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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.

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    $\begingroup$ Related and possibly duplicate: Defining four-vectors in General Relativity? $\endgroup$ – John Rennie Mar 22 '18 at 16:12
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    $\begingroup$ 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
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    $\begingroup$ "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
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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.

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

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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.

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  • $\begingroup$ 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

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