• #Newton's laws and relativity
  • #Third law and simultaneity
• #Work energy theorem :-
• #Transformation of forces :-
• #Proper force / Minkowski force :-

STR | PDF | << prev | next >>

1. Newton's first law was built into principles of reletivity
2. Newton's 2^nd Law remains valid as long as we use reletivistic momentum \[ \left\{ \vec{F} = \frac{d\vec{p}}{dt} \right\} \qquad \text{where} \quad \vec{p} = \gamma m\vec{u} \]
3. Newtons third law is not consistent with reletivity.

(Action \(\neq\) reaction) & are not simultaneous in different frames.

{ Third law is incompatible with reletivity of simultanity }

Side note in the margin:

1. inertia \(\to\) mass
2. inertia \(\sim\) momentum \[ 3)\quad \vec{p} \;=\; \gamma m\vec{v} \;=\; \frac{m\vec{v}}{\sqrt{1-v^2/c^2}} \]

   \begin{tikzpicture}[scale=1.0]
     \draw[->] (0,0) -- (0,3.0) node[above] {\small $\gamma$};
     \draw[->] (0,0) -- (6.4,0) node[right] {\small \% the speed of light};
     \foreach \y/\l in {0.35/1,0.70/2,1.05/3,1.40/4,1.75/5,2.10/6,2.45/7,2.80/8}
        { \draw (-0.08,\y) -- (0.08,\y); \node[left] at (-0.10,\y) {\tiny \l}; }
     \foreach \x/\l in {0.64/10,1.28/20,1.92/30,2.56/40,3.20/50,3.84/60,4.48/70,5.12/80,5.76/90,6.40/100}
        { \draw (\x,-0.08) -- (\x,0.08); \node[below] at (\x,-0.10) {\tiny \l}; }
     \draw[very thick] plot[smooth] coordinates
       {(0,0.35) (0.64,0.352) (1.28,0.357) (1.92,0.367) (2.56,0.382)
        (3.20,0.404) (3.84,0.438) (4.48,0.490) (5.12,0.583) (5.44,0.663)
        (5.76,0.803) (6.08,1.130) (6.24,1.520) (6.34,2.230) (6.38,2.800)};
     \draw[->] (5.30,2.10) -- (5.62,0.85);
     \node[draw,anchor=south west,align=left] at (4.55,2.15)
          {\tiny \textbf{90\% SPEED OF LIGHT}\\ \tiny $\gamma = 2.294$};
   \end{tikzpicture}
   

/[A screenshot of a gamma-vs-speed plot was pasted into the notebook at this point; the sketch above is a redrawn stand-in.]/

\begin{tikzpicture}
  \node at (0,0.45) {$A$};
  \node at (2.0,0.45) {$B$};
  \draw (0,0.1) -- (2.0,0.1);
  \node[below] at (1.0,0.12) {\small $\Delta x$};
\end{tikzpicture}

\[ \left. \begin{aligned} \vec{F}_{AB} &= F(t)\\ \vec{F}_{BA} &= -F(t) \end{aligned} \right\} \quad \text{In frame } S \]

But In frame \((S')\) :-

\[ \vec{F}\,'_{AB} \;=\; F'(t_1') \] \[ \text{then} \qquad \vec{F}\,'_{BA} \;\neq\; -F'(t_1') \;=\; F'(t_2') \] \[ \text{as} \qquad \Delta t' = \left( \Delta t - \frac{u\Delta x}{c^2} \right)\gamma \] \[ \Delta t' \;=\; -\left( \frac{u\Delta x}{c^2} \right)\gamma \] \[ t_2' - t_1' \;=\; -\,\frac{u\Delta x}{c^2}\,\gamma \] \[ \Rightarrow \quad \left\{ F'_{AB} \neq -F'_{BA} \right\} \quad \text{due to forces felt at different times } t_1' \& t_2'. \]

Note :-

Third law is only valid for \(\Delta x = 0\) \(\Rightarrow\) \((\Delta t' = \Delta t = 0)\)

