Showing that Lorentz transformations are the only ones possible

micromass

OK, I'll try to type out the proof for you in this special case.

Ah, but this is far more general since it deals with arbitrary fields and stuff. The proof will probably be significantly harder than the [itex]\mathbb{R}[/itex] case.

I don't think you can use the fundamental theorem to prove that [itex]\mathbb{R}[/itex] has only automorphism. I agree the author makes you think that. But what he actually wants to do is prove that the only line preserving maps [itex]\mathbb{R}^n\rightarrow\mathbb{R}^n[/itex] are the affine maps. The fundamental theorem deals with semi-affine maps: so there is an automorphism of the field. So in order to prove the case of [itex]\mathbb{R}^n[/itex] he needs a lemma that states that there is only one automorphism on [itex]\mathbb{R}[/itex]. It is not a result that (I think) follows from the fundamental theorem.

That said, the proof that [itex]\mathbb{R}[/itex] has only one automorphism is not very hard. Let [itex]\sigma:\mathbb{R}\rightarrow \mathbb{R}[/itex] be an automorphism. So:

• [itex]\sigma[/itex] is bijective
• [itex]\sigma(x+y)=\sigma(x)+\sigma(y)[/itex]
• [itex]\sigma(xy)=\sigma(x)\sigma(y)[/itex]

So [itex]\sigma(0)=\sigma(0+0)=\sigma(0)+\sigma(0)[/itex], so [itex]\sigma(0)=0[/itex].
Likewise, [itex]\sigma(1)=\sigma(1.1)=\sigma(1)\sigma(1)[/itex], so [itex]\sigma(1)=1[/itex] (unless [itex]\sigma(1)=0[/itex] which is impossible because if injectivity).

Take [itex]n\in \mathbb{N}[/itex]. Then we can write [itex]n=\sum_{k=1}^n 1[/itex]. So
[tex]\sigma(n)=\sigma\left(\sum_{k=1}^n 1\right)=\sum_{k=1}^n \sigma(1)=\sum_{k=1}^n 1=n[/tex]

Now, we know that [itex]0=\sigma(0)=\sigma(n+(-n))=\sigma(n)+\sigma(-n)[/itex]. It follows that [itex]\sigma(-n)=\sigma(n)[/itex].

So we have proven that [itex]\sigma[/itex] is fixed on [itex]\mathbb{Z}[/itex].

Take [itex]p\neq 0[/itex]. Then [itex]1=\sigma(1)=\sigma(p\frac{1}{p})= \sigma(p)\sigma(\frac{1}{p})=p\sigma(\frac{1}{p})[/itex]. So [itex]\sigma(1/p)=1/p[/itex].
So, for [itex]q,p\in \mathbb{Z}[/itex] with [itex]p\neq 0[/itex]: [itex]\sigma(p/q)=\sigma(p)\sigma(1/q)=p/q[/itex]. So this proves that [itex]\sigma[/itex] is fixed on [itex]\mathbb{Q}[/itex].

Take [itex]x>0[/itex] in [itex]\mathbb{R}[/itex]. Then there exists a unique [itex]y\in \mathbb{R}[/itex] with [itex]y^2=x[/itex]. But then [itex]\sigma(y)^2=\sigma(x)[/itex]. It follows that [itex]\sigma(x)>0[/itex].
Take [itex]x<y[/itex] in [itex]\mathbb{R}[/itex]. Then [itex]x-y>0[/itex]. So [itex]\sigma(x-y)>0[/itex]. Thus [itex]\sigma(x)<\sigma(y)[/itex]. So [itex]\sigma[/itex] preserves the ordering.

Assume that there exists an [itex]x\in \mathbb{R}[/itex] such that [itex]\sigma(x)\neq x[/itex]. Assume (for example), that [itex]\sigma(x)<x[/itex]. Then there exists a [itex]q\in \mathbb{Q}[/itex] such that [itex]\sigma(x)<q<x[/itex]. But since [itex]\sigma[/itex] preserves orderings and rationals, it follows that [itex]\sigma(x)>q[/itex], which is a contradiction. So [itex]\sigma(x)=x[/itex].

This proves that the identity is the only automorphism on [itex]\mathbb{R}[/itex].

