# Anamitra Palit on the mathematics of General Relativity

Discussion in 'Pseudoscience' started by Anamitra Palit, Dec 18, 2020.

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3. ### Anamitra PalitRegistered Senior Member

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Last edited: Jan 13, 2021
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7. ### Anamitra PalitRegistered Senior Member

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Last edited: Jan 13, 2021

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9. ### Anamitra PalitRegistered Senior Member

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Important revisions have been made in "the Extra Bit"

https://drive.google.com/file/d/1V_Ms1FfeiQKhMNk9lfqhxqr3GAcguBN5/view?usp=sharing

Relevant material in Latex:
Metric

$$c^2d\tau^2=c^2dt^2-dx2-dy^2-dz^2$$ (1)
$$c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2$$
$$c^2=c^2{v_t}^2-{v_x}^2-{v_y}^2-{v_z}^2$$(2)

We consider two proper velocities on the same manifold

$$c^2=c^2{v_{1t}}^2-{v_{1x}}^2-{v_{1y}}^2-{v_{1z}}^2$$(3.1)

$$c^2=c^2{v_{2t}}^2-{v_{2x}}^2-{v_{2y}}^2-{v_{2z}}^2$$(3.2)

Adding (3.1) and (3.2) we obtain

$$2c^2=c^2\left({v_{1t}}^2+{v_{2t}}^2\right)-\left({v_{1x}}^2+{v_{2x}}^2\right)-\left({v_{1y}}^2+{v_{2y}}^2\right)-\left({v_{1z}}^2+{v_{2z}}^2\right)$$

$$2c^2=c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2-2v_1\dot v_2$$

$$2c^2+2v_1\dot v_2=c^2\left(v_{1t}+v_{2t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-left(v_{1z}+v_{2z}\right)^2$$

Since v1.v2>=c^2 we have

$$c^2\left({v_{1t}}+{v_{2t}}\right)^2-\left({v_{1x}}+{v_{2 x}}\right)^2-\left({v_{1y}}+{v_{1y}}\right)^2-\left({v_{1z}}+{v_{1z}}\right)^2\ge 4c^2$$(4)

$$\left(v_1+v2)\dot (v_1+v_2)\right) \ge 4c^2$$(5)

If $v_1+v_2$ is a proper velocity then

$$c^2=c^2\left(v_{1t}+v_{2t})\right)^2-\left(v_{1x}+v_{2x})\right)^2-\left(v_{1y}+v_{2y})\right)^2-\left(v_{1z}+v_{2z})\right)^2$$ (6)

$$c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+2v_1.v_2$$

$$c^2=c^2+c^2+2v_1.v_2$$(7)

Therefore

$$v_1.v_2\le -½ c^2$$(8)

which is not true since

$$v.v=c^2$$

Therefore $$v_1+v_2$$ is not a four vector if $$v_1$$ and $$v_2$$ are four vectors

Again if $$v_1-v_2$$ is a four vector then

$$c^2=c^2\left(v_{1t}-v_{2t})\right)^2-\left(v_{1x}-v_{2x})\right)^2-\left(v_{1y}-v_{2y})\right)^2-\left(v_{1z}-v_{2z})\right)^2$$ (9)

$$c^2=c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2+ c^2v_{1t}^2-v_{1x}^2-v_{1y}^2-v_{1z}^2-2v_1.v_2$$
$$c^2=c^2+c^2-2v_1.v_2$$
$$½ c^2=v_1.v_2$$ (10)

But the above formula is not a valid one. Given two infinitesimally close four velocities their difference is not a four velocity. Therefore the manifold has to be a perforated one. The manifold indeed is a mesh of worldlines and each world line is a train of proper velocity four vectors as tangents. A particle moves along a timelike path and therefore each point on it has a four velocity as a tangent representing the motion. The manifold is discrete and that presents difficulty an impossibility to be precise with procedure like differentiation.

