Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I don't understand the following statement in Landau & Lifshitz, Classical Theory of Fields, p.5:

$ds$ and $ds'$ are infinitesimals of same order. [...] It follows that $ds^2$ and $ds'^2$ must be proportional to each other: $$ds^2 = a \, ds'^2.$$.

I don't get why the proportionality applies, and why does it apply to the squares of the infinitesimals.

share|improve this question
1  
It would be useful to provide the context. –  leongz Jan 4 '13 at 2:15

2 Answers 2

up vote 1 down vote accepted

First, Landau and Lifshitz stated that $ds$ and $ds'$ approach zero simultaneously, so that there is some hidden variable $x$ such that, \begin{equation} \lim_{x\to 0} ds(x) =0 \end{equation} and \begin{equation} \lim_{x\to 0} ds'(x) =0, \end{equation} assuming and $ds$ and $ds'$ are continuous functions of $x$.

Next, the two are infinitesimals of the same order since the two inertial frames $K$ and $K'$ are equivalent. The frame $K'$ (in which the interval $ds'$ is measured) moves relative to the frame $K$. Suppose $ds'$ is an infinitesimal of greater order than $ds$, i.e., according the the reference given in the above answer, \begin{equation} \lim_{x\to 0} \frac{ds'(x)}{[ds(x)]^n} = A,\quad A\neq 0,\quad n>1, \end{equation} where $A$ can depend only on the magnitude of the relative velocity, not its direction and certainly not the coordinates, for reasons related to homogeneity of space and time and isotropy of space. Since $K$ is also moving relative to $K'$ and the principle of relativity holds, by symmetry one ought to have \begin{equation} \lim_{x\to 0} \frac{ds(x)}{[ds'(x)]^n} = A,\quad A\neq 0,\quad n>1, \end{equation} which is absurd. Hence $ds$ and $ds'$ have to be infinitesimals of the same order.

share|improve this answer

If $\lim_{x\rightarrow 0}\frac{\alpha(x)}{\beta(x)}=A$ ($A$ is a number different from zero), then the functions $\alpha(x)$ and $\beta(x)$ are called infinitesimals of the same order [1].

The proportionality at $x\rightarrow 0$ should be obvious from this.

[1] http://www.math24.net/infinitesimals.html

share|improve this answer
    
I still don't understand why "$ds$ and $ds'$ are infinitesimals of the same order". –  becko Mar 2 '13 at 19:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.