STR | PDF | << prev
we have already defined some relativistic quantities.
\((a^\mu)\) :- It has 4 components and transform like position vector \((x^\mu)\) under lorentz Transformation
\[
x^\mu \;=\; (x^0,\ x^1,\ x^2,\ x^3)
\]
Where
\[
\left.
\begin{aligned}
\bar{x}^\mu &= \Lambda^\mu_{\ \nu}\, x^\nu\\
\text{So}\quad \bar{a}^\mu &= \Lambda^\mu_{\ \nu}\, a^\nu
\end{aligned}
\right\}
\quad
\begin{aligned}
&\text{Lorentz transformations of any}\\
&\text{four vector } a^\mu \text{ and position}\\
&\text{4-vector } x^\mu .
\end{aligned}
\]
We can define \(\Delta x^\mu = x_A^\mu - x_B^\mu\) as displacement 4-vector.
(\(x_A^\mu\) is the four-vector of event \(A\), \(x_B^\mu\) the position 4-vector of event \(B\).)
It also transform like those coordinates under LT.
\[
\left\{\; \overline{\Delta x^\mu} \;=\; \Lambda^\mu_{\ \nu}\, \Delta x^\nu \;\right\}
\]
Proper velocity 4-vector \((\eta^\mu)\) & proper velocity \((\vec\eta\,)\) :-
To have a vector we need to devide \(\Delta x^\mu\) by a scalar
here \(\Delta\tau\) is invarient under different frames so.
we define
\[
\eta^\mu \;=\; \frac{\Delta x^\mu}{\Delta\tau} \qquad \text{or} \qquad \eta^\mu \;=\; \frac{dx^\mu}{d\tau}
\]
here \(\eta^\mu\) is 4-velocity vector. It should also transform like coordinates under L.T
\[
\bar\eta^\mu \;=\; \Lambda^\mu_{\ \nu}\, \eta^\nu
\qquad
\begin{aligned}
d\tau &= \text{proper time}\\
d\tau &= dt\sqrt{1-u^2/c^2}
\end{aligned}
\]
\[
\left( \text{Here } \vec\eta = \text{spatial part of } \eta^\mu \;=\; \frac{\vec{u}}{\sqrt{1-u^2/c^2}} \;=\; \text{proper velocity (3-D vector)} \right)
\]
Remember, all 4-vectors should by defination transform like coordinates under Lorentz transformations (change of frames).
Where as, the ordinary velocity vector should transform in different way from one frame to another
\begin{tikzpicture}
% S frame axes
\draw[->] (0,0) -- (0,3.0) node[above] {$y$};
\draw[->] (0,0) -- (4.6,0) node[right] {$x$};
\draw[->] (0,0) -- (-1.5,-1.5) node[below left] {$z$};
\node[above left] at (0.05,3.0) {$S$};
% velocity vector at a point
\draw[->] (1.1,0.9) -- (2.15,2.15);
\node[above] at (2.15,2.15) {$\vec{V}$};
\draw[->] (1.1,0.9) -- (1.1,1.95);
\node[left] at (1.05,1.6) {$V_y$};
\draw[->] (1.1,0.9) -- (2.25,0.9);
\node[right] at (2.25,0.9) {$V_x$};
\draw[->] (1.1,0.9) -- (0.35,0.2);
\node[below] at (0.35,0.15) {$V_z$};
% S' frame
\draw (3.1,2.7) -- (3.1,1.15) -- (5.6,1.15);
\node[above] at (3.1,2.75) {$S'$};
\draw[->] (3.1,2.05) -- (3.75,2.05) node[right] {$u$};
\end{tikzpicture}
let us study motion of a particle travelling with velocity \(\vec{V}\) in frame \(S\)
Find \(\vec{V}\) in frame \(S'\).
