- #1
jk22
- 729
- 24
Suppose the system of equations (coming from invariance of the wave equation) :
$$B=-vE\\A^2-B^2/c^2=1\\E^2-c^2D^2=1\\AD=EB/c^2\\B=vA\\AE-BD=1$$
If one adds a lightspeed movement like
$$A=a+f\\B=b-cf\\D=d+h\\E=e-ch$$
Then solving equ 1 for f gives
$$f=(b+ve-vch)/c$$
Equ 4 for h implies
$$h=(ve^2+ac^2d+bcd+vecd)/(vec-ac^2-bc+vc^2d)$$
And finally equ 3 for e yields
$$vce^3\\+e^2(ac^2-vc^2d+bc+2vc^2d)\\+e(vc+vc^3d^2+2c(ac^2d+bcd+c^3vd^2))\\+(-ac^4d^2-bc^3d^2+vc^4d^3-ac^2-bc+vc^2d+2ac^4d^2+2bc^3d^2)=0$$
Hence this should make cubic roots appear in the solving of the system of equation.
Is it a mistake or is it possible to make a cubic come out of a quadratic system of equations ?
$$B=-vE\\A^2-B^2/c^2=1\\E^2-c^2D^2=1\\AD=EB/c^2\\B=vA\\AE-BD=1$$
If one adds a lightspeed movement like
$$A=a+f\\B=b-cf\\D=d+h\\E=e-ch$$
Then solving equ 1 for f gives
$$f=(b+ve-vch)/c$$
Equ 4 for h implies
$$h=(ve^2+ac^2d+bcd+vecd)/(vec-ac^2-bc+vc^2d)$$
And finally equ 3 for e yields
$$vce^3\\+e^2(ac^2-vc^2d+bc+2vc^2d)\\+e(vc+vc^3d^2+2c(ac^2d+bcd+c^3vd^2))\\+(-ac^4d^2-bc^3d^2+vc^4d^3-ac^2-bc+vc^2d+2ac^4d^2+2bc^3d^2)=0$$
Hence this should make cubic roots appear in the solving of the system of equation.
Is it a mistake or is it possible to make a cubic come out of a quadratic system of equations ?