{bmatrix} g_{00}(x^{\mu})&g_{01}(x^{\mu})&g_{02}(x^{\mu})&g_{03}(x^{\mu})\\ g_{10}(x^{\mu})&g_{11}(x^{\mu})&g_{12}(x^{\mu})&g_{13}(x^{\mu})\\ g_{20}(x^{\mu})&g_{21}(x^{\mu})&g_{22}(x^{\mu})&g_{23}(x^{\mu})\\ g_{30}(x^{\mu})&g_{31}(x^{\mu})&g_{32}(x^{\mu})&g_{33}(x^{\mu})\\ \end{bmatrix} $$

With this manifold framework, we can give a sufficiently general description of spacetime:

A spacetime is a 4-dimensional manifold $\mathcal{M}$ with a pseudo-riemannian metric $g_{\mu \nu}$: $$ (\mathcal{M},g_{\mu \nu}) $$

IV) Spacetime: Merging the intuitive idea with math

So, spacetime is the stage of special relativity and general relativity. It tells you which events are in your future, in your past and the ones which you cannot access in a sufficiently small proper time (the time of the clock in your hand, the time of the observer at rest on his own reference system, in general a tetrad). Spacetime is also a geometrical 4-dimensional entity which tells you that because of the Lorentz signature you need to give spatial and temporal directions and, of course, the geometry isn't Euclidean anymore.

  • 8
    $\begingroup$ Good information, but I'm not sure this is an answer "a layman can understand". $\endgroup$ Commented Sep 24, 2019 at 21:16
  • $\begingroup$ I've read too fast the question. $\endgroup$
    – M.N.Raia
    Commented Sep 24, 2019 at 21:58
  • 2
    $\begingroup$ Just to add to this: Lorentz transformations (rotations + velocity differences) are exactly those transformations that preserve the norm. Especially the velocity differences (Lorentz boosts) can give an intuition on time dilation when interpreted on differentials. Moreover, you forget to mention that to interpret the norm, one has to measure in natural units, otherwise some c's are missing. $\endgroup$ Commented Sep 24, 2019 at 23:06

Essentially, the dimension of a space is the number of numbers you need to specify a point in it.

  • The Earth's surface is two-dimensional, because you need to specify longitude and latitude.
  • The set of possible electromagnetic field values is six-dimensional, because you need to specify $E_x$, $E_y$, $E_z$, $B_x$, $B_y$, and $B_z$.
  • The set of possible $1000 \times 1000$ RGB images is $3000000$ dimensional, because you need to specify R, G, and B color values for each of $1000000$ pixels.

The point is that "dimension" mathematically need not have anything to do with space or spacetime. Asking if "the fourth dimension is time" is kind of like asking if "the derivative is force". No, the mathematical derivative is just that, mathematical. The statements only make sense if you're more specific: the time derivative of momentum is force, and the fourth dimension of spacetime is time.

Stated this way, it's also clear that "spacetime" as a concept has nothing to do with relativity. It takes four numbers to specify a position and a time in relativity, but it also did in Newton's day. The only difference is that in relativity, spacetime can also be given a nice mathematical structure as a whole, as a Lorentzian manifold, which is why we talk about it more. But structures like this are also not special to relativity, as Newtonian spacetime can be given a Newton-Cartan geometry.


Yes, it (time not spacetime) is a mathematical dimension. To describe an event, you need three numbers to define its position, and a fourth number to describe the time at which it occurs. The time dimension works a bit differently than the spatial one when we have to talk about "distances" between events, though. We're used to describing the between two points in space (denoted by the subscripts 1 & 2) as $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$. In spacetime, the time component is present in that sum, except with a relative minus sign (and a factor corresponding to the speed of light, which we often set to 1): $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 - (t_2 - t_1)^2$.

  • 4
    $\begingroup$ Spacetime is not a dimension. It has dimension. It's a mathematical model based on a four-dimensional manifold. $\endgroup$ Commented Sep 24, 2019 at 15:57
  • $\begingroup$ Are you picking at this answer or at the phrasing of the question, @SolomonSlow? It does ask if "spacetime" is a 4th dimension at one point, but, taken as a whole, it seems likely that the middle part of the question (which is most of it) about time being a 4th dimension is what was meant. $\endgroup$
    – Brick
    Commented Sep 24, 2019 at 17:52
  • 1
    $\begingroup$ @Brick, I don't see anything wrong with the answer apart from its first sentence. The OP asked, "Is spacetime a dimension?" The answer to that question is "no," but the first sentence in this answer says, "yes." I did not downvote the answer because it's mostly true, but I also did not upvote it because it doesn't speak to the root of the OP's misunderstanding. It doesn't help the OP to understand what "dimension" actually means. I wish I could answer the question myself, but I don't have time right now to give it the attention it deserves. $\endgroup$ Commented Sep 24, 2019 at 18:13
  • $\begingroup$ I agree that part could be cleaner, @SolomonSlow. $\endgroup$
    – Brick
    Commented Sep 24, 2019 at 18:17

Spacetime has 4 dimensions: 3 spatial and 1 temporal. The temporal (time) dimension is treated slightly different from the spatial dimensions. A 4th spatial dimension is something else, which can easily be modeled mathematically but has not been proven to be relevant in describing the reality we live in.

