Question

Check Eq. 12.29, using Eq. 12.27. [This only proves the invariance of the scalar product for transformations along the x direction. But the scalar product is also invariant under rotations, since the first term is not affected at all, and the last three constitute the three-dimensional dot product a-b . By a suitable rotation, the x direction can be aimed any way you please, so the four-dimensional scalar product is actually invariant under arbitrary Lorentz transformations.]

Short answer

It is proved that $-{\overline{a}}^0{\overline{b}}^0+{\overline{a}}^1{\overline{b}}^1+{\overline{a}}^2{\overline{b}}^2+{\overline{a}}^3{\overline{b}}^3=-a^0{b}^0+a^1{b}^1+a^2{b}^2+a^3{b}^3$.

Step-by-step solution

Expression for the four-dimensional scalar product:

Using the equation , write the equation for the four-dimensional scalar product.

$\mathbf{-}{\overline{\mathit{a}}}^{\mathbf{0}}{\overline{\mathit{b}}}^{\mathbf{0}}\mathbf{+}{\overline{\mathit{a}}}^{\mathbf{1}}{\overline{\mathit{b}}}^{\mathbf{1}}\mathbf{+}{\overline{\mathit{a}}}^{\mathbf{2}}{\overline{\mathit{b}}}^{\mathbf{2}}\mathbf{+}{\overline{\mathit{a}}}^{\mathbf{3}}{\overline{\mathit{b}}}^{\mathbf{3}}\mathbf{=}\mathbf{-}{\mathit{a}}^{\mathbf{0}}{\mathit{b}}^{\mathbf{0}}\mathbf{+}{\mathit{a}}^{\mathbf{1}}{\mathit{b}}^{\mathbf{1}}\mathbf{+}{\mathit{a}}^{\mathbf{2}}{\mathit{b}}^{\mathbf{2}}\mathbf{+}{\mathit{a}}^{\mathbf{3}}{\mathit{b}}^{\mathbf{3}}$ …… (1)

Prove the equation of the four-dimensional scalar product:

From the equation 12.27 , the values of ${\overline{\mathrm{a}}}^0,{\overline{\mathrm{a}}}^1,{\overline{\mathrm{a}}}^2\mathrm{and}{\overline{\mathrm{a}}}^3$are given as:

$$\begin{gathered} {\overline{a}}^0=\gamma \left(a^0-\beta a^1\right) \\ {\overline{a}}^1=\gamma \left(a^1-\beta a^0\right) \\ {\overline{a}}^2=a^2 \\ {\overline{a}}^3=a^3 \end{gathered}$$

Similarly, for the values of ${\overline{\mathrm{b}}}^0,{\overline{\mathrm{b}}}^1,{\overline{\mathrm{b}}}^2\mathrm{and}{\overline{\mathrm{b}}}^3$and :

$$\begin{gathered} {\overline{b}}^0=\gamma \left(b^0-\beta b^1\right) \\ {\overline{b}}^1=\gamma \left(b^1-\beta b^0\right) \\ {\overline{b}}^2=b^2 \\ {\overline{b}}^3=b^3 \end{gathered}$$

Solve L.H.S. of the given equation.

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}=-{\overline{a}}^0{\overline{b}}^0+{\overline{a}}^1{\overline{b}}^1+{\overline{a}}^2{\overline{b}}^2+{\overline{a}}^3{\overline{b}}^3 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\left\{\begin{array}{c}\left[\begin{array}{cc}-\left(\gamma \left(\begin{array}{c}{a}^0-\beta a^1\end{array}\right)\right)& \left(\gamma \left(\begin{array}{c}{b}^0-\beta b^1\end{array}\right)\right)\end{array}\right]+\\ \left[\begin{array}{cc}\left(\gamma \left(\begin{array}{c}\begin{array}{c}{a}^1-\beta a^0\end{array}\end{array}\right)\right)& \left(\begin{array}{c}\gamma \left(\begin{array}{c}{b}^1-\beta b^0\end{array}\right)\end{array}\right)\end{array}\right]+a^2{b}^2+a^3{b}^3\end{array}\right\} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}=-{\gamma }^2\left[\left(\begin{array}{c}{a}^0-\beta a^1\end{array}\right)\left(\begin{array}{c}\begin{array}{c}{b}^0-\beta b^1\end{array}\end{array}\right)\right]+{\gamma }^2\left[\begin{array}{cc}\left(\begin{array}{c}\begin{array}{c}{a}^1-\beta a^0\end{array}\end{array}\right)& \left(\begin{array}{c}\begin{array}{c}{b}^1-\beta b^0\end{array}\end{array}\right)\end{array}\right]+a^2{b}^2+a^3{b}^3 \end{gathered}$$

On further solving,

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}=\left\{-{\gamma }^2\left[\begin{array}{c}{a}^0{b}^0-a^0\beta b^1-\beta a^1{b}^0+{\beta }^2{a}^1{b}^1-a^1{b}^1+\\ a^1\beta b^0+\beta a^0{b}^1-{\beta }^2{a}^0{b}^0\end{array}\right]+a^2{b}^2+a^3{b}^3\right\} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}=-{\gamma }^2{a}^0{b}^0\left(1-{\beta }^2\right)+{\gamma }^2{a}^1{b}^1\left(1-{\beta }^2\right)+a^2{b}^2+a^3{b}^3 \end{gathered}$$

Substitute${\mathrm{\gamma }}^2\left(\begin{array}{c}1-{\mathrm{\beta }}^2\end{array}\right)=1$ in the above equation.

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}=-a^0{b}^0\left(1\right)+a^1{b}^1\left(1\right)+a^2{b}^2+a^3{b}^3 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}=-a^{\mathit{0}}{b}^{\mathit{0}}+a^{\mathit{1}}{b}^{\mathit{1}}+a^{\mathit{2}}{b}^{\mathit{2}}+a^{\mathit{3}}{b}^{\mathit{3}} \end{gathered}$$

So, it is found as:

$\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}$

Therefore, it is proved that$-{\overline{a}}^0{\overline{b}}^0+{\overline{a}}^1{\overline{b}}^1+{\overline{a}}^2{\overline{b}}^2+{\overline{a}}^3{\overline{b}}^3=-a^0{b}^0+a^1{b}^1+a^2{b}^2+a^3{b}^3$.