Now, for automorphisms on [itex]\mathbb{C}[/itex]. Let [itex]\tau[/itex] be a continuous automorphism on [itex]\mathbb{C}[/itex]. Completely analogously, we prove that [itex]\tau[/itex] is fixed on [itex]\mathbb{Q}[/itex]. Since [itex]\tau[/itex] is continuous and since [itex]\mathbb{Q}[/itex] is dense in [itex]\mathbb{R}[/itex], it follows that [itex]\tau[/itex] is fixed on [itex]\mathbb{R}[/itex].

Now, since [itex]i^2=-1[/itex]. It follows that [itex]\tau(i)^2=-1[/itex]. So [itex]\tau(i)=i[/itex] or [itex]\tau(i)=-i[/itex]. In the first case [itex]\tau(a+ib)=\tau(a)+\tau(i)\tau(b)=a+ib[/itex]. In the second case: [itex]\tau(a+ib)=a-ib[/itex].
So there are only two automorphisms on [itex]\mathbb{C}[/itex].

We don't really need projective spaces. We can prove the result without referring to it. But the result is often stated in this form because it is more general.
Also, one of the advantages of projective spaces is that [itex]\varphi(\mathbf{x})=\frac{A\mathbf{x}+B}{C\mathbf{x}+D}[/itex] is everywhere defined, even if the denominator is 0 (in that case, the result will be a point at infinity).

DrGreg

Maybe I need to spell this bit out. I think if T is continuous and your Step 3 is true and [itex]K = \mathbb{R}[/itex] then you can prove [itex]T(a\mathbf{x})=aT(\mathbf{x})[/itex] as follows.

It's clearly true for a = 2 (put x=y in step 3).

By induction it's true for any integer a (y = (a-1)x).

By rescaling it's true for any rational a.

By continuity of T and density of [itex]\mathbb{Q}[/itex] in [itex]\mathbb{R}[/itex] it's true for all real a.

Fredrik

Thank you micromass. That was exceptionally clear. I didn't even have to grab a pen. This saved me a lot of time.

Interesting idea. Thanks for posting it. I will however still be interested in a proof that doesn't rely on the assumption that T is continuous.

micromass

Here is a proof for the plane. I think the same method of proof directly generalizes to higher dimensions, but it might get annoying to write down.

DEFINITION: A projectivity is a function [itex]\varphi[/itex] on [itex]\mathbb{R}^2[/itex] such that


[tex]\varphi(x,y)=\left(\frac{Ax+By+C}{Gx+Hy+I},\frac{Dx+Ey+F}{Gx+Hy+I}\righ t)[/tex]

where [itex]A,B,C,D,E,F,G,H,I[/itex] are real numbers such that the matrix

[tex]\left(\begin{array}{ccc} A & B & C\\ D & E & F\\ G & H & I\end{array}\right)[/tex]

is invertible. This invertible-condition tells us exactly that [itex]\varphi[/itex] is invertible. The inverse is again a perspectivity and its matrix is given by the inverse of the above matrix.

We can see this easily as follows:
Recall that a homogeneous coordinate is defined as a triple [x:y:z] with not all x, y and z zero. Furthermore, if [itex]\alpha\neq 0[/itex], then we define [itex][\alpha x: \alpha y : \alpha z]=[x:y:z][/itex].

There exists a bijection between [itex]\mathbb{R}^2[/itex] and the homogeneous coordinates [x:y:z] with nonzero z. Indeed, with (x,y) in [itex]\mathbb{R}^2[/itex], we can associate [x:y:1]. And with [x:y:z] with nonzero z, we can associate (x/z,y/z).

We can now look at [itex]\varphi[/itex] on homogeneous coordinates. We define [itex]\varphi [x:y:z] = \varphi(x/z,y/z)[/itex]. Clearly, if [itex]\alpha\neq 0[/itex], then [itex]\varphi [\alpha x:\alpha y:\alpha z]=\varphi [x:y:z][/itex]. So the map is well defined.

