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Possible mistake in an article (rotations and boosts). 
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#1
Jan713, 07:50 PM

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PF Gold
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This is a linear algebra question, but it's about an article about Minkowski spacetime, so I think it's appropriate to post it here. The article is The rich structure of Minkowski space by Domenico Giulini. The detail I'm asking about is at the top of page 16.
The article is describing the 3+1dimensional version of the "nothing but relativity" argument that's been discussed here in a few threads recently. (The idea is to prove that the group of functions that make a coordinate change from one global inertial coordinate system to another, is either the group of Galilean boosts or the Lorentz group. So at the start, we do not assume that spacetime is Minkowski spacetime. We just assume that spacetime is some structure with underlying set ℝ^{4}). The article assumes that the group has two subgroups, one corresponding to rotations, and one corresponding to boosts. The rotations are 4×4 matrices $$R(D)=\begin{pmatrix}1 & 0\\ 0 & D\end{pmatrix},$$ where the zeroes are a 3×1 matrix and a 1×3 matrix, and D is a member of SO(3). The boosts can be expressed as a function of velocity, and the relationship between boosts and rotations is assumed to be $$B(Dv)=R(D)B(v)R(D^{1}).$$ Now the author claims that by chosing v to be a multiple of e_{1}, and D to be an arbitrary rotation around the 1 axis, we can see that $$B(v)=\begin{pmatrix}A & 0\\ 0 & \alpha I\end{pmatrix},$$ where A is a 2×2 matrix, I is the 2×2 identity matrix, and ##\alpha## is a real number. This result looks wrong to me. I want to know if I'm missing something. So here's my argument: First write $$B(v)=\begin{pmatrix}K & L\\ M & N\end{pmatrix}.$$ We have $$R(D)=\begin{pmatrix}I & 0\\ 0 & D'\end{pmatrix},$$ where I is the 2×2 identity matrix and D is a member of SO(2). So $$R(D^{1})=\begin{pmatrix}I & 0\\ 0 & D'^{1}\end{pmatrix}$$ and \begin{align}\begin{pmatrix}K & L\\ M & N\end{pmatrix} &=B(v)=B(Dv)=R(D)B(v)R(D^{1})\\ &=\begin{pmatrix}I & 0\\ 0 & D'\end{pmatrix}\begin{pmatrix}K & L\\ M & N\end{pmatrix} \begin{pmatrix}I & 0\\ 0 & D'^{1}\end{pmatrix} =\begin{pmatrix}K & LD'^{1}\\ D'M & D'ND'^{1}\end{pmatrix} \end{align} Now it's easy to see that L=M=0. For example, we have M=D'M for all D' in SO(2), and if we e.g. choose D' to be a rotation by ##\pi/2##, we can easily see that M=0. However, the same choice of D in the equation for N yields that N is of the form $$N=\begin{pmatrix}a & b\\ b & a\end{pmatrix},$$ and this is a number times a member of SO(2), but that member doesn't have to be the identity. And I don't think that there's a way to get b=0 by choosing another D', because ##D'ND'^{1}## will just be (a number times) a product of 3 rotations by angles θ,λ,θ that add up to λ, and this turns the equation ##N=D'ND'^{1}## into N=N, which tells us nothing. 


#2
Jan813, 02:44 AM

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I think you're right  the paper is missing something. This type of derivation usually proceeds by assuming that the spatial axes of both frames (anchored at the common spatiotemporal origin) are adjusted to coincide.
So spatial isotropy is used twice  in different ways: firstly to get rid of L and M completely, by insisting that those 2D rotations must be an automorphism of the group. But then, one says that we can multiply the B(v) matrix by a matrix consisting of 1 and ##N^{1}##, and silently redefine the product matrix as B(v) .... :) This is what one means by "adjusting the transverse axes to coincide in both frames". 


#3
Jan813, 11:04 AM

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Thank you. By the way, if we're allowed to choose D' as a reflection instead of a (proper) rotation, we can easily get the result that N is a number times the identity matrix. But the author doesn't seem to assume any kind of reflection invariance.



