# Four Dot Products and of Momenta

Discussion in 'Physics & Math' started by Anamitra Palit, Jan 11, 2021.

1. ### Anamitra PalitRegistered Member

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We deduce in the paper the following results
v1.v2>=c^2
[v1 and v2 are four velocities]
p1.p2>=m1m2c^2
[p1 and p2 are four velocities]

By applying the reversed Cauchy Schwarz Inequality we may arrive directly at the same results
Let's consider the reversed Cauchy Schwarz inequality.

c^2 t1 t2-x1 x2 -y1 y2-z1 z2>=Sqrt[c^2t1^2-x1^2-y1^2-z1^2]Sqrt[c^2t2^2-x2^2-y2^2-z2^2]

The equality sign holds when (t1,x1,y1,z1) and (t2,x2,y2,z2) are identical vectors

Replacing x^i b y dx^i we obtain

c^2 dt1 dt2-dx1dx2 -dy1 dy2-dz1 dz2>=Sqrt[c^2dt1^2-dx1^2-dy1^2-dz1^2]Sqrt[c^2dt2^2-dx2^2-dy2^2-dz2^2]
A paper on the Reversed Cauchy Schwarz Inequality:

Wikipedia Link on the Cauchy Schwarz Inequality:
https://en.wikipedia.org/wiki/Minkowski_space#Norm_and_reversed_Cauchy_inequality

Dividing both sides by dtau^2 we obtain

Four dot product v1.v2>=c^2
Multiplying both sides by m1m2[m1 and m2 being rest masses] we obtain,
m1v1.mv2>=m1 m2c^2
or,p1.p2>=m1m2 c^2

3. ### James RJust this guy, you know?Staff Member

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And so? What's the point?

5. ### Anamitra PalitRegistered Member

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The formulas v1.v2>=c^2,
p1.p2 >m1m2c^2 are important just like v,v=c^2

But alas..
c^2dtau^2=c^2dt^2-dx^2-dy^2-dz^2

c^2=c^2[dt/dtau]^2-[dx/dtau]^2-[dy/dtau]^2-[dz/dtau]^2

c^2=c^2v_t^2-v_x^2-v_y^2-v_z^2 (1)

v_i are proper speeds and as such they can exceed the speed of light without hurting or violating Special Relativity

For two proper velocities v1 and v2at the same point of the manifold.Since tensors are additive we have

c^2=c^2(v1_t+v2_t)^2-(v1_x+v2_x)^2-(v1_y+v2_y)^2-(v1_z+v2_z)^2(2)

or,c^2=c^2v1_t^2-v1_x^2-v1_y^2-v1_z^2+c^2v2_t^2-v2_x^2-v2_y^2-v2_z^2+2v1.v2

or, c^2=c^2+c^2+2v1.v2

v1.v2=-c^2 (3)

By calculations we have arrived at an untenable result.

An analogous result may be obtained in the General Relativity context

7. ### Anamitra PalitRegistered Member

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(in continuation)
Indeed v1,v2 and v=v1+v2 all satisfy(1) and hence heir existence is certified by Special relativity or even by General Relativity for that matter.

An analogous result may be obtained in the General Relativity context

c^2dtau^2=c^2 g_tt dt^2-g_xx dx^2-g_yy dy^2-g_zz dz^2 (4)

We consider transformations g_tt dt^2=dT^2, g_xxdx^2=dX^2, g_yydy^2=dY^2,g_zzdz^2=dZ^2 (5)

Local or even transformations over infinitesimally small regions would suffice.

Equations (4) and (5) combined gives us the flat space time metric[mathematical form of it]

c^2dtau^2=c^2 dT^2- dX^2- dY^2- dZ^2 (6)

All conclusions we made earlier follow.