Actually, our [itex]\varphi[/itex] is actually just matrix multiplication:

[tex]\varphi[x:y:z] = \left(\begin{array}{ccc} A & B & C\\ D & E & F\\ G & H & I\end{array}\right)\left(\begin{array}{c} x\\ y \\ z\end{array}\right)[/tex]

Now we see clearly that [itex]\varphi[/itex] has an inverse given by

[tex]\varphi^{-1} [x:y:z] = \left(\begin{array}{ccc} A & B & C\\ D & E & F\\ G & H & I\end{array}\right)^{-1}\left(\begin{array}{c} x\\ y \\ z\end{array}\right)[/tex]




LEMMA: Let x,y,z and t in [itex]\mathbb{R}^2[/itex] be four distinct points such that no three of them lie on the same line. Let x',y',z',t' in [itex]\mathbb{R}^2[/itex] also be four points such that no three of them lie on the same line. There exists a projectivity [itex]\varphi[/itex] such that [itex]\varphi(x)=x^\prime[/itex], [itex]\varphi(y)=y^\prime[/itex], [itex]\varphi(z)=z^\prime[/itex], [itex]\varphi(t)=t^\prime[/itex].

We write in homogeneous coordinates:
[tex]x=[x_1:x_2:x_3],~y=[y_1:y_2:y_3],~z=[z_1:z_2:z_3],~t=[t_1:t_2:t_3][/tex]

Since [itex]\mathbb{R}^3[/itex] has dimension 3, we can find [itex]\alpha,\beta,\gamma[/itex] in [itex]\mathbb{R}[/itex] such that

[tex](t_1,t_2,t_3)=(\alpha x_1,\alpha x_2,\alpha x_3)+(\beta y_1,\beta y_2,\beta y_3)+ (\gamma z_1, \gamma z_2,\gamma z_3)[/tex].

The vectors [itex](\alpha x_1,\alpha x_2,\alpha x_3), (\beta y_1,\beta y_2,\beta y_3), (\gamma z_1, \gamma z_2,\gamma z_3)[/itex] form a basis for [itex]\mathbb{R}^3[/itex] (because of the condition that not three of x,y,z or t is on one line).

We can do the same for the x',y',z',t' and we again obtain a basis [itex](\alpha^\prime x_1^\prime,\alpha^\prime x_2^\prime,\alpha^\prime x_3^\prime), (\beta^\prime y_1^\prime,\beta^\prime y_2^\prime,\beta^\prime y_3^\prime), (\gamma^\prime z_1^\prime, \gamma^\prime z_2^\prime,\gamma^\prime z_3^\prime)[/itex] such that

[tex](t_1^\prime, t_2^\prime,t_3^\prime)=(\alpha^\prime x_1^\prime,\alpha^\prime x_2^\prime,\alpha^\prime x_3^\prime)+(\beta^\prime y_1^\prime,\beta^\prime y_2^\prime,\beta^\prime y_3^\prime)+(\gamma^\prime z_1^\prime, \gamma^\prime z_2^\prime,\gamma^\prime z_3^\prime)[/tex]


By linear algebra, we know that there exists an invertible matrix T that sends the bases on each other. This implies directly that the associated projectivity sends x to x', y to y' and z to z'.
Since
[tex](t_1,t_2,t_3)=(\alpha x_1,\alpha x_2,\alpha x_3)+(\beta y_1,\beta y_2,\beta y_3)+ (\gamma z_1, \gamma z_2,\gamma z_3)[/tex]
we get after applying T that

[tex]T(t_1,t_2,t_3)=(\alpha^\prime x_1^\prime,\alpha^\prime x_2^\prime,\alpha^\prime x_3^\prime)+(\beta^\prime y_1^\prime,\beta^\prime y_2^\prime,\beta^\prime y_3^\prime)+(\gamma^\prime z_1^\prime, \gamma^\prime z_2^\prime,\gamma^\prime z_3^\prime)[/tex]

and thus [itex]T(t_1,t_2,t_3)=(t_1^\prime,t_2^\prime, t_3^\prime)[/itex]. Thus the projectivity also sends t to t'.



THEOREM Let [itex]U\subseteq \mathbb{R}^2[/itex] be open and let [itex]\varphi:U\rightarrow \mathbb{R}^2[/itex] be injective. Assume that [itex]\varphi[/itex] sends lines to lines, then it is a projectivity.

