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Old Sep 3rd 2018, 05:27 PM   #11
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So "born" here was essentially "supported"? In English that would be "borne".
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Old Sep 5th 2018, 06:16 PM   #12
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$\displaystyle \ \ \ \ \ $Part2 The relationship of the rest mass line density along AB between $\displaystyle K$and $\displaystyle K’$

$\displaystyle \ \ \ \ \ $ In $\displaystyle K$, an isolated rod AB rotates around its mid point O at uniform angular velocity $\displaystyle ω$, keeps the linear state. For actual rod AB, there is micro motion on molecular level. For each point on AB, In the process of simplifying and establishing a mathematical model, we do not consider this micro motion. The rod AB is an ideal system without any relative movement between the components relative to $\displaystyle K$. the distance from point O to this point measured in $\displaystyle K$ is constant; denote it is $\displaystyle r$. Such as for point D,$\displaystyle r_D$ is a constant. For each point on OA, its equation of motion is $\displaystyle x=r\cos(ωt)$, $\displaystyle y=r\sin(ωt)$ . For each point on OB, its equation of motion is $\displaystyle x=r\cos(ωt+π)$, $\displaystyle y=r\sin(ωt+π)$. Rod AB is a Born rigid body.
$\displaystyle \ \ \ \ \ $The distribution of material and potential energy on the rod AB along the length is symmetry about mid point O and it does not change with time, measured in $\displaystyle K$. Rod AB has no thickness along $\displaystyle θ$ direction, if transform $\displaystyle K$ from rectangular coordinate system to polar coordinate system. That is to say consider rod AB as a one-dimensional rod on $\displaystyle x-y$ plane relative to$\displaystyle K$.

$\displaystyle \ \ \ \ \ $We consider: A rod is static in an inertial reference system. When the rod is free, it has a rest mass. When the rod is stretched, it has another rest mass. The rest mass of the rod when it is stretched is greater than the rest mass of the rod when it is free. When the rod is static at different states such as free or stretched, the potential energy of the particles is different.

$\displaystyle \ \ \ \ \ $In $\displaystyle K$, static auxiliary circle D and M, take O as their center of their circle, intersecting rod OA at point D and M, intersecting rod OB at point E and N. Here $\displaystyle r_D$and $\displaystyle r_M$ can take any value from 0 to$\displaystyle r_A$.
$\displaystyle \ \ \ \ \ $The inertial reference frame $\displaystyle K’$ moves relative to $\displaystyle K$ at speed $\displaystyle v$. Relative to $\displaystyle K’$, the auxiliary circle D and M appear as ellipses D and M, and they move at speed $\displaystyle -v$.

$\displaystyle \ \ \ \ \ $Relative to$\displaystyle K’$, rod AB can't keep the linear state at all times. see Part1

$\displaystyle \ \ \ \ \ $The auxiliary circle in $\displaystyle K$, is an auxiliary ellipse with uniform translational velocity in $\displaystyle K’$. For confirmed the circle D and circle M in $\displaystyle K$, they are ellipse D and ellipse M in $\displaystyle K’$.

$\displaystyle \ \ \ \ \ $In $\displaystyle K$, at time $\displaystyle t$, for each position on OA, denote $\displaystyle ρ_α(t,r)$ as the rest mass line density at the position, thereinto $\displaystyle r $ is the distance from point O to this position. $\displaystyle ρ_α(t,r)$ is equal to the rest mass of the segment at position $\displaystyle r $ on OA divided by the length of this segment. (the length of this segment at the time $\displaystyle t$ need to be infinitesimal). $\displaystyle ρ_α(t,r)$ is a function only relative to $\displaystyle r $, $\displaystyle ρ_α(t,r)$= $\displaystyle ρ(r)$. Denote $\displaystyle ρ_β(t,r)$ as the rest mass line density for each position on OB. Because the mass distribution of AB is symmetry about mid point O, $\displaystyle ρ_β(t,r)$ =$\displaystyle ρ_α(t,r)$= $\displaystyle ρ(r)$.

$\displaystyle \ \ \ \ \ $The function's subscript $\displaystyle α$ indicates that the function is valid for the point on OA. The function's subscript $\displaystyle β$ indicates that the function is valid for the point on OB.

$\displaystyle \ \ \ \ \ $In $\displaystyle K’$, at time $\displaystyle t’$, for each position on AB, use $\displaystyle l’ $ to represent the curve length from point O to this position along AB.

$\displaystyle \ \ \ \ \ $In $\displaystyle K’$, at time $\displaystyle t’$, for each point on OA, denote $\displaystyle l’$as the curve length from point O to this point along curve OA. Such as $\displaystyle l_D ({t}’)’$ is the curve length from point O to this position of point $\displaystyle D$ along AB at time $\displaystyle t’$ in $\displaystyle K’$. It depends on the time $\displaystyle t’$ and the position of this point on OA. Because $\displaystyle r$ represents the position of a point on OA, and $\displaystyle r$ is a constant for each confirmed point; so for each point on OA, $\displaystyle l’$ can be considered as a function of $\displaystyle t’$and $\displaystyle r$, remember it as $\displaystyle l’$=$\displaystyle f_α (t’,r)’$. For example, at the moment $\displaystyle {t_1}’$, the length of the curve OD is $\displaystyle l_D ({t_1}’)’$=$\displaystyle f_α ({t}’,r)’$, here the parameter of $\displaystyle f_α ({t}’,r)’$ is $\displaystyle t’={t_1}’$,$\displaystyle r=r_D $ ($\displaystyle r_D $is a constant).

