Wikipedia says, "In special relativity, four-momentum is the generalization of the classical three-dimensional momentum to four-dimensional space-time. Momentum is a vector in three dimensions; similarly four-momentum is a four-vector in space-time."

Does this mean that four-momentum is the type of vector resulting from three spatial dimensions being placed, as a whole, into a greater level of momentum? As in three dimensions traveling along a fourth axis (time), correct? And this extra dimensional relationship is entirely relativistic, correct (I apologize if I am not making myself clear, I am very fascinated with relativity, spending hours on Wikipedia trying to understand it, but I need a teacher to walk me through a few things that confuse me)? What I am trying to say in that last statement I will hopefully make clear in the following example:

A train is traveling along it's track so smoothly that there is no way for its passengers to tell that they are in motion. A man in the train tosses a ball up in the air and catches it. From the perspective of the man, the ball has gone straight up and down. The path of that ball relative to the inside of the train can be calculated using classical three dimensional momentum. However, relative to someone living outside the train, the ball has traveled in an arc, not a straight line. The balls paths (up and down) relative to the inside of the train was a path that could be described on one axis. But, that same path relative to the outside of the train requires two axes to describe (up and down and side to side), and an extra vector. So the dimensional path of the ball went from one (a line) to two (a plane), and held a new vector given to it by the trains momentum, just by taking our perspective out of the train.

Does this accurately depict relative levels of momentum?

If the answer is yes, then here is where my confusion begins. I can take that same basic scenario and apply it to a man on earth who tosses a ball straight into the air. Relative to the man, the ball went up and down. Relative to the sun, the earth is moving so the ball traveled in an arc (well, the earth is also rotating so that gives the path of the ball another vector and its curvature greater complexity). Yet the sun is in motion, so relative to the space the sun travels through the ball moves in a sort of partial helix. Yet the system of motion containing the sun is contained by an even greater system of motion, and so on, adding further levels of relativistic spatial dimensions to the balls momentum. Will not the relative momentum of the ball quickly reach and exceed special relativity's "4-vector" as we continue to place our perspective into the greater levels of gravity that smaller systems of motion are always contained by?

So my questions is: How is it that while observing the universe realistically, Einsteins equations for space-time use four-momentum, and not rather five or six-momentum, seven, or even greater levels of spatial dimensions? What am I not seeing?

My first thought was that Einstein was describing the motion of the entire universe through time, and not the motion of bodies through space, but I then discovered that this is not entirely true as Einstein showed that these two descriptions are inseparable, as time is dynamically linked to motion itself.

My only idea is that the train scenario I described is somehow disconnected from what the building of multiple vectors actually is, and in order to understand relativity I would be incredibly grateful to anyone who could complete my understanding of the above "ball in train" depiction.

  • 1
    $\begingroup$ Possible duplicates of title question (v3): physics.stackexchange.com/q/10651/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Sep 8, 2013 at 22:43
  • $\begingroup$ No, my question is very different from the one you brought up, as I am not concerned with the formation of bodies in higher spatial planes. Also non of the answers to that question bridge the gap I am looking to close in my understanding of space-time. $\endgroup$
    – user29372
    Commented Sep 8, 2013 at 22:54
  • $\begingroup$ Mathematicians tend to define the idea of a vector in terms of the algebraic operations like vector addition. Physicists tend to define them instead in terms of their transformation properties. For 3-vectors that means how they behave under rotation. A playing card's orientation isn't a vector because it's unchanged under a 180-degree rotation. For 4-vectors it means how they behave under both rotations and Lorentz boosts. When we say momentum is a 4-vector, we mean that it transforms under a Lorentz boost in the same way as a spacetime displacement $(\Delta t,\Delta x,\Delta y,\Delta z)$. $\endgroup$
    – user4552
    Commented Sep 8, 2013 at 23:30

6 Answers 6


All four dimensions are present in both examples. All that you mean when you say that space-time is four dimensional is that you need four numbers to describe when and where an event happens.

The path of a particle is a string of such events--the ball is one inch above my hand, the ball is two inches above my hand, the ball is at it's peak, it's two inches above my hand, etc.

What is novel about describing the two reference frames that you do is that in one, the ball only travels vertically and in time, while in the second, it also has a horizontal component to its motion.

