# Anamitra Palit on the mathematics of General Relativity

Moderator note: Posts in this thread, above this one, were merged from a number of separate threads on related or similar topics, started by the same poster.

Mathematical Theorem
If
$\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\ge 0$
then either
$\left(a_1b_1-a_2b_2\right)\ge \sqrt{a_1^2-a_2^2}\sqrt{b_1^2-b_2^2}$(1.1)
or
$\left(a_1b_1-a_2b_2\right)\le -\sqrt{a_1^2-a_2^2}\sqrt{b_1^2-b_2^2}$(1.2)
Proof
$\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_1b_2-a_2b_1\right)^2$
$\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\ge 0$
$\left(a_1b_1-a_2b_2\right)^2\ge \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)$
Next we prove the reversed Cauchy Schwarz inequality:
If
$\left(a_1^2-a_2^2-a_3^2-......-a_n^2\right)\left(b_1^2-b_2^2-b_3^2-....-b_n^2\right)\ge 0$
Then
$\left(a_1b_1-a_2b_2-a_3b_3-....-a_nb_n \right)\ge \\ \sqrt{a_1^2-a_2^2-a_3^2-.....a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-....-b_n^2}$ (2.1)
or
$\left(a_1b_1-a_2b_2-a_3b_3-....-a_nb_n\right)\le \\ -\sqrt{a_1^2-a_2^2-a_3^2-....a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-....-b_n^2}$ (2.2)
Proof: Applying the Cauchy Schwarz inequality we have,
$\left(a_2b_2+a_3b_3+....+a_nb_n\right)^2\ge \left(a_2^2+a_3^2+....+a_n^2\right)\left(a_2^2+a_3^2+....+a_n^2\right)$
$\left[\frac{a_2 b_2+a_3b_3+...a_nb_n}{\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}}\right]^2\le 1$
$\Rightarrow -1 \le \frac{a_2 b_2+a_3b_3+...a_nb_n}{\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}}\le 1$
$\Rightarrow \frac{a_2 b_2+a_3b_3+...a_nb_n}{\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}}=\cos \theta$ (3.1)
$\Rightarrow a_2b_2+a_3b_3+...a_nb_n=\\ \sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}\cos \theta$
Since the square roots to the right of the last equation are positive quantities,
$\Rightarrow a_1b_1-a_2b_2-a_3b_3-...-a_nb_n=\\ a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}\cos \theta$(3.2)
But
$a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}\cos \theta\ge \\ a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}$(3.3)
Using our mathematical theorem,
$a_1 b_1-\sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2} \ge \\ \sqrt{a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2}$|(4)
provided
$\left(a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2\right)\left(b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2\right)\ge 0$
Therefore, subject to the above criterion,
$\Rightarrow \left( a_1 b_1-a_2b_2-a_3b_3-...-a_nb_n\right)^2\ge\\ \left(a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2\right)\left(b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2\right)$(5)
$\Rightarrow \left( a_1 b1-a_2b_2-a_3b_3-...-a_nb_n\right)\ge\\ \sqrt{a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2}\sqrt {b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2}$
or
$\Rightarrow \left( a_1 b1-a_2b_2-a_3b_3-...-a_nb_n\right)\le \\ -\sqrt{a_1^2-a_2^2-a_3^2-a_4^2-...-a_n^2}\sqrt {b_1^2-b_2^2-b_3^2-b_4^2-...-b_n^2}$(6)
Wikipedia link for the reversed Cauchy Schwarz inequality[in relation to relativity]
https://en.wikipedia.org/wiki/Minkowski_space#Norm_and_reversed_Cauchy_inequality
Keeping in the mind
$\left(c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2\right)\left(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2\right)\ge 0$
we have by the reversed Cauchy Schwarz inequality
$v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2}\sqrt(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2$(7)
$v_1.v_2\ge c^2$ (8)
Metric
$c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2$(9)
$\Rightarrow c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2$(10)
Differentiating both sides of the above with respect to proper time we obtain,
$c^2\frac{dt}{d\tau}\frac{d^2t}{d\tau^2}-\frac{dx}{d\tau}\frac{d^2x}{d\tau^2}-\frac{dy}{d\tau}\frac{d^2y}{d\tau^2}-\frac{dz}{d\tau}\frac{d^2z}{d\tau^2}=0$(11)
$\Rightarrow v.a=0$(12)