$\displaystyle \ \ \ \ \ $In $\displaystyle K’$, at time $\displaystyle t’$, for each position on OA, denote $\displaystyle ρ_α(t’,l’) ’$ as the rest mass line density at the position when the curve length from point O to this position is $\displaystyle l’$ measured in $\displaystyle K’$ at the time $\displaystyle t’$. $\displaystyle ρ_α(t’,l’) ’$ is equal to the rest mass of the segment at position $\displaystyle l’$ on OA divided by the length of this segment at the time $\displaystyle t’$. (the length of this segment at the time $\displaystyle t’$ need to be infinitesimal). $\displaystyle ρ_α(t’,l’) ’$ takes $\displaystyle t’$ and $\displaystyle l’$ as its parameters, $\displaystyle t’$ is the time on$\displaystyle K’$ and $\displaystyle l’ = f_α ({t}’,r)’ $ is the curve length from point O to this position along curve OA measured in $\displaystyle K’$ at the time $\displaystyle t’$. Then $\displaystyle ρ_α(t’,l’) ’$= $\displaystyle ρ_α(t’, f_α ({t}’,r)’ ) ’$. So relative to $\displaystyle K’$, for each position on OA, the rest mass line density also can be considered as it takes the time $\displaystyle t’$on $\displaystyle K’$ and the distance $\displaystyle r $ from point O to this position measured in $\displaystyle K$as parameters. Remember $\displaystyle g_α ({t}’,r)’= ρ_α(t’, f_α ({t}’,r)’ ) ’$.

$\displaystyle \ \ \ \ \ $In $\displaystyle K’$, at time $\displaystyle t’$, for each point on OB, denote $\displaystyle l’$as the curve length from point O to this point along curve OB, $\displaystyle l’$=$\displaystyle f_β(t’,r)’$. For example, at the moment $\displaystyle {t_1}’$, the length of the curve OE is $\displaystyle l_E ({t_1}’)’$=$\displaystyle f_β({t}’,r)’$, here the parameter of $\displaystyle f_β({t}’,r)’$ is $\displaystyle t’={t_1}’$,$\displaystyle r=r_E $.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’$, at time $\displaystyle t’$, for each position on OB, denote $\displaystyle ρ_β(t’,l’) ’$ as the rest mass line density at the position when the curve length from point O to this position is $\displaystyle l’$ measured in $\displaystyle K’$ at the time $\displaystyle t’$. Then $\displaystyle ρ_β(t’,l’) ’$= $\displaystyle ρ_β(t’, f_β({t}’,r)’ ) ’$. So relative to $\displaystyle K’$, for each position on OB, the rest mass line density also can be considered as it takes the time $\displaystyle t’$on $\displaystyle K’$ and the distance $\displaystyle r $ from point O to this position measured in $\displaystyle K$as parameters. Remember $\displaystyle g_β({t}’,r)’= ρ_β(t’, f_β({t}’,r)’ ) ’$.

$\displaystyle \ \ \ \ \ $Use $\displaystyle m_{sum\ 0 \ DM } (t)$ to represent the sum of rest masses of section $\displaystyle DM$ in $\displaystyle K$. ($\displaystyle m_{sum\ 0 \ DM }(t)$ is not equal to $\displaystyle m_{ 0 DM }(t)$, the invariant mass of section $\displaystyle DM$.)
$\displaystyle \ \ \ \ \ $Use $\displaystyle {m_{sum\ 0 \ DM } (t’)} ’ $ to represent the sum of rest masses of section $\displaystyle DM$ at time $\displaystyle t’$ in $\displaystyle K ’$.

$\displaystyle \ \ \ \ \ $$\displaystyle m_{sum\ 0 \ OD }(t) = \int_{0} ^{ r_D } {ρ_α(t,r)} \, dr $=$\displaystyle \int_{0} ^{ r_D } {ρ (r)} \, dr $
$\displaystyle \ \ \ \ \ $So denote $\displaystyle m_{sum\ 0 \ OD }(t)$=$\displaystyle m_{sum\ 0 \ OD } $

$\displaystyle \ \ \ \ \ $$\displaystyle {m_{sum\ 0 \ OD } ({t}’)} ’ = \int_{ 0} ^{ l_D(t’) } {ρ_α({t}’,l’) ’} \, dl’ $ , $\displaystyle l’$ is a function of $\displaystyle {t}’$ and $\displaystyle r$, $\displaystyle l’ = f_α ({t}’,r)’$. Make variable substitution for the integral $\displaystyle l’ →r$,
$\displaystyle \ \ \ \ \ $$\displaystyle {m_{sum\ 0 \ OD } ({t}’) } ’ =\int_{ 0} ^{ r_D } {ρ_α({t}’, f_α ({t}’,r)’) ’ \frac {\partial f_α ({t}’,r)’} {\partial r}} \, dr $ =$\displaystyle \int_{ 0} ^{ r_D } {g_α({t}’, r) ’ \frac {\partial f_α ({t}’,r)’} {\partial r}} \, dr $

$\displaystyle \ \ \ \ \ $The same $\displaystyle {m_{sum\ 0 \ OE } ({t}’ )} ’ $ =$\displaystyle \int_{ 0} ^{ r_E } {g_β({t}’, r) ’ \frac {\partial f_β({t}’,r)’} {\partial r}} \, dr $