\[
\begin{aligned}
\vec{V}\Big)_{S} &= \vec{V} = (V_x,\ V_y,\ V_z)\\
\vec{V}\Big)_{S'} &= V' = (V_x',\ V_y',\ V_z')
\end{aligned}
\]
\[
V_x' \;=\; \frac{\Delta x'}{\Delta t'} \;=\; \frac{(\Delta x - u\Delta t)\,\gamma}{\left( \Delta t - \dfrac{u\Delta x}{c^2} \right)\gamma} \;=\; \frac{\left( \dfrac{\Delta x}{\Delta t} - u \right)}{\left( 1 - \dfrac{u}{c^2}\dfrac{\Delta x}{\Delta t} \right)} \;=\; \frac{V_x - u}{1 - \dfrac{uV_x}{c^2}}
\]
\[
V_y' \;=\; \frac{\Delta y'}{\Delta t'} \;=\; \frac{\Delta y}{\left( \Delta t - \dfrac{u\Delta x}{c^2} \right)\gamma} \;=\; \frac{\dfrac{\Delta y}{\Delta t}\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{u}{c^2}\dfrac{\Delta x}{\Delta t} \right)} \;=\; \frac{V_y\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{uV_x}{c^2} \right)}
\]
\[
V_z' \;=\; \frac{\Delta z'}{\Delta t'} \;=\; \frac{\Delta z}{\left( \Delta t - \dfrac{u\Delta x}{c^2} \right)\gamma} \;=\; \frac{\left( \dfrac{\Delta z}{\Delta t} \right)\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{u}{c^2}\dfrac{\Delta x}{\Delta t} \right)} \;=\; \frac{V_z\sqrt{1-u^2/c^2}}{\left( 1 - \dfrac{uV_x}{c^2} \right)}
\]
here `\(u\)' is \(\left|\text{Relative velocity}\right|\) of \(S'\) w.r.t. \(S\)
We define energy-momentum 4vector (or) Momentum 4-vector as
\[
p^\mu \;=\; m_0\eta^\mu \;=\; m_0\frac{dx^\mu}{d\tau}
\]
\[
\text{here} \quad p^0 c = E \qquad \text{and} \qquad \vec{p} = \gamma m\vec{u} = \frac{m\vec{u}}{\sqrt{1-u^2/c^2}}
\]
(\(E\) is the Relativistic Energy of a particle of mass \(m_0\); \(\vec{p}\) the Relativistic momentum of particle moving with ordinary velocity \(\vec{u}\).)
\[
E \;=\; p^0 c \;=\; c\,m_0\eta^0 \;=\; c\,m_0\frac{dx^0}{d\tau} \;=\; \frac{m_0c^2}{\sqrt{1-u^2/c^2}}
\]
\[
\text{So,} \qquad E = \frac{m_0c^2}{\sqrt{1-u^2/c^2}} \qquad \& \qquad \vec{p} = \frac{m_0\vec{u}}{\sqrt{1-u^2/c^2}}
\]
We can thus express momentum 4 vector \(p^\mu\) as
\[
p^\mu \;=\;
\begin{pmatrix} p^0\\ p^1\\ p^2\\ p^3 \end{pmatrix}
\;=\;
\begin{pmatrix} p^0\\ \vec{p} \end{pmatrix}
\;=\;
\left( \frac{m_0c}{\sqrt{1-u^2/c^2}},\ \frac{m_0\vec{u}}{\sqrt{1-u^2/c^2}} \right)
\]
This 4-vector must also transform like coordinates under L.T.
\[
\bar{p}^\mu \;=\; \Lambda^\mu_{\ \nu}\, p^\nu
\]
\[
\text{i.e.} \quad
\left\{
\begin{aligned}
\bar{p}^0 &= (p^0 - \beta p^1)\,\gamma\\
\bar{p}^1 &= (p^1 - \beta p^0)\,\gamma\\
\bar{p}^2 &= p^2\\
\bar{p}^3 &= p^3
\end{aligned}
\right\}
\quad
\begin{aligned}
&\text{For motion of frame } S' \text{ along}\\
&x\,x' \ \underline{\text{axis}}\ \text{with speed } u\\
&\gamma = \frac{1}{\sqrt{1-u^2/c^2}}
\end{aligned}
\]
In every closed system Total relativistic energy and momentum are conserved.
Invarient - Same value in different frames of reference
Conserved - Same value before and after some process.
