# 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
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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

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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]

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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
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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.

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5 hours ago, Anamitra Palit said:

Adding (3.1) and (3.2) we obtain

(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
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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.

$$c^2=c^2 v_t^2-v_x^2-v_y^2-v_z^2$$

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

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.

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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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• 2 weeks later...

The current form of the  article in Latex

[It might be necessary to refresh the page for viewing the ormulas and the equations]

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
$\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)=\left(a_1b_2-a_2b_1\right)^2$

$\left(a_1b_1-a_2b_2\right)^2-\left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)\ge 0$

$\left(a_1b_1-a_2b_2\right)^2\ge \left(a_1^2-a_2^2\right)\left(b_1^2-b_2^2\right)$

The last inequality implies (1.1) or (1.2)[subject to the initial criterion]
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)

$\Rightarrow a_2b_2+a_3b_3+...a_nb_n=\\ \sqrt{a_2^2+a_3^2+a_4^2+...+a_n^2}\sqrt{b_2^2+b_3^2+b_4^2+...+b_n^2}\cos \theta$(3.2)
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)
$2c^2=c^2\left(v_{1t}^2+v_{2t}^2\right)-\left(v_{1x}^2+v_{2x}^2\right)-\left(v_{1y}^2+v_{2y}^2\right)-\left(v_{1z}^2+v_{2z}^2\right)$
$2c^2=c^2\left(v_{1t}+v_{1t}\right)^2-\left(v_{1x}+v_{2x}\right)^2-\left(v_{1y}+v_{2y}\right)^2-\left(v_{1z}+v_{2z}\right)^2-2v_1.v_2$(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, are indicative of  a subtle issue of  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.

Edited by Anamitra Palit
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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:product of two square roots, a positive number

For X>=0

$X\ge X \cos \theta$

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

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

$X=\sqrt{a_2^2+a_3^2+.....+a_n^2}\sqrt{b_2^2+b_3^2+.....+b_n^2}$

NB:

$v_1.v_2\le -c^2$ will not apply since $v_1.v_1=c^2>0$

[We can make v2=v1]

Edited by Anamitra Palit
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3 hours ago, Anamitra Palit said:

One has to think of equations(24.1),(24.2) and (24.3) in the mathematical sense regardless of the physics involved with them.

Please also explain why this is posted in relativity section if your mathematics is not about physics.

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The section of my second  last post from after equation (17.2) and before the section " The Sum and  Difference of Two Proper Velocities " has been revised

We restate (13),(14),(15) ..(17.1) and then proceed..

$v.v= c^2$(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. 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)
$\Rightarrow c^2\ge cc'\Rightarrow c\ge c'$
That is c'=||v-ka||<=c (A)
If ka could be a proper velocity then ||v-ka||^2<=2c^2 [we have had this earlier from the second last post] from  The Sum and Difference of Two  Proper Speeds.

$||v-ka||^2\le 2c^2$ (B) if ka is a proper speed
We show that ka can indeed be a proper speed:

Norm square of the acceleration vector

$c^2a_t^2-a_x^2-a_y^2-a_z^2=\bar c^2$

c bar has the unit of acceleration

$m^2\left(c^2a_t^2-a_x^2-a_y^2-a_z^2\right)=m^2\bar c^2$
m in the above equation has the dimension of time and we make the magnitude of m unity [in the concerned units]. Thus we have
$c^2a_t^2-a_x^2-a_y^2-a_z^2=\bar c^2$
Next we multiply both sides of the above equation by dimensionless  k^2 so that k* c bar is equal to c. Now we have
$c^2k^2a_t^2-k^2a_x^2-k^2a_y^2-k^2a_z^2=k^2\bar c^2$

$c^2\left(k a_t\right)^2-\left(k a_x\right)^2-\left(k a_y\right)|^2 -\left(k a_z\right)^2=c^2$
Therefore ka is a candidate for a proper velocity.But (A) and (B) contradict each other

Edited by Anamitra Palit
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On 1/27/2021 at 3:18 AM, Anamitra Palit said:

Therefore

v.(vka)=c2

(15)
We could expect from the last equation

(vka)

Um... what?!

