Equating these elements and rearranging gives: γ ( v ′ ) γ ( v ) − v δ ( v ′ ) γ ( v ) = − v ′ γ ( v ′ ) δ ( v ) + γ ( v ′ ) γ ( v ) {\displaystyle \gamma (v')\gamma (v)-v\delta (v')\gamma (v)=-v'\gamma (v')\delta (v)+\gamma (v')\gamma (v)\,} v δ ( v ′ ) γ ( v ) = v ′ γ ( v ′ ) δ ( v ) {\displaystyle v\delta (v')\gamma (v)=v'\gamma (v')\delta (v)\,} δ ( v ) v γ ( v ) = δ ( v ′ ) v ′ γ ( v ′ ) . {\displaystyle {\frac {\delta (v)}{v\gamma (v)}}={\frac {\delta (v')}{v'\gamma (v')}}.\,} The denominator will be nonzero for nonzero v, because γ(v) is always nonzero; γ 2 + v δ γ = 1 {\displaystyle \gamma ^{2}+v\delta \gamma =1} .
Bu unsurlar denk ve yeniden düzenlenerek verilir: γ ( v ′ ) γ ( v ) − v δ ( v ′ ) γ ( v ) = − v ′ γ ( v ′ ) δ ( v ) + γ ( v ′ ) γ ( v ) {\displaystyle \gamma (v')\gamma (v)-v\delta (v')\gamma (v)=-v'\gamma (v')\delta (v)+\gamma (v')\gamma (v)\,} v δ ( v ′ ) γ ( v ) = v ′ γ ( v ′ ) δ ( v ) {\displaystyle v\delta (v')\gamma (v)=v'\gamma (v')\delta (v)\,} δ ( v ) v γ ( v ) = δ ( v ′ ) v ′ γ ( v ′ ) . {\displaystyle {\frac {\delta (v)}{v\gamma (v)}}={\frac {\delta (v')}{v'\gamma (v')}}.\,} Payda sıfır için sıfır olmayan v için sıfır olmayacak,çünkü γ(v) her zaman sıfırdan farklı ; γ 2 + v δ γ = 1 {\displaystyle \gamma ^{2}+v\delta \gamma =1} .