Quantity
Invarient
Conserved
mass
yes
no
Energy
no
yes
electric charge
yes
yes
Velocity
no
no
\(a^\mu a_\mu\)
yes
?
let us find the Invarient scalar product of 4-momentum vector
\[
\begin{aligned}
I \;=\; p^\mu p_\mu &= -(p^0)^2 + \vec{p}\cdot\vec{p}\\
&= -\left( \frac{mc}{\sqrt{1-u^2/c^2}} \right)^2 + \left( \frac{m\vec{u}}{\sqrt{1-u^2/c^2}}\cdot\frac{m\vec{u}}{\sqrt{1-u^2/c^2}} \right)\\
&= \frac{-m^2c^2}{1-u^2/c^2} + \frac{m^2u^2}{1-u^2/c^2}\\
&= -\,\frac{m^2c^2(c^2-u^2)}{(c^2-u^2)} \;=\; -m^2c^2
\end{aligned}
\]
\[
\text{i.e.} \qquad p^\mu = \left( p^0,\ \vec{p} \right) \;=\; \left( \frac{E}{c},\ \vec{p} \right)
\]
\[
p_\mu = \left( -p_0,\ \vec{p} \right) \;=\; \left( -E/c,\ \vec{p} \right)
\]
\[
I \;=\; p^\mu p_\mu \;=\; -\frac{E^2}{c^2} + p^2 \;=\; -m^2c^2
\]
\[
\Rightarrow \qquad \boxed{\; E^2 - p^2c^2 \;=\; m^2c^4 \;}
\]
We can define idea of Invarient mass as
\[
m \;=\; \frac{-p^\mu p_\mu}{c^2}
\qquad
\left( \begin{aligned}&\text{But this is not in general}\\ &\text{Sum of Individual mass}\end{aligned} \right)
\]
Coordinate of any event \(= X = (ct,\ x,y,z) = (x^0, x^1, x^2, x^3) = x^\mu\)
\[
\text{under} \ \underline{\text{L.T.}} \qquad \bar{x}^\mu = \Lambda^\mu_{\ \nu} x^\nu
\]
\[
\text{i.e.} \quad
\begin{aligned}
\bar{x}^1 &= (x^1 - \beta x^0)\,\gamma\\
\bar{x}^0 &= (x^0 - \beta x^1)\,\gamma
\end{aligned}
\]
we have defined the invarient
\[
\left(
\begin{aligned}
x^\mu x_\mu &= X\cdot X = x^2 = -(x^0)^2 + \vec{r}\cdot\vec{r} = -(x^0)^2 + r^2\\
\Delta x^\mu \Delta x_\mu &= (\Delta x)^2 = -(\Delta x^0)^2 + (\Delta x^1)^2 + (\Delta x^2)^2 + (\Delta x^3)^2
\end{aligned}
\right)
\]
\[
\text{Similarly} \qquad p^\mu p_\mu \;=\; \bar{p}^\mu \bar{p}_\mu \qquad (\text{invarient})
\]
let me prove it using tensor notation.
\(p^\mu\) transform like coordinates under LT
\[
\left\{ \bar{p}^\mu = \Lambda^\mu_{\ \nu}\, p^\nu \right\} \qquad \left\{ \bar{p}_\mu = \Lambda^{-1\,\mu}_{\ \ \ \nu}\, p_\nu \right\}
\]
\[
\Rightarrow \quad \bar{p}^\mu \bar{p}_\mu \;=\; \Lambda^\mu_{\ \nu}\, p^\nu\, \Lambda^{-1\,\mu}_{\ \ \ \nu}\, p_\nu \;=\; \left( \Lambda^\mu_{\ \nu} \Lambda^{-1\,\mu}_{\ \ \ \nu} \right) p^\mu p_\mu
\]
\[
\boxed{\; \bar{p}^\mu\, \bar{p}_\mu \;=\; p^\mu p_\mu \;}
\]
\[
\begin{aligned}
\bar{p}^0 &= (p^0 - \beta p^1)\,\gamma\\
\bar{p}^1 &= (p^1 - \beta p^0)\,\gamma\\
\bar{p}^2 &= p^2\\
\bar{p}^3 &= p^3
\end{aligned}
\quad \Rightarrow \quad
\begin{aligned}
-\bar{p}_0 &= (-p_0 - \beta p_1)\,\gamma \quad \Rightarrow \quad \bar{p}_0 = (p_0 + \beta p_1)\,\gamma\\
\bar{p}_1 &= (p_1 + \beta p_0)\,\gamma\\
\bar{p}_2 &= p_2\\
\bar{p}_3 &= p_3
\end{aligned}
\]
\[
\Lambda^\mu_{\ \nu} \;=\;
\begin{pmatrix}
\gamma & -\beta\gamma & 0 & 0\\
-\beta\gamma & \gamma & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}
\]
We could have defined Invarient quantity \(I = a^\mu b_\mu\) as in both ways.
\[
\left.
