Every time I try to learn special relativity the statement is made that ds^2 = ds'^2. This time I am trying to derive it. And so here is where I get stuck. (I use 'D' as the Greek capital "delta".)(adsbygoogle = window.adsbygoogle || []).push({});

Accepting Postulate 2 ('c' is constant in all inertial frames), and that the front of a flash of light travels with speed v = c in the original S frame, then:

v_{x}^{2}+ v_{y}^{2}+ v_{z}^{2}= c^{2}

To an inertial observer in the S' frame, then they write the equation as:

v'_{x}^{2}+ v'_{y}^{2}+ v'_{z}^{2}= c^{2}

or, in terms of the coordinates:

(Dx/Dt)^{2}+ (Dy/Dt)^{2}+ (Dz/Dt)^{2}= c^{2}

and

(Dx'/Dt')^{2}+ (Dy'/Dt')^{2}+ (Dz'/Dt')^{2}= c^{2}

Since Dt is the same interval for all three components of speed in S,

and Dt' is the same interval for all three components of speed in S',

then we can write:

(Dx)^{2}+ (Dy)^{2}+ (Dz)^{2}= c^{2}Dt^{2}

and

(Dx').^{2}+ (Dy')^{2}+ (Dz')^{2}= c^{2}Dt'^{2}

Subtracting the "cdt" terms from their respective equations this yields:

c^{2}Dt^{2}- (Dx)^{2}- (Dy)^{2}- (Dz)^{2}= 0

and

c^{2}Dt'^{2}- (Dx')^{2}- (Dy')^{2}- (Dz')^{2}= 0

Here is the exact place where I get stuck. How can I equate these? (Yes, they are both zero, but is that a valid argument? Are any two things, which are equal to zero, themselves equal? E.g. does: 1 - cos^{2}- sin^{2}= Ds^{2}??)

Thanks in advance for any help you may offer.

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# Show that ds^2 = ds'^2

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