# Definable relationships between 4 variables with 2 equations

Tags:
1. Nov 22, 2015

### Darren73

Hi,

I saw a derivation in a book and I don't see the logical connection. Suppose $1.\ \ a=b \text{ and } 2.\ \ x=y$

Then $3.\ \ a-b= \lambda (x-y)$ makes sense to me since $0=\text{Anything}⋅0$

However they said "similarly" $4.\ \ a+b= \mu (x+y)$, and this I don't understand. To me $a+b= 2a \text{ or }2b$ and $x+y=2x\text{ or }2y$, so unless there was a defined linear relationship such as $a \propto x \text{ or }y$, OR $b \propto x \text{ or } y$, then I don't see how they could propose equation 4. to be necessarily true.

2. Nov 22, 2015

### axmls

It would help to get some context as to what the derivation was and what it was deriving, and in particular, what this relationship was used to prove. To me, it's just the statement: the quantity $a+b$ is related to the quantity $x + y$ by some constant $\mu$. Clearly $\mu \neq 0$ since $a + b \neq 0$ and $x + y \neq 0$.

3. Nov 22, 2015

### Darren73

This is taken from "The Einstein Theory of Relativity" by Lillian R. Lieber. It is supposed to be a high school introduction to relativity using simple algebra and calculus. From page 45 in the "First Paul Dry Books Edition, 2008".

In it we are supposing that there are two frames of reference (FORs), called K and K' moving with relative velocity v along a single dimension x. Within the K FOR it defines x = ct where as in the K' FOR x'=ct'. Earlier in the book it settled why x and x' are different within different FORs (because light signals used to measure distance depend on your relative velocity). And so is says that since x = ct and x' = ct' then we can get equations 3. and 4. that I posted above. From the physics of the situation maybe the necessary linear relationship is available, but if so it is not mentioned in the book.