We can of course assume that U contains an equilateral triangle ABC. Let P be the centroid of ABC.
By the previous lemma, there exists a projectivity [itex]\psi[/itex] such that [itex]\psi(\varphi(A))=A, ~\psi(\varphi(B))=B, ~\psi(\varphi(C))=C, ~\psi(\varphi(P))=P[/itex]. So we see that [itex]\sigma:=\psi\circ\varphi[/itex] sends lines to lines and that [itex]\sigma(A)=A,~\sigma(B)=B,~\sigma(C)=C,~\sigma(P)=P[/itex]. We will prove that [itex]\sigma[/itex] is the identity.

HINT: look at Figure 2.1, p.19 of the Mccallum paper.

Define E the midpoint of AC. Then E is the intersection of AC and PB. But these lines are fixed by [itex]\sigma[/itex]. Thus [itex]\sigma(E)=E[/itex]. Let D be the midpoint of BC and F the midpoint of AB. Likewise follows that [itex]\sigma(D)=D[/itex] and [itex]\sigma(F)=F[/itex].

Thus [itex]\sigma[/itex] preserves the verticles of the equilateral traingles AFE, FBD, DEF and EDC. Since [itex]\sigma[/itex] preserves parallelism, we see easily that [itex]\sigma[/itex] preserves the midpoints and centroids of the smaller triangles. So we can subdivide the triangles in even smaller triangles whose vertices are preserved. We keep doing this process and eventually we find a set S dense in the triangle such that [itex]\sigma[/itex] is fixed on that dense set. If [itex]\sigma[/itex] were continuous, then [itex]\sigma[/itex] is the identity on the triangle.

To prove continuity, we show that certain rhombuses are preserved. Look at Figure 2.3 on page 20 of McCallum. We have shown that the vertices of arbitrary triangles are preserved. Putting those two triangles together gives a rhombus. We will show that [itex]\sigma[/itex] sends the interior of any rhombus ABCD into the rhombus ABCD. Since the rhombus can be made arbitrarily small around an arbitrary point, it would follow that [itex]\sigma[/itex] were continuous.

By composing with a suitable linear map, we restrict to the following situation:

LEMMA: Let A=(0,0), B=(1,0), C=(1,1) and D=(0,1) and let [itex]\Sigma[/itex] be the square ABCD. Suppose that [itex]\sigma:\Sigma\rightarrow \mathbb{R}^2[/itex] sends lines to lines and suppose that [itex]\sigma[/itex] is fixed on A,B,C and D. Then [itex]\sigma(\Sigma)\subseteq \Sigma[/itex].

Take S on CB. We can make a construction analogous to 2.4 p.22 in MCCallen. So we let TS be horizontal, TU have slope -1 and VU be vertical. We define Q as the intersection of AS and VU. If S has coordinates [itex](1,s)[/itex] for some s. Then we can easily check that Q has coordinates [itex](s,s^2)[/itex]. In particular, Q lies in the upper half-plane (= everything about AB).

Since S in CB and since C and B are fixed. We see that [itex]\sigma(S)\in CB[/itex]. Let's say that [itex]\sigma(S)=(1,t)[/itex] for some t. The line TS is a horizontal and [itex]\sigma[/itex] maps this to a horizontal. So [itex]\sigma(T)[/itex] has the form (0,t). The line TU has slope -1. So [itex]\sigma(U)[/itex] has the form (t,0). Finally, it follows that [itex]\sigma(Q)[/itex] has the form [itex](t,t^2)[/itex]. In particular, [itex]\sigma(Q)[/itex] is in the upper half plane.

So we have proven that if S is on CB, then they ray AS emanating from A is sent into the upper half plane. Let P be an arbitrary point in the square, then it is an element of a ray AS for some S. This ray is taken to the upper half plane. So [itex]\sigma(P)[/itex] is in the upper half plane.

So this proves that the square ABCD is sent by [itex]\sigma[/itex] into the upper half plane. Similar constructions show that the square is also sent to the lower half plane, the left and right half planes. So taking all of these things together: ABCD is sent into ABCD. This proves the lemma.

So, right now we have shown that [itex]\sigma[/itex] is the identity on some small equilateral triangle in [itex]U[/itex]. So [itex]\varphi[/itex] is a projectivity on some small open set [itex]U^\prime[/itex] of U (namely on the interior of the triangle). We prove now that [itex]\varphi[/itex] will be a projectivity on entire U.

Around any point P in U, we can find some equilateral triangle. And we proved for such triangles that [itex]\varphi[/itex] is a projectivity and thus analytic. The uniqueness of analytic continuation now proves that [itex]\varphi[/itex] is a projectivity on entire U.

