Showing that Lorentz transformations are the only ones possibleby bob900 Tags: lorentz, showing, transformations 

#91
Nov2212, 03:30 AM

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PF Gold
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#92
Nov2212, 04:04 AM

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Suppose that X=ℝ^{2}. If T:X→X is a bijection that takes straight lines to straight lines and 0 to 0, then T is linear.For this theorem, the steps are as follows: 1. If K and L are two different lines through 0, then T(K) and T(L) are two different lines through 0. 2. If K and L are two parallel lines, then T(K) and T(L) are two parallel lines. 3. For all x,y such that {x,y} is linearly independent, T(x+y)=Tx+Ty. (This is done by considering a parallelogram as you suggested). 4. For all vectors x and all real numbers a, T(ax)=aTx. (Note that this result implies that T(x+y)=Tx+Ty when {x,y} is linearly dependent). The strategy for step 4 is as follows: Let x be an arbitrary vector and a an arbitrary real number. If either x or a is zero, we have T(ax)=0=aTx. If both are nonzero, we have to be clever. Since Tx is on the same straight line through 0 as T(ax), there's a real number b such that T(ax)=bTx. We need to prove that b=a. Let B be the map ##t\mapsto tx##. Let C be the map ##t\mapsto tTx##. Let f be the restriction of T to the line through x and 0. Define ##\sigma:\mathbb R\to\mathbb R## by ##\sigma=C^{1}\circ f\circ B##. Since $$\sigma(a)=C^{1}\circ f\circ B(a) =C^{1}(f(B(a)) =C^{1}(T(ax)) =C^{1}(bTx)=b,$$ what we need to do is to prove that σ is the identity map. Berger does this by proving that σ is a field isomorphism. Since both the domain and codomain is ℝ, this makes it an automorphism of ℝ, and by the lemma that micromass proved so elegantly above, that implies that it's the identity map. 



#93
Nov2212, 05:15 AM

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#94
Nov2212, 07:05 PM

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Your posts deserve to be polished and turned into a library item, so I'll mention a couple of minor typos I noticed: Also, 



#95
Nov2312, 02:38 AM

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Just out of curiosity, do people use the term "line" for curves that aren't straight? Do we really need to say "straight line" every time?




#96
Nov2312, 06:25 AM

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A perspectivity is something slightly different. I don't know why I used that term... 



#97
Nov2312, 11:43 AM

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EDIT: If you're interested, you can see the definitions here. 



#98
Nov2412, 12:20 AM

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I think I have completely understood how to prove the following theorem using the methods described in Berger's book.
If ##T:\mathbb R^2\to\mathbb R^2## is a bijection that takes lines to lines and 0 to 0, then ##T## is linear.I have broken it up into ten parts. Most of them are very easy, but there are a few tricky ones. Notation: If L is a line, then I will write TL instead of T(L).
I won't explain all the details of part 8, because they require a diagram. But I will describe the idea. If you want to understand part 8 completely, you need to look at the diagrams in Berger's book. Notation: I will denote the line through x and y by [x,y].




#99
Nov2512, 03:45 PM

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This is a very interesting thread. Sorry I'm late to the conversation. I appreciate all the contributions. But I'm getting a little lost.
The question of the OP was asking about what kind of transformation keeps the following invariant: [tex]c^2t^2  x^2  y^2  z^2 = 0[/tex] [tex]c^2t'^2  x'^2  y'^2  z'^2 = 0 [/tex] But Mentz114 in post 3 interprets this to means that the transformation preserves dt'2 + dx'2 = dt2 + dx2. And Fredrik in post 8 interprets this to mean If Λ is linear and g(Λx,Λx)=g(x,x) for all x∈R4, then Λ is a Lorentz transformation. And modifies this in post 9 to be If Λ is surjective, and g(Λ(x),Λ(y))=g(x,y) for all x,y∈R4, then Λ is a Lorentz transformation. Are these all the same answer in different forms? Or is there a side question being addressed about linearity? Thank you. 



