# How to prove a 4D vector is a 4-Vector?

This is a fairly open ended question.

Given a set of 4 Components, that is, a 4D Vector, what is the process for determining rather or not it is a "4-Vector" as defined in special relativity? I want to know the general method for answering this question. It is still unclear to me what additional constraints make 4-vectors "special".

• If a vector is a vector if and only if it satisfies all properties of a vector, then a 4-vector... – FGSUZ Jun 7 at 22:20
• 4-vectors must transform with the Lorentz transformation matrix between reference frames. – gented Jun 7 at 22:26

## 3 Answers

It is a 4-vector if and only if the components change by a Lorentz transformation when you pass from the coordinate system of one inertial frame to another.

However, if all you have is the 4 components, then you do not know how it transforms: you need further information.

The further information can come in several forms. Here are some that would suffice. The first three are mathematical, the fourth is physical.

1. You can write an expression showing that your 4-component object can be written in terms of something else that you already know to be a 4-vector. e.g. four-velocity and four momentum: $$u^a = dx^a/d\tau, \;\;\;\;\; p^a = m u^a$$ if we already know $$dx^a$$ is a 4-vector and $$d\tau$$ and $$m$$ are invariants, then we have 4-vector $$u^a$$ and $$p^a$$.

2. Item 1 can be expanded more generally by using the quotient rule. If you have a valid tensorial expresson (i.e. one that obeys the rules of tensor analysis) in which everything else except your 4-component object is a tensor of some rank or other, and you know the expression holds in all coordinate systems, then your object is a 4-vector.

3. You write your components in some frame of your choosing, and then you simply announce that it is a 4-vector by definition. The trouble with this method is that you may thus obtain a 4-vector of limited use in the rest of physics. However it can work out nicely sometimes. An example is the flux 4-vector $$j^a$$. There is a natural frame to pick: the rest frame of a fluid element. You give to $$j^a$$ the components $$(\rho_0 c,0,0,0)$$ in that frame where $$\rho_0$$ is proper density. Then you announce it is a 4-vector. Then your next job is to find out the physical meaning of the components in a general frame. In this example there is a natural physical meaning if it is the flux of something such as electric charge which is itself a scalar invariant. See a good textbook for more information. Actually this 4-vector can also be written $$\rho_0 u^a$$ so that is another way to prove it is a 4-vector, but the job of working out the physical significance of the components remains.

4. You have a physical argument to show that the quantities you are considering are indeed the components of a 4-vector because they must transform the right way. For example, you can, with a bit of care, argue that the phase $$\phi$$ of an oscillation of a wave will be Lorentz-invariant (and therefore a tensor of rank zero). You then show that this same phase can be written $$\phi = k^\mu x_\mu$$ where $$x^a$$ is a position 4-vector. You may then deduce that $$k^a$$ is a 4-vector.

A set of 4 components, that is a 4-dimensional vector, is a 4-vector if it is transformed as the position 4-vector between reference frames.

Related : Transformation of 4−velocity.

$$k$$-vectors represent physically relevant things: displacement, velocity, orientation, current, force, et al. What physically relevant thing does your list of four components represent? I ask, because whether you have a 4-vector depends on how the physically relevant thing you are representing transforms.

There is no way to go from a list of four numbers to the physically relevant thing it transforms, so there is no way to start from a list of four numbers and discern whether the physically relevant thing it represents transforms as a $$4$$-vector. This transformation property is a property of the physically relevant thing, not of the list of numbers.