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Lorentz transformation derivation (probably simple calculus)

Technically this is for physics but I think my problem is weak calculus so I'd appreciate some help please.

This is what I have in my notes:

c2dt2dx2=c2dt2dx2c^2dt^2-dx^2=c^2dt'^2-dx'^2

Then using the following;

t=Dx+Et[br]x=Ax+Bt[br]y=y[br]z=z[br]t'=Dx+Et[br]x'=Ax+Bt[br]y'=y[br]z'=z[br]

somehow the next line reads
c2dt2dx2=c2(D2dx2+E2dt2+2DEdxdt)(A2dx2+B2dt2+2ABdxdt)c^2dt^2-dx^2=c^2(D^2dx^2+E^2dt^2+2DEdxdt)-(A^2dx^2+B^2dt^2+2ABdxdt)

I follow the rest of the derivation but basically don't get how

dt2=D2dx2+E2dt2+2DEdxdtdt'^2=D^2dx^2+E^2dt^2+2DEdxdt
and
dx2=A2dx2+B2dt2+2ABdxdtdx'^2=A^2dx^2+B^2dt^2+2ABdxdt

Is there some differentiation procedure I'm forgetting for differentiating:
t=Dx+Et[br]x=Ax+Btt'=Dx+Et[br]x'=Ax+Bt ?

Thanks.
Reply 1
Original post by specrel
Technically this is for physics but I think my problem is weak calculus so I'd appreciate some help please.

This is what I have in my notes:

c2dt2dx2=c2dt2dx2c^2dt^2-dx^2=c^2dt'^2-dx'^2

Then using the following;

t=Dx+Et[br]x=Ax+Bt[br]y=y[br]z=z[br]t'=Dx+Et[br]x'=Ax+Bt[br]y'=y[br]z'=z[br]

somehow the next line reads
c2dt2dx2=c2(D2dx2+E2dt2+2DEdxdt)(A2dx2+B2dt2+2ABdxdt)c^2dt^2-dx^2=c^2(D^2dx^2+E^2dt^2+2DEdxdt)-(A^2dx^2+B^2dt^2+2ABdxdt)

I follow the rest of the derivation but basically don't get how

dt2=D2dx2+E2dt2+2DEdxdtdt'^2=D^2dx^2+E^2dt^2+2DEdxdt
and
dx2=A2dx2+B2dt2+2ABdxdtdx'^2=A^2dx^2+B^2dt^2+2ABdxdt

Is there some differentiation procedure I'm forgetting for differentiating:
t=Dx+Et[br]x=Ax+Btt'=Dx+Et[br]x'=Ax+Bt ?

Thanks.


It's not really differentiation, it's about manipulating small quantities or "differentials". If t' = Dx + Et then dt' = Ddx + Edt then you just square this using the binomial theorem. Similiarly for dx'.
Reply 2
I feel so stupid.

Thanks a lot.

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