∑ F ⃗ = m a ⃗ ; \sum{\vec{F}}=m\vec{a}; ∑ F = m a ;
vector statementsum of force statementdoes not make distinction between object at rest andobject moving at constant velocity . V = 0 ≡ V = const \begin{matrix}V=0&\equiv&V=\text{const}\end{matrix} V = 0 ≡ V = const
e.g.
S : x y -frame S: xy\text{-frame} S : x y -frame
Q1 : In the classroom, fixed to the surface of the earth, are we on an inertial frame?
A1 : No. Because the earth is spinning.
Q2 : Do we see something spinning, because the earth is spinning?
A2 : No. (except coriolis effect )
in general
∇ × B ⃗ − 1 c ∂ E ⃗ ∂ t = 4 π c j ⃗ ∇ ⋅ E ⃗ = 4 π ρ ∇ × E ⃗ + 1 c ∂ B ⃗ ∂ t = 0 ⃗ ∇ ⋅ B ⃗ = 0 \begin{align*} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align*} ∇ × B − c 1 ∂ t ∂ E ∇ × E + c 1 ∂ t ∂ B ∇ ⋅ B = c 4 π j ∇ ⋅ E = 0 = 0 = 4 π ρ
v ′ = v − V , { v 1 = v 1 ′ + V ; v 2 = v 2 ′ − V ; \begin{align*} v'&=v-\mathbf{V},&& \begin{cases} v_1=v_1'+\mathbf{V}; \\ v_2=v_2'-\mathbf{V}; \end{cases} \end{align*} v ′ = v − V , { v 1 = v 1 ′ + V ; v 2 = v 2 ′ − V ;
m 1 v 1 + m 2 v 2 = ( m 1 + m 2 ) v f ; ⟨ v = v ′ + V ⟩ m 1 ( v 1 ′ + V ) + m 2 ( v 2 ′ + V ) = ( m 1 + m 2 ) ( v f ′ + V ) ; m 1 v 1 ′ + m 2 v 2 ′ + ( m 1 + m 2 ) V = ( m 1 + m 2 ) ( v f ′ ) + ( m 1 + m 2 ) V ) ; m 1 v 1 ′ + m 2 v 2 ′ = ( m 1 + m 2 ) v f ′ \begin{align*} m_1v_1+m_2v_2&=(m_1+m_2)v_f;&&\langle v=v'+\mathbf{V} \rangle\\ m_1(v_1'+\mathbf{V})+m_2(v_2'+\mathbf{V})&=(m_1+m_2)(v_f'+\mathbf{V});\\ m_1v_1'+m_2v_2'+(m_1+m_2)\mathbf{V}&=(m_1+m_2)(v_f')+(m_1+m_2)\mathbf{V});\\ m_1v_1'+m_2v_2'&=(m_1+m_2)v_f' \end{align*} m 1 v 1 + m 2 v 2 m 1 ( v 1 ′ + V ) + m 2 ( v 2 ′ + V ) m 1 v 1 ′ + m 2 v 2 ′ + ( m 1 + m 2 ) V m 1 v 1 ′ + m 2 v 2 ′ = ( m 1 + m 2 ) v f ; = ( m 1 + m 2 ) ( v f ′ + V ) ; = ( m 1 + m 2 ) ( v f ′ ) + ( m 1 + m 2 ) V ) ; = ( m 1 + m 2 ) v f ′ ⟨ v = v ′ + V ⟩
∴ S ′ : m 1 v 1 ′ + m 2 v 2 ′ = ( m 1 + m 2 ) v f ′ conservation of momentum \underset{\text{conservation of momentum}}{\therefore{S'}: m_1v_1'+m_2v_2'=(m_1+m_2)v_f'} conservation of momentum ∴ S ′ : m 1 v 1 ′ + m 2 v 2 ′ = ( m 1 + m 2 ) v f ′
v ′ = v − V ; d x ′ d t ′ = d x d t − V ; ⟨ d t ′ = d t ⟩ d x ′ = d x − V d t ; x ′ = x − V d t ; \begin{align*} v'&=v-\mathbf{V};\\ \frac{dx'}{dt'}&=\frac{dx}{dt}-\mathbf{V};&&\langle dt'=dt \rangle\\ dx'&=dx-\mathbf{V}dt;\\ x'&=x-\mathbf{V}dt; \end{align*} v ′ d t ′ d x ′ d x ′ x ′ = v − V ; = d t d x − V ; = d x − V d t ; = x − V d t ; ⟨ d t ′ = d t ⟩