  • 1
    $\begingroup$ How is it treated differently? $\endgroup$
    – BMF
    Commented Sep 24, 2019 at 16:29
  • $\begingroup$ @BMF It has the opposite sign of the other three dimensions in the metric signature. $\endgroup$ Commented Sep 24, 2019 at 17:58
  • 6
    $\begingroup$ I think your answer could benefit by explaining the differences between space and time dimensions, instead of just stating they're different. $\endgroup$
    – BMF
    Commented Sep 24, 2019 at 21:00

IMHO, it depends on the context. Minkowski's geometry of space-time is not Euclidean. So while spacetime is a type of mathematical 4 dimension geometry, not all mathematical 4 dimension geometries have time as the 4th dimension. In fact, math (and engineering and finance) has no issues dealing with n-space, where n is as big as needed.

Here is a representation of a 4-dimension cube being projected in 3D I stole from wikipedia's entry on 4D space:


  • 8
    $\begingroup$ That's actually a 4D cube being projected in 2D :P $\endgroup$
    – Luaan
    Commented Sep 25, 2019 at 12:14
  • 2
    $\begingroup$ It's actually a 4D cube projected onto 3D, which is then projected onto 2D. $\endgroup$ Commented Sep 26, 2019 at 22:12


There is a straightforward extension of 3D geometry to higher dimensions, where all the normal Euclidean axioms apply. This is the world where we draw tesseracts and spinning toruses as shown in other answers. This is what you are most likely referring to when you say "the mathematical 4th dimension". In this Euclidean 4-dimensional space, all of the dimensions are alike, and the normal Pythagorean theorem works there -- the distance between two points is the square root of the sum of the squares of the distance along each dimension:

$$\begin{eqnarray*} d&=&\sqrt{x^2+y^2+z^2+w^2}\\ d^2&=&x^2+y^2+z^2+w^2\end{eqnarray*}$$

The Pythagorean theorem is equivalent to the parallel postulate, meaning that if you assume the Pythagorean theorem as an axiom, you can prove the parallel postulate from it. Therefore, the Pythagorean theorem in a sense is an indicator of whether your space is Euclidean or not.

In Minkowski geometry (the geometry underlying special relativity), the Pythagorean theorem does not hold. Since one of the dimensions is timelike, you end up with the following formula for distance between two points (or in spacetime vocabulary, the interval between two events):

$$\begin{eqnarray*} d&=&\sqrt{x^2+y^2+z^2-t^2}\\ d^2&=&x^2+y^2+z^2-t^2\end{eqnarray*}$$

This implies that if the time interval between two events is longer, the total interval between them is shorter. If the time interval is long enough, it makes the squared interval $d^2$ negative, and therefore the total interval is imaginary. We talk about a spacelike interval, or proper distance, when $d$ is real, and a timelike interval, or proper time, if $d$ is imaginary.

Minkowski geometry is flat, and yet not Euclidean, since the Pythagorean theorem doesn't hold in general. Any 3D slice perpendicular to the $t$ axis is Euclidean, but the whole space is not. This implies that the parallel postulate isn't true either in general in Minkowski geometry.

Note that the $t$ axis is not a set direction, like "true north" or something like that. Each observer thinks that there is a unique $t$ axis, but if each observer were to point along their $t$ axis, they might not point parallel to each other.

Everything I say here is implied in the mathematics of Minkowski geometry, but in my opinion, when special relativity is taught or especially discussed in popular science presentations, too much emphasis is placed on "spacetime is fourth dimensional, where time is a dimension just like space since time can be 'rotated' into space". Not enough emphasis is placed on "The time dimension is not just like the space dimensions. It can rotate into space, but according to different rules."

"Rotation" of a Minkowski space with one spacelike and one timelike dimension

"Rotation" of a Minkowski space with one spacelike and one timelike dimension. In some sense, the green lines stay perpendicular no matter the rotation. Stolen from Wikipedia.

