Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

My favorite way of deriving the Lorentz transformation is to start from symmetry principles (an approach originated in Ignatowsky 1911; cf. Pal 2003), and one of my steps is to prove a lemma stating that the Lorentz transformation has to preserve volume in spacetime, i.e., in fancy language, it has to have a Jacobian determinant of 1. My visual proof of the lemma (for 1+1 dimensions) is given at the link above, in a figure and its caption.

Intuitively, the idea behind the proof is that it would be goofy if a boost could, say, double the area, since then what would a boost in the opposite direction do? If it's going to undo the first boost, it has to halve the area, but then we'd be violating parity symmetry. The actual proof is actually a little more complicated than this, though. If a rigorous proof could be this simple, I'd actually be satisfied and say the result was so simple and obvious that we should call it a day. But in fact if you want to get it right it becomes a little more involved, as shown at the link where I give the actual proof. One way to see that the intuitive argument I've given here above doesn't quite suffice is that it seems to require that area in the $x$-$y$ plane be preserved under a boost in the $x$ direction, and that's not true.

This is different from the approach based on Einstein's 1905 axiomatization. In that approach, you derive the Lorentz transformation and then prove the unit Jacobian as an afterthought.

Although my proof of the unit Jacobian from symmetry principles works, I've always felt that there must be some deeper reason or better physical interpretation of this fact. Is there?

Takeuchi 2010 says on p. 92:

This conservation of spacetime area maintains the symmetry between [...] frames, since each is moving at the exact same speed when observed from the other frame, and ensures that the correspondence between the points on the two diagrams is one-to-one."

This seems clearly wrong to me, since you can have one-to-one functions that don't preserve area. (Takeuchi is writing for an audience of liberal arts students.)

Mermin has a very geometrical pedagogy that he's honed over the years for a similar audience, and he interprets space-time intervals as areas of "light rectangles" (Mermin 1998). Since he uses the 1905 Einstein axiomatization, conservation of area, which is equivalent to conservation of spacetime intervals, comes as an afterthought.

One thing that bugs me is that the area-preserving property holds for Galilean relativity (and my proof of it holds in the Galilean case without modification). So any interpretation, such as Mermin's light rectangles, that appeals specifically to something about SR seems unsatisfying. The notion of area here is really just the affine one, not the metrical one. (In Galilean relativity we don't even have a metric.)

Another way of getting at this is that if you start with a square in the $x$-$t$ plane and apply a Lorentz transformation, it becomes a parallelogram, and the factors by which the two diagonals change are the forward and backward Doppler shifts. Conservation of area then follows from the fact that these Doppler shifts must be inverses of one another.

Laurent 2012 is an unusual coordinate-free presentation of SR. He interprets $\epsilon_{abcd}U^aB^bC^cD^d$ as a 3-volume measured by observer whose normalized velocity vector is $U$. Restricting to 1+1 dimensions, $\epsilon_{ab}U^aB^b$ is the length of vector $B$ according to this observer. He gives the example of associating affine volume with the number of radioactive decays in that volume. The implications of this are vague to me.

What is the best way of interpreting the preservation of spacetime volume by Lorentz transformation?

Please do not reply with answers that start from the known form of the Lorentz transformation and calculate the Jacobian determinant to be 1. I know how to do that, and it's not what I'm interested in.

W.v. Ignatowsky, Phys. Zeits. 11 (1911) 972

Bertel Laurent, Introduction to spacetime: a first course on relativity

Mermin, "Space-time intervals as light rectangles," Amer. J. Phys. 66 (1998), no. 12, 1077; the ideas can be found at links from http://people.ccmr.cornell.edu/~mermin/homepage/ndm.html , esp. http://www.ccmr.cornell.edu/~mermin/homepage/minkowski.pdf