  • $\begingroup$ Yes, though to the man outside the train, the ball took a longer length of time to complete its path than it did inside the train. So in reality the second frame of reference is a bent version of the first frame. When my perspective "bends" from the first to the second, the balls path bends along with it, and lengthens, and time lengthens as well. To the outside man, the inside man lives in a state of "bent perspective." Where the inside man used 4 numbers to chart a point along the balls path, will not the outside man perceive that point as bent and thus require a greater coordinate to chart? $\endgroup$
    – user29372
    Commented Sep 9, 2013 at 2:44
  • $\begingroup$ @user29372: no, they will both use four numbers to chart the path of the ball. All that will happen is that they will percieve the space and time coordinates that they use to be "mixed" with respect to each other, when they compare notes. $\endgroup$ Commented Sep 9, 2013 at 3:01
  • $\begingroup$ I'm trying to conceptualize what "mixed" might look like. For example: If I take a warped mirror and view the reflection of a dot I have drawn on a piece of paper, where the mirror is flat the dot will appear as a point. If I move the reflection to a specific place in the mirror, the dot is reflected back to me as a line. In which case, to graph my dot on the mirror, I must drawn a line from the point where it begins to the point where it ends. Does the bending of space-time never behave as the light in the bent mirror, and when you say "mixed" are we perhaps closely describing the same thing? $\endgroup$
    – user29372
    Commented Sep 9, 2013 at 5:15
  • $\begingroup$ @user29372: no. There is no change of dimensionality. There are just time dialation and length contraction effects. $\endgroup$ Commented Sep 9, 2013 at 12:16
  • $\begingroup$ @user29372. Contractions, not contradictions. $\endgroup$ Commented Sep 9, 2013 at 17:19

Note that in you train-ball example, it might happen that in a particular frame, the motion of an object is restricted to, for example, a straight line, and that in another frame, the motion is more complicated, but there is no frame in which the ball's motion suddenly becomes "spatially $n$-dimensional" for some $n>3$. Changing frames just means that the motion of the ball through 3-dimensional space will look differently 3-dimensional. Said in another way,

The number of dimensions sufficient to describe the spatial position of a particle according to any single inertial observer is precisely three.

When one transitions to relativity, an analogous statement is true;

The number of dimensions sufficient to describe the spacetime position of a particle according to any single inertial observer is precisely four.

As for the four-momentum and other such vectors, it may be the case that in a particular frame, one of the components of the vector vanishes; frame-changing alters the values of vector components in general, but it never magically adds on an extra component. The fundamental 4-dimensional description remains sufficient for all observers.


To rephrase what the other answers are already saying in a different way:

Space and time together were always 4 dimensional, even before relativity. It was understood that there were 3 space dimensions and 1 time dimension.

However, pre-SR physics assumed that time just ticked along in a universal manner for all locations and all reference frames the same way. As a result, objects could be said to be "moving through time" and have a "time coordinate" for when something happened to them, but this information could often be neglected. That is, switching reference frames would alter the spatial description of events but never the temporal description, so why worry about it?

When relativity came along, the key change in thinking was not just including time in descriptions of things. Rather it was how to transform one 4-vector into another when shifting reference frames.

  • You could do the naive thing and just do a linear transformation of the spatial components according to the new frame's velocity relative to the old, leaving the time untouched. This would be a Galilean transformation.
  • Or you could employ a more complicated rule -- specifically that of the Lorentz transformation -- which inherently mixes up space and time.

It turns out nature follows the second choice, and that accurately describes what happens. Without all 4 dimensions being used, something is often missing.


Your question might be answered by pure algebra: the relative movement of an n-dimensional object point $y$ with respect to an n-dimensional reference frame $x$ in its most general form is described by the bilinear outer product (the Jacobian) of an n-dimensional differential operator with the object point's coordinates: $\frac{d}{dx} \otimes y$. This Jacobian is an $n\times n$ matrix, but it should be part of the same n-dimensional space as the vectors, otherwise the movement of a movement would no longer be a movement. We thus need a space wherein a bilinear complete vector product exists, such that $\frac{d}{dx} \otimes y = z$, wherein $x$, $y$, and $z$ are vectors of the same space, and wherein the length of the product of two vectors equals the product of the lengths of the vectors. As A. Hurwitz showed in 1898, there are only four possible spaces with these properties, noteworthy the real numbers $\mathbb{R}$ (trivial case); the complex numbers $\mathbb{C}$; the quaternions $\mathbb{H}$, and the octonions $\mathbb{O}$. Except for the real numbers $\mathbb{R}$, the other spaces have negative metrics: $\mathbb{C} (+1,-1)$; $\mathbb{H} (+1,-1,-1,-1)$; $\mathbb{O} (+1,-1,-1,-1,-1,-1,-1,-1)$. The metric of the quaternion space $\mathbb{H}$ is the metric of space-time. Special Relativity follows immediately out of this metric, as well as Maxwell's equations and the wave nature of any displacement in this space (foundations of quantum mechanics). I have written a small paper on this: "De la réalité des nombres", Bulletin de la société Fribourgeoise des Sciences Naturelles, 103 (2014), p. 83-90; online under:


P.S.: One must then assume, of course, that the luminiferous ether exists, but that Michelson & Moorley's experiment could not detect any ether wind, because it was conceived for a 3-dimensional Euclidian space with metric $(+1,+1,+1)$. The ether wind in $(+1,-1,-1,-1)$ space shows rather up as electric and magnetic fields! There are more forces in particle physics than the mere electromagnetic one, but spacetime at these small dimensions is likely to be 8-dimensional, with metric $(+1,-1,-1,-1,-1,-1,-1,-1)$. The question why the additional dimensions and forces do not show up at the macroscopic scale is out of the present scope.

