Einstein needs to show his work

[rhenretta]

I'm reading Einstein's book on relativity (aptly named Relativity), and trying to follow his "simple" derivation of the Lorentz transformation in Appendix 1. However, since Einstein feels it acceptable to skip over important steps, it leaves me to figure them out, but I am lost on one step...

The relevant equations are as follows (keeping the equation numbers consistent with his)

Eq 5:
$$x' = ax - bct$$

$$ct' = act - bx$$

Eq 6:
$$v=\frac{bc}{a}$$

He claims:

But if the snapshot is taken from K'(t'=0), and if we eliminate t from the equations (5), taking into account the expression (6) we obtain:

$$x' = a(1-\frac{v^2}{c^2})x$$

I tried following the steps expressed:

First, substituting t'=0 into (5):

$$x' = ax - bct$$

$$0 = act - bx$$

Now, to eliminate t, I can only see to do this with equation 6 since

$$v = \frac{x}{t}$$

$$t = \frac{x}{v}$$

This leaves me with:

$$x' = ax - bc\frac{x}{v}$$

$$0 = ac\frac{x}{v} - bx$$

It is at this point I am at a loss where to go next to obtain his result.

 

[George Jones]

$$0 = act - bx$$

Solve for $t$ in the above, stick the result in the $x'$ equation of (5), and then use (6).

 

[rhenretta]

Following that, I get:

$$t=\frac{bx}{ac}$$

Substituting in eq 5 (and reducing for simplicity - ironic I am not showing my work):

$$x' = x(a-\frac{b^2}{a})$$

I'm trying to use eq 6 by solving it for b:

$$b = \frac{av}{c}$$

substituting b:

$$x' = x(a-\frac{( \frac{av}{c})^2}{a})$$

Do a little reduction: (bear with me, doing the work via the forums)
$$x' = x(a (\frac{a}{a}) -\frac{( \frac{av}{c})^2}{a})$$

$$x' = x(\frac{a^2}{a} -\frac{( \frac{av}{c})^2}{a})$$

$$x' = x(\frac{\frac{a^2v^2}{c^2} - a^2}{a})$$

$$x' = x(\frac{\frac{a^2v^2}{c^2} - a^2 * \frac{c^2}{c^2}}{a})$$

$$x' = x(\frac{\frac{a^2v^2}{c^2} - \frac{a^2c^2}{c^2}}{a})$$

$$x' = x(\frac{\frac{a^2v^2 - a^2c^2}{c^2}}{a})$$

$$x' = x(\frac{\frac{a^2(v^2 - c^2)}{c^2}}{a})$$

$$x' = x(\frac{a^2(v^2 - c^2)}{c^2} * \frac{1}{a})$$

$$x' = x(\frac{a(v^2 - c^2)}{c^2})$$

$$x' = a(\frac{v^2 - c^2}{c^2})x$$

Getting close, but here is why my algebra skills fail me. How does this reduce down to:

$$x' = a(1-\frac{v^2}{c^2})x$$

 

[rhenretta]

Note, I fixed my math above (I think), though since I reused my last post it could be cached. Shift-F5 will fix that (or ctl f5, or shift refresh, or whatever that keystroke is)

 

[George Jones]

How did you go from

$$x' = x(\frac{a^2}{a} -\frac{( \frac{av}{c})^2}{a})$$

to

$$x' = x(\frac{\frac{a^2v^2}{c^2} - a^2}{a})$$

 

[rhenretta]

crap, I even double checked my work before submitting...

$$x' = (\frac{a^2 - \frac{a^2v^2}{c^2}}{a})x$$

$$x' = (\frac{a^2 * \frac{c^2}{c^2} - \frac{a^2v^2}{c^2}}{a})x$$

$$x' = (\frac{\frac{a^2c^2}{c^2} - \frac{a^2v^2}{c^2}}{a})x$$

$$x' = (\frac{\frac{a^2(c^2 - v^2)}{c^2}}{a})x$$

$$x' = (\frac{a^2(c^2 - v^2)}{c^2} * \frac{1}{a})x$$

$$x' = (\frac{a(c^2 - v^2)}{c^2})x$$

$$x' = a(\frac{c^2 - v^2}{c^2})x$$

I feel sufficiently stupid now, especially since I must be forgetting a reduction rule in algebra still. I don't see how that reduces down farther.

 

[JesseM]

$$x' = a(\frac{c^2 - v^2}{c^2})x$$

I feel sufficiently stupid now, especially since I must be forgetting a reduction rule in algebra still. I don't see show that reduces down farther.

The reduction rule you're missing is just that $$\frac{x + y}{z} = \frac{x}{z} + \frac{y}{z}$$. Apply that to the fraction in your equation above and you get Einstein's equation.

 

[cshum00]

$$x' = a(\frac{c^2 - v^2}{c^2})x$$

I feel sufficiently stupid now, especially since I must be forgetting a reduction rule in algebra still. I don't see how that reduces down farther.

Anyone can forget things unless you were being sarcastic.
$$x' = a(\frac{c^2}{c^2} - \frac{v^2}{c^2})x$$

$$x' = a(1 - \frac{v^2}{c^2})x$$

 

[Doc Al]

$$x' = a(\frac{c^2 - v^2}{c^2})x$$

I feel sufficiently stupid now, especially since I must be forgetting a reduction rule in algebra still. I don't see how that reduces down farther.

Just separate the terms in that fraction, dividing each term in the numerator by the denominator. In general:

$$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$$

 

[rhenretta]

oh, of course, thank you all of you. And, no I was not being sarcastic, I am tough on myself. Expecting perfection is the only way to achieve it.