You already showed on another thread that for two four velocities $$u$$ and $$v$$ their inner product satisfies $$u.v\geq c^2$$, with equality iff $$u=v$$. So what you should expect from (15) is that $$v-ka$$ is not a four velocity, except in the case that $$ka=0$$. In fact, a moment's thought would tell you that $$v-ka$$ isn't even necessarily timelike. Furthermore, you already know that $$a$$ is spacelike, so your $$m$$ must be imaginary and cannot be 1 as you so blithely assert. This whole line of argument is utterly incoherent.

If you want to see the effect of four acceleration on a four velocity you need to calculate $$a(\tau)$$ or some other parameter, then integrate $$v(\tau)=v(0)+\int_0^\tau a(\tau')d\tau'$$ (or whatever parameter you prefer to $$\tau$$). Simply adding a multiple of the four acceleration and hoping does not work.

Edited by Kino
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3 hours ago, Kino said:

Furthermore, you already know that a is spacelike, so your m must be imaginary and cannot be 1 as you so blithely assert.

Correcting myself here: $$m$$ can be anything it likes because it does exactly nothing. However if you insist that $$k^2\bar c^2=c^2$$, then noting that $$\bar c^2$$ is negative since $$a$$ is spacelike, then $$k$$ is imaginary.

Edited by Kino
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13 hours ago, Kino said:

If you want to see the effect of four acceleration on a four velocity you need to calculate a(τ) or some other parameter, then integrate v(τ)=v(0)+τ0a(τ)dτ (or whatever parameter you prefer to τ ). Simply adding a multiple of the four acceleration and hoping does not work.

I’ve also pointed this out to him already on his “Norm of 4-Acceleration” thread, and gave the explicit expressions for the necessary transforms of x and ct, but he ignored it completely. In fact, so far he has ignored pretty much everything that has been pointed out to him - it doesn’t seem like there is any genuine openness to feedback. It feels more like a personal blog to me.

Just for the casual reader’s reference - in the special case of motion starting from rest at the origin, with uniform acceleration, the integral becomes straightforward, and the expressions evaluate to:

$v( \tau ) =c \tanh\left(\frac{a\tau }{c}\right)$
$v(t) =\frac{at}{\sqrt{1+\left(\frac{at}{c}\right)^{2}}}$

Which is, of course, hyperbolic motion, as we would expect in a Minkowski spacetime.

Edited by Markus Hanke
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To add to the panoply of excellent comments by Kino and Markus, you cannot expect an arbitrary linear combination of 4-vectors to be a physically significant 4-vector. Both vectors must be timelike $$\left(u_{\left(i\right)}^{0}\right)^{2}-\boldsymbol{u}_{\left(i\right)}\cdot\boldsymbol{u}_{\left(i\right)}\geq0$$ and orthochronous $$u_{\left(1\right)}^{0},\:u_{\left(2\right)}^{0}>0$$. Also, the resulting 4-vector must be normalised to $$c^2$$. In that sense, when you're working with 4-velocities, you're not working on a plain linear space --Minkowski space--, but in some kind of "unitary quotient of it." This distinction is referred-to in physics by means of the buzzwords "on-shell" and "off-shell." Adding vectors off-shell can lead you to vectors on-shell, and vice versa.

This point has arisen before --Ghideon has been particularly persistent.

Off the top of my head, you can derive a common (CoM) 4-velocity for 2 material particles moving every which way by calculating the common 4-momentum, and then dividing by $$m_1+m_2$$ --which are relativistic invariants. Another thing you could do is calculate the centre-of-energies motion and then impose that it be normalised as to become a physical 4-vector. One last thing you could do is use Einstein's addition of velocities --which doesn't involve the masses--, to obtain a physical 4-vector, by multiplying by the appropriate observer-dependent factor as to obtain a 4-vector. I don't know. I'm just trying to help you so that your effort is not in vain.

So far, it is in vain, simply because you're not distinguishing with any care what's on-shell and what's off-shell.

Edited by joigus
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I don't know how to interpret the fact that the OP has decided not to address my comments at all.

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8 hours ago, joigus said:

I don't know how to interpret the fact that the OP has decided not to address my comments at all.

If you look carefully, you’ll notice that he hasn’t really addressed most comments made so far. He either ignores them, or just repeats earlier posts.

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11 hours ago, joigus said:

I don't know how to interpret the fact that the OP has decided not to address my comments at all.