Last edited: Jan 14, 2021
10. ### Anamitra PalitRegistered Senior Member

Messages:
102
[It may be necessary to refresh the page for proper viewing]
We start with the norm of proper velocity[metric signature(+,-,-,-)]
$$c^2=c^2 v_t^2-v_x^2-v_y^2-v_z^2$$

$$c^2=c^2\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dt}{d\tau}\right)^2-\left(\frac{dt}{d\tau}\right)^2$$

Differentiating with respect to propertime,

\begin {equation}c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}-\frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0\end {equation} (1)

$$\Rightarrow v.a=0$$(2)

We choose k such that [k] =T so that ka has the dimension of velocity

We have,

$$\Rightarrow v.ka=0$$(3)

We have had earlier, $$v.v=c^2$$(4)

From (3) and (4)

$$\Rightarrow v. \left(v-ka\right)=c^2$$ (5)

By adjusting the value [but maintaining its dimension as that of time] we always do have equation (5)

If $\left(v-ka\right)=v'$ is a proper velocity then we have $v.v'=c^2$ in opposition to $v.v'>=c^2$

If $\left(v-ka\right)=v'$ is a not a proper velocity then
$$c^2\left(\frac{dt'}{d\tau'}\right)^2-\left(\frac{dx'}{d\tau'}\right)^2-\left(\frac{dy'}{d\tau'}\right)^2-\left(\frac{dz'}{d\tau'}\right)^2=c'^2 \ne c^2$$

We have from the reversed Cauchy Schwarz inequality,

\begin{array}{l}\left(c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\right)^2\ge\\ \left(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\right)\left(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\right)\end{array}(6)

or,$$\left(c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\right)^2\ge c^2c'^2$$ (7)

$$c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\ge cc'$$

or,

$$c^2 \frac{dt}{d\tau}\frac{dt'}{d\tau'}-\frac{dx}{d\tau}\frac{dx'}{d\tau'}-\frac{dy}{d\tau}\frac{dy'}{d\tau'}-\frac{dz}{d\tau}\frac{dz'}{d\tau'}\le -cc'$$

v.v'>=cc' or v.v'<=-cc'

But v.v'=c^2. Therefore the solution is c'=c .We have v' is a proper velocity.||But we assumed /postulated at the very outset that v' is not a proper velocity.

Last edited: Jan 15, 2021
11. ### Anamitra PalitRegistered Senior Member

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102
[One may require to refresh the page for proper viewing]
Norm of Four Acceleration

Four Acceleration
$$\left(c\frac{d^2 t}{d\tau^2},\frac{d^2 x}{d\tau^2},\frac{d^2 y}{d\tau^2}, \frac{d^2 z}{d\tau^2}\right)$$ (1)
Let
$$c^2N=c^2\left(\frac{d^2 t}{d\tau^2}\right)^2-\left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2$$ (2)
We consider the metric
$$c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2$$ (4)
\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
(5)
Differentiating (5) with respect to proper time we have,
$c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0$ (6)
By applying the Cauchy Schwarz inequality we have,
$\left(\frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}+\frac{d y}{d\tau}\frac{d^2y}{d \tau^2}+\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}\right)^2 \\ \ge \left(\left(\frac{d x}{d\tau}\right)^2+\left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau} \right)^2\right)\left(\left(\frac{d^2 x}{d \tau^2}\right)^2+\left(\frac{d^2 y}{d \tau^2}\right)^2+\left(\frac{d^2 z}{d \tau^2}\right)^2\right)$(7)
or,
$\left(c^2\left(\frac{d t}{d \tau}\right)^2-c^2\right)\left(c^2\left( \frac {d^2 t}{d \tau^2}\right)^2-c^2N\right) \ge\left( c^2 \frac {d^2 t}{d\tau^2}\right)^2\left(c^2\frac{d t}{d\tau}\right)^2$
$\left(\left(\frac{dt}{d \tau}\right)^2-1\right)\left(\left( \frac {d^2 t}{d \tau^2}\right)^2-N\right) \ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2$ (8)
$\left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2$
$\Rightarrow-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge 0$ (9)
$N\left(1-\left(\frac{dt}{d\tau}\right)^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2$
$N\left(1-\gamma^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2$(10)
The right side of (10) is always positive or zero. Therefore the left side is also positive or zero. Therefore N<=0 since gamma[Lorentz factor] is positive[>= unity]. N cannot be positive unless the particle is moving uniformly.
If N is negative then from (1) we have
$c^2\left(\frac{d^2 t}{d\tau^2}\right)^2\le \left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2$
For a particle at rest (spatially) and N<0,
$\left(\frac{d^2 t}{d\tau^2}\right)^2\le 0$(11)
Equation (11) will not hold, the left side being a [perfect square and hence positive or zero]unless
$$\frac{d^2 t}{d\tau^2}=0$$
that is unless $$\frac{dt}{d\tau}=constant \Rightarrow \gamma=constant$$
that is unless the particle is moving with a constant velocity. An accelerating particle will not cater to N<0.
For N=0 we have from (10)
$\left(\frac{d^2x}{d\tau^2}\right)^2\le 0$(12)
Equation (12) is not a valid on unless the particle moves with a constant velocity.
We conclude that the norm square of the acceleration vector c^2N cannot be positive, negative or zero unless the particle is moving uniformly that is unless it moves with a constant velocity