\[ W \;=\; \int \vec{F}\cdot d\vec{x} \qquad (\text{holds reletivistically}) \] \[ W \;=\; \int \frac{d\vec{p}}{dt}\cdot dx \;=\; \int \frac{d\vec{p}}{dt}\cdot\frac{d\vec{x}}{dt}\,dt \;=\; \int \left( \frac{d\vec{p}}{dt}\cdot\vec{v} \right) dt \] \[ \left\{ \text{here } \vec{v} \text{ is velocity of particle under force } \vec{F} \right\} \] \[ \frac{d\vec{p}}{dt}\cdot\vec{v} \;=\; \frac{d}{dt}\left( \frac{m\vec{v}}{\sqrt{1-v^2/c^2}} \right)\cdot\vec{v} \;=\; m\left( \frac{\dfrac{d\vec{v}}{dt}\sqrt{1-v^2/c^2} + \dfrac{\vec{v}\,(v/c^2)}{\sqrt{1-v^2/c^2}}\dfrac{dv}{dt}}{\left( 1-v^2/c^2 \right)} \right)\cdot\vec{v} \] \[ = m\left( \frac{\vec{a}\left( 1 - v^2/c^2 \right) + \vec{v}\,\dfrac{v}{c^2}\dfrac{dv}{dt}}{\left( 1-v^2/c^2 \right)^{3/2}} \right)\cdot\vec{v} \] \[ = m\,\frac{\vec{v}\cdot\dfrac{d\vec{v}}{dt}\left( \dfrac{1-v^2}{c^2} \right) + \dfrac{v}{c^2}\left( \vec{v}\cdot\vec{v} \right)\dfrac{dv}{dt}}{\left( 1-v^2/c^2 \right)^{3/2}} \] \[ = m\,\frac{\vec{v}\cdot\dfrac{d\vec{v}}{dt} \;-\; \dfrac{v^2}{c^2}\dfrac{dv}{dt} \;+\; \dfrac{v}{c^2}\,v\,\dfrac{dv}{dt}}{\left( 1-v^2/c^2 \right)^{3/2}} \]

(the last two terms cancel)

\[ \left. \begin{aligned} \frac{d}{dt}(v^2) &= \frac{d}{dt}\left( \vec{v}\cdot\vec{v} \right) = 2\vec{v}\cdot\frac{d\vec{v}}{dt}\\ \text{also,}\quad \frac{d}{dt}(v^2) &= 2v\frac{dv}{dt} \end{aligned} \right\} \Rightarrow \left( v\frac{dv}{dt} = \vec{v}\cdot\frac{d\vec{v}}{dt} \right) \] \[ \frac{d\vec{p}}{dt}\cdot\vec{v} \;=\; \left[ m\left( \frac{d\vec{v}}{dt} \right) \frac{1}{\left( 1-v^2/c^2 \right)^{3/2}} \right]\cdot\vec{v} \;=\; \left( \frac{m\vec{v}}{\left( 1-\dfrac{v^2}{c^2} \right)^{3/2}} \right)\cdot\frac{d\vec{v}}{dt} \] \[ = \frac{d}{dt}\left( \frac{mc^2}{\sqrt{1-v^2/c^2}} \right) \] \[ \boxed{\; \frac{d\vec{p}}{dt}\cdot\vec{v} \;=\; \frac{dE}{dt} \;} \] \[ \left( W \;=\; \int \frac{dE}{dt}\,dt \;=\; \Delta E \;=\; E_f - E_i \right) \] \[ \text{If rest mass energy is constant} \quad \left( W = \Delta \text{K.E} \right) \]

Since force is \(\dfrac{d\vec{p}}{dt}\) it transforms like improper velocity or normal velocity

If \((F_x, F_y, F_z)\) are forces in \(S\) frame

and \((F_x', F_y', F_z')\) are forces in \(S'\) frame

then

\begin{tikzpicture}
  \draw (0,1.9) -- (0,0) -- (2.4,0);
  \node[above right] at (0,1.9) {$S$};
  \draw[->] (0,1.55) -- (0,1.9);
  \draw[->] (0.55,0.35) -- (1.05,1.05);
  \node[above] at (1.05,1.05) {$\vec{v}$};
  \draw (2.9,1.9) -- (2.9,0.9) -- (4.9,0.9);
  \node[above] at (2.9,1.9) {$S'$};
  \draw[->] (2.9,1.35) -- (3.5,1.35) node[right] {$u$};
  \draw[->] (2.6,0.05) .. controls (1.6,-0.35) and (0.9,-0.05) .. (0.62,0.28);
  \node[anchor=west] at (2.7,0.05) {velocity of particle};
  \node[anchor=west] at (5.6,0.05) {and $\beta = u/c$};
\end{tikzpicture}