\begin{aligned}
I &= a^\mu b_\mu = -(a^0b^0) + \vec{a}\cdot\vec{b}\\
\text{or}\quad I &= -a^\mu b_\mu = (a^0b^0) - \vec{a}\cdot\vec{b}
\end{aligned}
\right\}
\quad
\begin{aligned}&\text{Both are correct and in}\\ &\text{use in literature.}\end{aligned}
\]
For any two four vectors \(A\) & \(B\).
\[
\begin{aligned}
A^\mu &= (A^0,\ \vec{A}\,)\\
B^\mu &= (B^0,\ \vec{B}\,)
\end{aligned}
\]
\[
\begin{aligned}
I &= A^\mu B_\mu \;=\; -A^0B^0 + \vec{A}\cdot\vec{B} \;=\; \text{Invarient}\\
\text{or}\quad \underline{I} &= -A^\mu B_\mu \;=\; A^0B^0 - \vec{A}\cdot\vec{B} \;=\; \text{Invarient}
\end{aligned}
\]
\[
\text{For special case } \vec{A} = \vec{B},\qquad I = A^\mu A_\mu = -(A^0)^2 + |A|^2 = \text{Invarient}
\]
\[
\text{Similarly for } A^\mu = p^\mu \quad \Rightarrow \quad I = (p^0)^2 - p^2 = m^2c^2
\]
So Four vector of any kind has its \((\text{norm})^2 = a^\mu a_\mu\) which is Invarient under LT (under different frames it is same).
\begin{tikzpicture}
\draw (0,2.1) -- (0,0) -- (3.4,0);
\node[above right] at (0,2.1) {$S$};
\draw[dashed] (1.15,1.85) -- (1.15,0.75);
\node[above] at (1.15,1.9) {$S'$};
\fill (1.15,1.15) circle (1.5pt);
\node[left] at (1.10,1.15) {$m$};
\draw[->] (1.25,1.15) -- (1.95,1.15) node[right] {$u$};
\end{tikzpicture}
\[
p^\mu\Big)_{S} \;=\; \left( \frac{E}{c},\ \vec{p} \right)
\]
\[
\bar{p}^\mu\Big)_{S'} \;=\; \left( \frac{mc}{\sqrt{1-u^2/c^2}},\ \frac{m\vec{u}}{\sqrt{1-u^2/c^2}} \right)
\qquad
\left\{ \begin{aligned}&u = \text{velocity of}\\ &\text{particle}\end{aligned} \right.
\]
(the spatial part goes to \(0\) in the particle's own frame)
\[
\bar{p}^\mu \;=\; \left( mc,\ 0 \right)
\]
\[
\overline{p^\mu}\cdot\overline{p}_\mu \;=\; m^2c^2 \qquad \text{Which is Same}
\]
\[
p^\mu p_\mu \;=\; \frac{E^2}{c^2} - p^2 \;=\; m^2c^2
\]
Shortcut for \(p^\mu p_\mu = -m^2c^2\)
\begin{tikzpicture}
\draw[->] (0,0) -- (0,1.5);
\draw[->] (0,0) -- (1.7,0);
\draw[->] (0,0) -- (-0.8,-0.8);
\fill (0.85,1.15) circle (1.4pt);
\node[above] at (0.85,1.22) {\small $m$};
\draw[->] (0.95,1.15) -- (1.55,1.15) node[right] {\small $v$};
\end{tikzpicture}
In frame of particle
\[
\begin{aligned}
p &= \left( p^0,\ \vec{p} \right)\\
&= \left( \frac{mc}{\sqrt{1-u^2/c^2}},\ \frac{m\vec{u}}{\sqrt{1-u^2/c^2}} \right) \quad (u \to 0)\\
&= \left( mc,\ 0 \right)
\end{aligned}
\]
\[
\boxed{\; p\cdot p \;=\; m^2c^2 \;}
\]
For any particle of energy \(E\) & momentum \(\vec{p}\)
\[
p \;=\; \left( p^0,\ \vec{p} \right) \;=\; \left( \frac{E}{c},\ \vec{p} \right)
\qquad
\left\{
\begin{aligned}
E &= \frac{mc^2}{\sqrt{1-u^2/c^2}}\\
\vec{p} &= \frac{m\vec{u}}{\sqrt{1-u^2/c^2}}
\end{aligned}
\right.