$\displaystyle \ \ \ \ \ $Rod AB is a Born rigid body.
$\displaystyle \ \ \ \ \ $$\displaystyle {m_{sum\ 0 \ OD } ({t}’ )} ’={m_{sum\ 0 \ OD } }(t) $

$\displaystyle \ \ \ \ \ $$\displaystyle \int_{ 0} ^{ r_D } {g_α({t}’, r) ’ \frac {\partial f_α ({t}’,r)’} {\partial r}} \, dr $=$\displaystyle \int_{ 0} ^{ r_D } ρ(r) dr $, and because of the arbitrariness of the auxiliary circle D, there is
$\displaystyle \ \ \ \ \ $$\displaystyle {g_α({t}’, r) ’ \frac {\partial f_α ({t}’,r)’} {\partial r}}$=$\displaystyle ρ(r) $
$\displaystyle \ \ \ \ \ $The same, there is $\displaystyle {g_β({t}’, r) ’ \frac {\partial f_β({t}’,r)’} {\partial r}}$=$\displaystyle ρ(r) $

$\displaystyle \ \ \ \ \ $This is the relationship of the rest mass line density along AB between $\displaystyle K$ and $\displaystyle K′$.

$\displaystyle \ \ \ \ \ $Set $\displaystyle β{ (u{ (t, r ) }) } = \frac {u{ (t, r ) }} {c}$. (It isn’t function's subscript $\displaystyle β$) (It isn’t $\displaystyle β=\frac v {c}$)
$\displaystyle \ \ \ \ \ $Set $\displaystyle γ{ (u{ (t, r ) }) } = \frac 1 {\sqrt {1 - \frac {{u { (t, r ) }}^2} {c^2}}} = \frac 1 {\sqrt { 1- {β({ u(t, r )} ) }^2}}$.
$\displaystyle \ \ \ \ \ $Set $\displaystyle {β{ ({u { (t’,l’)}}’) } }= \frac {{u { (t’,l’)}}’} {c}$.
$\displaystyle \ \ \ \ \ $Set $\displaystyle {γ{ ({u { (t’,l’)}}’)} } = \frac 1 {\sqrt { 1- {{β({ u(t’,l’) }’ ) } } ^2}}$.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’$, Momentum of any section JK is $\displaystyle {p_{ JK }(t’) }’ $ = $\displaystyle \int_{{ l_J(t’) } } ^{ { l_K(t’) } } {γ { ({u { (t’,l’)}}’) } } ρ(t’,l’) ’ {u{ (t’,l’) }}’ {d l’ }$
$\displaystyle \ \ \ \ \ $When section JK is on OA , such as section DM.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’$, Momentum of section DM is $\displaystyle {p_{ DM }(t’) }’ $ = $\displaystyle \int_{{ l_D(t’) } } ^{ { l_M(t’) } } {γ { ({u { (t’,l’)}}’) } } ρ_α(t’,l’) ’ {u_α{ (t’,l’) }}’ {d l’ }$
$\displaystyle \ \ \ \ \ $$\displaystyle {γ ({u { (t’,l’)}}’) } {u {(t’,l’) }}’ $ will be calculate at the next part .