Next we consider the following two equations
$v.v= c^2$(13)
$v.ka =0$(14)
Therefore
$v.\left(v-ka\right) = c^2$ (15)
We could expect from the last equation
$\left(v-ka\right)$
to be a proper velocity. Then norm^2=c^2. But we can vary k and make the norm different from c^2 so that v-ka is not a proper velocity.
Let
$\left||v-ka\right||=\bar c^2 \ne c^2$ (16)
We adjust the value of k so that norm^2 of v-ka is positive but diffeent from c^2. This permits the application of the reversed Cauchy Schwarz inequality
From (a) by applying reversed Cauchy Schwarz we have for v and v-ka
$v.\left(v-ka\right) \ge cc'$(17.1)
or
$v.\left(v-ka\right) \le -cc'$ (17.2)
But c' is adjustable. Equation (A) fails
The Sum and Difference of Two Proper Speeds
$c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2$(18.1)
$c^2=c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2$ (18.2)
Adding the last two equations
$2c^2=c^2\left(v_{1t}^2+v_{2t}^2\right)-\left(v_{1x}^2+v_{2x}^2\right)-\left(v_{1y}^2+v_{2y}^2\right)-\left(v_{1z}^2+v_{2z}^2\right)$
$2c^2=c^2\left(v_{1t}+v_{1t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2-2v_1.v_2$(19)
$2c^2+2v_1.v_2=c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2$
$c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2 \ge 4c^2$(20)
If v1 and v2 are four velocities then
$\left ||v_1+v_2 \right||^2\ge 4c^2 \ne c^2$(21)
Thus the sum of two proper velocities is not a proper velocity since the norm fails to match. In a vector space it is not necessary that all vectors should have the same norm. Nevertheless in the physical or in the intuitive sense it is quite uncanny!
The Difference of Two Proper Speeds:
$2c^2=c^2\left(v_{1t}^2-v_{2t}^2\right)-\left(v_{1x}^2-v_{2x}^2\right)-\left(v_{1y}^2-v_{2y}^2\right)-\left(v_{1z}^2-v_{2z}^2\right)$
$2c^2=c^2\left(v_{1t}^2+v_{2t}^2\right)-\left(v_{1x}^2+v_{2x}^2\right)-\left(v_{1y}^2+v_{2y}^2\right)-\left(v_{1z}^2+v_{2z}^2\right)-2v_2.v_2$
$2c^2=c^2\left(v_{1t}-v_{1t}\right)^2-\left(v_{1x}-v_{2x}\right)^2-\left(v_{1y}-v_{2y}\right)^2-\left(v_{1z}-v_{2z}\right)^2+2v_1.v_2-2v_2.v_2$
$2c^2-2v_1.v_2+2v_2.v_2=c^2\left(v_{1t}-v_{2t}\right)^2-\left(v_{1x}-v_{2x}\right)^2-\left(v_{1y}-v_{2y}\right)^2-\left(v_{1z}-v_{2z}\right)^2$
$c^2\left(v_{1t}-v_{2t}\right)^2-\left(v_{1x}-v_{2x}\right)^2-\left(v_{1y}-v_{2y}\right)^2-\left(v_{1z}-v_{2z}\right)^2 \le 2c^2$(22)
If v1 and v2 are four velocities then
$\left ||v_1-v_2 \right||^2\le 2c^2 \ne c^2$(23)
Reversing the sign of v2 in (19) we obtain (23)
We may solve the following three equations
$c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2$(24.1)
$c^2=c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2$(24.2))
$c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2$(24.3)
We have from the last three equations,
$v_1.v_2 \le -\frac{1}{2}c^2$(24.4)
Equation (24.3)assumes the sum of two proper speeds to be a proper speed.
The interesting point is that solutions do exist for these equation: some of the components are not real.If the three equations are viewed mathematically regardless of any physics they imply some sort of a contradiction.equation (24.4) stands in opposition to equation (8)
We may repeat the mathematical theorem and the derivation of the Cauchy |Schwarz inequality considering ai and bj to be complex numbers with
$a_1 b_1-a_2 b_2-a_3b_3-....-a_nb_n$
and
$\left(a_1^2-a_2^2-a_3^2.......-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.......-b_n^2\right)$
as real with
$\left(a_1^2-a_2^2-a_3^2.......-a_n^2\right)\left(b_1^2-b_2^2-b_3^2.......-b_n^2\right)\ge 0$
Existence of solutions to (21.1),(24.2) and (24.3) or equivalently those to (24.1),(24.2) and (24.4) ,irrespective of the nature of the solutions being real or complex, corresponds to a subtle contradiction revealed by the simultaneous validity of equations (8) and (24.4).[We may solve (24.1),(24.2) and (24.4) directly ]
One has to think of equations(24.1),(24.2) and (24.3) in the mathematical sense regardless of the physics involved with them.