TrickyDicky

Nice proof!
If I understand it correctly this proves that the most general transformations that take straight lines to straight lines are the linear fractional ones.
To get to the linear case one still needs to impose the condition mentioned above about the continuity of the transformation, right?
Classically(Pauli for instance) this was done just assuming the euclidean (minkowskian) space as the underlying geometry.

Fredrik

It's sufficient to assume that the map that takes straight lines to straight lines is defined on the entire vector space, rather than a proper subset. It's not necessary to assume that the map is continuous. (If you want the map to be linear, rather than linear plus a translation, you must also assume that it takes 0 to 0).

Fredrik

I've been examining the proof in Berger's book more closely. (Change the .se to your own country domain if the url is giving you trouble). His strategy is very close to yours, but there's a clever trick at the end that allows us to drop the assumption of continuity. Consider the following version of the theorem:

Suppose that X=ℝ². 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 non-zero, 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.

TrickyDicky

What I meant is that one must impose that the transformation must map finite coordinates to finite coordinates, which I think is equivalent to what you are saying here.

strangerep

Thank you Micromass.
Your posts deserve to be polished and turned into a library item, so I'll mention a couple of minor typos I noticed:
Even though this is a synonym, I presume it should be "projectivity", since that's the word you used earlier.

Also,

Fredrik

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?

micromass

Ah yes, thank you!! It should indeed be projectivity.
A perspectivity is something slightly different. I don't know why I used that term...

lugita15

Yes, at least historically line was just used to mean any curve. I think Euclid defined a line to be a "breadthless length", and defined a straight line to be a line that "lies evenly with itself", whatever that means.

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

Fredrik

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).

1. If K is a line through 0, then so is TK.
2. If K,L are lines through 0 such that K≠L, then TK≠TL. (Note that this implies that if {x,y} is linearly independent, then so is {Tx,Ty}).
3. If K is parallel to L, then TK is parallel to TL.
4. For all x,y such that {x,y} is linearly independent, T(x+y)=Tx+Ty.
5. If x=0 or a=0, then T(ax)=aTx.
6. If x≠0 and a≠0, then there's a b such that T(ax)=bTx. (Note that this implies that for each x≠0, there's a map σ such that T(ax)=σ(a)Tx. The following steps determine the properties of σ for an arbitrary x≠0).
7. σ is a bijection from ℝ² into ℝ².
8. σ is a field homomorphism.
9. σ is the identity map. (Combined with 5-6, this implies that T(ax)=aTx for all a,x).
10. For all x,y such that {x,y} is linearly dependent, T(x+y)=Tx+Ty.

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].

1. Since T takes lines to lines, TK is a line. Since T0=0, 0 is on TK.
2. Suppose that TK=TL. Let x be an arbitrary non-zero point on TK. Since x is also on TL, T^-1(x) is in both K and L. But this implies that T^-1(x)=0, which contradicts that x≠0.
3. If K=L, then obviously TK=TL. If K≠L, then, they are either parallel or intersect somewhere, and part 2 tells us that they don't intersect.
4. Let x,y be arbitrary vectors such that {x,y} is linearly independent. Part 2 tells us that {Tx,Ty} is linearly independent. Define
   K=[0,x] (This is the range of ##t\mapsto tx##).
   L=[0,y] (This is the range of ##t\mapsto ty##).
   K'=[x+y,y] (This is the range of ##t\mapsto y+tx## so this line is parallel to K).
   L'=[x+y,x] (This is the range of ##t\mapsto x+ty## so this line is parallel to L).
   Since x+y is at the intersection of K' and L', T(x+y) is at the intersection of TK' and TL'. we will show that Tx+Ty is also at that intersection. Since x is on L', Tx is on TL'. Since L' is parallel to L, TL' is parallel to TL (the line spanned by {Ty}). These two results imply that TL' is the range of the map B defined by B(t)=Tx+tTy. Similarly, TK' is the range of the map C defined by C(t)=Ty+tTx. So there's a unique pair (r,s) such that T(x+y)=C(r)=B(s). The latter equality can be written as Ty+rTx=Tx+sTy. This is equivalent to (r-1)Tx+(1-s)Ty=0, and since {Tx,Ty} is linearly independent, this implies r=s=1. So T(x+y)=B(1)=Tx+Ty.
   
