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# On Four Velocity and Four Momentum

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In relation to the last post

We may always choose the eight unknowns  unknowns:v1_i and v2_j with each i and j=1,2,3,4 , in such a manner that the next three equations hold

c^2=c^2v1_t^2-v1_x^2-v1_y^2-v1_z^2

c^2=c^2v2_t^2-v2_x^2-v2_y^2-v2_z^2+2v1.v2

and c^2=c^2(v_1+v2_t)^2_-(v1_x-v2_x)^2-(v1_y-v2_y)^2-(v1_z-v2_z)^2

They will lead to 2v1 .v2<=-c^2

Edited by Anamitra Palit

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On 1/10/2021 at 11:33 AM, Anamitra Palit said:

I will write in Latex at my earliest opportunity

May I suggest that you focus on that opportunity?

16 hours ago, Anamitra Palit said:

In the mean time you may consider the following

c^2dtau^2=c^2dt^2-dx^2-dy^2-dz^2

Why should every interested member have to parse your posts to participate in the discussion when you could provide a readable format in a quick one-time effort?

On 1/11/2021 at 3:33 PM, Anamitra Palit said:

This paper dos not indicate at any error in theory.

By "This paper" you mean the attachment named "Time Conflict 5.pdf"? Please note that you have attached several papers.

Edited by Ghideon
grammar

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Time Conflict 5 does not indicate any error in the theory. The paper on Four Velocity and Four Momentum on modification indicates error in theory. "The Extra Bit " by itself is indicative of an error

Edited by Anamitra Palit

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Important revisions have been implemented in the file on Four velocity and Four momenta. The revised file has been uploaded.[pl see the section "The Extra Bit" for the revision]

Edited by Anamitra Palit

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8 hours ago, Anamitra Palit said:

Important revisions have been implemented in the file on Four velocity and Four momenta.

But your initial issues are still in the revised pdfs you keep attaching. I’ll try to provide a simple comparing example to highlight an issue in case you did not grasp @joigus and @Markus Hanke's excellent explanations:

For arbitrary numbers a and b we know that $| b-a | \geq 0$ (1)

Lets take x and y such that x-y=0 (2)

You seem to argue that inserting x and y constrained by (2) into equation (1) means $| x-y | \geq 0$ and hence a possibility that $| x-y | > 0$. But that is of course not true because x and y are constrained. What is true for arbitrary numbers a and b is not true for x and y. Just as your inequality; it holds for arbitrary numbers but not for four-vectors.

Edited by Ghideon
spelling

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Important revisions/corrections have been made with the 'Extra Bit'.Relevant portion has been provided in Latex. The entire file[revised file] has been uploaded.

Metric

$$c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2$$ (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

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

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

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

Since $$v1.v2\ge c^2$$ we have

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

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

If $v_1+v_2$ is a proper  velocity then

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

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

$$c^2=c^2+c^2+2v_1.v_2$$

Therefore

$$v_1.v_2\le -½ c^2$$

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

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

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

$$c^2=c^2+c^2-2v_1.v_2$$

$$½ c^2=v_1.v_2$$

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 tangent representing the motion. The manifold is discrete and that presents difficulty with procedure like differentiation.

Edited by Anamitra Palit

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5 hours ago, Anamitra Palit said:

Adding (3.1) and (3.2) we obtain

Please explain how that addition is valid, the components you add are not arbitrary numbers.

(Also, the paper still contains the issues previously mentioned)

Edited by Ghideon

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5 hours ago, Anamitra Palit said:

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 tangent representing the motion. The manifold is discrete and that presents

difficulty with procedure like differentiation

Ok. There's a couple of things that you're seeing there that are already taken care of in the mainstream formalism. And have been mentioned repeatedly.

One of them is that for any physical 4-vector the pseudo-norm with signature (+,-,-,-) must be positive --causality. The other one that I at least forgot to mention --it's possible that either @Ghideon and/or @Markus Hanke have mentioned it and I've missed it--, is that the naught component must be positive. Those are called orthochronous 4-vectors. So you cannot add two arbitrary 4-vectors and hope that that has any physical meaning. When you add two orthochronous 4-vectors, you obtain another orthochronous 4-vector.

Also, the orthochronous character ( $$v^0 > 0$$ ) is split in mutually disconnected subsets of the Minkowski space, so the idea that you suggest that there is a "perforated" structure is actually a misguided intuition. The reason is that both the sign of the Minkowski pseudometric and the zero component are continuous and differentiable function of their arguments:

$f\left(\alpha^{0},\alpha^{1},\alpha^{2},\alpha^{3},\beta^{0},\beta^{1},\beta^{2},\beta^{3}\right)=\alpha\cdot\beta$

$g\left(\alpha\right)=\alpha^0$

So you can't have either changes in sign of the product, nor changes in the sign of any component for arbitrarily close 4-velocities, as you posit.

Edited by joigus
minor addition

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Ghideon "Please explain how that addition is valid, the components you add are not arbitrary numbers."

The squares of the components have been added. Its true that these numbers (squares of the components)are not independent. They are related by (3.1) and by(3.2). That does not matter.We can always add the two  sides of a pair of equations.

Next Issue[Independent of reply to Ghideon]

For proper viewing one may have to refresh the page.

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 do have, $$v.v=c^2$$(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 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 that is v' is a proper velocity.||But we assumed /postulated at the very outset that v' is not a proper velocity.

Edited by Anamitra Palit

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23 minutes ago, Anamitra Palit said:

The squares of the components have been added. Its true that these numbers (squares of the components)are not independent. They are related by (3.1) and by(3.2). That does not matter.We can always add the two  sides of a pair of equations.

Yes, mathematics allows for addition of equations. That does not imply you can draw conclusions about physics from that addition without further clarifications.

Naive comparison: 0.5e (half the elementary positive charge) is mathmenatically valid but it does not follow that there exists an elementary particle with that charge or that standard model have issues.

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