Last edited:
Equation (7) of the last post has been rewritten[it could not be parsed in the last post and there was no time left for editing]:

$v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2\sqrt(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2$

For writing (3.3) in the last post we have the following considerations

For X>=0,

X>=X cos theta

Therefore,

a1 b1+X>=a1b1+X cos theta

or,

a1b1-X cos theta>=a1 b1-X

X representing the product of two square roots is a positive number.

For X>=0

$X\ge X \cos \theta$

$a_1 b_1+X\ge a_1b_1+X\cos\theta$

$\Rightarrow a_1b_1-X\cos\theta\ge a_1b_1-X$
$X=\sqrt{a_1^2-a_2^2-a_3^2-.....a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-.....b_n^2}$
Equation (7) has been rewritten:

$v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2\sqrt(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2$

For writing (3.3) in the last post we have the following considerations

For X>=0,

X>=X cos theta

Therefore,

a1 b1+X>=a1b1+X cos theta

or,

a1b1-X cos theta>=a1 b1-X

Xroduct of two square roots, a positive number

For X>=0

$X\ge X \cos \theta$

$a_1 b_1+X\ge a_1b_1+X\cos\theta$

$\Rightarrow a_1b_1-X\cos\theta\ge a_1b_1-X$

Last edited:
Equation (7) of the last post has been rewritten[it could not be parsed in the last post and there was no time left for editing]:

$v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2\sqrt(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2$

For writing (3.3) in the last post we have the following considerations

For X>=0,

X>=X cos theta

Therefore,

a1 b1+X>=a1b1+X cos theta

or,

a1b1-X cos theta>=a1 b1-X

X representing the product of two square roots is a positive number.

For X>=0

$X\ge X \cos \theta$

$a_1 b_1+X\ge a_1b_1+X\cos\theta$

$\Rightarrow a_1b_1-X\cos\theta\ge a_1b_1-X$
$X=\sqrt{a_1^2-a_2^2-a_3^2-.....a_n^2}\sqrt{b_1^2-b_2^2-b_3^2-.....b_n^2}$
Equation (7) has been rewritten:

$v_1.v_2=cv_{1t}cv_{2t}-v_{1x}v_{2x}-v_{1y}v_{2y}-v_{1z}v_{2z}\ge \\ \sqrt{c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2\sqrt(c^2v_{2t}^2-v_{2x}^2-v_{2y}^2-v_{2z}^2}=c^2$

For writing (3.3) in the last post we have the following considerations

For X>=0,

X>=X cos theta

Therefore,

a1 b1+X>=a1b1+X cos theta

or,

a1b1-X cos theta>=a1 b1-X

Xroduct of two square roots, a positive number

For X>=0

$X\ge X \cos \theta$

$a_1 b_1+X\ge a_1b_1+X\cos\theta$

$\Rightarrow a_1b_1-X\cos\theta\ge a_1b_1-X$

Fine , Now in words .