Also note that while Minkowski geometry is used in special relativity, it is not the same as special relativity. Minkowski geometry can be studied as an object in its own right, without even thinking about such concepts as "observers" or "objects", just "points", "lines", etc. It would be valid to study this geometry even if our real world didn't follow special relativity. Imagine a geometry where two of the dimensions are timelike, IE a longer interval in either of those tow directions results in a shorter interval. We would have:


(look up Dichronauts by Greg Egan)

This could be studied with just as much mathematical rigor as Minkowski geometry, even though I don't believe there is anywhere in the universe that follows that geometry. It wouldn't be as useful, since we tend to study math that models the world as we see it, but you could make definitions and prove theorems with equal rigor to 3D Euclidean space or Minkowski spacetime. Likewise imagine a 92-dimensional space where 47 of the dimensions are timelike. This is more dimensions than even the string theorists believe in, but would be perfectly valid to study.


I'm going to be pedantic here and nitpick your wording because I hear this quite often, and it seems to confuse people.

Time is not THE fourth dimension. Time (in at least some models) is A dimension, which may or may not correspond to the fourth dimension in any given context.

If we're talking about three spatial dimensions, then we add time as a dimension, it makes natural sense to call it the "fourth" dimension. But we could just as easily start with a one-spatial-dimension analog, add time as the "second" dimension, then add two more spatial dimensions as the "third" and "fourth" dimensions.

It's also worth noting that our reality is not intrinsically three or four dimensional. Dimensions also include things like energy level, brightness, gravity field strength, gravity field gradient, etc.

Human vision, for example, is six dimensional, with two discrete channels. The eye records brightness for each color channel (red, green and blue, three dimensions) at each receptor, keeping track of the vertical and horizontal positions of the receptor (fourth and fifth dimensions), across time (sixth dimension).

Having two channels (one for each eye) allows us to localize objects more readily, and is sometimes considered an extra dimension (for example, when a movie is played in "3D" but is just two 2D images, or "3D" audio formats that are only planar), but I don't think it really qualifies mathematically.

Random Trivia

The first reference I can find to time being considered a dimension (it's even labeled a "fourth" dimension) is from circa 17511, by Jean-Baptiste le Rond d'Alembert 2.

I'm told some butterflies have 15-dimensional color space they use to identify specific flower types and rival butterflies, which would give them 18-dimensional vision. But I've never asked a butterfly to find out for myself. 3

1 From d'Alembert and the Fourth Dimension by Rosine G. Van Oss via Science Direct. https://www.sciencedirect.com/science/article/pii/0315086083900071
2 A Wikipedia article, Jean le Rond d'Alembert. https://en.wikipedia.org/wiki/Jean_le_Rond_d%27Alembert
3 From a 2016 article posted on Frontiers in Ecology and Evolution, Extreme Spectral Richness in the Eye of the Common Bluebottle Butterfly, Graphium sarpedon. https://www.frontiersin.org/articles/10.3389/fevo.2016.00018/full


Pick any massive object in the world, like a football ball, or whatever you want. Now, think what was this object's history: which position $x$ has it had throughout its existence? Now, you can tag each point in this object's history with the time you read on the world's clock, or better, any clock you have with you, at the moment you saw the object in some place. Then you can write down the times and the places $(t,x)$ showing the object's history. At 5:00 pm the ball was on Brazil's net, at 5:01 pm was on the pitch's center spot, and at 5:02 pm it was on Brazil's net again. If, instead of time, you were using the space to tag each point in the ball's history, you'll soon run into a problem as you'll probably see the ball going over and over to the same spot. You have lots of $t$'s for only one $x$, and you don't even have the same amount of $t$'s for every $x$, while using the time-tag you neatly have one and only one $t$ for $every$ point in the history. Choosing the time tag has the mathematical advantage of allowing us to use a function $x(t)$ to describe the object's history. Then the time dimension is a natural independent variable we can always choose, because we know every measurement will always have only one $t$ associated with it. This doesn't happen with any spatial dimension, so time is not on an equal footing with space.


Layman answer:

Three dimensions are not enough to describe things that vary in time. Mathematically, you need a fourth dimension.
You can 'measure' that fourth dimension with a clock.

It is analogous to:

Two dimensions are not enough to describe how things like birds move. You need a third dimension.
You can 'measure' that third dimension with an altimeter.

In the case of the fourth dimension, the unit of measurement is the second. In the case of altitude the unit is the millibar or hectopascal.

In both cases there is a conversion factor from the unit of measurement to the units of the other axes.

In the altitude case, the conversion factor is a real number, for time it is an imaginary number.


No, spacetime is not the "usual" euclidean mathematical space people imply when they talk about the 4th dimension. The easiest way to see that is that in usual euclidean space the distance $s$ between two point can never be negative: $$ s^2 = x^2 + y^2 + z^2 + w^2 $$ In spacetime the distance between two point can be negative, as can be seen from the formula to compute the distance between two spacetime events: $$ s^2 = x^2 + y^2 + z^2 - c^2 t^2 $$ The minus sign in front of the last term is what makes all the difference.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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