# Proving invariance of $ds^2$ from the invariance of the speed of light

I've started today the book of Landau and Lifshitz Vol.2: The Classical Theory of Fields $\S 2$. They start from the invariance of the speed of light, express it as the fact that $$c^2(\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2=0$$ is preserved when we change inertial frame, so they consider $$ds^2=c^2dt^2-dx^2-dy^2-dz^2,$$ and say

We have observed that if $ds=0$ in one frame then $ds'=0$ in another frame. But $ds$ and $ds'$ are infinitesimal of the same order. So it follows that $ds^2$ and $ds'^2$ have to be proportional that is $ds^2=ads'^2$...

and he goes on to prove that $a=1$.

How to translate this argument in a rigorous one? I'm really interested in this, both to understand this deduction and also to be able in future to make similar ones.

• So you mean that the passage "So it follows that $ds^2$ and $ds'^2$ have to be proportional.." is inaccurate? – user35543 Dec 9 '13 at 13:34
• @user35543: that is not the case, see my answer below. – Kyle Kanos Dec 9 '13 at 13:47
• I think I misunderstood something that Ben said. The post that I was thinking of is physics.stackexchange.com/q/12435 – jinawee Dec 9 '13 at 15:05

It is nothing but a problem with real quadratic forms. You have a pair of vectors $v,v' \in R^4$ with, respectively, components $(\Delta t, \Delta x, \Delta y, \Delta z)$ and $(\Delta t', \Delta x', \Delta y', \Delta z')$. Actually these components describe the same vector in spacetime (describing the difference of events) but referring to two different reference frames.

Next you consider two quadratic forms $$ds^2(v,v)= c^2\Delta t^2 - \Delta x^2 - \Delta y^2 -\Delta z^2$$ and $$ds'^2(v',v')= c^2\Delta t'^2 - \Delta x'^2 - \Delta y'^2 -\Delta z'^2\:.$$ Since there is a linear transformation connecting the two reference frames (which will be proven to be the Lorentz transformation!) we also have $v'= Av$, where $A$ is some non-singular $4 \times 4$ real matrix independent from $v$, we can also write: $$ds'^2(v',v')= ds'^2(Av,Av)$$ We finally know, from physics, that: $$ds'^2(v',v') =0 \quad \mbox{if and only if}\quad ds^2(v,v) =0\:,$$ that is $$ds'^2(Av,Av) =0 \quad \mbox{if and only if}\quad ds^2(v,v) =0\:.$$ In other words: the quadratic forms $ds^2(\:,\:)$ and $ds'^2(A\cdot\:,A\cdot)$ have the same zeros.

A theorem of quadratic forms states that:

THEORM. Two real quadratic forms on a real $n$ dimensional vector space, respectively associated to the symmetric matrix $$\eta = diag(1,-1, \ldots, -1)$$
and to the symmtric matrix $\eta'$, have the same zeros if and only if they are proportional, i.e.,
$$\eta' = c \eta \quad \mbox{for some c\in \mathbb R \setminus\{0\}}\:.$$

Applying the reuslt to our case we have that there is $a\neq 0$ with $$ds^2(v,v)= ads'^2(Av,Av)\quad \mbox{for every v\in R^4}.$$ I do not have Ladau's book, so I do not know how it is subsequently proved therein that $a=1$. I think that some physical symmetry argument is exploited in addition to the invariance of $c$.

1. PROOF of the Theorem.

Due to the form of $\eta$, we have that $U^t\eta U =0$ if and only if $U_V^0 = \pm |V|$ and $U_V^{i}= V_{i}$ for $V \in \mathbb R^{n-1}$ and $i=1,\ldots,n-1$, where $|V| = \sqrt{V_1^2+\cdots+V_{n-1}^2}$. Our hypotheses on $\eta'$ can therefore be made explicit as follows. $$U_V^t\eta'U_V=0 \quad \forall V \in \mathbb R^{n-1}\:.$$ In components (latin indices are summed from $1$ to $n-1$) $$\eta'_{00}|V|^2 \pm \sum_i \eta'_{0i} |V| V_i + \sum_{ij} \eta'_{ij} V_iV_j=0\:.$$ We henceforth use the sign $+$, the other case can be treated similarly. It must be $$\eta'_{00}|V|^2 + \sum_i \eta'_{0i} |V| V_i + \sum_{ij} \eta'_{ij} V_iV_j=0\quad \forall V \in \mathbb R^{n-1}\:.\tag{1}$$ Since (1) holds for $V$ and $-V$, we immediately have that $$\sum_i \eta'_{0i} |V| V_i=0 \quad \forall V \in \mathbb R^{n-1}\:.$$ Arbitrariness of $V$ easily implies that $\eta'_{0i}=0$ for all $i=1,\ldots, n-1$.

