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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? $\endgroup$ Mar 22, 2018 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, 2018 at 17:26
  • $\begingroup$ Yes I calculated according to the metric of the Minkowski space. $\endgroup$
    – Keith
    Mar 22, 2018 at 18:43
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    $\begingroup$ "invariant inner product" Perhaps you mean "invariant norm-square" or "invariant inner product with itself". $\endgroup$ Mar 23, 2018 at 0:13
  • $\begingroup$ Yes that is what I mean $\endgroup$
    – Keith
    Mar 23, 2018 at 19:13

4 Answers 4

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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 $(t,x,y,z)$. $\endgroup$ Mar 27, 2019 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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My answer is implicitly contained in CR Drost's answer, but I'm mainly going for a clean mathematical exposition.

The mathematical definition of a 4-vector is quite straightforward: it's an element of a 4-dimensional vector space $V$ which is equipped with a Minkowski metric $\eta$ (i.e. a bilinear, real-valued function with a $(-,+,+,+)$ or a $(+,-,-,-)$ signature). In this sense, $v = (1, 2, 3, 4)$ is a 4-vector in $\mathbb R^{1 + 3}$, with Minkowski norm $\eta(v, v) = -1^2 + 2^2 + 3^2 + 4^2 = 28$.

However, issues start to arise in physics because people like to define things in terms of what they can measure. This means that most definitions of real quantities are as components in a particular basis, which is to say a choice of $xyz$-axes and a direction of time (i.e. a velocity). If we think about regular force, for example, it's defined as follows:

In any given basis, the force on a particle is defined to be the derivative of its momentum with respect to time in the basis.

The point is that this doesn't just define a 4-vector in one basis. Instead, it defines a 4-vector for every basis at once, and there's nothing compelling this definition to make them all the same. The vectors certainly aren't the same for this definition; due to relativistic effects, time derivatives are different in each frame. As such, we're dealing with a different vector in each basis, which is both exceptionally unnatural and difficult to deal with.

When defining vectors based on their coordinates in a basis, therefore, we prefer to have definitions which don't change as you change the basis. This is what a physicist means when they say something like "such-and-such isn't a 4-vector because it doesn't transform properly under Lorentz transformations": they're really saying that a certain basis-dependent definition doesn't happen to give the same results in each basis.

Let's make this rigorous! In Minkowski space, we only really care about basis transformations which preserve $\eta$ (since these correspond to rotations and Lorentz boosts). For this reason, we only need to verify that the components $a^\mu$ of a vector $a = a^\mu e_\mu$ change properly under this group of transformations, called $\text{SO}(1, 3)$ (the Lorentz group). If the basis elements $e_\mu$ transform with $\text{SO}(1, 3)$, we need the components $a^\mu$ to transform against $\text{SO}(1, 3)$, to keep the product $a = a^\mu e_\mu$ the same. This is what is meant by covariance and contravariance.

You don't need to think about any of this, on the other hand, if you make sure all of your definitions are in terms of quantities which don't change with basis transformations. Think about the proper velocity: $$ u(\tau) = \frac{d}{d\tau} x(\tau), $$ where $x: \mathbb R \to V$ is the position of the particle as a function of the proper time. Or the relativistic force: $$ f(\tau) = \frac{d}{d\tau} (m u(\tau)), $$ where $m$ is the invariant mass of the particle (alternatively $f = dL$ if you know Lagrangian mechanics). Or the four-current density: $$ J = \rho_0 u, $$ where $\rho_0$ is the proper charge density. Note that components don't pop up anywhere in these definitions; they are all made from scalars and vectors, which don't change from basis to basis. Physicists call such definitions of vectors manifestly covariant, since we don't even need to check that the components transform correctly: we already know, because that's how vectors work in the first place!

This is not to say that there is no place for component-wise manipulations and definitions. The components of a vector are, as I said earlier, the things we can actually measure, and it would make things much harder if we didn't have rules for translating between coordinate systems. Moreover, definitions are often much harder to make if we can only look at relativistic invariants. For example, the usual way of introducing the electromagnetic field strength tensor is by combining elements of electric and magnetic fields into a matrix, and showing that it behaves correctly under Lorentz transformation. Any method of introducing it without component-wise combinations requires a "leap of faith," which can be offputting for new students.


So, if you want the physicist's definition of a 4-vector, it's a set of four components that transform contravariantly with respect to the Lorentz group $\text{SO}(1, 3)$. But if you want a more honest definition, it's just an element of a 4D vector space. Transformation laws only need to be considered when you're giving a vector to each basis, since you need to transform between each basis to compare each vector.

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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, 2019 at 20:21

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