Math Is Fun Forum

LQ

Real Member

Registered: 2006-12-04

Posts: 1,285

Solving a new coordinate system. Is my derivation valid.

I have: (x^2 + y^2 + z^2)(1-(v/c)^2) + (ct)^2 = roomtime distance squared = D^2

I divide with t^2 I get:

v^2(1 - (v/c)^2) + c^2 = roomtime speed squared = w^2

now i draw a root:

√(v^2(1 - (v/c)^2 + c^2)

now I draw out v^2 beyond the root:

v√((1 - (v/c)^2 + (c/v)^2)

xc*√((1 - (x)^2 + (1/x)^2)

c*√((x^2 - (x)^4 + 1) = w

(w/c)^2 = x^2 - x^4 + 1

T - T^2 + (1 - (w/c)^2) = 0

T^2 - T - (1 - (w/c)^2) = 0

1/2 +/-√(1/4+1 - (w/c)^2) = T = (v/c)^2 = 1/2 +/-√(5/4 + (w/c)^2)

((v/c)^2 - 1/2)^2 = 1/4 + 1 - (w/c)^2

v^4/c^4 - (v/c)^2 = 1-(w/c)^2

v^4/c^2 - v^2 - 1 = w^2

v^4 - (vc)^2 - c^2- (wc)^2 = 0

c^2/2 +/-√(c^4/4+c^2 + (wc)^2 = v^2

(v^2 - c^2/2)^2 = c^4/4 + c^2 + (wc)^2

v^4 + c^4/4 - (vc)^2 = c^4/4 + c^2 + (wc)^2

v^4 - (vc)^2 = c^2 + (wc)^2

v^2(v^2-c^2) = c^2(1 + w^2)

(v/c)^2(v^2 - c^2) - 1 = w^2

+/-√(v^4/c^2 - v^2 - 1) = w

mw = +/-m√(v^4/c^2 - v^2 - 1)

Last edited by LQ (2010-05-23 21:51:07)

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bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Solving a new coordinate system. Is my derivation valid.

Hi LQ;

Right here:

v^4/c^4 - (v/c)^2 = 1-(w/c)^2

looks like you multiplied both sides by c^2. You forgot to multiply the 1 by c^2.

v^4/c^2 - v^2 - 1 = w^2      Not right.

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