(1) can be re-written as follows $$\sum_{ij} (\eta'_{00}\delta_{ij}+\eta'_{ij}) V_iV_j=0\quad \forall V \in \mathbb R^{n-1}\:.\tag{2}$$ Since $\eta'_{ij}=\eta'_{ji}$, using $V=X+Y$ and next $V=X-Y$, (2) yields $$\sum_{ij} (\eta'_{00}\delta_{ij}+\eta'_{ij}) X_iY_j=0\quad \forall X,Y \in \mathbb R^{n-1}\:.\tag{3}$$ In other words the matrix inside the scalar product above is the zero matrix, $$\eta'_{00}\delta_{ij}+\eta'_{ij}=0\:.$$ Summing up $$\eta'= \eta'_{00} diag (1, -1, \cdots, -1)\:.\tag{4}$$ Notice that $\eta'_{00}$ cannot vanish otherwise the zeros of $\eta'$ would fill the whole vector space, differently form the zeros of $\eta$, but we know that the set of zeros must coincide by hypotheses. (4) is the thesis if $c:= \eta'_{00}$. QED

2. Evidently, making use of Sylvester's theorem, the statement of the proved theorem implies an apparently stronger form of it like this.

THEORM. Consider two real quadratic forms on a real $n$ dimensional vector space, respectively associated to the symmetric matrices $\eta$ and $\eta'$, where $\eta$ has signature $+,-,\ldots, -$.

$\eta$ and $\eta'$ have the same zeros if and only if they are proportional

• Ok seen in this way is very very clear! Thank you very much. As i can see the fundamental point that i was missing was the linearity of the transformation that reduces it to a quadratic forms problem. So as a final question: when Landau says that the 2 infinitesimal have the same order is he essentially saying the fact that there will be your linear transformation A(indeed two infinitesimal of the same order are connected by linear transformation)? – user35543 Dec 9 '13 at 14:02
• Yes, I think so in the general case. Actually I was supposing to deal with the elementary case of (the construction of) special relativity, where the transformations between reference frames are supposed to (or proved to, depending on physical requirements) be linear. I do not like Landau's books because sometimes, like in this case, there is not a clear distinction between physical facts and mathematical results. – Valter Moretti Dec 9 '13 at 14:08
• -1. The question asked is "How is it proved that $a=1$?" and your answer is "I don't know." – Kyle Kanos Dec 9 '13 at 14:18
• No, the question was "How to translate this argument in a rigorous one?" – Valter Moretti Dec 9 '13 at 14:24
• @VM9: You are correct, I misread the question. If you make a correction, I can change my downvote to an upvote. – Kyle Kanos Dec 9 '13 at 14:43

In general, uniform motion in one reference frame implies uniform motion in a different reference frame. Suppose that frame $K'$ is moving at a constant velocity $\mathbf{v}$ relative to frame $K$. The transformation from frame $K$ to $K'$ must be linear, so it must be true that $$ds'^2=a\,ds^2\tag{1}$$ where $a$ depends on the relative motion of $K$ and $K'$, but not on the 4-coordinates themselves1! We are pretty much left with $a\to a(\mathbf{v})$ (this means $a$ is a function of the velocity vector $\mathbf{v}$).

But $\mathbf{v}$ introduces a direction to the transformation, which means that the orientation of the two frames would matter. But since $ds^2$ depends on the quadratic components, then we must have that $a\to a(v)$ ($a$ is a function of the velocity's magnitude).

Now comes the fun part: the inverse transformation is obtained from the forward transformation by changing the sign of the velocity vector. Thus, we see that $$ds^2=a(v)\,ds'^2\tag{2}$$ Comparing (1) and (2), $$ds^2=a(v)ds'^2=\frac{ds'^2}{a(v)}$$ and clearly the only way for this to occur is $a(v)=1$.

1. Allowing $a$ to depend on the coordinates would mean that space-time is not homogeneous.

• Thank you very much for the answer. Also here i can see that the main point is linearity, that, if guess correctly, was hidden in the Landau observation that the 2 infinitesimal have the same order(maybe also with the assumption of homogeneity of space and time as you pointed out). – user35543 Dec 9 '13 at 14:03
• Yes, linearity is a key to the proof. I'd recommend picking up a 2nd reference book for when Landau appears a little vague. – Kyle Kanos Dec 9 '13 at 14:15
• "and clearly the only way for this to occur is $a(v) = 1$" Well, no: what about $a(v) = -1$? In $1+1$ dimensions, taking the $-1$ makes a mathematically consistent linear transformation that happens to switch timelike and spacelike four-velocities, which would be physically strange. With more dimensions, things become inconsistent if the coordinates are restricted to be real. But either way, one needs at least a little bit of consideration to exclude $a(v) = -1$. – Stan Liou Dec 9 '13 at 14:16
• -1 because the question was "How to translate this argument (that in the box) in a rigorous one?" and your statement that $ds^2 = a ds'^2$ is consequence of linearity of the transformation between $K$ and $K'$ is wrong (without using the theorem on quadratic forms I mentioned). If it were correct, it would be the same as saying that, if $g$ is the metric, $A^\mu_\alpha A^\nu_\beta g_{\mu\nu}=a g_{\alpha\beta}$ for every linear transformation $A$, that is false. – Valter Moretti Dec 9 '13 at 14:41
• @StanLiou: We can exclude $a=-1$ because two successive transformations must give the same result as a single transformation, such that $a^2=a$ which can only be satisfied with $a=1$. – Kyle Kanos Dec 9 '13 at 14:46