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In this writing we dive a highly restrictive relation for the metric coefficients in Genera Relativity
We start with the the formula
\[\sum g^{\alpha\beta}g _{\alpha\beta}=4\]    (1)
Summation extends over alpha and beta

In the orthogonal systems (1) reduces to

\[g_{00}g^{00}+ g_{11}g^{11}+ g_{22}g^{22}+ g_{33}g^{33}=4\] (1')
Applying the reversed Cauchy Schwarz inequality to (1')we have,
\[\left[ g_{00}g^{00}- g_{11}\left(-g^{11}\right)- g_{22}\left(-g^{22}\right)- g_{33}\left(-g^{33}\right)\right]^2\\ \ge\left(g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right)\left(\left|(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right)]\] (2)
Using (1) and (2) we obtain,
\[ 16\ge \left(g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right)\ \left[\left(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right]\] (3)
Since for orthogonal coordinate systems we have the formula[1]

\[g_{\alpha \alpha}g^{\alpha\alpha}=1\]
with no summation on alpha in the above
\[\Rightarrow g_{\alpha \alpha}=\frac{1}{g^{\alpha\alpha}}\]

we now have,
\[ \left [g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right]\\ \left[\left(\frac{1}{g_{00}}\right)^2-\left(\frac{1}{g_{11}}\right)^2-\left(\frac{1}{g_{22}}\right)^2-\left(\frac{1}{g_{33}}\right)^2\right]\le 16\]
(4)

Equation (4) is highly restrictive.

The Reversed Cauchy Schwarz Inequality:
We consider 4  real numbers
\[a_1,a_2\]and \[b_1,b_2\]
We have

\[\left(a_1b_1-a_2b_2\right)^2- \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_ab_2-a_2b_1\right)^2\ge 0\]
\[\left(a_1b_1-a_2b_2\right)^2\ge \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\]

\[\left(a_1b_1-a_2b_2\right)^2\ge \left|a_1^2-a_2^2\right|\left|b_1^2-b_2^2\right|\]  (5)

Next we take 2n real numbers a1,a2...an and b1,b2..bn
By applying the Cauchy Schwarz inequality we have,
[\left(a_2b_2+a_3b_3....+a_nb_n\right)^2\le\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)\] 
\[\frac{\left(a_2b_2+a_3b_3....+a_nb_n\right)^2}{\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)}\le 1\]
\[-1\le \frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}\le 1\]
Therefore
\[\frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}=\cos \theta\]

\[\Rightarrow  a_2b_2+a_3b_3....+a_nb_n=\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta  \]
\[a_1 b_1-a_2b_2-a_3b_3....-a_nb_n= \\ a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\] (6)
Applying (5) on the right side of (6) we obtain,
\[ a_1 b_1-a_2b_2-a_3b_3....-a_nb_n  \ge\\ \sqrt{\left|a_1^2-a_2^2-a_3^2.....-a_n^2\right|}\sqrt{\left|b_1^2-b_2^2-b_3^2.....-b_n^2\right|}\]
Or

\[-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\\ \le a_1 b_1-a_2b_2-a_3b_3....-a_nb_n \]
Therefore

\[\left(a_1 b_1-a_2b_2-a_3b_3....-a_nb_n\right)^2 \ge \left|a_1^2-a_2^2-a_3^2.....-a_n^2\right| \\ \times \left|b_1^2-b_2^2-b_3^2.....-b_n^2\right|\]
References
1. Spiegel M,THeory and Problems of Vector Analysis with an Introduction to Tensor Analysis Schaum's series, McGraw-Hill Book Company,Singapore, 1974,Chapter8; Tensor Analysis, problem 48b,p194
 

 

Edited January 18, 20214 yr by Anamitra Palit

Revised Post:

[One may have to refresh the page to view the formulas/equations]
In this writing we dive a highly restrictive relation for the metric coefficients in General Relativity
We start with the the formula
\[\sum g^{\alpha\beta}g _{\alpha\beta}=4\]   (1)
Summation extends over alpha and beta.

In orthogonal systems (1) reduces to

\[g_{00}g^{00}+g_{11}g^{11}+g_{22}g^{22}+g_{33}g^{33}=4\]  (1' )

Applying the reversed Cauchy Schwarz inequality we have,
\[\left[ g_{00}g^{00}- g_{11}\left(-g^{11}\right)- g_{22}\left(-g^{22}\right)- g_{33}\left(-g^{33}\right)\right]^2\\ \ge\left(g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right)\left(\left(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right)\] (2)
Using (1) and (2) we obtain,
\[ 16\ge \left|g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right|\\ \left|\left(g^{00}\right)^2-\left(g^{11}\right)^2-\left(g^{22}\right)^2-\left(g^{33}\right)^2\right|\] (3)
Since for orthogonal coordinate systems we have the formula[1]

\[g_{\alpha \alpha}g^{\alpha\alpha}=1\]
with no summation on alpha in the above
\[\Rightarrow g_{\alpha \alpha}=\frac{1}{g^{\alpha\alpha}}\]

we now have,
\[ \left[g_{00}^2-g_{11}^2-g_{22}^2-g_{33}^2\right]\\ \left[\left(\frac{1}{g_{00}}\right)^2-\left(\frac{1}{g_{11}}\right)^2-\left(\frac{1}{g_{22}}\right)^2-\left(\frac{1}{g_{33}}\right)^2\right]\le 16\]
(4)
Equation (4) is highly restrictive.

On The Reversed Cauchy Schwarz Inequality:
We consider 4  real numbers
\[a_1,a_2\]and \[b_1,b_2\]
We have

\[\left(a_1b_1-a_2b_2\right)^2- \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_ab_2-a_2b_1\right)^2\ge 0\]
\[\left(a_1b_1-a_2b_2\right)^2\ge \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\] (5)

Next we take 2n real numbers a1,a2...an and b1,b2..bn
By applying the Cauchy Schwarz inequality we have,
\[\left(a_2b_2+a_3b_3....+a_nb_n\right)^2\le\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)\] 
\[\frac{\left(a_2b_2+a_3b_3....+a_nb_n\right)^2}{\left(a_2^2+a_3^2+.....a_n^2\right)\left(b_2^2+b_3^2+.....b_n^2\right)}\le 1\]
\[-1\le \frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}\le 1\]
Therefore
\[\frac{a_2b_2+a_3b_3....+a_nb_n}{\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}}=\cos \theta\]

\[\Rightarrow  a_2b_2+a_3b_3....+a_nb_n=\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta  \]
\[a_1 b_1-a_2b_2-a_3b_3....-a_nb_n= \\ a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\] (6)
\[\left(a_1 b_1-a_2b_2-a_3b_3....-a_nb_n\right)^2= \\ \left(a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\right)^2\] (7)
Applying (5) on the right side of (7) we obtain,
\[\left(a_1b_1-\sqrt{a_2^2+a_3^2+.....a_n^2}\sqrt{b_2^2+b_3^2+.....b_n^2}\cos \theta\right)^2\ge \\ \left(a_1^2-a_2^2-a_3^2.....-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.....-b_n^2\right)\] (8)

From (7) and (8) we have,

\[ \left (a_1 b_1-a_2b_2-a_3b_3....-a_nb_n\right)^2  \ge\\ \left(a_1^2-a_2^2-a_3^2.....-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.....-b_n^2\right)\]References
1. Spiegel M,Theory and Problems of Vector Analysis with an Introduction to Tensor Analysis Schaum's series, McGraw-Hill Book Company,Singapore, 1974,Chapter8; Tensor Analysis, problem 48b,p194
 

Edited January 18, 20214 yr by Anamitra Palit

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