#4
Jan813, 06:36 PM

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Possible mistake in an article (rotations and boosts).



#5
Jan913, 12:58 AM

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what's the idea behind it? take all Lie groups with 4dim. rep. and "derive" in some way that spacetime symmetry is Lorentz symmetry? this is strange for several reasons:
 it misses Poincare invariance  it misses lessons from GR where some/all these symmetries become local gauge symmetries  it misses diffeomorphism invariance  there is no mathematical reason to single out SO(3,1), so everything is contained in additional assumptions; are there any guiding principles not already using the result SO(3,1)? 


#6
Jan913, 01:48 AM

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I guess you didn't follow the previous thread(s) where Fredrik and I discussed this subject. Anyway, the focus is on deriving the maximal group (of coordinate transformations) that preserves inertial motion (i.e., unaccelerated observers) from the relativity principle alone, i.e., without the light principle. It doesn't miss Poincare invariance. GR is not relevant here since we're only talking about inertial motion. [Edit: the older thread is: http://www.physicsforums.com/showthread.php?t=651640 ] 


#7
Jan913, 08:13 AM

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The idea is to see what the principle of relativity says about theories of physics in which space and time are represented by ##\mathbb R^4## equipped with global inertial coordinate systems. (Yes, these are theories that are less sophisticated than GR, but the point isn't to find exciting new theories. It's to explain what the less sophisticated ones have in common, and to show the power of the principle of relativity). No assumptions are made about those coordinate systems other than that the functions (permutations of ##\mathbb R^4##) that change coordinates from one global inertial coordinate system to another have properties that are suggested by the principle of relativity, and the idea that an inertial coordinate system takes the world line of a nonaccelerating particle to a straight line. The most important assumption is that these functions form a group. The claim is that a small set of additional "inspired by the principle of relativity" assumptions imply that this group is either the Galilean group or (isomorphic to) the Poincaré group.
One of the assumptions is that these functions take straight lines to straight lines. (This one isn't inspired by the principle of relativity. Instead it should be thought of as part of what we mean by "inertial coordinate system"). This implies that if T is such a function, there's a linear ##\Lambda## and a vector y such that ##T(x)=\Lambda x+y##. (To prove this is by far the hardest part of the argument). The most important of the other assumptions is that the set of all of these functions form a group. Because of the above, the set of linear functions in that group is a subgroup. What we are talking about is how to find that subgroup. 


