Lorentz transformation aims at an invariant speed of light for all reference frames.

Speed only involves x and t. Therefore it's reasonable to understand the Lorentz transform can act on spacetime four vector (ct,x1,x2,x3)

But speed does not directly involves other physical quantities such as E field, B field, current density and so on. My questions is why do we need these four vector covariant under Lorentz transform?

For example:

$x_{\mu }x^{\mu}=const$ directly involves space time

$\partial_{\mu }{\color{Red} j^{\mu }}=0$ indirectly involves space time

$\partial_{\mu }\partial ^{\mu }{\color{Red}A^{\nu} }= \frac{4\pi}{c}{\color{Red} j^{\nu }}$ does not involve space time (if only judge from this equation, we can conclude that A and j obeys Lorentz transform rule)

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    $\begingroup$ The vectors themselves may not depend on space/time, but their components depend on your choice of coordinates. $\endgroup$ – Javier May 17 '14 at 3:21
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    $\begingroup$ Rethink please: in terms of three vectors. Is the electric field a three vector? etc? Why? because it has three components E_x,E_y,E_z in euclidean space. You just add an E_t for the pseudo-euclidean space where the Lorenz transformations hold, instead of the Galilean ones. en.wikipedia.org/wiki/Galilean_transformation $\endgroup$ – anna v May 17 '14 at 3:26
  • $\begingroup$ "Rotation in $xy$-plane only involves $x$ and $y$. Therefore, it's reasonable to understand the rotation transformation can act on a spatial three-vector. But rotation doesn't directly involve other physical quantities like velocity or current density. So, why do we need these three-vectors to be covariant under rotation?" $\endgroup$ – Muphrid May 17 '14 at 3:47
  • $\begingroup$ @annav Having three components isn't enough of a justification to be a three-vector. A column of three scalars doesn't represent a three-vector, for example. $\endgroup$ – Ruslan May 17 '14 at 7:39
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    $\begingroup$ @Ruslan If you define it to, then it does. The electric field is a three vector by definition in euclidean space a four vector in lorenz pseudoeuclidean space $\endgroup$ – anna v May 17 '14 at 7:51

Other 4-vectors not necessarily true

To be sure, if it is a four-vector, it necessarily transforms by a Lorentz transformation in the way as the displacement four-vector.

If it doesn't transform by a Lorentz transformation in this way, it is necessarily not a four-vector.

Now, imagine a scalar field $\phi(x^{\alpha})$ defined on spacetime; a rule that assigns a number to each event. A scalar field on spacetime is Lorentz invariant.

It's trivial to construct a four-vector field from this scalar field as so

$$v^{\alpha} = \partial^{\alpha}\phi(x^{\alpha})$$


$$\partial^\alpha \ = \left(\frac{1}{c} \frac{\partial}{\partial t}, -\nabla \right)$$

And there you have it - an infinity of four-vectors, one at each event in spacetime, that aren't obviously involving speed.


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