12. ### Anamitra PalitRegistered Senior Member

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102
In Latex:
[One may require to refresh the page for proper viewing]
The difference of proper speeds is not a proper speed
Indeed
$c^2=c^2v_{1t}^2-v_{1x}^2--v_{1y}^2-v_{1z}^2$
$c^2=c^2v_{2t}^2-v_{2x}^2--v_{2y}^2-v_{2z}^2$
By adding the last two equations we obtain
$2c^2=c^2\left(v_{1t}^2+v_{2t}^2\right)-\left(v_{1x}^2+v_{2x}^2\right)-\left(v_{1y}^2+v_{2y}^2\right)-\left(v_{1z}^2+v_{2z}^2\right)$
$2c^2=c^2\left(v_{1t}-v_{2t}\right)^2-\left(v_{1x}-v_{2x}\right)^2-\left(v_{1y}-v_{2y}\right)^2-\left(v_{1z}-v_{2z}\right)^2+2v_1.v_2$
Since v1.v2>=c^2 [earlier postings] we obtain
[Norm(v1-v2)]<=0 that is $\left(v_1-v_2\right).\left(v_1-v_2\right)<=0$
That the difference or the sum of two proper speeds is not a proper speed[in that their norms are not equal to that of a proper speed] is a bit uncanny...
The difference of proper speeds is related to acceleration...
Four Acceleration
$$\left(c\frac{d^2 t}{d\tau^2},\frac{d^2 x}{d\tau^2},\frac{d^2 y}{d\tau^2}, \frac{d^2 z}{d\tau^2}\right)$$ (1)
Let
$$c^2N=c^2\left(\frac{d^2 t}{d\tau^2}\right)^2-\left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2$$ (2)
We consider the metric
$$c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2$$ (4)
\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
(5)
Differentiating (5) with respect to proper time we have,
$c^2\frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}-\frac{dy}{d\tau}\frac{d^2 y}{d \tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}=0$ (6)
By applying the Cauchy Schwarz inequality we have,
$\left(\frac{dx}{d\tau}\frac{d^2 x}{d \tau^2}+\frac{d y}{d\tau}\frac{d^2y}{d \tau^2}+\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2}\right)^2 \\ \ge \left(\left(\frac{d x}{d\tau}\right)^2+\left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau} \right)^2\right)\left(\left(\frac{d^2 x}{d \tau^2}\right)^2+\left(\frac{d^2 y}{d \tau^2}\right)^2+\left(\frac{d^2 z}{d \tau^2}\right)^2\right)$(7)
or,
$\left(c^2\left(\frac{d t}{d \tau}\right)^2-c^2\right)\left(c^2\left( \frac {d^2 t}{d \tau^2}\right)^2-c^2N\right) \ge\left( c^2 \frac {d^2 t}{d\tau^2}\right)^2\left(c^2\frac{d t}{d\tau}\right)^2$
$\left(\left(\frac{dt}{d \tau}\right)^2-1\right)\left(\left( \frac {d^2 t}{d \tau^2}\right)^2-N\right) \ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2$ (8)
$\left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge \left( \frac {d^2 t}{d\tau^2}\right)^2\left(\frac{dt}{d\tau}\right)^2$
$\Rightarrow-N\left(\frac{dt}{d \tau}\right)^2-\left( \frac {d^2 t}{d \tau^2}\right)^2+N\ge 0$ (9)
$N\left(1-\left(\frac{dt}{d\tau}\right)^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2$
$N\left(1-\gamma^2\right)\ge \left( \frac {d^2 t}{d \tau^2}\right)^2$(10)
The right side of (10) is always positive or zero. Therefore the left side is also positive or zero. Therefore N<=0 since gamma[Lorentz factor] is positive[>= unity]. N cannot be positive unless the particle is moving uniformly.
If N is negative then from (1) we have
$c^2\left(\frac{d^2 t}{d\tau^2}\right)^2\le \left(\frac{d^2 x}{d\tau^2}\right)^2-\left(\frac{d^2 y}{d\tau^2}\right)^2-\left( \frac{d^2 z}{d\tau^2}\right)^2$
For a particle at rest (spatially) and N<0,
$\left(\frac{d^2 t}{d\tau^2}\right)^2\le 0$(11)
Equation (11) will not hold, the left side being a [perfect square and hence positive or zero]unless
$$\frac{d^2 t}{d\tau^2}=0$$
that is unless $$\frac{dt}{d\tau}=constant \Rightarrow \gamma=constant$$
that is unless the particle is moving with a constant velocity. An accelerating particle will not cater to N<0.
For N=0 we have from (10)
$\left(\frac{d^2x}{d\tau^2}\right)^2\le 0$(12)
Equation (12) is not a valid on unless the particle moves with a constant velocity.
We conclude that the norm square of the acceleration vector c^2N cannot be positive, negative or zero unless the particle is moving uniformly that is unless it moves with a constant velocity