#8
Jan913, 09:42 AM

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The task is easier when we take spacetime to be ℝ^{2} instead of ℝ^{4}. I will show one way to find the subgroup that consists of linear proper orthochronous coordinate transformations. I'm confident about the calculation, but I'd like to discuss the assumptions. I'll explain what I want to discuss later, in another post.
First note that if $$\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix},$$ then $$\Lambda^{1}=\frac{1}{adbc}\begin{pmatrix}d & b\\ c & a\end{pmatrix}.$$ The velocity associated with ##\Lambda## can be determined by examining what ##\Lambda^{1}## does to a point on the time axis. I'm using the convention that the time coordinate is the upper one. Since $$\Lambda^{1}\begin{pmatrix}1\\ 0\end{pmatrix}=\begin{pmatrix}(\Lambda^{1})_{00}\\ (\Lambda^{1})_{10}\end{pmatrix},$$ the velocity of ##\Lambda## is $$\frac{(\Lambda^{1})_{10}}{(\Lambda^{1})_{00}}=\frac{c}{d}=\frac{\Lambda_{10}}{\Lambda_{11}}.$$ Similarly, the velocity of ##\Lambda^{1}## is c/a. Let G be a nontrivial subgroup of GL(ℝ^{2}) such that the following statements are true.
Let ##\Lambda\in G## be arbitrary. Assumption 3 says that $$\frac{c}{a}=v(\Lambda^{1})=v(\Lambda)=\frac{c}{d}.$$ So if c≠0, then d=a. However, if c=0, then ##v(\Lambda)=0##, and now assumption 4 says that ##\Lambda=I##, which implies that d=a=0. So we always have d=a. Define ##\gamma=a##, ##v=c/a## (with apologies for giving the symbol v a second meaning) and ##\alpha=b/a##. Note that assumption 3 implies that ##v=v(\Lambda)## (where the v on the left is the new one, and the v on the right is the old one). We have $$\Lambda=a\begin{pmatrix}1 & b/a\\ c/a & d/a\end{pmatrix}=\gamma\begin{pmatrix}1 & \alpha\\ v & 1\end{pmatrix}.$$ For all ##\Lambda',\Lambda''\in G##, $$G\ni \Lambda'\Lambda'' =\gamma'\gamma''\begin{pmatrix}1 & \alpha'\\ v' & 1\end{pmatrix}\begin{pmatrix}1 & \alpha''\\ v'' & 1\end{pmatrix} =\gamma'\gamma''\begin{pmatrix}1\alpha'v'' & \alpha''+\alpha'\\ v'v'' & v'\alpha''+1\end{pmatrix}.$$ This implies that ##1\alpha'v''=v'\alpha''+1##. Note that since this holds for all ##\Lambda''## including one with ##v''\neq 0##, it implies that if ##v'=0##, then ##\alpha'=0##. We will use this in a minute. Let ##\Lambda',\Lambda''## be arbitrary members of ##G## such that ##v'\neq 0## and ##v''\neq 0##. Now the result ##1\alpha'v''=v'\alpha''+1## implies that $$\frac{\alpha''}{v''}=\frac{\alpha'}{v'}.$$ This means that ##\alpha'/v'## has the same value for all ##\Lambda'\in G##. Denote this value by K. We have ##\alpha'=Kv'## for all ##\Lambda'\in G##. So $$\Lambda=\gamma\begin{pmatrix}1 & Kv\\ v & 1\end{pmatrix},\quad\Lambda^{1}=\frac{1}{\gamma(1Kv^2)}\begin{pmatrix}1 & Kv\\ v & 1\end{pmatrix}.$$ Assumption 5 tells us that $$\gamma=\frac{1}{\gamma(1Kv^2)}.$$ Note that if K>0, this implies that ##1Kv^2>0##. If we define ##c=1/\sqrt{K}##, this is equivalent to ##v\in(c,c)##. Assumption 1 implies that ##\gamma>0##, so we have $$\gamma=\frac{1}{\sqrt{1Kv^2}}.$$ When K=0, ##\Lambda## is a proper and orthochronous Galilean boost. When ##K=1##, ##\Lambda## is a proper and orthochronous Lorentz boost. A few things that I didn't show here: We must have K≥0, because K<0 contradicts that G is a group or that all its members are orthochronous. For all K≥0, let ##G_K## be a set whose members are all the "Lambdas" of the form found above, with all the velocities that are consistent with the value of K. Then ##G_K## is a group that satisfies the assumptions. For all K>0, ##G_K## is isomorphic to ##G_1##, i.e. the restricted Lorentz group. 


#9
Jan913, 12:40 PM

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Here's what I want to discuss:
A. It's not clear to me what assumptions about space we're using here. Just reflection invariance? Are we also using translation invariance? Consider two identical point particles with different constant velocities in otherwise empty (1dimensional) space. Is it obvious that if the velocity of A in a coordinate system comoving with B is v, then the velocity of B in a coordinate system comoving with A is v if the systems have the same orientation and v if they have opposite orientations? B. If we consider the group that includes reflections of space or time instead of its restricted subgroup, we will get results like d=±a instead of d=a, and these extra minus signs will make the calculation awkward. So maybe the easiest way to find the group is to start by defining the timereversal operator T and the parity operator P, and prove that any member is equal to a product of T's and P's and a member of the restricted subgroup. Then we find the restricted subgroup as above. C. I'm undecided. I would say that the argument doesn't help us understand Galilean spacetime, and that it doesn't help us understand Minkowski spacetime. It only helps us understand how the two can be viewed as different versions of the same thing. I think that's interesting, but the argument would be much cooler if a person who doesn't know SR could have come up with it, and I doubt it. Edit: When I referenced "assumption 4" in three different places under C above, I really meant "assumption 5". I'm sorry about that. 