Palash B. Pal, "Nothing but Relativity," http://arxiv.org/abs/physics/0302045v1

Takeuchi, An Illustrated Guide to Relativity

share|improve this question
1  
I'm not sure what it is you're asking for. It is simply mathematical fact that a Minkowski spacetime has its own special orthogonal group, which has as group members both spatial rotations and Lorentz boosts. To me, it is a physical statement to say that Minkowski spacetime can be used as a model for the real world, that objects must follow worldlines in this spacetime and, as such, the operations of the special orthogonal group are relevant and have corresponding physical interpretations. Are you asking why we should seek out such special orthogonal operators in the first place? –  Muphrid Aug 24 '13 at 3:43
1  
@Murphid: It is simply mathematical fact that... Here you're assuming that the Lorentz transformations have already been established. My interest in this fact is that in my preferred way of developing the Lorentz transformations, I use it as a way of deriving the Lorentz transformations in the first place. Essentially, what I'm asking is a "why" question, and "why" questions are always tricky, because the answer can be different depending on what you take as a prior assumption. To me, it is a physical statement to say that... Here you're stating your prior assumptions. Mine differ. –  Ben Crowell Aug 24 '13 at 4:11
1  
Read with interest your question and some of the texts linked to. I now realize my favorite way of explaining Special Relativity (representing material particles in 1+1D by zig-zag-ing light rays and using area preservation of the 'light rectangles') is by no means original. Need to get access to Takeuchi's book and Mermin's article. - On your question: a non area-preserving transformation would render the observations of two observers in relative motion no longer mutually interchangeable. –  Johannes Aug 24 '13 at 4:19
2  
@Muphrid: Pinging you because Ben Crowell spelt your name wrongly so you were probably not notified. –  Dimensio1n0 Aug 24 '13 at 4:35
3  
This question seems tangentially related to physics.stackexchange.com/questions/31534/… , in that an answer to one of them might shed some light on the other. –  Nathaniel Aug 24 '13 at 4:45
show 6 more comments

1 Answer

First, you have to distinguish mathematical proofs from physics "proofs". You can't really rigorously prove statements about the real world at the same level of rigor that one has in maths.

The unimodularity (Jacobian = 1 or -1, to agree with the usual definitions of the adjective) of the Lorentz transformations is trivial to prove in maths, whatever definition of the Lorentz transformations we adopt.

Something else is the "proof" that the transformation responsible for changing the frame in physics is unimodular. Such a "proof" has to accept some physical assumptions that are ultimately justified empirically.

The usual "proof" is simple and you sketched it at the beginning.

In the empty space, the transformation $B$ (boost) from an observer at rest to an observer moving by the speed $v$ in the $z$ direction is given by $$ (t,x,y,z)\mapsto (t',x',y',z')$$ Because of time- and space-translational symmetry, this map has to be linear (with a possible shift, i.e. an inhomogeneous term, generalizing Lorentz to Poincaré transformations). Because it is linear, its Jacobian is constant, so it is just one number-valued function of $v$.

By using the rotational symmetry, one can write $$ B(-v) = R_\pi B(v) R_\pi $$ In words, the boost by the opposite speed may be obtained by rotating the system by $\pi$ around e.g. $x$-axis, boosting it by $+v$, and rotating back. The Jacobian is a determinant so for the product transformation above, it's just the product of the determinants and the determinants for the rotations are one (one could have also used parity so that the signs from these two determinants would cancel as well but I chose to avoid negative-determinant transformations).

It follows that $$\det B(-v) = \det B(v)$$ i.e. the determinant is an even function of $v$. At the same moment, the boost by $-v$ is nothing else than the transformation switching back to the original frame, i.e. $$B(-v) = B(v)^{-1}$$ which also means (in combination with the previous identity, to eliminate $\det B(-v)$) $$\det B(v) = \det B(v)^{-1}$$ and the determinant is therefore $\pm 1$. Because it's $1$ for $v=0$ (identity) and it is a continuous function, we must have $\det B(v)=1$ for all $v$.

share|improve this answer
    
Well, it is true for Galilean transformations also. In fact, looking at a boost in the $z$ axis, linearity and rotational invariance implies a transformation belonging to a continuous 1-parameter subgroup of $SL(2,R)$. They are known and listed. The distinction between these different subgroups have to be done with other supplementary basic axioms, and with physical principles (for instance the energy must be positive or the speed of light is always the same) –  Trimok Aug 24 '13 at 9:05
    
Nice. This is similar to but different from my proof, which involves doing a boost of a square, dissecting the resulting parallelogram into small squares, and then doing the reverse boost on the squares. –  Ben Crowell Aug 24 '13 at 13:04
    
one could have also used parity [rather than rotations]. An advantage of using parity is that the argument is then valid in 1+1 dimensions. Time reversal also works. –  Ben Crowell Aug 24 '13 at 21:16
    
Agreed, @Ben. The advantage of using rotation is that the theory may be parity-violating. And even in 1+1 dimensions, one may involve the 180-degree-like rotations directly although the rotations by less degrees aren't doable. Such 180-degree rotations appear in the CPT theorem - even though they are disconnected from the identity. –  Luboš Motl Aug 25 '13 at 8:00
    
Hi Luboš Motl, I think that in your last sentence you meant “because it's $1$ for $v=0$...” –  pppqqq Jan 16 at 21:55
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.