Last edited: Jan 16, 2021
13. ### James RJust this guy, you know?Staff Member

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35,782
Anamitra Palit:

It is not clear what you want to discuss in this thread.

You have not given any indication as to why you believe the results you have obtained are important, or what they imply.

This is a discussion forum. Do you have questions for our members?

If you're merely looking for a place to publish your work, why don't you submit it to a peer-reviewed scientific journal?

14. ### Anamitra PalitRegistered Senior Member

Messages:
102
On the Riemann Curvature Tensor
Linkhttps://drive.google.com/file/d/1C2-ru6uuDIw9u_e4HQ1bdwa01oKysQ-B/view?usp=sharing
Material in Latex[It might be necessary to refresh the page for viewing the formulas and the equations]
The paper establishes mathematically that he Riemann Tensor is a zero tensor.
$R^{\mu\nu}_{\quad\gamma\delta}=g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}$ (1)
Interchanging the dummy indices alpha and beta we have,
$R^{\mu\nu}_{\quad\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}$(2)
Therefore,
$g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}$ (3)
But
$R_{\beta\alpha\mu\nu}=-R_{\alpha\beta\mu\nu}$ (4)
Thus we have
$g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=-g^{\beta\mu}g^{\alpha\nu}R_{\alpha\beta\gamma\delta}$ (5)
$R_{\alpha\beta\gamma\delta}\left[g^{\alpha\mu}g^{\beta\nu}+g^{\beta\mu}g^{\alpha\nu}\right]=0$ (6)
With (6) alpha and beta are dummy indices;others are free indices.Treating the Riemann tensor components as variables[unknowns], there are sixteen of them for the sixteen alpha beta combinations any given gamma , delta and mu, nu combinations.Each equation has gamma,delta,mu and nu as constant quantities. In total we have 256 equations[four values for each gamma,delta,mu and nu].Since these equations are of homogeneous nature,
$R_{\alpha\beta\gamma\delta}=0$ (7)
In the orthogonal system of coordinates we obtain from (6) for distince mu and nu
$R_{\mu\nu\gamma\delta}\left[g^{\mu\mu}g^{\nu\nu}+g^{\nu\mu}g^{\mu \nu}\right]=0$ (8)
In (8) we have considered different values of mu and nu as well as different values for gamma and delta. One should also take note of the fact that there is no summation on mu and on nu.
$R_{\mu\nu\gamma\delta}g^{\mu\mu}g^{\nu\nu}=0$ (9)
In (8) also there is no summation on mu and on nu.
Since g_mu mu not equal to 0 and g_nu nu not equal to zero we have for unequal mu ,nu and unequal gamma delta,
$R_{\mu\nu\gamma\delta}=0$ (10)
If the Riemann curvature tensor is a zero tensor then the Ricci tensor is also a zero tensor. The Ricci scalar becomes a zero valued scalar

Last edited: Jan 20, 2021
15. ### James RJust this guy, you know?Staff Member

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35,782
Clearly, you must have made an error somewhere.

16. ### QuarkHeadRemedial Math StudentValued Senior Member

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1,717
No he hasn't, all his so-called equalities are false, starting with the ridiculous equation (1)

Everything else is wrong, and we end up with this pearl:

with respect to orthogonal coordinates the curvature field is zero i.e. flat space is flat.

Well tickle my tits till Tuesday, who would have thought it

17. ### exchemistValued Senior Member

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10,288
It's curried Reiku, in my opinion.

Please Register or Log in to view the hidden image!

May not be Gareth himself, but if not actually him then someone with a very similar mental condition, i.e. on the, ahem, spectrum.

