# Why define four-vectors to be quantities that transform only like the position vector transforms?

A four-vector is defined to be a four component quantity $A^\nu$ which transforms under a Lorentz transformation as $A^{\mu'} = L_\nu^{\mu'} A^\nu$, where $L_\nu^{\mu'}$ is the Lorentz transformation matrix, which including boosts, rotations and compositions. (In other words, as the components of a position vector $(x0,x1,x2,x3)$would transform).

The useful property of four-vectors is claimed to be that if two four-vector expressions are equal in one frame, then they will be equal in all frames :

$A^\mu = B^\mu \Leftrightarrow A^{\mu'} = B^{\mu'}$

and therefore we can express laws of physics in terms of four vectors, because they remain invariant in all frames.

But this property will be true even for four component quantities that transform (across reference frames) as $A^{\mu'} = T_\nu^{\mu'} A^\nu$, where $T$ is any transformation matrix (not necessarily a Lorentz one). As long as we can find a $T$ that will describe how the quantity's components transform, we can apply that T to both sides of an equality.

So why require (i.e. define) four-vectors to only be quantities that transform under a Lorentz matrix?

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– Ben Crowell Jun 1 '13 at 19:54

dayareishq has captured the basic idea as it is typically portrayed in differential geometry (and general relativity, as an application of that discipline). But you would be right to think "vectors are partial derivatives? that makes no sense!" Because it doesn't. Nevertheless, this identification is pervasive in differential geometry; it's something you have to get used to...or find a good and solid alternative to (which do exist).

The transformation law for vectors follows from drawing curves in spacetime. Given a curve $c(\lambda)$, transforming this under an arbitrary, smooth, differentiable transformation $f(x) = x'$ yields $c'(\lambda) = (f \circ c)(\lambda)$. You then find that $dc'/d\lambda = \underline f(dc/d\lambda)$, where $\underline f$ is the Jacobian. This is a basic application of the chain rule, and that is all the mathematical content of dayareishq's equation $v^{\mu'} = v^\mu \partial x^{\mu'}/\partial x^\mu$. Lorentz boosts and rotations just obey the simple idea that, as linear operators, they are equal to their own Jacobians.

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Agreed, this is perhaps easier to digest for a new user. I must admit, I myself, do not understand very well the idea that "vectors are partial derivatives." My guess is that defining things in that manner makes the Jacobian transformation law fall out quite naturally, all while keeping the definitions very "mathy" (defining everything to be maps). – childofsaturn May 31 '13 at 1:24
I've seen it suggested that someone took Cartan's idea that a tangent vector $e_\alpha = \partial x/\partial x^\alpha$ and realized that the quantity was not well-defined because $x$ describes a point, but if you just took the partial derivative, you could finagle it into working. The alternative I spoke of takes the opposite tack: let $x$ be a vector in some infinite-dimensional vector space (though the set of points defining a manifold generally won't form a vector space), and then the derivative operation is well-defined. – Muphrid May 31 '13 at 1:28

Ultimately, you make a totally legitimate point; we very well could have defined the term "four-vector" to refer to a type of object that transforms in a different way, but we make the particular definition that we do because it is useful to have a term that refers things that transform like spacetime positions when you change frame. Here are two reasons why:

Fact 1. Given any two four vectors $A^\mu$ and $B^\nu$, the quantity $g_{\mu\nu}A^\mu B^\nu$ is invariant under a change of frame.

Notice that this would not have been true unless $A^\mu$ and $B^\nu$ were four-vectors because the proof of this fact relies on the metric being preserved by Lorentz transformations, and not by other arbitrary things. Here is another reason why the definition is useful

Fact 2. Lots of really useful and physically significant quantities happen to be four-vectors. Take, for example, $J^\mu$ and $A^\mu$ (the current and vector potential) in electromagnetism.

Having said all of this, however, note that there are tons of other quantities that do not transform as four-vectors when one changes frame. In fact, given any representation $\rho$ of the Lorentz group, one often encounters quantities $Q$ that transform as $$Q' = \rho(\Lambda) Q$$ For example, there are objects called Weyl spinors that transform as $$\psi' = \rho_\mathrm{weyl}(\Lambda)\psi$$ when one transforms between frames.

The upshot of all of this is the following

Upshot. Lorentz 4-vectors are not special. However, since every change of reference frame can be associated with a Lorentz transformation, every quantity that you want to transform between frames must necessarily transform in a way the depends, in some way or another, on the Lorentz transformation between the frames. This leads us to not only define four-vectors, but a host of other objects that have specified transformation laws under changes of frame and to give them special names. Doing this is useful because such objects appear all over the place in physics, and we can prove useful properties about objects with certain transformation behaviors.

