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Active and passive Lorentz transformation

  1. Mar 21, 2013 #1
    Physics books rarely make the distinction between active or passive Lorentz transformations. The usual Lorentz transformations of the spacetime coordinates in two different inertial frames seem to me to be passive transformations, because by definition passive transformations are coordinates transformations; but, we also say that spacetime coordinates transform as a 4-vector (a 4-vector is by definition a collection of entries that transform just as the coordinates do) and active transformations are those transformations that act on the vectors. Grouping the four spacetime coordinates in a 4-vector seems to blur the distinction between active and passive Lorentz transformations. 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)? A Lorentz transformations act on Minkowki spacetime M, whose elements are spacetime events, right? So, where do general 4-vector live? And "how" do Lorentz transformations act on them?

    To sum it all up, the physically obvious concept of "change of reference frame" would be called active or passive by a mathematician?
     
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  3. Mar 21, 2013 #2

    WannabeNewton

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    Where did you ever see space - time coordinates being represented as vectors?

    Let ##(M.g_{ab})## be a space - time and ##p\in M## an event. There exists a chart ##(U,\varphi)## containing ##p##. The local coordinate representation or coordinate of ##p## is ##\varphi(p) = (x^0,x^1,x^2,x^3)\in \mathbb{R}^{4}##. Just as in smooth manifold texts, in GR one usually blurs the distinction between the event, which is an element of the space - time, and its coordinate representation in euclidean space.

    A tangent 4 - vector at ##p\in M## is by definition some ##v\in T_p(M)## (where ##T_p(M)## is the tangent space to ##M## at the event ##p##); an event in space - time is obviously not such a thing. Your definition of a 4 - vector makes no sense: how can a set of coordinate functions "transform", in the sense that components of vectors transform, when they are the things that are in the definition of the transformation in the first place?

    Let ##\Lambda :(\mathbb{R}^{4},\eta _{ab})\rightarrow (\mathbb{R}^{4},\eta '_{ab})## be a lorentz transformation; we know, of course, that it is an isometry of minkowski space - time. It's action on minkowski space - time is simply ##p\rightarrow \Lambda (p),p\in \mathbb{R}^{4}##. It's action on a vector is ##v\rightarrow \Lambda _*(v),v\in T_p(\mathbb{R}^{4})## where ##\Lambda _{*}## is the pushforward (using the notation of Lee). In coordinates, this is simply ##v'^{i} = \Lambda ^{i}_jv^j## where ##\Lambda ^{i}_j## is the matrix representation of ##\Lambda ## in whatever coordinate basis you are using for ##T_p(M)##. Furthermore, because ##\Lambda ## is an isometry as mentioned earlier, ##\Lambda ^{*}\eta' _{ab} = \eta _{ab}## where ##\Lambda ^*## is the pullback.
     
  4. Mar 21, 2013 #3
    Hi WN, I'm not sure what you are arguing here. See for instance this from WP:"Four-position: A point in Minkowski space is called an "event" and is described in a standard basis by a set of four coordinates ... These coordinates are the components of the position four-vector for the event."
     
  5. Mar 21, 2013 #4

    WannabeNewton

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    Well firstly, the description (I'm assuming this means coordinate representation) of an event has nothing to do with a basis, it is simply the 4 - tuple gotten from applying a coordinate map to the event. Secondly, the WP article is doing it for a very very special case in which the events in minkowski space are represented by cartesian coordinates. Even in ##\mathbb{R}^{2}## if I'm in a suitable open subset ##U\subset \mathbb{R}^{2}## and have a polar coordinate chart ##(U,\varphi )## then the coordinate representation of a point will be ##\varphi(p) = (r,\theta )##. It is basic calc 3 to know that these are not components of a position vector in such a case.
     
  6. Mar 21, 2013 #5
    We are using different languages here. I was thinking of spacetime simply as a linear space and it is in this context that the terminology active vs passive transformations arises. Also, I was secretly thinking of a vector as its components, which is nonsense in your differential geometry language.

    When you change coordinate neighbourhood around a point P, the coordinates of P, x_U, (the coordinate functions evaluated at P in the first coordinate chart) are related to the coordinates of P, x_V, (the coordinate functions evaluated at P in the second coordinate chart) just as the components of a tangent vector at that point are related in the two coordinate neighbourhoods (this transformation is simply given by the jacobian of the functions that connect the new and the old coordinates).
     
