The spacetime of SR can be defined in at least three different ways, as a vector space, as an affine space, or as a smooth manifold. This give us three slightly different versions of the theory, but they're only different in the sense that the assumptions that define the theory are different. They all make the same predictions about results of experiment.
The definition of "four-vector" and "Lorentz transformation" depends on this choice. Let's go with the simplest option. We define spacetime as the set ℝ
4 with the standard vector space structure, and the bilinear form g defined by
$$g(x,y)=x^T\eta y$$ for all ##x,y\in\mathbb R^4##. The world line of a particle is a curve ##C:(a,b)\to \mathbb R^4##. The particle's four-velocity is just KC', where C' is the derivative of C, and K>0 is such that
$$g(KC'(t),KC'(t))=-1.$$ So for any point p on the world line, the four-velocity at p is KC'(t), where t is the unique real number in the interval (a,b) such that C(t)=p. Note that both C(t) and KC'(t) are members of ##\mathbb R^4##.
A Lorentz transformation is a linear transformation ##\Lambda:\mathbb R^4\to\mathbb R^4## such that ##g(\Lambda x,\Lambda x)=0## for all ##x\in\mathbb R^4##. It can be shown that ##\Lambda## is a Lorentz transformation if and only if ##\Lambda^T\eta\Lambda=\eta##.
These are some things I've said about active and passive transformations in other posts: (I'm using the convention to not write any summation sigmas, since we can remember to always sum over those indices that appear twice).
Fredrik said:
The Wikipedia article uses rotations of ##\mathbb R^2## to illustrate what they mean by active and passive transformations. For example, ##x\mapsto Rx## is considered an active transformation (by the rotation R) of the components of x, while a passive transformation of the components of x is the transformation ##x_i\mapsto x_i'## where the ##x_i'## are defined by ##x=x_ie_i=x_i'Re_i##. It's not hard to turn this into a formula for ##x_i'##. First let's expand ##Re_i## in basis vectors:
$$x_i'Re_i=x_i'(Re_i)_je_j=x_i'R_{ji}e_j.$$ This is equal to ##x_je_j##, so we must have
$$x_j=R_{ji}x_i'.$$ Multiply by ##(R^{-1})_{kj}## (and sum over j).
$$(R^{-1})_{kj}x_j=\delta_{ki} x_i'=x_k'.$$ So we have
$$x_i'=(R^{-1})_{ij}x_j$$ for the passive transformation, and
$$x_i'=R_{ij}x_j$$ for the active transformation. This makes "passive transformations" of component matrices a pretty useless concept in my opinion. A passive transformation (of the component matrix) by R is just an active transformation by R-1.
Fredrik said:
If you need a more intuitive way to think about these things, I suggest that you think of an active transformation by a rotation matrix R as a physical rotation by R (say a counterclockwise rotation by an angle of π/4) of the object on which we're going to do measurements, and the corresponding passive transformation by R as a physical rotation by R-1 (a clockwise rotation by π/4) of the labratory around the object (while the object is held fixed relative to the Earth).
In both cases, there's a physical change. The point is that the changes are equivalent, a far as physics experiments are concerned (unless of course we're doing experiments with something like a compass needle; in those cases, you have to imagine these things taking place in intergalactic space or something).
In the passive case, the orientation of the object relative to the Earth (or some other fixed stuff outside the laboratory) doesn't change. But we would still change our description of its orientation, if we describe it relative to the walls of the laboratory (the new basis vectors). In the active case, our description of the orientation of the object relative to the walls changes in exactly the same way as in the passive case.
Edit: Note that this last bit is consistent with what I've been saying about active and passive transformations of a coordinate 4-tuple ##(x_\mu)_{\mu=0}^3##:
Active transformation by ##\Lambda##: ##x^\mu\to\Lambda^\mu{}_\nu x^\nu##.
Passive transformation by ##\Lambda##: ##x^\mu\to(\Lambda^{-1})^\mu{}_\nu x^\nu##.
Passive transformation by ##\Lambda^{-1}##: ##x^\mu\to\Lambda^\mu{}_\nu x^\nu##. (This is the same as the active transformation by ##\Lambda##, as suggested by the informal argument above).
Note in particular the suggestion about how to think about active and passive transformations, and the fact that an active transformation by ##\Lambda## is a passive transformation by ##\Lambda^{-1}##.
SpinorData said:
Also, what do we mean precisely when we say that energy and momentum have the same transformation property of the coordinates (i.e. form a 4-vector)?
My experience is that the people who use this terminology don't fully understand it themselves. That's how confusing it is. I think the terminology is also completely useless. The idea is that if we associate a 4-tuple of real numbers
with each ordered basis for ##\mathbb R^4##, then you can think of the 4-tuple associated with one basis as a "transformed" version of the 4-tuple associated with another basis. Let ##\langle e_\mu\rangle_{\mu=0}^3## be an arbitrary ordered basis. Let ##\Lambda## be an arbitrary Lorentz transformation. Since ##\Lambda## is invertible, ##\langle \Lambda e_\mu\rangle_{\mu=0}^3## is an ordered basis too. Now define ##e_\mu'=\Lambda e_\mu##, and denote the 4-tuples associated with these two ordered bases by ##(v^0,v^1,v^2,v^3)## and ##(v'^0,v'^1,v'^2,v'^3)## respectively. If ##v'^\mu=(\Lambda^{-1})^\mu{}_\nu v^\nu## where ##(\Lambda^{-1})^\mu{}_\nu v^\nu## is row ##\mu##, column ##\nu## of the standard matrix representation of ##\Lambda^{-1}##, then it makes some sense to call the association of 4-tuples of real numbers with ordered bases a
contravariant 4-vector.
Unfortunately the people who use this terminology have the nasty habit of calling the specific 4-tuple ##(v^0,v^1,v^2,v^3)## a 4-vector. This doesn't make any kind of sense. If anything should be described as a 4-vector because if its transformation properties, it's the function that associates 4-tuples with ordered bases.
It's easy to see that each member of ##\mathbb R^4## can be used to define a contravariant 4-vector in this sense. Let ##v\in\mathbb R^4## be arbitrary. For each ordered basis ##\langle e_\mu\rangle_{\mu=0}^3##, there's a unique 4-tuple of real numbers ##(v^0,v^1,v^2,v^3)## such that ##v=v^\mu e_\mu##. This is an association of 4-tuples with ordered bases. Now you just need to verify that if we define ##e_\mu'## and ##v'^\mu## as above, then ##v'^\mu=(\Lambda^{-1})^\mu{}_\nu v^\nu##.