\]
\[
p\cdot p \;=\; m^2c^2 \;=\; \frac{E^2}{c^2} - p^2
\]
\[
\Rightarrow \qquad \boxed{\; E^2 \;=\; p^2c^2 + m^2c^4 \;}
\]
\[
\text{For a photon} \quad m_0 = 0 \quad \Rightarrow \quad \left\{\, E = pc \,\right\}
\]
\[
p^\mu \;=\;
\begin{pmatrix} p^0\\ p^1\\ p^2\\ p^3 \end{pmatrix}
\;=\;
\begin{pmatrix} p^0\\ \vec{p} \end{pmatrix}
\;=\;
\begin{pmatrix} mc\big/\sqrt{1-u^2/c^2}\\ m\vec{u}\big/\sqrt{1-u^2/c^2} \end{pmatrix}
\]
\[
\text{with} \qquad p^\mu p_\mu \;=\; -m^2c^2
\]
\[
\text{Where} \qquad p_\mu \;=\; \left( -p_0,\ \vec{p} \right)
\]
(Q.1) Write four vector for photon.
Ans )
\[
p^\mu \;=\; (p^0,\ p^1,\ p^2,\ p^3)
\]
\[
p^\mu \;=\; m\,\eta^\mu \;=\; m\,\frac{dx^\mu}{d\tau}
\qquad
\text{where} \quad x^\mu =
\begin{pmatrix} x^0\\ x^1\\ x^2\\ x^3 \end{pmatrix}
= \begin{pmatrix} ct\\ x\\ y\\ z \end{pmatrix}
\]
\[
\begin{aligned}
&= \frac{m}{\sqrt{1-u^2/c^2}} \begin{pmatrix} c\\ \vec{u} \end{pmatrix}\\
&= \begin{pmatrix} \dfrac{mc}{\sqrt{1-u^2/c^2}}\\ \dfrac{m\vec{u}}{\sqrt{1-u^2/c^2}} \end{pmatrix}
\end{aligned}
\]
\[
\text{for photon} \qquad p^\mu = \left( \frac{E}{c},\ \vec{p} \right) \qquad \frac{E^2}{c^2} - p^2 = m_0^2c^2 = 0
\]
\[
\Rightarrow \quad \left( \frac{E}{c} = p \right)
\]
\[
\left\{ \text{So} \quad p^\mu = \left( \frac{E}{c},\ \vec{p} \right) \quad \text{such that} \quad E = pc \right\}
\]
(Q.1)
\begin{tikzpicture}
% photon
\draw (0,0.55) -- (0.18,0.75) -- (0.36,0.35) -- (0.54,0.75) -- (0.72,0.35) -- (0.90,0.55);
\draw[->] (0.90,0.55) -- (1.25,0.55);
\node[above] at (0.45,0.80) {\small $W$};
% target mass
\draw (2.1,0.55) circle (0.17);
\fill (2.1,0.55) circle (0.17);
\node[above] at (2.1,0.78) {\small $m$};
% arrow
\draw[->] (3.0,0.55) -- (5.1,0.55);
% product
\draw (6.0,0.55) circle (0.17);
\draw[->] (6.25,0.55) -- (6.95,0.55) node[right] {\small $v$};
\node[above] at (6.0,0.78) {\small $m'$};
\node[right] at (7.6,0.55) {(Find $m'$ \& $v$)};
% labels below
\draw[->] (0.45,0.05) -- (0.45,-0.45);
\node[below] at (0.45,-0.50) {\small $p_1^\mu$};
\draw[->] (2.10,0.05) -- (2.10,-0.45);
\node[below] at (2.10,-0.50) {\small $p_2^\mu$};
\draw[->] (6.00,0.05) -- (6.00,-0.45);
\node[below] at (6.00,-0.50) {\small $p_3^\mu$};
\end{tikzpicture}
\[
p_1^\mu + p_2^\mu \;=\; p_3^\mu
\]
\[
\left( \frac{E}{c},\ \vec{p} \right) + \left( mc,\ 0 \right) \;=\; \left( \frac{m'c}{\sqrt{1-v^2/c^2}},\ \frac{m'v}{\sqrt{1-v^2/c^2}} \right)
\]
\[
\frac{E}{c} + mc \;=\; \frac{m'c}{\sqrt{1-v^2/c^2}}
\]
\[
\& \qquad p \;=\; \frac{m'v}{\sqrt{1-v^2/c^2}}
\]
\[