  7. Mar 21, 2013 #6
    Unless the events are considered the elements of a space like Minkowski considered as a vector space [itex]ℝ^4[/itex].
    Right, and it's this special case that the OP seems to be talking about. That was my point.

    EDIT: I see the OP replied at the same time I did, and I seemed to be reading him right.
     
  8. Mar 21, 2013 #7

    WannabeNewton

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    Well let's be careful here. This works for minkowski spacetime, not all spacetimes. What is your definition of a passive vs. active transformation in this special context (there is a notion of passive vs active in the more general context of arbitrary space - times but that requires diff geo).
     
  9. Mar 21, 2013 #8
    Hey, now you are talking buddy! :wink:
     
  10. Mar 21, 2013 #9

    Fredrik

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    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 ℝ4 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).
    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}##.

    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##.
     
  11. Mar 21, 2013 #10

    haushofer

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    I would say passive. Changing frames of reference means changing coordinates without moving the actual event. Mathematically such a transformation is a function from one chart to another chart.

    A diffeomorphism is what mathematicians would call an active transformation, I guess. That's a function from the manifold to the manifold, which actually moves the event (spacetime point).
     
  12. Mar 22, 2013 #11
    Thank you all for your answers, especially Fredrick (only thing, you have a typo in your definition of Lorentz transformation: I think you mean g(\Gamma x, \Gamma x)=g(x,x)). I have an additional question: the only definition I know of Lorentz transformations is as linear transformations in minkowski space, they are basically the same as orthogonal transformations in euclidean space, but now the signature of the metric is slighty different and all the consequences; in more general spacetimes, I thought Lorentz transformations are defined only between tangent spaces at a given event. Is there a generalization of lorentz transformations as diffeomorphism between general spacetime manifolds? Giving me some good reference on the subject would be good
     
  13. Mar 22, 2013 #12

    haushofer

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    Yes. In that case you can make the LT's general functions of your spacetime coordinates and let them act on curved indices instead of flat ones. What this physically means however is not clear to me. In Van Proeyen's notes on supergravity something similar is treated, because in obtaining GR by gauging the Poincare algebra you impose a conventional constraint to rewrite tangent space translations into general coordinate transformations and LT's with field-dependent parameters. The resulting algebra is also called a soft algebra.
     
  14. Mar 22, 2013 #13

    Fredrik

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    Oops. Yes, of course. Don't know how that happened.

    The concept of "isometry" can certainly be generalized. If M and N are smooth manifolds, and N has a metric g, a diffemorphism ##\phi:M\to N## can be used to define a metric on M called the pullback metric. It's denoted by ##\phi^*g##. If M=N, and ##\phi^*g=g##, ##\phi## is said to be an isometry. The set of all isometries of the metric is a group. When we define the spacetime of SR as a smooth manifold, the Poincaré group can be defined as the group of isometries of the metric. See posts #5 and #6 here for definitions and stuff. Sorry about the ugly layout in post #6. I think I just copied and pasted from something I had written for my personal notes and made the bare minimum of changes to get the code to work here. Hm, I see that I'm referring to something denoted by G in that post, but I don't see a definition of G. G is the group of isometries. P is the Poincaré group, defined in a way that's similar to how I tried to define the Lorentz group in this thread.
     
  15. Mar 22, 2013 #14

    atyy

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    I think the tetrad transformations are gauge transformations, so they would be "passive".

    http://casa.colorado.edu/~ajsh/phys5770_10/grtetrad.pdf [Broken]
    "Gauge transformations are transformations of the coordinates or tetrad. Such transformations do not change the underlying spacetime.

    Quantities that are unchanged by a coordinate transformation are coordinate gauge-invariant. Quantities that are unchanged under a tetrad transformation are tetrad gauge-invariant. For example, tetrad tensors are coordinate gauge-invariant, while coordinate tensors are tetrad gauge-invariant.

    Tetrad transformations have the 6 degrees of freedom of Lorentz transformations, with 3 degrees of freedom in spatial rotations, and 3 more in Lorentz boosts. General coordinate transformations have 4 degrees of freedom. Thus there are 10 degrees of freedom in the choice of tetrad and coordinate system. The 16 degrees of freedom of the vierbein, minus the 10 degrees of freedom from the transformations of the tetrad and coordinates, leave 6 physical degrees of freedom in spacetime, the same as in the coordinate approach to GR, which is as it should be."
     
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