Equation (7) of the last post has been rewritten[it could not be parsed in the last post and there was no time left for editing]:

v1.v2=cv1tcv2t&#x2212;v1xv2x&#x2212;v1yv2y&#x2212;v1zv2z&#x2265;c2v1t2&#x2212;v1x2&#x2212;v1y2&#x2212;v1z2(c2v2t2&#x2212;v2x2&#x2212;v2y2&#x2212;v2z2=c2" role="presentation" style="font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative; display: table-cell !important; width: 10000em !important;">v1.v2=cv1tcv2t−v1xv2x−v1yv2y−v1zv2z≥c2v21t−v21x−v21y−v21z(√c2v22t−v22x−v22y−v22z−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=c2v1.v2=cv1tcv2t−v1xv2x−v1yv2y−v1zv2z≥c2v1t2−v1x2−v1y2−v1z2(c2v2t2−v2x2−v2y2−v2z2=c2

For writing (3.3) in the last post we have the following considerations

For X>=0,

X>=X cos theta

Therefore,

a1 b1+X>=a1b1+X cos theta

or,

a1b1-X cos theta>=a1 b1-X

X representing the product of two square roots is a positive number.

For X>=0

X&#x2265;Xcos&#x2061;&#x03B8;" role="presentation" style="display: inline; font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">X≥XcosθX≥Xcos⁡θ

a1b1+X&#x2265;a1b1+Xcos&#x2061;&#x03B8;" role="presentation" style="display: inline; font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">a1b1+X≥a1b1+Xcosθa1b1+X≥a1b1+Xcos⁡θ

&#x21D2;a1b1&#x2212;Xcos&#x2061;&#x03B8;&#x2265;a1b1&#x2212;X" role="presentation" style="display: inline; font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">⇒a1b1−Xcosθ≥a1b1−X⇒a1b1−Xcos⁡θ≥a1b1−X

X=a12&#x2212;a22&#x2212;a32&#x2212;.....an2b12&#x2212;b22&#x2212;b32&#x2212;.....bn2" role="presentation" style="display: inline; font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">X=a21−a22−a23−.....a2n−−−−−−−−−−−−−−−−−√b21−b22−b23−.....b2n−−−−−−−−−−−−−−−−√X=a12−a22−a32−.....an2b12−b22−b32−.....bn2

Equation (7) has been rewritten:

v1.v2=cv1tcv2t&#x2212;v1xv2x&#x2212;v1yv2y&#x2212;v1zv2z&#x2265;c2v1t2&#x2212;v1x2&#x2212;v1y2&#x2212;v1z2(c2v2t2&#x2212;v2x2&#x2212;v2y2&#x2212;v2z2=c2" role="presentation" style="font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative; display: table-cell !important; width: 10000em !important;">v1.v2=cv1tcv2t−v1xv2x−v1yv2y−v1zv2z≥c2v21t−v21x−v21y−v21z(√c2v22t−v22x−v22y−v22z−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=c2v1.v2=cv1tcv2t−v1xv2x−v1yv2y−v1zv2z≥c2v1t2−v1x2−v1y2−v1z2(c2v2t2−v2x2−v2y2−v2z2=c2

For writing (3.3) in the last post we have the following considerations

For X>=0,

X>=X cos theta

Therefore,

a1 b1+X>=a1b1+X cos theta

or,

a1b1-X cos theta>=a1 b1-X

Xroduct of two square roots, a positive number

For X>=0

X&#x2265;Xcos&#x2061;&#x03B8;" role="presentation" style="display: inline; font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">X≥XcosθX≥Xcos⁡θ

a1b1+X&#x2265;a1b1+Xcos&#x2061;&#x03B8;" role="presentation" style="display: inline; font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">a1b1+X≥a1b1+Xcosθa1b1+X≥a1b1+Xcos⁡θ

&#x21D2;a1b1&#x2212;Xcos&#x2061;&#x03B8;&#x2265;a1b1&#x2212;X" role="presentation" style="display: inline; font-style: normal; line-height: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">⇒a1b1−Xcosθ≥a1b1−X

Fine , Now in words .

Why words ? Because words translates the mathematical , into the physical . To have a mathematical thought is one thing , for the theory to be real is translated into words , words describe the physical , hence prove that the theory is correct or good , or not , or inbetween ( some right , some wrong ) .