# Simple Derivation (1D) Lorentz Transformation

1. Sep 19, 2013

### phatgandy

Appendix 1 - simple Lorentz transformation derivation found at - http://www.bartleby.com/173/a1.html

Given in equation (3)

(x'-ct') = Y(x-ct) [Y = const.]

by rearrangement, it yields,

(x'-ct')/(x-ct) = Y.

But it is stated that both (x-ct) and (x'-ct') are zero, so we have "zero over zero" - a mathematical no-no.
There is surely no way something so trivial could of been over-looked when proof-reading, so I ask where it is I am going wrong in this derivation? Also, is it such the case when considering the 3D model?

Many thanks.

2. Sep 19, 2013

### The_Duck

Nowhere in the text do I see the expression (x'-ct')/(x-ct), so what is the problem exactly? Einstein never divides by zero in the course of his derivation.

3. Sep 19, 2013

### PAllen

I don't see any such division in the referenced link.

The overall argument is what I would call a 'motivation' not a strict derivation. The gist is: start with something we know is true for light (x-ct = 0; x' - ct' = 0) and note that if something more general were true for all coordinates, this fact about light would follow. There is no proof given that this is the only possible assumption. Over the years, there has been an 'industry' of papers examining what are the truly minimal but sufficient assumptions to derive the Lorentz transform.

Any way, he notes that if x'-ct' = γ (x-ct) always, then the required equations for light would follow.

Last edited: Sep 19, 2013
4. Sep 19, 2013

### stevendaryl

Staff Emeritus
$x$ and $t$ are not ALWAYS going to satisfy $x - ct = 0$. $x$ is the location of some event, and $t$ is the time of that event. $x$ and $t$ can be anything. Similarly, $x'$ and $t'$ can be absolutely anything. But in the special case in which $x - ct = 0$, we know that $x' - ct' = 0$, as well.

So we look for a relationship between the coordinates $(x,t)$ and the coordinates $(x',t')$ that ALWAYS holds, but has the implication that IF $x-ct = 0$, THEN $x' - ct' = 0$.

So we're looking for a relationship between the coordinates (a linear relationship, specifically) that has this implication. The following relationship works:

$x - ct = Y (x' - ct')$

It's supposed to hold for all values of $x$ and $t$, not just in the special case $x - ct = 0$.

5. Sep 20, 2013

### phatgandy

The_Duck & PAllen, I obtained the division by simple rearrangement of x−ct=Y(x′−ct′), which is totally reasonable.

Either way, my point was the statement reads 0 = Y*0 or by rearrangement 0/0 = Y

I still cannot entirely get my head around the validity of the equation in this form, when regarding the special case of x-ct=0, due to the fact that it represents 0 = Y*0, and this indeed is our basis (or so I have seen thus far).

Could you offer any more wise words?

6. Sep 20, 2013

### The_Duck

You are not allowed to do the rearrangement you are trying to do. Consider the equation

0 = 5*0.

It is a true equation, and there is nothing wrong with it. Indeed, if you multiply five by zero you get zero. Few arithmetic problems are simpler. Now, you might feel tempted to divide both sides by zero and get

0/0 = 5.

If you ever feel this temptation, you must resist it. You are not allowed to divide by zero, and the resulting equation does not make sense. The fact that this second equation is nonsense does not in any way diminish the validity of the original equation 0 = 0*5.

To reiterate, suppose in the course of some algebraic manipulations you obtain the equation

0 = Y*0.

This is a true equation for any value of Y, and there is nothing wrong with it. Again you must resist the temptation to divide by zero and rewrite the equation as Y = 0/0. The equation 0 = Y*0 is trivially true. The equation Y = 0/0 is meaningless.