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Right, I see.. It is just that in some textbooks I saw statements like "If a statement has any chance of being true in all frames, it must involve only 4-vectors." [Intro to Classical Mechanics by Morin, page 644]. That is what I found confusing. – Tim May 31 '13 at 23:40
@Tim Yeah. That statement is not true in general. Lorentz-covariance can be achieved through objects with all sorts of Lorentz transformation properties. Take, for example, the Dirac Equation which is written in terms of spinors en.wikipedia.org/wiki/…. – joshphysics Jun 2 '13 at 18:19
In $Q'=\rho(\Lambda)Q$, must the function $\rho$ be a Lorentz invariant if a physical law exists involving these quantities? – Physiks lover Aug 16 '15 at 2:32
@Physicslover I'm not quite sure what you mean in this context. Could you explain what it means for a representation of the Lorentz group to be Lorentz-invariant? – joshphysics Aug 16 '15 at 3:42
Every frame uses the same Lorentz transformation when transforming events with the same lab space-time coordinates to other frames with the same relative velocity $v$. It's a transformation that doesn't change upon boosting to another frame, making it some kind of "Lorentz invariant". Maybe this isn't appropriate then? – Physiks lover Aug 16 '15 at 21:18

Only Lorentz transformations correspond to the geometric transformation which are associated with physical rotations and velocity change, and they connect the frames in which you compute the proper times and leghts as $\sqrt{\text dt^2-\text d \vec x^2}$. The metric components involved in computing $x^2=x_\mu x^\mu=\eta_{\mu\nu} x^\mu x^\nu$ are merely factors $\pm 1$ if $x$ are the cooridnates in an inertial frame, or any frame which is obtained from that frame via a Lorentz transformation (which we again call inertial frame).

We don't require that you only transform them in a certain way in general, you can do any transformation you want. If your cat walks next to you with a certain speed, and if you want to see how the world looks from her point of view, you gotta do a boost transformation of your coordinates. And Lorentz figured that the Lorentz transformations do that more accurately than the Galilean boost. That doesn't mean you can't do a Galilean transformation to describe the world from another perspective in which the ongoing value of the time coordinate doesn't match up with your cats perception of current time (the error will not be big).

If you know how position changes and you do a transformation of an equation involving it, then the transformation law of other geometric quantities, like velocities, co-vectors which eat velocities, or any form on the cotangent space (like the electromagnatic field strengh tensor) are induced by consistency requiremt and the claim that a principle of relativity should hold.

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The reason you're asking this question is because defining a vector or a more general tensor as "a quantity that transforms a certain way" is not very conceptually illuminating, but is often done to avoid introducing some slightly formal mathematics. Here's a better way to proceed:

Let $M$ be a manifold (just think of Minkowski space as a simple example). We define a vector $v$ at a point $p\in M$ to be a linear map that takes a function $f \in C^{\infty}(M)$ (all smooth real valued functions on $M$) to a real number $c\in \mathbb{R}$, that also obeys the Leibniz rule:

$$v(fg) = f(p)v(g) + g(p)v(f).$$

Now for each point $p \in M$, this set forms a vector space, called the tangent space $T_p M$. An "$n$-vector" then (where $n$ is the dimension of our manifold), is simply a smooth vector field, i.e for each point $p\in M$, it gives us a vector living in $T_pM$. It turns out that the set of partial derivative operators $\{\partial_{\mu}\}$ for any valid coordinate system $\{x^{\mu}\}$, forms a basis for our tangent space, so that any 4-vector can be written as $$v = v^{\mu}\partial_{\mu},$$ where $v^{\mu}$ are real-valued functions. However, in another coordinate system we may also write this as $$v = v^{\mu'}\partial_{\mu'}.$$ Of course, these two vectors are equal: $$v^{\mu}\partial_{\mu} = v^{\mu'}\partial_{\mu'}.$$

From this we see that the new coordinates in terms of the old are given by

$$v^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^{\mu}}v^{\mu}.$$

So in general, for any vector (field), under a change of coordinates, the change in components is given by the matrix of partial derivatives. For the special case of change of coordinates by Lorentz transformations, the matrix of partials are the corresponding matrices of the Lorentz boosts (not very different, since the Lorentz transformations are linear).

Thus, making this definition you see that your $T$ is only allowed to be the matrix of partial derivatives, which in your particular case, happens to be $L^{\mu'}_{\mu}$, and not any other arbitrary transformation.

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As noticed below, a general formula for vector transformations, under coordinate transformations, is $A^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^{\mu}}A^{\mu}.$

And it work perfectly fine in the special case of Lorentz transformations.

But laws of physics can not only be expressed in term of vectors, you could use also tensors, spinors, etc.

The point is that you must equal 2 quantities that transform in the same way, under coordinate transformations.

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I am reading Schutz on the subject of vectors, and I believe that he says that; given a manifold with two sets of coordinate systems that are related through a linear transformation; a pair of sets of numbers (with the same dimension as the manifold) is a vector if one set of numbers is related to the other set of numbers through the linear transformation.

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## protected by Qmechanic♦May 11 at 14:43

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