Last edited by Liuxinhua; Sep 5th 2018 at 11:51 PM.
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Old Sep 5th 2018, 11:05 PM   #13
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Part3 Velocity character of each position on $\displaystyle AB$ in $\displaystyle K’ $
$\displaystyle \ \ \ \ \ $ In $\displaystyle K$, an rod $\displaystyle AB$ rotates around its mid point $\displaystyle O$ at uniform angular velocity $\displaystyle ω$, keeps the linear state.
$\displaystyle \ \ \ \ \ $For each point on $\displaystyle AB$, the distance from point $\displaystyle O$ to this point measured in $\displaystyle K$ is constant; denote it is $\displaystyle r$. Such as for point $\displaystyle D$,$\displaystyle r_D$ is a constant. For each point on $\displaystyle OA$, its equation of motion is $\displaystyle x=r\cos(ωt)$, $\displaystyle y=r\sin(ωt)$ . For each point on $\displaystyle OB$, its equation of motion is $\displaystyle x=r\cos(ωt+π)$, $\displaystyle y=r\sin(ωt+π)$.
$\displaystyle \ \ \ \ \ $In $\displaystyle K$, at time $\displaystyle t$, the coordinates of point $\displaystyle D$ on $\displaystyle OA$ is$\displaystyle (r_D\cos(ωt), r_D \sin(ωt),0,t) $; the coordinates of point $\displaystyle E$ on $\displaystyle OB$ is $\displaystyle (r_E\cos(ωt+π), r_E \sin(ωt+π),0,t )$; the velocity of point $\displaystyle D$ is $\displaystyle u_ α(t, r_D) $; the velocity of point $\displaystyle E$ is $\displaystyle u_β(t, r_E) $.
$\displaystyle \ \ \ \ \ $The inertial reference frame $\displaystyle K’$ moves relative to $\displaystyle K$at speed $\displaystyle v$.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’ $, at time $\displaystyle t ’$, for each point on $\displaystyle OA$, its velocity is $\displaystyle u_ α(t’, l’)’ $, thereinto $\displaystyle l’=f _ α(t’, r)’$; for each point on $\displaystyle OB$, its velocity is $\displaystyle u_β(t’, l’)’ $, thereinto $\displaystyle l’=f _β(t’, r)’$. $\displaystyle l’ $ represent the curve length from point $\displaystyle O$ to this position along AB. Such for position $\displaystyle D$, its velocity $\displaystyle u_ α(t’, l_D(t’)’)’ $is $\displaystyle u_ α(t’, f _ α(t’, r_D)’)’ $.
$\displaystyle \ \ \ \ \ $For point on$\displaystyle AB$, $\displaystyle {γ (u(t’,l’) )} $=$\displaystyle \frac 1 {\sqrt {1- \frac { u ({ t}’, l’) ’ ^2} {c^2} }} $ . Here $\displaystyle l’$=$\displaystyle {l (t’,r) }’ $.
$\displaystyle \ \ \ \ \ $For point on $\displaystyle OA$, $\displaystyle γ({u_ α (t’,l’) }’ )$=$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {α} ({ t}’, l’) ’ ^2} {c^2} }} $ .
$\displaystyle \ \ \ \ \ $$\displaystyle {γ ( {u_ α ({ t_1}’,l’) }’ )} $=$\displaystyle γ ( {u_ α ({ t_1}’, {l_D}’ }’ )$=$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {D} ({ t_1}’) ’ ^2} {c^2} }} $ =$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {α} ({ t_1}’, {l_D}’) ’ ^2} {c^2} }} $. Here $\displaystyle {l_D}’$=$\displaystyle {l (t’,r_D) }’ $.
$\displaystyle \ \ \ \ \ $For point on $\displaystyle OB$, $\displaystyle γ ( {u_β({ t}’,l’) }’ ) $=$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {β} ({ t}’, l’) ’ ^2} {c^2} }} $ .
$\displaystyle \ \ \ \ \ $Set time $\displaystyle {t_1}’=0$ in $\displaystyle K’ $.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’ $, at time $\displaystyle {t_1} ’$, the event $\displaystyle E_{E { t_1}’} $ occurs at the coordinates of point $\displaystyle E( {x_{E1}} ’, {y_{E1}} ’ , 0 , { t_1}’ )$. Using Lorentz transformation between $\displaystyle K $ and $\displaystyle K’ $, translate coordinates $\displaystyle E( {x_{E1}}’, {y_{E1}}’ , 0 , { t_1}’ )$ into the coordinates in $\displaystyle K $. The event$\displaystyle E_{E { t_1}’} $ occurs at coordinates $\displaystyle (r_E\cos(ω{ t_{E\ one } }+π), r_E \sin(ω{ t_{E\ one } }+π),0,t_{E\ one })$ , the coordinates of point $\displaystyle E$ at time $\displaystyle t_{E\ one }$ in $\displaystyle K $.