In Minkowski space, I know that there are some vectors such as the ordinary velocity that are not proper 4-vectors.

But what is the exact definition of a 4-vector? For any fixed numbers, say 1,2,3,4, does $(1,2,3,4)$ become a 4-vector in Minkowski space with the invariant inner product 28? I am confused.

  • 1
    $\begingroup$ Related and possibly duplicate: Defining four-vectors in General Relativity? $\endgroup$ – John Rennie Mar 22 '18 at 16:12
  • 1
    $\begingroup$ Note that "28" is not an inner product, i.e. it doesn't obey the inner product axioms. In special relativity, the inner product we use is the Minkowski metric. $\endgroup$ – knzhou Mar 22 '18 at 17:26
  • $\begingroup$ Yes I calculated according to the metric of the Minkowski space. $\endgroup$ – Keith Mar 22 '18 at 18:43
  • 1
    $\begingroup$ "invariant inner product" Perhaps you mean "invariant norm-square" or "invariant inner product with itself". $\endgroup$ – dmckee --- ex-moderator kitten Mar 23 '18 at 0:13
  • $\begingroup$ Yes that is what I mean $\endgroup$ – Keith Mar 23 '18 at 19:13

In Euclidean space, we can define vector as an object which transforms in a specific way under rotation.

To define vector in special relativity, we use Lorentz transformation instead of rotation. (Actually, Lorentz transformation is a kind of rotation in 4-dimensional space,)

Suppose that the events of stationary observer $O$ are given by $(t,x,y,z)$. Consider another frame $O'$ which moves along the x-axis with velocity $v$ and whose events are given by $(t',x',y',z')$. The Lorentz transformation between the two observers is:

$$t'=\gamma(t-vx/c^2),\ x'=\gamma(x-vt),\ y'=y,\ z'=z $$

From this, we can conclude that $(t,x,y,z)$ is a 4-vector.

Here is another example : Electromagnetic four potential is given by $$ A_\mu=(\phi/c,A_x,A_y,A_z) $$

If this is a 4-vector, it must obey the Lorentz transformation rule, so that

$$ \phi'/c=\gamma(\phi/c-vA_x/c^2),\ A_x'=\gamma(A_x-v\phi/c),\ A_y'=A_y,\ A_z'=A_z $$

This conclusion can be derived by classical Electrodynamics.

  • 1
    $\begingroup$ More exactly: $(ct,x,y,z)$ is a 4-vector, not $(t,x,y,z)$. $\endgroup$ – Thomas Fritsch Mar 27 '19 at 22:11

So I can confirm that if you are looking at some Minkowski space, $(1, 2, 3, 4)$ is a 4-vector in that space. In addition if you have a particle and you see it having some velocity components $v_{x,y,z}$, there is a 4-vector with components $(0, v_x, v_y, v_z)$, another with components $(1, v_x, v_y, v_z)$, another with components $(2, v_x, v_y, v_z)$ -- one can pick nits here about units but one cannot deny that basically for any numbers $a,b,c,d$ there is some 4-vector in some basis with components $(a,b,c,d)$.

When we say that something “is a 4-vector” we are not speaking of that, per se. I would say, it is a syntactic statement and not a semantic statement. It has to do with the syntax of the expression that you are using to build the thing, and whether that syntax refers to a consistent geometric entity after coördinate transforms.

So for example when you deal with general relativity you will learn that there is a thing called a Christoffel symbol, and it is calculated from certain things about the metric tensor, which is a tensor, but that we proudly tell our students, “The Christoffel symbol $\Gamma^a_{bc}$ is not a tensor.” This is the same as whether something is a vector or is not a vector, it just has more indexes to fuss over.

The thing is, there surely is a tensor which has the components that the Christoffel symbol has. So what we mean when we say that the Christoffel symbol is not a tensor is this:

  • You might use different coördinates to describe the space, say transformed by a Lorentz boost.
  • In those different coördinates you might calculate a new Christoffel symbol at this point.
  • You might also take the tensor that you assembled from the Christoffel symbol and transform it into the new coördinates.
  • But, crucially, these two approaches give you two different objects. The answers that you get when you do these two operations are not identical.

Therefore “Christoffel symbol” is something like a title of nobility or something: much like how, if you translate yourself forwards or backwards in time, the title ‘Earl of Grantham’ might refer to a fundamentally different human being; so, too, if you look at the world in a different way, the Christoffel symbol might be embodied by a different fundamental geometric object in your manifold.

The same is true of, say, $(0, v_x, v_y, v_z)$. That is certainly a four-vector that exists in my space, but if I transform this by a Lorentz boost I will in general find some $(a,b,c,d)$ where $a \ne 0$, and then this will not be the exact same vector I assemble if I look at the same particle's new velocity components $v_{x,y,z}'$ and assemble the four-vector $(0, v_x', v_y', v_z'),$ because that first component will be nonzero.

By contrast if I assemble the four-velocity, which we’d say is a four-vector, $$v^\mu = \frac{1}{\sqrt{1 - (v_x^2 + v_y^2 + v_z^2)/c^2}} (c, v_x, v_y, v_z),$$what makes that thing a four-vector is that those two approaches give you the same object: you can use other Lorentz-transformed coördinates and calculate a velocity in those coördinates and assemble this four-vector from those new components, or you can just transform this four-vector, and you will find out that both of these approaches have given the same geometrical object.

In a more pithy phrasing,

A vector is anything that transforms like a vector. A tensor is anything that transforms like a tensor.

This is not circular: we have model examples like position vectors to tell us how such vectors transform, so that “transforms like a vector” is actually an independent criterion by which we can evaluate whether a proposed method of assembling together numbers into a vector really constitutes a vector. So in 2D space, my shopping list whereby I need one apple and two grapefruits can hypothetically be abused to allow me to assemble a ‘vector’ $(1, 2)$ to describe this shopping list, but that shopping list is ‘not a vector’ because if I rotate space by 45 degrees I do not find that I suddenly need $3 \sqrt{1/2}$ apples and $\sqrt{1/2}$ grapefruits.


Don't think about the 4-vector's definition, think about how these objects are transformed. A four-vector is a simple four ranged covariant tensor, mathematically it is no more than that. The special characteristic is that some of them as four-velocity, position or potential are transformed between inertial frames by Lorentz transformation tensor. This is defined in a metric space which metric is just the minkowski metric.

  • $\begingroup$ For whatever it's worth, I never understood tensors from this idea of "think about how they transform". It only made sense once I learned how to think of them in a basis-independent way. $\endgroup$ – DanielSank Mar 29 '19 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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