  • $\begingroup$ Welcome on Physics SE and thank you for your answer :) For future reference, the editor of PSE accepts TeX-Code (inline with opening and ending $ or in an additional line with $$) $\endgroup$
    – Sanya
    Commented Aug 7, 2016 at 16:27
  • $\begingroup$ I feel honered to texify your high quality answer. The $\frac{d}{dx}xy$ part seemed a little bit cloudy to me, I am not sure it was your intent to write. $\endgroup$
    – peterh
    Commented Aug 7, 2016 at 17:02
  • $\begingroup$ @peterh The first differential expression should be an outer product, and the second one a cross product (I do not know how to type it in here): (d/dx) 'outer product' (y); and (d/dx) 'cross product' (y) = (z). Thanks for editing! Edgar Mueller $\endgroup$ Commented Aug 7, 2016 at 20:35
  • $\begingroup$ @EdgarMueller I hope you didn't go away - I had to google for that. This is what you wanted to write? What is an "n dimensional object point"? $\endgroup$
    – peterh
    Commented Aug 9, 2016 at 21:33

@JerrySchirmer: Im almost there. My problem is that I have a hard time grasping something if I can't conceptualize it, though maybe this can not be conceptualized, just accepted, in which case I thank you for your patience. The problem I'm facing now in my mis-leading "conceptual" mind is that when you say "dilations, and length contractions" I can't help thinking that in reality there are so many other things happening to time than dilations and contractions. When we observe the warping of space-time we see both space and time being bent out of shape in fantastic ways. Time is not only dilated, but "smashed," and "warped," and curved. In fact, in Einstein's definition of gravity space-time "swirls" towards the center of large masses, because the mass is literally warping both time and space. By definition that is a three dimensional process.

To describe a point on a line, we need one number. To describe a point on a curve we need two numbers. To describe a point on a "curving curve" like a helix we need three numbers (one each for the components of hight, width, and depth). If the path is traveling on a one dimensional track of time, in other words along a "line" of time, we would need a fourth number to describe its position in time. But if the path is curving through time, then there is also a horizontal component to time, so how is it will we not need 2 numbers to record a point on that arcing track? Again, I apologize for stubborn imagination.

Einstein talks of the curvature of time. Here is my main desire: Please, just explain to me how it is possible to chart a point on a curving path using one number.

Is there a way for me to understand such a process conceptually, or must I just accept it?

But also, as Einstein's definition of gravity shows, mass warps both space and time three dimensionally. Did he not imagine that the curving track of time was also simultaneously bending? meaning it has a vertical component, a horizontal component, and a depth component, as it is warped by gravity. so to chart a point in time we must have 3 extra numbers. 3 for the position in space, and three for the position in time. So I have no problem leaving behind the idea of 5 or greater dimensions, though is it then wrong for me to think of 3D systems of gravity contained within greater 3D systems of gravity while containing smaller 3D systems, when thinking of the planets in a solar system in a galaxy?

Though again, my main desire is for this question to be answered:

If time contains curvature, and if a curving path requires at least two numbers to chart as it has a vertical component and a horizontal component, then how is time charted with one number?

P.S. I wanted to use the comment section to respond but there was not enough room.

  • $\begingroup$ Einstein's definitions do not show that mass (and energy and momentum and stress) "warps both space and time three dimensionally", rather they show that mass warps spacetime in a four dimensional sense. Also, a point on a curve or a "curbing curve" can still be described by a single number, such as the arclength. Consider a tape-measure twisted in space. $\endgroup$ Commented Feb 16, 2015 at 14:05

Lets take the very beginning of your question.

Just as the three dimension vector locating a position in space, $(x,y,z)$ is generalised to a 4 dimensional vector $(x, y, z, ct)$ that includes time, we can generalise most of our 3 dimension vectors to 4 vectors.

The three dimensions of momentum become $(p_x, p_y, p_z, E/c)$. The forth component of this vector is energy. The force vector becomes 4 vector with power as the 4th component.

Angular momentum is tricky. When we generalise it, instead of a 4 vector, we get a 4x4 matrix with 6 unique entries. There are the original 3 angular momentum components, and 3 new guys.

We get 4x4 matrix any time you take a 'cross product' of two 4 vectors.

The 3 components of the electric field and the 3 components of the magnetic field combine in a similar 4x4 matrix. So do electric and magnetic dipoles.


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