#10
Jan913, 05:16 PM

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That's indeed used as an assumption in most published treatments I've seen, but it's unnecessary. If one merely assumes that there is some parameter ##v^## that corresponds to the inverse transformation, then it possible to deduce (in this case) that ##v^ =  v##. One need simply multiply the forward and inverse transformations, insist that the result is the identity, and then equate coefficients of like powers of the coordinates in the resulting equation to get several constraints, one of which turns out to be ##v^ =  v##. But (potentially) more useful is to see how the group theoretic approach can reveal unexpected invariant constant(s) in more general cases. 


#11
Jan913, 06:21 PM

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1. This one is just saying that I have decided to find the restricted (=proper and orthochronous) subgroup first. 2. Every ##\Lambda## has a finite velocity. 3. If the velocity of ##\Lambda## is v, then the velocity of ##\Lambda^{1}## is v. 4. The only ##\Lambda## with velocity 0 is the identity. 5. Every ##\Lambda## has the same 00 (top left) component as its inverse. 


#12
Jan913, 06:54 PM

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##\begin{pmatrix}K & L\\ M & N\end{pmatrix} = \begin{pmatrix}K & LD'^{1}\\ D'M & D'ND'^{1}\end{pmatrix}## must hold for arbitrary ##D'## in SO(2). You've just chosen particular values, which does not tell us anything. The way to proceed from here is to note that ##L=LD'^{1}## is satisfied for arbitrary ##D'## in SO(2) only for ##L=0##, since in particular the equivalence after rotation by π requires ##L=L##. Likewise for ##M##. On the other hand, ##N = D'ND'^{1}## does require that ##N## is a multiple of the unit matrix, as the article says. This follows from Schur's lemma, since ##N## commutes with every group representative of SO(2) in an irreducible representation. So the author's statement is correct. When you said that a rotation by π/2 requires: ##N=\begin{pmatrix}a & b\\ b & a\end{pmatrix}## which is not a multiple of the unit matrix, that is strictly true—but you are not taking into account all the other restrictions imposed by the other possible choices of rotation angle. 


#13
Jan913, 07:58 PM

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I did realize that these equalities hold for arbitrary D' in SO(2). That implies that they hold for the specific choices of D' that make the calculations easy. So let's write $$L=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$$ and choose D' to be a a rotation by ##\pi/2##. We have $$\begin{pmatrix}a & b\\ c & d\end{pmatrix}=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}=\begin{pmatrix}b & a\\ d & c\end{pmatrix},$$ and now we immediately see that a=b, b=a, c=d, and d=c. These results imply that a=b=c=d=0, and therefore that L=0. $$N=\begin{pmatrix}a & b\\ b & a\end{pmatrix}=s\begin{pmatrix}\frac a s & \frac b s\\ \frac{b}{s} & \frac a s\end{pmatrix}.$$ In particular, this holds when ##s=\sqrt{a^2+b^2}##. With this choice of s, the righthand side above is s times a matrix whose rows are orthonormal. This matrix is in SO(2), so it's a rotation by some angle λ. So we can write ##N=sR(\lambda)##, where ##R(\lambda)## denotes a rotation by λ. D' is a rotation by an arbitrary angle ##\theta##. We are free to choose that angle as we see fit, but no matter what we choose, we will always have $$N=D'ND'^{1}=R(\theta)sR(\lambda)R(\theta)=sR(\theta+\lambda\theta)=sR(\lambda)=N.$$ My interpretation of this was that no choice of θ can give me additional information about the components of N. Maybe that's just the wrong conclusion. I looked at the Wikipedia article on Schur's lemma (link) and if I understand it correctly, it does imply that this N is a multiple of the identity. Edit: I've been playing around with Wolfram Alpha a bit. It's pretty cool that I can check my (partial) result for N simply by typing
$$\begin{pmatrix}a & b\\ b & a\end{pmatrix}$$ every time, and this tells me nothing. I also see that Wikipedia requires the vector space to be complex. These things have brought me back to thinking that Giulini got that detail wrong in his article. Maybe Schur's lemma simply doesn't apply here? 


