Simplest way of deriving Lorentz transforms

[RobikShrestha]

Is there a way to derive Lorentz transforms by knowing just Galilean/Newtonian relativity (adding up velocities) and Pythagoras theorem? I do not understand the principle of homogeneity or any higher mathematics. I can derive t' = gamma * t by using common sense and Pythagoras theorem. I am looking for such easy way to derive Lorentz transforms. I am really new at 'relativity' please help!

 

[HallsofIvy]

Here's a cute method for deriving the length equations. Imagine a person swimming in a river with current v m/s. The person himself can swim at c m/s. First, he swims a length l downstream and the same length upstream. His speed, relative to the shore as he swims downstream, is c+v and the time required to swim downstream a length l is
[tex]\frac{l}{c+ v}[/tex].
His speed, relative to the shore as he swims upstream, is c- v and the time required to swim upstream a length l is
[tex]\frac{l}{c- v}[/tex]

so his total time is
[tex]\frac{l}{c+ v}+ \frac{l}{c- v}= \frac{l(c- v)+ l(c+ v)}{c^2- v^2}= \frac{2lc}{c^2- v^2}= \frac{2}{c}\frac{l}{1- \left(\frac{v}{c}\right)^2}[/tex]

Now, he swims a distance l' across the river and back. Of course, to do that, he must angle slightly upstream so that the vector addition of his speed and the river's will keep him moving across the river. If he take time t, then our vector diagram is a right triangle with hypotenuse of length ct, one leg of length cv, and the other of length l'. By the Pythagorean theorem, [itex](ct)^2= (vt)^2+ l'^2[/itex]. Then [itex](c^2- v^2)t= l'^2[/itex] so that
[tex]t^2= \frac{l'^2}{c^2- v^2}[/tex]
and
[tex]t= \frac{l'}{\sqrt{c^2- v^2}}= \frac{1}{c}\frac{l'}{\sqrt{1- \left(\frac{v}{c}\right)^2}}[/tex]

That is one way across. Coming back will take exactly the same time and so the total time will be
[tex]\frac{2}{c}\frac{l'}{\sqrt{1- \left(\frac{v}{c}\right)^2}}[/tex].

Now, suppose these two swims take exactly the same time. That is,
[tex]\frac{2}{c}\frac{l}{1- \left(\frac{v}{c}\right)^2}= \frac{2}{c}\frac{l'}{\sqrt{1- \left(\frac{v}{c}\right)^2}}[/tex]

Then we must have
[tex]l= l'\sqrt{1- \left(\frac{v}{c}\right)^2}[/tex]
and
[tex]l'= \frac{l}{\sqrt{1- \left(\frac{v}{c}\right)^2}}[/tex].

Now, do you see how that relates to the Michaelson-Moreley experiment and length contraction?

 

[RobikShrestha]

Thanks. Now I have another question: the above derivation assumes the person to be a point. I want to assume the person as a 'length' i.e. I want to keep track of both points of the person (head and legs). This I believe should directly prove length contraction: length2 = length1 / gamma. I can derive t' or l' for an event crossing my frame of reference at the origin. But what if it crosses at x=xo and travels with v velocity? I am struggling to derive the transforms. WHAT I REALLY WANT TO DERIVE is: t' = gamma * (t + x * v / (c*c)). I am really struggling with that.