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’ $, at time $\displaystyle { t_1}’$, the event $\displaystyle E_{D { t_1}’} $ occurs at the coordinates of point $\displaystyle D( {x_{D1}}’, {y_{D1}}’ , 0 , { t_1}’ )$. Translate $\displaystyle D( {x_{D1}}’, {y_{D1}}’ , 0 , { t_1}’ )$ into the coordinates in $\displaystyle K $. The event$\displaystyle E_{D { t_1}’} $ occurs at coordinates $\displaystyle (r_D\cos(ω{ t_{D\ one }}), r_D \sin(ω{ t_{D\ one } }),0,t_{D\ one })$, the coordinates of point $\displaystyle D$ at time $\displaystyle t_{D\ one }$ in $\displaystyle K $.
$\displaystyle \ \ \ \ \ $Set time $\displaystyle t_{three}=\frac π {2ω}$ in $\displaystyle K $.
$\displaystyle \ \ \ \ \ $In $\displaystyle K $, at time $\displaystyle t_{three}$, the event $\displaystyle E_{E\ t_{three} } $occurs at the coordinates of point $\displaystyle E( x_{E t_{three}}, y_{E t_{three}} , 0 , t_{three} )$, here $\displaystyle x_{E t_{three}}=0 $. Translate coordinates $\displaystyle E( x_{E t_{three}}, y_{E t_{three}} , 0 , t_{three} )$ into the coordinates in $\displaystyle K’ $. The event$\displaystyle E_{E\ t_{three} } $ occurs at the coordinates of point $\displaystyle E$ at time $\displaystyle {t_3}’=γ{\frac π {2ω}} $ in $\displaystyle K’ $. Thereinto $\displaystyle γ=\frac 1 { \sqrt {1-\frac {v^2} {c^2}}}$
$\displaystyle \ \ \ \ \ $In $\displaystyle K $, at time $\displaystyle t_{three}$, the event $\displaystyle E_{D\ t_{three} } $occurs at the coordinates of point $\displaystyle D( x_{D t_{three}}, y_{D t_{three}} , 0 , t_{three} )$, here $\displaystyle x_{D t_{three}}=0 $. Translate coordinates $\displaystyle D( x_{D t_{three}}, y_{D t_{three}} , 0 , t_{three} )$ into the coordinates in $\displaystyle K’ $. The event$\displaystyle E_{D\ t_{three} } $ occurs at the coordinates of point $\displaystyle D$ at time $\displaystyle {t_3}’ $ in $\displaystyle K’ $ too.
$\displaystyle \ \ \ \ \ $In $\displaystyle K $, at time $\displaystyle t_{D\ one }$, the coordinates of point $\displaystyle D$ is$\displaystyle (r_D\cos(ωt_{D\ one }), r_D \sin(ωt_{D\ one }),0,t_{D\ one })$, the velocity of point $\displaystyle D$ is $\displaystyle u_ α(t_{D\ one }, r_D) $=$\displaystyle (-r_Dω \sin(ωt_{D\ one }),r_Dω \cos(ωt_{D\ one }),0)$. In $\displaystyle K’ $, at time $\displaystyle { t_1}’$, the velocity of point $\displaystyle D$ is $\displaystyle u_ D({ t_1}’)’= u_ α({ t_1}’, l _ D(t_1 ’)’)’ $
$\displaystyle \ \ \ \ \ $$\displaystyle u_ {Dx} (t_{D\ one }) $=$\displaystyle u_ {αx} (t_{D\ one }, r_D) $= $\displaystyle -r_Dω \sin(ωt_{D\ one })$
$\displaystyle \ \ \ \ \ $$\displaystyle u_ {Dy} (t_{D\ one }) $=$\displaystyle u_ {αy} (t_{D\ one }, r_D) $= $\displaystyle r_Dω \cos(ωt_{D\ one })$
$\displaystyle \ \ \ \ \ $$\displaystyle u_ {Dx} ({ t_1}’) ’ $=$\displaystyle u_ {αx} ({ t_1}’, l_D ({ t_1}’) ’)’ $=$\displaystyle \frac { u_ {Dx} (t_{D\ one })-v } { 1-\frac v {c^2} { u_ {Dx} (t_{D\ one })}} $ $\displaystyle =\frac {-r_Dω \sin(ωt_{D\ one }) -v} {1+\frac v {c^2} r_Dω \sin(ωt_{D\ one })} $
$\displaystyle \ \ \ \ \ $$\displaystyle u_ {Dy} ({ t_1}’) ’ $=$\displaystyle u_ {αy} ({ t_1}’, l_D ({ t_1}’) ’)’ $=$\displaystyle \frac 1 {γ(v)}\ \frac {u_ {Dy} (t_{D\ one }) } { 1-\frac v {c^2} { u_ {Dx} (t_{D\ one })}} $ $\displaystyle = \frac 1 {γ(v)}\ \frac { r_Dω \cos(ωt_{D\ one }) } {1+\frac v {c^2} r_Dω \sin(ωt_{D\ one })} $
$\displaystyle \ \ \ \ \ $$\displaystyle \sqrt {1- \frac { u_ {D} ({ t_1}’) ’ ^2} {c^2} } = \sqrt {1- \frac { u_ {Dx} ({ t_1}’) ’ ^2+ u_ {Dy} ({ t_1}’) ’ ^2} {c^2} } $=$\displaystyle \frac {\sqrt {( c^2- {r_ D}^2 ω^2) (c^2- v^2) } } {{(1-\frac v {c^2} u_ {Dx} (t_{D\ one }) ) }c^2 } $=$\displaystyle \frac {\sqrt {( c^2- {r_ D}^2 ω^2) (c^2- v^2) } } { ({1+\frac v {c^2} r_Dω \sin(ωt_{D\ one })} ){c^2} }$