#14
Jan913, 09:06 PM

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Yes, it looks like you're right. I'm not exactly sure how Schur's lemma requires the complex field as it seems pretty generally stated. But clearly something is wrong. Either way, I think it's a fair wager that this is exactly the mistake the author made.



#15
Jan913, 09:42 PM

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OK, thank you LastOneStanding.
Back to the discussion about the assumptions that go into the "nothing but relativity" argument... I have realized that one thing that I've been taking for granted is questionable. Does the set of orthochronous transformations really form a subgroup? How do we even define "orthochronous" here? How about this? ##T:\mathbb R^n\to\mathbb R^n## is said to be orthochronous if ##(T(x))_0>(T(y))_0## for all ##x,y\in\mathbb R^n## such that ##x_0>y_0##. I don't see how this implies that the set of orthochronous maps is closed under composition. This is a bit tricky even when we know that the group is the Lorentz group. I did that proof here. Unfortunately it's an old thread with several LaTeX errors (which were unnoticeable before the upgrade to MathJax). Maybe there's no easier way to prove that there's a proper subgroup and an orthochronous subgroup, than to first determine the full group and then use that result to prove that the group has these subgroups. If that's the case, the list of assumptions will have to be changed. Edit: The calculation that attempts to find the full group right away is a lot uglier. But I just realized that we can appeal to the principle of relativity and make the existence of an orthochronous subgroup one of our assumptions. If Alice agrees with Bob about the temporal order of any two events, and Bob agrees with Charlie about the temporal order of any two events, then Alice should agree with Charlie about the temporal order of any two events. 


#16
Jan913, 10:55 PM

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#17
Jan913, 11:37 PM

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OK, here's another attempt. (Just a sketch. I'm leaving out some details). We're looking for a nontrivial group G that's a subgroup of GL(ℝ^{2}).
Assumption: There's a ##v:G\to\mathbb R## such that ##v(\Lambda)=\frac{\Lambda_{10}}{\Lambda_{11}}## for all ##\Lambda\in G##. This one implies that both ##\Lambda_{11}## and ##\Lambda_{00}## are nonzero. (The latter because ##\Lambda_{00}=(\det\Lambda)(\Lambda^{1})_{11}##). Assumptions: ##(\Lambda^{1})_{00}=\Lambda_{00}## and ##(\Lambda^{1})_{11}=\Lambda_{11}##. Let ##\Lambda\in G## be arbitrary, and denote its components by a,b,c,d, as in post #8. The assumptions imply that ##\det\Lambda=\pm 1## and that ##d=(\det\Lambda)a##. It's now trivial to prove that ##G_p##, defined by ##G_p=\{\Lambda\in G\det\Lambda>0\}##, is a subgroup of G. Assumption: The set ##G_o## defined by ##G_o=\{\Lambda\in G\Lambda\text{ is orthochronous}\}## is a subgroup of ##G##. The intersection of two subgroups is a subgroup, so the set H defined by ##H=G_o\cap G_p## is a subgroup. Let ##\Lambda## be an arbitrary member of H, and denote its components by a,b,c,d as above. We have d=a, and it's easy to show that a>0. (Use that ##\Lambda## is orthochronous). $$\Lambda=a\begin{pmatrix}1 & b/a\\ c/a & 1\end{pmatrix}.$$ We define ##\gamma=a## and ##\alpha=b/a##. Since $$\frac{c}{a}=\frac{c}{d}=v(\Lambda),$$ it's convenient to also define v=c/a. Now we have $$\Lambda=\gamma\begin{pmatrix}1 & \alpha\\ v & 1\end{pmatrix}.$$ From here, everything is the same as in post #8. We use that H is closed under matrix multiplication to prove that ##\alpha/v## has the same value for all ##\Lambda\in H## with v≠0. If we denote that value by K, we have ##\alpha=Kv##. Then we use the assumption about the 00 component to determine ##\gamma##. This time, I can prove ##v(\Lambda^{1})=v(\Lambda)## as a theorem, but only because I added another assumption, the one about the 11 component. Apparently this has also enabled me to drop the assumption that the only member of H with zero velocity is the identity. 


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