\text{from frame of } (m') \qquad \bar{p}_3^\mu \;=\; \left( m'c,\ 0 \right)
\]
\[
\bar{p}_3^\mu\cdot\bar{p}_{3\mu} \;=\; m'^2c^2 \;=\; (p_1+p_2)\cdot(p_1+p_2)
\]
\[
\bar{p}_3^\mu \bar{p}_{3\mu} \;=\; m'^2c^2 \;=\; \left( \frac{E}{c}+mc,\ \vec{p} \right)\cdot\left( \frac{E}{c}+mc,\ \vec{p} \right)
\]
\[
m'^2c^2 \;=\; \left( \frac{E}{c}+mc \right)^2 - p^2
\]
\[
p = \frac{E}{c} \quad \Rightarrow \quad (m'c)^2 \;=\; \frac{E^2}{c^2} + m^2c^2 + 2mE - \frac{E^2}{c^2}
\]
\[
(m'c)^2 \;=\; m^2c^2 + 2mE
\]
\[
m' \;=\; \sqrt{ m^2 + \frac{2mE}{c^2} }
\]
\[
\text{Where} \quad E = h\nu \quad \text{or} \quad \hbar\omega
\]
\[
\boxed{\; m' \;=\; \sqrt{ m^2 + \frac{2m\hbar\omega}{c^2} } \;}
\]
(Q.2)
\begin{tikzpicture}
\node at (0,0.55) {$p$};
\draw[->] (0.25,0.55) -- (1.25,0.55);
\node[below] at (0.80,0.50) {\small $u$};
\draw (2.2,0.55) circle (0.22);
\node at (2.2,0.55) {\small $p$};
\node[below] at (2.2,0.28) {\small Rest};
\draw[->] (2.9,0.55) -- (5.3,0.55);
\node[anchor=west] at (5.6,0.62) {$p + p + p + \bar{p}$};
\node[anchor=west] at (7.3,0.10) {\small $\hookrightarrow$ antiproton};
\end{tikzpicture}
with how much energy first proton should be imparted so that we have \(3p + \bar{p}\) moving together with some final velocity.
\[
p + p \;\longrightarrow\; \{ p+p+p+\bar{p} \} \;\longrightarrow\; V
\]
\[
\left( \frac{mc}{\sqrt{1-u^2/c^2}},\ \vec{p} \right) + \left( mc,\ 0 \right) \;=\; \left( \frac{4mc}{\sqrt{1-V^2/c^2}},\ \frac{4m\vec{V}}{\sqrt{1-V^2/c^2}} \right)
\]
\[
\text{or} \quad \left( \frac{E}{c},\ \vec{p} \right) + \left( mc,\ 0 \right) \;=\; \left( \frac{4mc}{\sqrt{1-V^2/c^2}},\ \frac{4m\vec{V}}{\sqrt{1-V^2/c^2}} \right) \;=\; p_f^\mu
\]
\[
\bar{p}_f^\mu\, \bar{p}_{f\mu} \;=\; p_f^\mu p_{f\mu} \qquad (\text{Invarient})
\]
So let me find this result from frame of C.O.M of All four particles.
\[
-\,\bar{p}_f^\mu\, \bar{p}_{f\mu} \;=\; \left( 4mc,\ 0 \right) \begin{pmatrix} 4mc\\ 0 \end{pmatrix} \;=\; 16m^2c^2
\]
\[
p_f^\mu p_{f\mu} \;=\; \left( \frac{E}{c}+mc,\ p \right) \begin{pmatrix} \frac{E}{c}+mc\\ p \end{pmatrix} \;=\; \left( \frac{E}{c}+mc \right)^2 - p^2
\]
\[
\Rightarrow \qquad \frac{E^2}{c^2} + m^2c^2 + 2mE - p^2 \;=\; 16m^2c^2 \qquad (1)
\]
Also from first particles
\[
p_1^\mu p_{1\mu} \;=\; I \;=\; -m^2c^2
\]
\[
-\frac{E^2}{c^2} + p^2 \;=\; -m^2c^2
\]
\[
\left( \frac{E^2}{c^2} - p^2 \;=\; m^2c^2 \right)
\]
eq^n (1) becomes,
\[
2m^2c^2 + 2mE \;=\; 16m^2c^2
\]
\[
\boxed{\; E \;=\; \frac{14m^2c^2}{2m} \;=\; 7mc^2 \;}
\]
STR | PDF | << prev