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I have a question about specifically whether the components of a 4-vector could depend on the position $x \in \mathcal{R}^4$, where I denote Minkowski space with $\mathcal{R}^4$. I know that the components of a 4-vector transform according to the following rule:

$(v^\prime)^i=\Lambda^i_j v^j$ with $v=v^ie_i = (v^\prime)^i e_i^\prime \in \mathcal{R}^4$ where $e_i, e_i^\prime$ are basis vectors along $x^i,(x^\prime)^i$ respectively and $\Lambda^i_j$ the component of a Lorentz transformation.

Considering $\mathcal{R}^n$ and some vector field $v=v^i \partial_i$ we could have that $v^i=v^i(x), x \in \mathcal{R}^n$, where I use the tangent basis $\{\partial_i\}$. For Minkowski space this basis is at every point the same so we can just use a standard basis.

Are four vectors always called vectors also when the components depend on position (like for example the position 4-vector) or are the components except for the position vector not a function of $x$? And are there examples of four-vectors (except position) where the components depend on the position?

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A 4-vector (or more generally 4-tensor) depending on the position variable is called a 4-tensor field. Those are very common objects.

For example the 4-potential $A^\mu$ is a tensor field, in Gaußian units it has the components $\big(A^\mu(x)\big) = \big(\phi(x), \vec A(x)\big)$, which depend on the space-time position. In the Lorenz gauge $\partial_\mu A^\mu = 0$ its equation of motion is $\partial^\nu\partial_\nu A^\mu = \frac{4\pi}{c} j^\mu$, where $j^\mu$, the 4-current density, is another tensor field.

As another example, consider a particle in some force field, the right hand side of the relativistic equation of motion will be a 4-vector whose components depend on the space time coordinates: $$ \dot p^\mu = K^\mu(x). $$

As a note of caution: Actually $j^\mu$ is a tensor density, which is a generalization of a tensor field. This gets relevant once you work in coordinates that are not orthonormal or in curved space-time (Wikipedia has more on electrodynamics in curved space time).

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  • $\begingroup$ Thank you very much. This makes it clear. I now understand that there are 4-vectors that can depend on the space time coordinates, and if this is the case we call them a 4-vector field (or tensor field). So still one question just to be sure that I understand it right. If one writes an equation like $(v^\prime)^i = \Lambda^i_j v^j$ they implicitly know whether $v^i=v^i(x)$ and if $v^i$ are constant then this still holds? An analogy that I could make is a constant function. One writes $f(x)$ for some function $f: \mathcal{R} \rightarrow \mathcal{R}$ with specifiying whether $f(x)$ is constant. $\endgroup$ – Dani Aug 13 '18 at 11:54
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    $\begingroup$ Yes, the transformation equations have the same form for 4-vectors and 4-vector fields (as soon as we leave Minkowski space time we can no longer speak about global 4-vectors, since the parallel transport of vectors along curves is no longer unique independently of the curve). $\endgroup$ – Sebastian Riese Aug 13 '18 at 12:07
  • $\begingroup$ Thanks Sebastian you did a great job explaning this material! $\endgroup$ – Dani Aug 13 '18 at 12:16
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The position relative to an arbitrary origin is not a four-vector, because a simple translation changes its length.

However, a displacement vector, pointing from one event to another, is a four-vector.

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    $\begingroup$ The 4-position relative to an arbitrary 4-origin is a 4-vector. $\endgroup$ – Frobenius Aug 13 '18 at 11:03

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