\[ (1)\quad F_x' \;=\; \frac{d\vec{p}_x}{dt'} \;=\; \frac{\left( dp_x - \beta\,dp^0 \right)\gamma}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{\left( \dfrac{dp_x}{dt} - \dfrac{\beta}{c}\dfrac{dE}{dt} \right)}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \;=\; \frac{\left( F_x - \dfrac{\beta}{c}\dfrac{dE}{dt} \right)}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \] \[ \boxed{\; F_x' \;=\; \frac{F_x - \dfrac{\beta}{c}\left( \vec{F}\cdot\vec{v} \right)}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \;} \qquad \begin{aligned} p^\mu &= (p^0, p^1, p^2, p^3)\\ \bar{p}^0 &= (p^0 - \beta p^1)\gamma\\ \bar{p}^1 &= (p^1 - \beta p^0)\gamma\\ \bar{p}^2 &= p^2\\ \bar{p}^3 &= p^3 \end{aligned} \] \[ (2)\quad F_y' \;=\; \frac{dp_y'}{dt'} \;=\; \frac{dp_y}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \] \[ \boxed{\; F_y' \;=\; \frac{F_y\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \;} \] \[ (3)\quad F_z' \;=\; \frac{dp_z'}{dt'} \;=\; \frac{dp_z}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{F_z\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \] \[ \boxed{\; F_z' \;=\; \frac{F_z\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{u v_x}{c^2} \right)} \;} \]

You can see that force here transforms like normal velocity.

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  \draw[->] (0,1.25) -- (0,1.6);
  \draw[->] (0,0.35) -- (-0.6,-0.25);
  \node[below left] at (-0.6,-0.25) {\small $S$};
  \fill (0.35,0.95) circle (1.2pt);
  \node[left] at (0.30,0.95) {\small $P$};
  \draw[->] (0.45,1.00) -- (0.95,1.25) node[right] {\small $v$};
  \draw[->] (0.35,0.10) -- (0.35,0.85);
  \node[anchor=west] at (0.45,0.05) {\small particle};
  \draw (2.5,1.6) -- (2.5,0.75) -- (3.8,0.75);
  \node[above] at (2.5,1.62) {\small $S'$};
  \draw[->] (2.5,1.15) -- (3.05,1.15) node[right] {\small $u$};
  \node[anchor=west] at (3.9,1.15) {\small Then $s$ frame};
\end{tikzpicture}

\[ V_x' \;=\; \frac{dx'}{dt'} \;=\; \frac{(dx - u\,dt)\,\gamma}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{(V_x - u)}{\left( 1 - \dfrac{uV_x}{c^2} \right)} \] \[ V_y' \;=\; \frac{dy'}{dt'} \;=\; \frac{dy}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{V_y\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{uV_x}{c^2} \right)} \] \[ \& \quad V_z' \;=\; \frac{dz'}{dt'} \;=\; \frac{dz}{\left( dt - \dfrac{u\,dx}{c^2} \right)\gamma} \;=\; \frac{V_z\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{uV_x}{c^2} \right)} \]

Note :- Throught the notes of relativity I have used Symbol \(u\)' for speed / velocity of another frame \(S'\) w.r.t \(S\). and \(v\)' represents velocity of objects moving in these frames.

let us define proper force

\[ K^\mu \;=\; \frac{dp^\mu}{d\tau} \qquad \left\{ p^\mu = m\,\eta^\mu = m\,\frac{dx^\mu}{d\tau} \right\} \] \[ \text{here} \quad K^0 \;=\; \frac{dp^0}{d\tau} \;=\; \frac{dp^0}{d\tau} \;=\; \frac{1}{c}\frac{dE}{d\tau} \qquad \left\{ K^\mu = \frac{dp^\mu}{d\tau} = m\,\frac{d^2x^\mu}{d\tau^2} \right\} \] \[ K^1 \;=\; \frac{dp^1}{d\tau} \;=\; \frac{dp_x}{d\tau} \] \[ K^2 \;=\; \frac{dp^2}{d\tau} \;=\; \frac{dp_y}{d\tau} \qquad \& \qquad K^3 \;=\; \frac{dp_z}{d\tau} \] \[ \text{here} \quad \left\{ \bar{K}^\mu \;=\; \Lambda^\mu_{\ \nu} K^\mu \right\} \quad \text{transforms like a four vector.} \] \[ K^\mu \;=\; \left( K^0,\ \vec{K} \right) \;=\; \left( \frac{1}{c}\frac{dE}{d\tau},\ \frac{d\vec{p}}{d\tau} \right) \qquad \left( \vec{p} = \frac{m\vec{v}}{\sqrt{1-v^2/c^2}} \right) \] \[ = \left( \frac{1}{c}\frac{dE}{d\tau},\ \frac{F}{\sqrt{1-v^2/c^2}} \right) \] \[ \boxed{\; K^\mu \;=\; \left( \frac{1}{c}\frac{dE}{d\tau},\ \gamma F \right) \;} \]