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Assume $\mathbf{x}$ to be a lightlike 4-vector in reference Frame $F$ and $\mathbf{x'}$ the corresponding vector in reference frame $F'$ where both frames share the same origin. Then due to the principle of relativity:

$$(x_0)^2-(x_1)^2-(x_2)^2-(x_3)^2 = (x_0')^2-(x_1')^2-(x_2')^2-(x_3')^2 = 0$$

According to Dodson/Poston (page 16), from this "it follows fairly easily that there is a positive number $a$ such that for any [vector] labelled $(x_0, x_1, x_2, x_3)$ by one system and $(x_0', x_1', x_2', x_3')$ by the other, not just [lightlike vectors], we have

$$\tag{*}(x_0)^2-(x_1)^2-(x_2)^2-(x_3)^2 = a((x_0')^2-(x_1')^2-(x_2')^2-(x_3')^2)$$"

I just changed the nomenclature a bit and set $c=1$ but otherwise cited the original text. The idea is to derive some basic property of Minkowski space / Lorentz transformations directly from the constancy of the speed of light.

I tried to restate the problem like this:

  1. Let $f: \mathbb{R}^4 \rightarrow \mathbb{R}: \mathbf{x} \mapsto (x_0)^2 - (x_1)^2 - (x_2)^2 - (x_3)^2$ (the Lorentz form)
  2. Let S be the set of all mappings $\mathbb{R}^4 \rightarrow \mathbb{R}^4$ such that for every $L \in S$, $L(0) = 0$ and $f(\mathbf{x}) = 0 \Rightarrow f(L(\mathbf{x})) = 0$ (constancy of the speed of light in different reference frames)
  3. It should then be possible to derive from (1) and (2) that for every $\mathbf{x} \in \mathbb{R}^4$ and a given $L \in S$: $$f(\mathbf{x}) = a f(L(\mathbf{x}))$$ with $a$ some constant positive real number.

EDIT: As has been pointed out in the comments, the restrictions I place on $L$ are not sufficient to derive $(*)$. I cannot think of more conditions that I could directly extract from the text, though. Anyone can help to come up with a minimum of conditions that would allow the deduction proposed by Dodson/Poston?

jerico
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    You must be missing some condition on $L$. By your definition, the map $\mathbf{x} \mapsto f(\mathbf{x}) \mathbf{x}$ belongs to $S$ but $f(L(\mathbf{x})) = f(\mathbf{x})^3$ – achille hui Mar 17 '13 at 23:20
  • Yes. You're right. I'll check this. – jerico Mar 17 '13 at 23:36
  • It seems a lot is missing, since the assumption about $L$ imposes no constraint at all on $L(\mathbf x)$ when $f(\mathbf x)\neq 0$. – Andreas Blass Mar 17 '13 at 23:41
  • @achillehui: I tried to extract more conditions from the text but I couldn't find more. I therefore cite the original text now so that hopefully someone can help me extracting the right conditions. :-( – jerico Mar 18 '13 at 00:26
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    It looks like $L : \mathbb{R}^4 \to \mathbb{R}^4$ should at least be a diffeomorphism fixing $0$, for it to be giving you a new reference frame. In fact, I wouldn't be surprised if $L$ really should be an invertible linear transformation. – Branimir Ćaćić Mar 18 '13 at 00:41
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    Judging from the sentences "We choose rectangular coordinates (x,y,z) if we can, though this is usually only locally and approximately possible in the General theory" and "in the limit of smallest" in the book. The intention is really for infinitesimal small $(x,y,z)$. One can probably assume $L$ is linear or close to linear.... – achille hui Mar 18 '13 at 00:41

1 Answers1

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Instead of the quadratic form $f$, use its related bilinear form $B({\mathbf x},{\mathbf y}) = x_0y_0 - x_1y_1 - x_2y_2 - x_3y_3$, which has more flexibility since it has two vector inputs instead of one.

Theorem. Let $B$ be a nondegenerate bilinear form on a finite-dimensional real vector space $V$. The following conditions on a linear transformation $L \colon V \rightarrow V$ are equivalent:

a) $B({\mathbf x},{\mathbf y}) = 0 \Longrightarrow B(L{\mathbf x},L{\mathbf y}) = 0$. That is, $L$ preserves orthogonality relative to $B$.

b) There is a real number $c$ such that $B(L{\mathbf x},L{\mathbf y}) = cB({\mathbf x},{\mathbf y})$ for all ${\mathbf x}$ and ${\mathbf y}$ in $V$.

Proof: See Theorem 3.20 at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/bilinearform.pdf, which is carried out with vector spaces over a general scalar field, and you can take the scalars to be the real numbers.

In this theorem $L$ is assumed to be linear, and that is missing from the setup in your problem. The geometric content of this theorem is that a linear transformation preserves orthogonality relative to $B$ exactly when it changes $B$-values by a uniform scaling factor. (Your $a$ is my $1/c$.)

To connect this theorem to your question, let $Q({\mathbf x}) = B({\mathbf x},{\mathbf x})$ be the quadratic form associated to $B$. (So $Q = f$ in your notation.) Then condition a implies that $Q({\mathbf x}) = 0 \Longrightarrow Q(L{\mathbf x}) = 0$ by using ${\mathbf y} = {\mathbf x}$ in a, but that implication about where $Q$ takes the value $0$ doesn't imply condition a in general.

Condition b in the theorem, however, is equivalent to $Q(L{\mathbf x}) = cQ({\mathbf x})$ for all ${\mathbf x}$ in $V$. In one direction, setting ${\mathbf y} = {\mathbf x}$ in condition b gives us $Q(L{\mathbf x}) = cQ({\mathbf x})$ for all ${\mathbf x}$. In the other direction, by the formula $$ B({\mathbf x},{\mathbf y}) = \frac{1}{2}(Q({\mathbf x} + {\mathbf y}) - Q({\mathbf x}) - Q({\mathbf y})), $$ if $Q(L{\mathbf x}) = cQ({\mathbf x})$ for all ${\mathbf x}$ then condition b is true by using $L{\mathbf x}$ in place of $\mathbf x$ and $L{\mathbf y}$ in place of $\mathbf y$ in the formula for $B({\mathbf x},{\mathbf y})$ in terms of values of $Q$; we need the additivity of $L$ for that, which is a special instance of being linear.

On a related note, a function $V \rightarrow V$ that preserves all $B$-values (not just $B$-value $0$) must be linear. Here is that result.

Theorem. Let $B$ be a nondegenerate bilinear form on a finite-dimensional vector space $V$. If $\sigma \colon V \rightarrow V$ satisfies $B(\sigma({\mathbf x}),\sigma({\mathbf y})) = B({\mathbf x},{\mathbf y})$ for all ${\mathbf x}$ and ${\mathbf y}$ in $V$, then $\sigma$ is linear and must be injective and surjective.

Proof: See my answer at Isometry without injection and surjection.

There is definitely no theorem about the condition $Q(\sigma{\mathbf x}) = Q({\mathbf x})$ for all ${\mathbf x}$ implying $\sigma$ is linear, since such a condition is satisfied if $\sigma$ is a random permutation in each level set of $Q$ (the sets where $Q$ takes a common value).

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