18. ### Anamitra PalitRegistered Senior Member

Messages:
102
[It might be necessary to refresh the page for viewing the formulas and the equations]
We consider the transformation of the rank two contravariant tensor:
$\bar A^{\mu \nu}=\frac{\partial \bar x^{\mu}}{\partial x^\alpha}\frac{\partial \bar x^{\nu}}{\partial x^\beta}A^{\alpha\beta}$(1)
Inverse transformation[for non singular transformations]
$A^{\alpha \beta}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\beta}}{\partial \bar x^\nu}\bar A^{\mu\nu}$(2)
For the diagonal components[alpha=beta]
$A^{\alpha \alpha}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\alpha}}{\partial \bar x^\nu}\bar A^{\mu\nu}$ (3)
We consider the situation where the off diagonal components of A[alpha not equal to beta] are all zero
The diagonal elements of A-bar are given by
$\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2 A^{\alpha\alpha}$ (4)
For off diagonal elements of A-bar are given by
'
$\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}$ (4’)
Subject to the situation that the off diagonal elements of A are all zero we consider the following three cases:
Case 1.
Assume A-bar^ mu nu=0for all μ≠ν for all barred reference frames[the general condition that off diagonal elements of A are zero continues to hold] we have
$\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}=0$ (4’’)
We have (4'') irrespective of the transformation elements[all reference frames being considered]. From (4’’) A-bar^mu mu=0.Again from (4’) and (4’’) |A-bar ^mu nu becomes zero for μ≠ν.
Then the tensor becomes null
Case 2
Off diagonal elements are non zero for all A-bar[that the off diagonal elements of A are all zero continues to hold]
$\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}$
$\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}$
$\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}$
$\Rightarrow\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2\delta^\alpha_\beta \delta^\alpha_\gamma=\frac{\partial \bar x^\mu} {\partial x^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}$
For β≠γ we have from (5)
$\frac{\partial \bar x^\mu}{\partial x ^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}=0$(6)
$\Rightarrow\frac{\partial \bar x^\mu}{\partial x^\beta}=0$or
$\frac{\partial \bar x^\mu}{\partial x ^\beta}=0$ (7)
$\Rightarrow \bar A^{\mu\nu}=0\Rightarrow A^{\alpha\beta}=0$ (8)
Equations represented by (7) and (8)are not true! We do have an enigma of a persistent nature.
[In fact (8) implies that the metric tensor is the null tensor;the Riemann tensor and the Ricci tensor are null tensors; the Ricci scalar is zero valued.]
Case 3
For A as per our initial postulation the off diagonal elements of A are all zero. For A-bar there we assume the existence of at least one mu ,nu pair in each barred frame such that A-bar^mu nu=0.These mu nu pairs may not be identical for all barred frames. Since we may have an infinitude of reference frames catering to our assumption it follows that for some specific (μ,ν)pair also A-bar mu mu is zero in an infinitely many frames of reference
$\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}$
Since A^m mu=0 for a specific pair in infinitely many frames,we have,
$\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}=0$(7)
For the same mu nu pair (7) we have an infinitude of equations like (7) whee the transformation elements are distinct. We have included infinitely many barred frames in equation (7)
$\Rightarrow A^{\alpha\alpha}=0\Rightarrow A=0$
(8)
Case 1 is an instance of case 3
Points to Observe:
1. Anti symmetric tensors transform to anti symmetric tensors in all frames of reference. Noe the null tensor is an antisymmetic tensor asides being a symmetric one.An arbitrary non zero[non null] tensor may be expressed as the sum of a symmetric an an antisymmetric tensor. Our analysis might be considered for an arbitrary tensor which is neither symmetric or antisymmetric.It reduces the null tensor making both the symmetric and the antisymmetric parts zero.
2.If a tensor has a component of identical value in all reference frames it has to be the null tensor.
3. Let us represent (1) in a standard matrix form
$\bar A=MAM^T$(9)
where M is the transformation matrix
For an arbitrary M even if M is non singular in nature, we will not necessarily have both A and A-bar as diagonal[both withy the off diagonal elements as zero].If one of A or A-bar is diagonal with the other non diagonal then A ,A-bar both become null tensors as we have already seen!
One may test the situation using a diagonal tensor with A and an arbitrary square matrix for M
For an arbitrary transformation matrix ,a diagonal tensor A does not necessarily produce a diagonal A-bar. Only by specific valid choice of the space time transformation can we have both sides of (10) or (1) for that matter, as diagonal[components zero for alpha not equal to beta]. An arbitrary[non singular ] transformation will produce from a diagonal tensor a non diagonal. It is always possible to have infinitely many transformation for which the off diagonal elements are non zero[diagonal elements of the tensor in some reference frame are assumed to be non zero]

Last edited: Jan 23, 2021
19. ### Anamitra PalitRegistered Senior Member

Messages:
102
QuarkHead:
Equation (1) and the facts ensuing from it correct; the only point is that $R^{\mu\nu}_{\quad \alpha\beta}$ is not formally used. Nevertheless raising and lowering of indices can always done.The antisymmetric property applied in the work [and notified through equation (4)] is a formal one.