$\displaystyle \ \ \ \ \ $$\displaystyle γ ( { u _D ( { t_1}’) }’ )$=$\displaystyle \frac 1 {\sqrt {1-β ( { u _D ( { t_1}’) }’ )^2} } $=$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {D} ({ t_1}’) ’ ^2} {c^2} }}$=$\displaystyle { ({1+{\frac v {c^2} } r_Dω \sin(ωt_{D\ one })}) } γ(r_ Dω) γ(v) $
$\displaystyle \ \ \ \ \ $$\displaystyle γ ( { u _D ( { t_1}’) }’ ) { u_ {Dx} ({ t_1}’) ’ } $=$\displaystyle \frac { u_ {Dx} ({ t_1}’) ’ } { \sqrt {1- \frac { u_ {D} ({ t_1}’) ’ ^2} {c^2} } } $= $\displaystyle - ({r_D ω \sin(ωt_{D\ one }) +v} )γ(r_ Dω) γ(v) $
$\displaystyle \ \ \ \ \ $Set $\displaystyle ω>0$, $\displaystyle c>v=ωr_A>0$.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’ $, at time $\displaystyle t’ $, for any point on $\displaystyle OA$, its coordinates is $\displaystyle ( x’, y’ , 0 , t ’ )$. Assume an event $\displaystyle E_{r \ t’} $ occurs at $\displaystyle ( x’, y’ , 0 , t ’ )$. Using Lorentz transformation, translate $\displaystyle ( x’, y’ , 0 , t ’ )$ into $\displaystyle ( x, y , 0 , t )$, the coordinates of event $\displaystyle E_{r \ t’} $ in $\displaystyle K $. The event $\displaystyle E_{r \ t’} $ occurs at $\displaystyle ( x, y , 0 , t )$in $\displaystyle K $. Here, $\displaystyle t’ $is a function of $\displaystyle t $ and $\displaystyle r $. Also it can be considered $\displaystyle t = t_α( t’ , r)$. Here are, $\displaystyle γ({u_α(t’,l’) }’ ) {u_{αx}{(t’,l’) }}’ $ =$\displaystyle u_{αx}({ t}’,l’) ’ / \sqrt {1- \frac { u_{α} ({ t_1}’ ,l’) ’ ^2} {c^2} }$= $\displaystyle -( {r ω \sin(ωt_α( t’ , r)) +v} ) γ(r_ Dω) γ(v) $ . Corresponding to the time $\displaystyle t’ ={ t_1}’=0$, in $\displaystyle K’ $, for any point on $\displaystyle OA$ except $\displaystyle O$, there are $\displaystyle \sin (ω t_α( t’ , r))>0$ (reason 1); Corresponding to the time $\displaystyle t’ ={ t_3}’= γ(v) \frac π {2ω} $ in $\displaystyle K’ $, for any point on $\displaystyle OA$ except $\displaystyle O$, there are $\displaystyle t_α({ t_3}’ , r) =t_{three} = \frac π {2ω} $, $\displaystyle \sin (ω t_α({ t_3}’ , r))=1$.
$\displaystyle \ \ \ \ \ $In $\displaystyle K’ $, at time $\displaystyle t’ $, for any point on $\displaystyle OB$, its coordinates is $\displaystyle ( x’, y’ , 0 , t ’ )$. Assume an event $\displaystyle E_{r \ t’} $ occurs at $\displaystyle ( x’, y’ , 0 , t ’ )$. Translate $\displaystyle ( x’, y’ , 0 , t ’ )$ into $\displaystyle ( x, y , 0 , t )$, the coordinates of event $\displaystyle E_{r \ t’} $ in $\displaystyle K $. The event $\displaystyle E_{r \ t’} $ occurs at $\displaystyle ( x, y , 0 , t )$in $\displaystyle K $. Here, $\displaystyle t = t_β( t’ , r)$. Here are, $\displaystyle γ({u_ β (t’,l’) }’) {u_ {βx}{(t’,l’) }}’ $=$\displaystyle u_{βx}({ t}’,l’) ’ / \sqrt {1- \frac { u_{β} ({ t_1}’ ,l’) ’ ^2} {c^2} }$= $\displaystyle -( {r ω \sin(ωt_β( t’ , r)+ π) +v} ) γ(r_ Dω) γ(v) $ . Corresponding to the time $\displaystyle t’ ={ t_1}’=0$, in $\displaystyle K’ $, for any point on $\displaystyle OB$ except $\displaystyle O$, there are $\displaystyle \sin (ω t_β( t’ , r) + π)>0$ (reason 2); Corresponding to the time $\displaystyle t’ ={ t_3}’=γ(v) \frac π {2ω} $ in $\displaystyle K’ $, for any point on $\displaystyle OB$ except $\displaystyle O$, there are $\displaystyle t_β({ t_3}’ , r) =t_{three} = \frac π {2ω} $, $\displaystyle \sin (ω t_β({ t_3}’ , r) + π)=-1$.
$\displaystyle \ \ \ \ \ $reason 1:
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle \begin{cases} {t_1}’=γ(v)\ (t- \frac v {c^2} x) \\
x= r \cos(ωt) \\
t= t_α({t_1}’ , r) \\
γ(v)= \frac 1 {\sqrt {1- \frac {v^2} {c^2} }} \\
{t_1}’=0 \end{cases}$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle t_α({t_1}’ , r)- \frac v {c^2} r \cos(ωt_α({t_1}’ , r))=0$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle ωt_α({t_1}’ , r)- \frac v {c^2}rω \ cos(ωt_α({t_1}’ , r))=0$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $For any point on $\displaystyle OA$ except $\displaystyle O$, $\displaystyle 0<r<r_A$, $\displaystyle 0<rω<c$,$\displaystyle 0<\frac v {c^2}rω<1$.( Because $\displaystyle ω>0$,$\displaystyle c>v=ωr_A>0$ )
