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".
 A: 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.
A: 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.


*

*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 $p^a$.

*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.

*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.

*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: $k$-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.