20. ### Anamitra PalitRegistered Senior Member

Messages:
102
In relation to my last comment
that $R^{\mu\nu}_{\quad\alpha\beta}$ is not formally used, is meant to imply that this tensor or $R^{\mu\nu}_{\quad\gamma\delta}$ for that matter are not conventionally used in literature . But there is really nothing wrong or incorrect with these concepts.They result from the raising of indices from $R_{\rho\sigma\alpha\beta}$ or $R_{\alpha\beta\gamma\delta}$
We have
$g^{\rho\mu}g^{\sigma\nu}R_{\rho\sigma\alpha\beta}=R^{\mu\nu}_{\quad\alpha\beta}$
$g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=R^{\mu\nu}_{\quad\gamma\delta}$
As mentioned in the last comment referring to equation (4) ,that is to
$R_{\beta\alpha\mu\nu}=-R_{\alpha\beta\mu\nu}$ is conventionally used and is quite frequent in the in its application.

Last edited: Jan 23, 2021
21. ### Anamitra PalitRegistered Senior Member

Messages:
102
Considering the advice of James R[Staff Member]and considering interactions from other forums on this subject:
The product of a symmetric and an antisymmetric tensor is the null tensor . In view of that (6) in the initial post is an identity.
Nevertheless we may consider the following:
First we recall equation (1) of the initial post:
$R_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0$(1)
A significant aspect of (1)- is that each equation holds for any (mu, nu) pair.
Keeping gamma and delta constant[and we maintain so throughout this post] we vary alpha and beta for each equation in (1). In order to obtain new equations of the type (1) we vary mu and nu There are 16 unknowns in R_αβγδ for the four alpha and the four beta while we have 16 equations[linear homogeneous equations ] in that mu and nu take on four values of each. Unless the determinant of the coefficient matrix
is zero the solutions to R_αβγδ are trivial in their nature. This cannot be violated[subject to a non zero determinant of the coefficient matrix] We have
$R_{\alpha\beta\gamma\delta}=0$ (2)
[The null tensor is always antisymmetric[ asides being symmetric]]
We may write (1) in the form
$\left(R_{\alpha\beta\gamma\delta}+R_{\beta\alpha\gamma\delta}\right)\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0$(3)
We have eight variables of the form
$R_{\alpha\beta\gamma\delta}+R_{\beta\alpha\gamma\delta}$
while the number of equations continue to remain sixteen.
Gamma and delta have been maintained constant throughout the writing as mentioned earlier.

Last edited: Jan 25, 2021
22. ### Anamitra PalitRegistered Senior Member

Messages:
102
In relation to my last post:
1. Keeping alpha beta same , for various selections of gamma and delta R_{alpha beta gamma delta will be the same:|R_{alpha beta gamma delta}=R_{alpha beta gamma', delta prime};gamma and delta do not figure into the coefficient matrix.
Example R 1234=R1232

2. If the determinant of the coefficient matrix is non zero(or zero) then it will remain non zero (or zero) irrespective of the choice of gamma and delta.. The coefficient matrix does not change even if the values of gamma and delta are changed.
The relevant equation is:
$R_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+g^{\beta\mu}g^{\alpha\nu}\right)=0$
If some R_{alpha beta gamma delta} is non zero for a particular choice of gamma and delta then the determinant of the coefficient matrix is zero.For any other combination of gamma and delta at least one R_{alpha beta gamma delta} will be non zero.

Last edited: Jan 25, 2021
23. ### Write4UValued Senior Member

Messages:
16,249
Riemann Geometry
https://en.wikipedia.org/wiki/Riemannian_geometry

I looked at some of the pictorial illustrations and got the thrust of the theory. Now I know the connection between the two terms.

Question: But what do the above mathematics illustrate other than the mathematics of the theory? Is there a problem with any of that?