Incidentally, there is one point to take note of:the Lorentz transformations follow from (6) in a unique manner [Reference; Steve Wienberg,Gravitation and Cosmology,Chapter 2:Special Relativity]

8. ### Anamitra PalitRegistered Member

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85
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 (1)

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

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 (3)
Equations (1),(2) and (3) are all certified by the relation
c^2=c^2v_t^2-v_x^2-v-y^2-v_z^2 which is equivalent to the Lorentz transformations
as stated earlier.
Equations (1),(2) and (3) are all certified by the relation
c^2=c^2v_t^2-v_x^2-v-y^2-v_z^2 which is equivalent to the Lorentz transformations
as stated earlier.

Last edited: Jan 12, 2021 at 8:39 AM
9. ### exchemistValued Senior Member

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What is it you want to discuss?

10. ### Michael 345New year. PRESENT is 71 years oldValued Senior Member

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The Inner Mind?

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12. ### Anamitra PalitRegistered Member

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Last edited: Jan 13, 2021 at 5:28 AM

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14. ### Anamitra PalitRegistered Member

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Last edited: Jan 13, 2021 at 1:16 PM

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16. ### Anamitra PalitRegistered Member

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Important revisions have been made in "the Extra Bit"

Relevant material in Latex:
Metric

\begin{equation}c^2d\tau^2=c^2dt^2-dx2-dy^2-dz^2 \end{equation} (1)
\begin{equation}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\end{equation}
\begin{equation}c^2=c^2{v_t}^2-{v_x}^2-{v_y}^2-{v_z}^2\end{equation}(2)

We consider two proper velocities on the same manifold

\begin{equation}c^2=c^2{v_{1t}}^2-{v_{1x}}^2-{v_{1y}}^2-{v_{1z}}^2\end{equation}(3.1)

\begin{equation}c^2=c^2{v_{2t}}^2-{v_{2x}}^2-{v_{2y}}^2-{v_{2z}}^2\end{equation}(3.2)

Adding (3.1) and (3.2) we obtain

\begin{equation}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)\end{equation}

\begin{equation}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\end{equation}

\begin{equation}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\end{equation}

Since v1.v2>=c^2 we have

\begin{equation}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\end{equation}(4)

\begin{equation}\left(v_1+v2)\dot (v_1+v_2)\right) \ge 4c^2\end{equation}(5)

If $v_1+v_2$ is a proper velocity then

\begin{equation}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\end{equation} (6)

\begin{equation}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\end{equation}

\begin{equation}c^2=c^2+c^2+2v_1.v_2\end{equation}(7)

Therefore

\begin{equation}v_1.v_2\le -½ c^2\end{equation}(8)

which is not true since

\begin{equation}v.v=c^2\end{equation}

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

\begin{equation}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\end{equation} (9)

\begin{equation}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\end{equation}
\begin{equation}c^2=c^2+c^2-2v_1.v_2\end{equation}
\begin{equation} ½ c^2=v_1.v_2\end{equation} (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 at 11:42 AM
17. ### Anamitra PalitRegistered Member

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[It may be necessary to refresh the page for proper viewing]
\begin{equation}c^2=c^2 v_t^2-v_x^2-v_y^2-v_z^2\end{equation}

\begin{equation}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\end{equation}

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)

\begin{equation}\Rightarrow v.a=0\end{equation}(2)

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

We have,

\begin{equation}\Rightarrow v.ka=0\end{equation}(3)

From (3) and (4)

\begin{equation}\Rightarrow v. \left(v-ka\right)=c^2\end{equation} (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
\begin{equation}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\end{equation}

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,\begin{equation}\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\end{equation} (7)

\begin{equation}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'\end{equation}

or,

\begin{equation}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'\end{equation}

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 at 8:27 PM
18. ### Anamitra PalitRegistered Member

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

Four Acceleration
\begin{equation}\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)\end{equation} (1)
Let
\begin{equation}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\end{equation} (2)
We consider the metric
\begin{equation}c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2 \end{equation} (4)
\begin{equation}\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
\end{equation} (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
\begin{equation}\frac{d^2 t}{d\tau^2}=0\end{equation}
that is unless \begin{equation}\frac{dt}{d\tau}=constant \Rightarrow \gamma=constant\end{equation}
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