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $There is only one solution $\displaystyle 0<x< \frac π 2$ satisfying for the equation $\displaystyle x-a \ cosx=0$, at condition $\displaystyle 0<a<1$.
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $So $\displaystyle 0<ωt_α({t_1}’ , r)< \frac π 2$, $\displaystyle \ sin(ωt_α({t_1}’ , r))>0$.
$\displaystyle \ \ \ \ \ $reason 2:
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $ $\displaystyle \begin{cases} {t_1}’=γ(v)\ (t- \frac v {c^2} x) \\
x= r \cos(ωt+π) \\
t= t_β({t_1}’ , r) \\
γ(v)= \frac 1 {\sqrt {1- \frac {v^2} {c^2} }} \\
{t_1}’=0 \end{cases}$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle t_β({t_1}’ , r)- \frac v {c^2}r \ cos(ωt_β({t_1}’ , r) +π)=0$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle ωt_β({t_1}’ , r)+ \frac v {c^2}rω \ cos(ωt_β({t_1}’ , r))=0$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $For any point on $\displaystyle OB$ except $\displaystyle O$, $\displaystyle 0<r<r_B$, $\displaystyle 0<rω<c$, $\displaystyle 0<\frac v {c^2}rω<1$.( Because $\displaystyle ω>0$,$\displaystyle c>v=ωr_B>0$ )
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $There is only one solution $\displaystyle -\frac π 2<x<0 $ satisfying for the equation $\displaystyle x+a \ cosx=0$, at condition $\displaystyle 0<a<1$.
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $So $\displaystyle -\frac π 2<ωt_β({t_1}’ , r)< 0$, $\displaystyle \ sin(ωt_β({t_1}’ , r) +π)>0$.
$\displaystyle \ \ \ $Summary of part 3
$\displaystyle \ \ \ \ \ $$\displaystyle {γ({u {(t’,l’) }}’) } {u {(t’,l’) }}’ $ has been calculate at part 3
$\displaystyle \ \ \ \ \ $For point on $\displaystyle AB$, $\displaystyle {γ (u(t’,l’) )} $=$\displaystyle \frac 1 {\sqrt {1- \frac { u ({ t}’, l’) ’ ^2} {c^2} }} $ . Here $\displaystyle l’$=$\displaystyle {l (t’,r) }’ $.
$\displaystyle \ \ \ \ \ $For point on $\displaystyle OA$, $\displaystyle γ({u_ α (t’,l’) }’ )$=$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {α} ({ t}’, l’) ’ ^2} {c^2} }} $ .
$\displaystyle \ \ \ \ \ $$\displaystyle {γ ( {u_ α ({ t_1}’,l’) }’ )} $=$\displaystyle γ ( {u_ α ({ t_1}’, {l_D}’ }’ )$=$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {D} ({ t_1}’) ’ ^2} {c^2} }} $ =$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {α} ({ t_1}’, {l_D}’) ’ ^2} {c^2} }} $. Here $\displaystyle {l_D}’$=$\displaystyle {l (t’,r_D) }’ $.
$\displaystyle \ \ \ \ \ $For point on $\displaystyle OB$, $\displaystyle γ ( {u_β({ t}’,l’) }’ ) $=$\displaystyle \frac 1 {\sqrt {1- \frac { u_ {β} ({ t}’, l’) ’ ^2} {c^2} }} $ .
$\displaystyle \ \ \ \ \ $Set $\displaystyle ω>0$, $\displaystyle c>v=ωr_A>0$.
$\displaystyle \ \ \ \ \ $Set time $\displaystyle {t_1}’=0$ in $\displaystyle K’ $.
$\displaystyle \ \ \ \ \ $$\displaystyle {t_3}’=γ(v){\frac π {2ω}} $ in $\displaystyle K’ $,here, $\displaystyle γ(v)= \frac 1 {\sqrt {1- \frac {v^2} {c^2} }} $
$\displaystyle \ \ \ \ \ $$\displaystyle {γ(u_α(t’,l’)’) } {u_{αx}{(t’,l’) }}’ $ =$\displaystyle \ \ \ \ \ $$\displaystyle u_α({ t}’ ,l’) ’ / \sqrt {1- \frac { u_α({ t}’ ,l’) ’ ^2} {c^2} }$= $\displaystyle - ({r ω \ sin(ωt_α(t’,r)) +v}) γ(r_ Dω) γ(v) $
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle \ sin(ωt_α({t_1}’ , r))>0$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle \ sin (ω t_α({ t_3}’ , r))=1$
$\displaystyle \ \ \ \ \ $$\displaystyle {γ(u_ β (t’,l’)’) }{u_ {βx} {(t’,l’) }}’ $ = $\displaystyle -( {r ω \sin(ωt_β(t’,r) +π) +v} ) γ(r_ Dω) γ(v) $
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle \ sin(ωt_β({t_1}’ , r) +π)>0$
$\displaystyle \ \ \ \ \ $$\displaystyle \ \ \ \ \ $$\displaystyle \ sin (ω t_β({ t_3}’ , r) + π)=-1$