#100
Nov2512, 10:26 PM

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There is another approach to relativity that's been discussed in a couple of other threads recently. In this approach, the speed of light isn't mentioned at all. (Note that the g in my theorems is the Minkowski metric, so the speed of light is mentioned there). Instead, we interpret the principle of relativity as a set of mathematically precise statements, and see what we get if we take those statements as axioms. The axioms are telling us that the set of functions that change coordinates from one inertial coordinate system to another is a group, and that each of them takes straight lines to straight lines. The problem I'm interested in is this: If space and time are represented in a theory of physics as a mathematical structure ("spacetime") with underlying set ℝ^{4}, then what is the structure? When ℝ^{4} is the underlying set, it's natural to assume that those functions are defined on all of ℝ^{4}. The axioms will then include the statement that those functions are bijections from ℝ^{4} into ℝ^{4}. (Strangerep is considering something more general, so he is replacing this with something weaker). The theorems we've been discussing lately tell us that a bijection ##T:\mathbb R^4\to\mathbb R^4## takes straight lines to straight lines if and only if there's an ##a\in\mathbb R^4## and a linear ##\Lambda:\mathbb R^4\to\mathbb R^4## such that ##T(x)=\Lambda x+a## for all ##x\in\mathbb R^4##. The set of inertial coordinate transformations with a=0 is a subgroup, and it has a subgroup of its own that consists of all the proper and orthochronous transformations with a=0. What we find when we use the axioms is that this subgroup is either the group of Galilean boosts and proper and orthochronous rotations, or it's isomorphic to the restricted (i.e. proper and orthochronous) Lorentz group. In other words, we find that "spacetime" is either the spacetime of Newtonian mechanics, or the spacetime of special relativity. Those are really the only options when we take "spacetime" to be a structure with underlying set ℝ^{4}. Of course, if we had lived in 1900, we wouldn't have been very concerned with mathematical rigor in an argument like this. We would have been trying to guess the structure of spacetime in a new theory, and in that situation, there's no need to prove that theorem about straight lines. We can just say "let's see if there are any theories in which Λ is linear", and move on. In 2012 however, I think it makes more sense to do this rigorously all the way from the axioms that we wrote down as an interpretation of the principle of relativity, because this way we know that there are no other spacetimes that are consistent with those axioms. 



#101
Nov2612, 02:25 AM

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#102
Nov2612, 04:19 AM

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Those formulas do imply that ##F=0\Leftrightarrow \dot v=0##.
$$\gamma=\frac{1}{\sqrt{1v^2}},\qquad m=\gamma m_0$$ $$\dot\gamma=\frac{1}{2}(1v^2)^{\frac{3}{2}}(2v\dot v)=\gamma^3v\dot v$$ $$\dot m=\dot\gamma m_0=\gamma^3v\dot v m_0$$ \begin{align} F &=\frac{d}{dt}(mv)=\dot m v+m\dot v=\gamma^3v^2\dot v m_0+\gamma m_0\dot v =\gamma m_0\dot v(\gamma^2v^2+1)\\ & =\gamma m_0\dot v\left(\frac{v^2}{1v^2}+\frac{1v^2}{1v^2}\right) =\gamma^3 m_0\dot v \end{align} A complete specification of a theory of physics must include a specification of what measuring devices to use to test the theory's predictions. In particular, a theory about space, time and motion must describe how to measure lengths. It's not enough to just describe a meter stick, because the properties of a stick will to some degree depend on what's being done to it. So the theory must also specify the ideal conditions under which the measuring devices are expected to work the best. It's going to be very hard to specify a theory without ever requiring that an accelerometer displays 0. I don't even know if can be done. So nonaccelerated motion is probably always going to be an essential part of all theories of physics. In all of our theories, motion is represented by curves in the underlying set of a structure called "spacetime". I will denote that set by M. A coordinate system is a function from a subset of M into ℝ^{4}. If ##C:(a,b)\to M## is a curve in M, U is a subset of M, and ##x:U\to\mathbb R^4## is a coordinate system, then ##x\circ C## is a curve in C. So each coordinate system takes curves in spacetime to curves in ℝ^{4}. If such a curve is a straight line, then the object has zero velocity in that coordinate system. If a coordinate system takes all the curves that represent nonaccelerating motion to straight lines, then it assigns a constant velocity to every nonaccelerating object. Those are the coordinate systems we call "inertial". There's nothing particularly oldfashioned about that. Edit: Fixed four (language/typing/editing) mistakes in the last paragrah. 



#103
Nov2612, 04:51 AM

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#104
Nov2612, 05:06 AM

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But if you follow this path it is completely superfluous to prove anything about mapping straight lines to straight lines to get the most general transformation that does that and once you have it restrict it to the linear ones with a plausible physical assumption, since you are already starting with linear transformations. 



#105
Nov2612, 06:22 AM

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#106
Nov2612, 06:46 AM

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#107
Nov2612, 06:55 AM

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Given that space, the transformations that leave inertial coordinates invariant in the sense of SR first postulate must automatically be linear transformations, do you agree? Maybe this is not as obvious to see as I think, but I I think it is correct. 



#108
Nov2612, 07:03 AM

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