He doesn't even seem to pay attention to his own posts, let alone anyone else's. See my comment about him proving that $$u.v> c^2$$ for distinct $$u$$, $$v$$ then asserting that $$v.x=c^2$$, with $$x\neq v$$, implies $$x$$ is a four velocity.

Edited by Kino
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On 1/30/2021 at 10:45 AM, Kino said:

He doesn't even seem to pay attention to his own posts, let alone anyone else's. See my comment about him proving that u.v>c2 for distinct u , v then asserting that v.x=c2 , with xv , implies x is a four velocity.

Yes, I've noticed that.

On 1/30/2021 at 7:55 AM, Markus Hanke said:

He either ignores them, or just repeats earlier posts.

Seems like all of our comments have gone unanswered. I think I've noticed a pattern though. After a couple of posts he even stops addressing the person. I wonder what that means...

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!

Moderator Note

Animitra Palit, too many of your posts have strayed away from mainstream physics, so I'm moving this to the Speculations section. You need to do a better job of addressing the concerns expressed by other members in the discussion, and support your assertions with mainstream science.

If you can't support your ideas here, the thread will be closed and you won't be allowed to bring it up again. The membership is willing to spend a great deal of time discussing this with you, but only if you are also willing to put in the work.

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• 1 month later...
On 1/29/2021 at 3:00 AM, Kino said:

Correcting myself here: m can be anything it likes because it does exactly nothing. However if you insist that k2c¯2=c2 , then noting that c¯2 is negative since a is spacelike, then k is imaginary.

The acceleration four vector is indeed space like or null[for uniform motion]

The norm of the acceleration four vector as suggested by Kino is negative[or zero , for uniform motion].There are different methods of proving the stated fact. One is presented here as a prelude to the point referred to finally.
$c^2d\tau ^2=c^2dt^2-dx^2-dy^2-dz^2$ (1.1)
$c^2=c^2 \left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2$  (1.2)
Differentiating both sides of (1.2) with respect to proper time tau, we obtain,
$c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2} - \frac{dx}{d\tau}\frac{d^2y}{d\tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2} =0$(3)
Next we transform to an inertial  frame of reference where the particle is momentarily[instantaneously] at rest [maintaining the acceleration of the original frame]
Equation (3) reduces to :
$c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}\frac{d^2 t}{d\tau^2}=0$
Since v.v=c^2 and v_x=v_y=v_z=0 implying gamma=1 that is dt/d tau=1,we have
$c^2 \frac{d^2 t}{d \tau^2}=0$(4)
From (4) we infer that a.a<=0 in the reference frame we have transformed to. From the invariance of the dot product we have
$a.a \le 0$ (5)
for all inertial frames of reference.
We consider one dimensional motion in the x-x’ direction for which a_yand a_z are zero [implying from the Lorentz transformations a’_y=a’_z=0 ]. The reference frames are translating in the x-x’ direction with a uniform speed v.The accelerating particle is also moving in the same direction.
a.a<=0 implies
$c^2a_t^2-a_x^2\le 0$ (6)
Again
$a_t=\frac{d^2 t}{d \tau^2}=\frac{d\gamma}{d \tau}=\gamma^3 v_x a_x \frac {1}{c^2}$
where,
$\gamma=\frac{1}{\sqrt{1-\frac{v_x^2}{c^2}}}$
We have,
$a_t= \gamma^3 v_x a_x \frac {1}{c^2}$(7)
From (6) and (7) we have
$\frac{1}{c^2} \gamma^6 v_x^2 a_x^2\le a_x^2$(8)
For non zero a_x,
$\frac{1}{c^2} \gamma^6 v_x ^2\le 1$(9)
For v_x tending to c the last equation fails.

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Posted (edited)

On 12/19/2020 at 2:15 PM, Markus Hanke said:

This is meaningless. If you are projecting a shorter 4-vector onto a longer one, the result can never exceed the length of the longer vector, that’s just basic geometry. In other words, the longest a projection can ever be is that of a 4-vector onto itself, which, in Minkowski spacetime, is thus -1. Since the inner product is invariant, this is true for all 4-vectors in all frames.

Consider a general 4-velocity of the type

uμ=(γ,γvx,γvy,γvz)

The inner product with itself is

uμuμ=γ2(1v2)=γ2/γ2=1

as expected. So your claim is wrong.

