Deriving Lorentz transformation part 2 | Special relativity | Physics | Khan Academy

Continuing the algebra to solve for the Lorentz factor.

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[Jorn]

great video's, however I find the constant changing of colours really distracting, I care about the math and physics, not the colour:)

keep up the great work

[lbqxylk]

Every derivation I see presupposes that the transformation equations are of a particular form. Even in Einstein's original paper, he asserts that the transformation must be linear because of homogeneity nature of space. Most derivations presume beforehand that the transformation equations are linear. It would be nice to see a derivation doesn't presuppose anything aside from the two postulates of SR.

[NITHINS]

Two succesive lorentz transformation in same direction?

[GaudioWind]

It seems to me that you derived a gama that works for time and space coordinates of points on the trajectory of a light speed moving object. Can we really be sure that this same gama would also work for slower speed x t coordinates?

[family-accountemail]

This is better han my lecturers explanation thanks

[takyc]

I just love how general this is, to be able to derive such a simple yet important formula

[Anders]

In the presentation only the principal (positive) square root of gamma is used. There is also a negative root. The only case for both roots being equal is zero. How is a mathematical division-by-zero fallacy avoided in the derivation of the Lorentz transformation?

[robertwilsoniii]

Notice at 6:00 he technically cheated and made an important algebraic mistake, one that many high school students make, so I want to clear this up so people don’t think it’s okay to do it.

You can’t divide out common terms across addition or subtraction in the numerator. You have to factor everything first, and then you’re only allowed to divide off terms that are strictly being multiplied together.

So, if you have to factor out t*t’ from the expression first, to get (c^2)(t)(t’) = ((gamma)^2)(t)(t’)(c^2-v^2) *first, * and then and only then, can you divide by (t)(t^2) to get c^2 = ((gamma)^2)(c^2-v^2).

Please, please, please don’t divide terms across addition without factoring first, to those tempted to do this by this video. He just got lucky that it worked; this is not actually a true statement in general in math.

[FatBitches]

i wish you would use specific numbers.