Last edited by Liuxinhua; Sep 6th 2018 at 07:43 AM.
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Old Sep 5th 2018, 11:45 PM   #14
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Part4 4-momentum of AB in $\displaystyle K’ $ at time $\displaystyle {t_1}’$ and $\displaystyle {t_{3}}’$
Calculate the 4 momentum of a rotating rod. The work has been divided into 4 parts.
$\displaystyle \ \ \ \ \ $This part is built on the basis of part2 part 3.

$\displaystyle \ \ \ \ \ $ In $\displaystyle K$, static auxiliary circle D and M, take O as their center of their circle, intersecting rod OA at point D and M, intersecting rod OB at point E and N. For confirmed auxiliary circle D and auxiliary circle M, $\displaystyle r_D= r_E$ $\displaystyle r_M= r_N$.
$\displaystyle \ \ \ \ \ $Now calculate 4-momentum of MD and NE on $\displaystyle x’ $ direction in $\displaystyle K’ $ at time$\displaystyle {t_1}’$ and $\displaystyle {t_{3}}’$.
$\displaystyle \ \ \ \ \ $$\displaystyle {p_{ MDx }({t_1}’) }’ $ = $\displaystyle \int_{{ l_M({t_1}’) } ’} ^{ { l_D({t_1}’) } ’} {γ(u_α{ ({t_1}’,l’) ’ } ) } ρ_α({t_1}’,l’) ’ {u_ {αx}{ ({t_1}’,l’) }}’ {d l’ }$
$\displaystyle \ \ \ \ \ $= $\displaystyle \int_{{ l_M({t_1}’) } ’} ^{ { l_D({t_1}’) }’ } ρ_α({t_1}’,l’) ’ ( {-r ω \sin(ωt_α( {t_1}’ , r)) -v} ) γ(rω) γ(v) {d l’ }$
$\displaystyle \ \ \ \ \ $Though $\displaystyle r$ is a constant of each point. But for determined $\displaystyle {t_1}’$, each $\displaystyle r$ corresponds to $\displaystyle l’$ one by one. So $\displaystyle r$ can be considered as a function of $\displaystyle {t_1}’$ and$\displaystyle l’$. And $\displaystyle l’$ also can be considered as a function of $\displaystyle {t_1}’$ and $\displaystyle r$.
$\displaystyle \ \ \ \ \ $Because $\displaystyle \ sin(ωt_α({t_1}’ , r))>0$
$\displaystyle \ \ \ \ \ $$\displaystyle {p_{ MDx }({t_1}’) }’ $<$\displaystyle \int_{{ l_M({t_1}’) }’ } ^{ { l_D({t_1}’) }’ } {( - v)ρ_α({t_1}’,l’) ’ } γ(rω) γ(v) {d l’ }$
$\displaystyle \ \ \ \ \ $$\displaystyle l’ $ is function of $\displaystyle {t_1}’$ and $\displaystyle r$. $\displaystyle {t_1}’$=0 is a constant. Make variable substitution$\displaystyle l’ →r$
$\displaystyle \ \ \ \ \ $$\displaystyle {p_{ MDx }({t_1}’) }’ $<$\displaystyle \int_{{ r_M } } ^{ { r_D } } (-v) { ρ_α({t_1}’, f_ α({t_1}’, r) ’) ’ } γ(rω) γ(v) \frac {\partial f_α({t_1}’,r)’} {\partial r} {d r}$
$\displaystyle \ \ \ \ \ $=$\displaystyle \int_{{ r_M } } ^{ { r_D } } (-v) { g_α({t_1}’, r) ’ } γ(rω) γ(v) \frac {\partial f_α({t_1}’,r)’} {\partial r} {d r}$
$\displaystyle \ \ \ \ \ $=$\displaystyle \int_{{ r_M } } ^{ { r_D } } (-v ) {ρ ( r) } γ(rω) γ(v) {d r}$
$\displaystyle \ \ \ \ \ $The same $\displaystyle {p_{ NEx }({t_1}’) }’ $<$\displaystyle \int_{{ r_N } } ^{ { r_E } } (-v) {ρ ( r) } γ(rω) γ(v) {d r}$
$\displaystyle \ \ \ \ \ $$\displaystyle {p_{ MDx }({t_3}’) }’ $=$\displaystyle \int_{{ r_M } } ^{ { r_D } } { ( -rω-v )ρ ( r) } γ(rω) γ(v) {d r}$
$\displaystyle \ \ \ \ \ $$\displaystyle {p_{ NEx }({t_3}’) }’ $=$\displaystyle \int_{{ r_N } } ^{ { r_E } } { ( rω-v )ρ ( r) } γ(rω) γ(v) {d r}$
$\displaystyle \ \ \ \ \ $For confirmed auxiliary circle D and auxiliary circle M, $\displaystyle r_D= r_E$ $\displaystyle r_M= r_N$.
$\displaystyle \ \ \ \ \ $$\displaystyle \int_{{ r_M } } ^{ { r_D } } ( -rω) { ρ ( r) } γ(rω) γ(v) {d r}$+$\displaystyle \int_{{ r_N } } ^{ { r_E } } { rωρ ( r) } γ(rω) γ(v) {d r}$=0
$\displaystyle \ \ \ \ \ $So $\displaystyle {p_{ MDx }({t_3}’) }’ $+$\displaystyle {p_{ NEx }({t_3}’) }’ $=$\displaystyle \int_{{ r_M } } ^{ { r_D } } ( -v ){ ρ ( r) } γ(rω) γ(v) {d r}$+$\displaystyle \int_{{ r_N } } ^{ { r_E } } ( -v ){ ρ ( r) } γ(rω) γ(v) {d r}$
$\displaystyle \ \ \ \ \ $Then $\displaystyle {p_{ MDx }({t_1}’) }’ $+$\displaystyle {p_{ NEx }({t_1}’) }’ $< $\displaystyle {p_{ MDx }({t_3}’) }’ $+$\displaystyle {p_{ NEx }({t_3}’) }’ $
$\displaystyle \ \ \ \ \ $ This may be a wrong result. But where is the mistake?
$\displaystyle \ \ \ \ \ $I really need your help. Thanks for this.
$\displaystyle \ \ \ \ \ $There is one point to note: the result is independent of the space-time bending.
$\displaystyle \ \ \ \ \ $Remember $\displaystyle k_{AB}$=$\displaystyle \frac {{p_{ BAx }({t_1}’) }’ -{p_{ BAx }({t_3}’) }’} {{p_{ BAx }({t_3}’) }’}$.
$\displaystyle \ \ \ \ \ $$\displaystyle k_{AB}$ is determined by parameters$\displaystyle v, ω,r_A.$
$\displaystyle \ \ \ \ \ $Rod PQ can replace the rod AB, The shape of rod PQ is the same as rod AB. The mass of rod PQ is $\displaystyle m$ times ($\displaystyle m$ can be 2, 4, 8... or 1/2, 1/4, 1/8..., any arbitrary value) the quality of rod AB (the proportion of mass distribution along length remains unchanged). But $\displaystyle k_{PQ}$=$\displaystyle k_{AB}$. The quality of AB is near to infinitesimal, so the result is independent of the space-time bending.

Last edited by Liuxinhua; Sep 6th 2018 at 07:25 AM.
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Old Sep 6th 2018, 12:05 AM   #15
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I haven't the vaguest notion of what you are trying to communicate. And, frankly, I haven't the slightest bit of desire to.

Please do me (and probably everyone else) the favor of telling us what your ideas are. We can fill in the Math later.

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Old Sep 6th 2018, 07:32 AM   #16
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Get a result, but it may be a wrong.
So, it need to sure :
The potential energy of born rigid body is a constant ?
If it is sure , $\displaystyle {m_{sum\ 0 \ OD } ({t}’ )} ’={m_{sum\ 0 \ OD } }(t) $
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