5. Let x be an arbitrary vector and a an arbitrary real number. If either of them is zero, we have T(ax)=0=aT(x).
   
6. Let x be non-zero but otherwise arbitrary. 0,x, and ax are all on the same line, K. So 0,x and T(ax) are on the line TK. This implies that there's a b such that T(ax)=bTx. (What we did here proves this statement when a≠0 and x≠0, and part 5 shows that it's also true when a=0 or x=0).
   
7. The map σ can be defined explicitly in the following way. Define B by B(t)=tx for all t. Define C by C(t)=tTx for all t. Let K be the range of B. Then the range of C is TK. Define ##\sigma=C^{-1}\circ T|_K\circ B##. This map is a bijection (ℝ→ℝ), since it's the composition of three bijections (ℝ→K→TK→ℝ). To see that this is the σ that was discussed in the previous step, let b be the real number such that T(ax)=bTx, and note that
   $$\sigma(a)=C^{-1}\circ T|_K\circ B(a) =C^{-1}(T(B(a))) =C^{-1}(T(ax)) =C^{-1}(bTx)=b.$$
8. Let a,b be arbitrary real numbers. Using the diagrams in Berger's book, we can show that there are two lines K and L such that (a+b)x is at the intersection of K and L. This implies that the point at the intersection of TK and TL is T((a+b)x)=σ(a+b)Tx. Then we use the diagram (and its image under T) to argue that T(ax)+T(bx) must also be at that same intersection. This expression can be written (σ(a)+σ(b))Tx, so these results tell us that
   $$(\sigma(a)+\sigma(b)-\sigma(a+b))Tx=0.$$ Since Tx≠0, this implies that σ(a+b)=σ(a)+σ(b). Then we use similar diagrams to show that σ(ab)=σ(a)σ(b), and that if a<b, then σ(a)<σ(b). (The book doesn't include a diagram for that last part, but it's easy to imagine one).
9. This follows from 8 and the lemma that says that the only automorphism of R is the identity.
10. Suppose that {x,y} is linearly dependent. Let k be the real number such that y=kx. Part 9 tells us that T(x+y)=T((1+k)x)=(1+k)Tx=Tx+kTx=Tx+T(kx)=Tx+Ty.

friend

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.

Fredrik

Those aren't interpretations of the original condition. I would interpret the OP's assumption as saying that g(Λx,Λx) for all x∈ℝ⁴ such that g(x,x)=0 (i.e. for all x on the light cone). This assumption isn't strong enough to to imply that Λ is a Lorentz transformation, so I described two similar but stronger assumptions that are strong enough. The two statements you're quoting here are theorems I can prove.

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 ℝ⁴, then what is the structure? When ℝ⁴ is the underlying set, it's natural to assume that those functions are defined on all of ℝ⁴. The axioms will then include the statement that those functions are bijections from ℝ⁴ into ℝ⁴. (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 ℝ⁴.

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.

Blackforest

OK. Thank you for all these explanations. But don't you think that the "obsession" with preservation of straight lines is entirely due to our false and old fashioned use of the definition of what an inertial observer is? What do I mean? Inertial observer is not = observer without acceleration, but = observer on which no force is acting. And this is not the same thing within a generalized theory of relativity where F = d(m. v)/dt = m. acceleration + dm/dt. speed => F = 0 is not acceleration = 0.

Fredrik

Those formulas do imply that ##F=0\Leftrightarrow \dot v=0##.

$$\gamma=\frac{1}{\sqrt{1-v^2}},\qquad m=\gamma m_0$$
$$\dot\gamma=-\frac{1}{2}(1-v^2)^{-\frac{3}{2}}(-2v\dot v)=\gamma^3v\dot v$$
$$\dot m=\dot\gamma m_0=\gamma^3v\dot v m_0$$
\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}{1-v^2}+\frac{1-v^2}{1-v^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 non-accelerated 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 ℝ⁴. 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 ℝ⁴. 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 non-accelerating motion to straight lines, then it assigns a constant velocity to every non-accelerating object. Those are the coordinate systems we call "inertial". There's nothing particularly old-fashioned about that.

Edit: Fixed four (language/typing/editing) mistakes in the last paragrah.