If this were true, the momentum of a photon would be zero. This is evidently false, as we know already from experiment and observation. Mathematically, you can show this in a similar manner as above.
So again, you are wrong.

I’ve already shown above that the inner product of a 4-velocity with itself is -1, so the magnitude of a 4-velocity in spacetime is always exactly c. Since the inner product is invariant, this is true in all frames, so it can’t be a function of the gamma factor. Also, if you look at this expression, you should notice immediately that the resultant magnitude of does not correspond to the gamma factor you are inputting, so the expression is meaningless.
You are wrong on this one, too.

Yes, because all three points you have presented contradict both basic maths, as well as observation in the real world. It’s simply wrong.

Forum expert Markus Hanke declined believing in the relation v1.v2>=c^2 that stems from the reversed Cauchy Schwarz inequality.In case he happens to believe in the stated relation my discussion has made a substantial  contributed to the forum and it[my discussion] is far from being of a speculative nature. Kino seems to have accepted this relation v1.v2>=c^2

[The relation v.v=c^2[gamma^2-1+1/gamma^2] is not a valid one . Nevertheless v1.v2>=c^2 is a valid formula]

On 12/21/2020 at 2:18 PM, Markus Hanke said:

You haven’t. You are simply repeating the same statements again and again.

You are right, this was my mistake, I actually thought of the photon’s 3-momentum (which is not zero!) when I wrote my post, whereas you were referring to the 4-momentum.

I have corrected a loose statement made by the forum expert. |his indeed has a non speculative contribution.

A writing that locates a contradiction in the exiting theory[formal theory]

The norm of the acceleration four vector  is negative[or zero , for uniform motion].There are different methods of proving the stated fact. One is presented here.
$c^2d\tau ^2=c^2dt^2-dx^2-dy^2-dz^2$ (1.1)
$c^2=c^2 \left(\frac{dt}{d\tau}\right)^2-\left(\frac{dx}{d\tau}\right)^2-\left(\frac{dy}{d\tau}\right)^2-\left(\frac{dz}{d\tau}\right)^2$  (1.2)
Differentiating both sides of (1.2) with respect to proper time tau, we obtain,
$c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}- \frac{dx}{d\tau}\frac{d^2 x}{d \tau^2} - \frac{dx}{d\tau}\frac{d^2y}{d\tau^2}-\frac{dz}{d\tau}\frac{d^2 z}{d \tau^2} =0$(3)
Next we transform to an inertial  frame of reference where the particle is momentarily[instantaneously] at rest [maintaining the acceleration of the original frame]
Equation (3) reduces to :
$c^2 \frac{dt}{d\tau}\frac{d^2 t}{d \tau^2}\frac{d^2 t}{d\tau^2}=0$
Since v.v=c^2 and v_x=v_y=v_z=0 implying gamma=1 that is dt/d tau=1,we have
$c^2 \frac{d^2 t}{d \tau^2}=0$(4)
From (4) we infer that a.a<=0 in the reference frame we have transformed to. From the invariance of the dot product we have
$a.a \le 0$ (5)
for all inertial frames of reference.
We consider one dimensional motion in the x-x’ direction for which a_yand a_z are zero [implying from the Lorentz transformations a’_y=a’_z=0 ]. The reference frames are translating in the x-x’ direction with a uniform speed v.The accelerating particle is also moving in the same direction.
a.a<=0 implies
$c^2a_t^2-a_x^2\le 0$ (6)
Again
$a_t=\frac{d^2 t}{d \tau^2}=\frac{d\gamma}{d \tau}=\gamma^3 v_x a_x \frac {1}{c^2}$
where,
$\gamma=\frac{1}{\sqrt{1-\frac{v_x^2}{c^2}}}$
We have,
$a_t= \gamma^3 v_x a_x \frac {1}{c^2}$(7)
From (6) and (7) we have
$\frac{1}{c^2} \gamma^6 v_x^2 a_x^2\le a_x^2$(8)
For non zero a_x,
$\frac{1}{c^2} \gamma^6 v_x ^2\le 1$(9)
For v_x tending to c the last equation fails.

There is no reason to believe that by discussion is of a speculative nature.Requesting the moderator to restore it to its normal status.

Edited by Anamitra Palit
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