Page:Grundgleichungen (Minkowski).djvu/11

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(14) \varrho'\mathfrak{w'_{v}}=\frac{\varrho\mathfrak{w_{v}}-\varrho q}{\sqrt{1-q^{2}}},\ \varrho'\mathfrak{w'_{\bar{v}}}=\varrho\mathfrak{w_{\bar{v}}},


(15) \begin{array}{c}
(\mathfrak{e}'+i\mathfrak{m}')_{\mathfrak{\bar{v}}}=\frac{(\mathfrak{e}+i\mathfrak{m}-i[\mathfrak{w},\ \mathfrak{e}+i\mathfrak{m}])_{\bar{v}}}{\sqrt{1-q^{2}}}\\
(\mathfrak{e}'+i\mathfrak{m}')_{\mathfrak{v}}=(\mathfrak{e}+i\mathfrak{m}-i[\mathfrak{w},\ \mathfrak{e}+i\mathfrak{m}])_{\mathfrak{v}}\end{array},

Then it follows that the equations I), II), III), IV) are transformed into the corresponding system with dashes.

The solution of the equations (10), (11), (12) leads to

(16) \mathfrak{r_{v}}=\frac{\mathfrak{r'_{v}}+qt'}{\sqrt{1-q^{2}}},\ \mathfrak{r_{\bar{v}}}=\mathfrak{r'_{\bar{v}}},\ t=\frac{q\mathfrak{r'_{v}}+t'}{\sqrt{1-q^{2}}}.

Now we shall make a very important observation about the vectors \mathfrak{w} and \mathfrak{w}'. We can again introduce the indices 1, 2, 3, 4, so that we write x'_{1},\ x'_{2},\ x'_{3},\ x'_{4} instead of x,' y,' z,' it' , and \varrho'_{1},\ \varrho'_{2},\ \varrho'_{3},\ \varrho'_{4} instead of \varrho'\mathfrak{w}'_{x'}\ \varrho'\mathfrak{w}'_{y'}\ \varrho'\mathfrak{w}'_{z'}\ i\varrho'. Like the rotation round the z-axis, the transformation (4), and more generally the transformations (10), (11), (12), are also linear transformations with the determinant +1, so that

(17) x^{2}_{1} + x^{2}_{2} + x^{2}_{3} + x^{2}_{4}, d. i. x^{2} + y^{2} + z^{2} - t^{2}

is transformed into

x^{'2}_{1} + x^{'2}_{2} + x^{'2}_{3} + x^{'2}_{4}, d. i. x'^{2} + y'^{2} + z'^{2} - t'^{2}.

On the basis of the equations (13), (14), we shall have


transformed into \varrho'(1-\mathfrak{w}'^{2}) or in other words,

(18) \varrho\sqrt{1-\mathfrak{w}^{2}},

is an invariant in a Lorentz-transformation.

  1. The brackets shall only summarize the expressions, which are related to the index, and [\mathfrak{w},\mathfrak{e}+i\mathfrak{m}] shall denote the vector product of \mathfrak{w} and +i\mathfrak{m}.