{ x ′ = x − V t ; y ′ = y ; z ′ = z ; t ′ = t ; \begin{cases} x'=x-\mathbf{V}t;\\ y'=y;\\ z'=z;\\ t'=t;\\ \end{cases} ⎩ ⎨ ⎧ x ′ = x − V t ; y ′ = y ; z ′ = z ; t ′ = t ;
∮ E ⃗ ⋅ d S ⃗ = Q enc ϵ 0 , ⟨ ϵ 0 = 1 0 − 9 36 π ≈ 8.85 ⟩ 0 − 12 permittivity of free space ⟩ \begin{align*} \oint{\vec{E}\cdot{d}\vec{S}}&=\frac{Q_{\text{enc}}}{\epsilon_0},&&\langle \underset{\text{permittivity of free space}}{\epsilon_0=\frac{10^{-9}}{36\pi}\approx8.85 \rangle0^{-12}}\rangle \end{align*} ∮ E ⋅ d S = ϵ 0 Q enc , ⟨ permittivity of free space ϵ 0 = 36 π 1 0 − 9 ≈ 8.85 ⟩ 0 − 12 ⟩
∮ E ⃗ ⋅ d S ⃗ = − d Φ B d t , ⟨ ϕ B = ∮ B ⃗ ⋅ d S ⃗ magnetic flux ⟩ \begin{align*} \oint{\vec{E}\cdot{d}\vec{S}}&=-\frac{d\Phi_{B}}{dt},&&&&\langle \underset{\text{magnetic flux}}{\phi_B=\oint{\vec{B}\cdot{d}\vec{S}}} \rangle \end{align*} ∮ E ⋅ d S = − d t d Φ B , ⟨ magnetic flux ϕ B = ∮ B ⋅ d S ⟩
∮ B ⃗ ⋅ d S ⃗ = μ 0 i through , ⟨ B : magnetic flux density μ permeability of free angle 0 = 4 π × 1 0 − 7 = 12.6 × 1 0 − 7 ⟩ \begin{align*} \oint{\vec{B}\cdot{d}\vec{S}} &= \mu_0i_{\text{through}}, &&\langle \begin{matrix}B:\text{magnetic flux density}\\ \underset{\text{permeability of free angle}} \mu_0=4\pi\times10^{-7}=12.6\times10^{-7} \end{matrix} \rangle \end{align*} ∮ B ⋅ d S = μ 0 i through , ⟨ B : magnetic flux density permeability of free angle μ 0 = 4 π × 1 0 − 7 = 12.6 × 1 0 − 7 ⟩
∮ B ⃗ ⋅ d S ⃗ = μ 0 [ i through + ϵ 0 d Φ E d t ] , where \begin{align*} \oint{\vec{B}\cdot{d}\vec{S}}&=\mu_0\left[i_{\text{through}}+\epsilon_0\frac{d\Phi_E}{dt}\right],&\text{where}\\ \end{align*} ∮ B ⋅ d S = μ 0 [ i through + ϵ 0 d t d Φ E ] , where
⟩ ϕ E = ∮ E ⃗ ⋅ d S ⃗ = ∮ E d A cos θ = E A = [ q ϵ 0 A ] A ⟩ \rangle \begin{align*} \phi_E&=\oint{\vec{E}\cdot{d}\vec{S}} =\oint{E\:dA\:\cos{\theta}}\\&=E\:A=\left[\frac{q}{\epsilon_0A}\right]A \end{align*} \rangle ⟩ ϕ E = ∮ E ⋅ d S = ∮ E d A cos θ = E A = [ ϵ 0 A q ] A ⟩
∮ E ⃗ ⋅ d S ⃗ = E ( x + Δ x , t ) h + 0 − E ( x , t ) h + 0 ; ⟨ E ( x + Δ x ) = E ( x ) + d E d x Δ x ⟩ = [ E ( x ) + d E d x Δ x ] h − E ( x , t ) h ; = d E d x Δ x h ; ‾ \begin{align*} \oint{\vec{E}\cdot{d}\vec{S}} &=E(x+\Delta{x},t)h+0-E(x,t)h+0;&&\langle E(x+\Delta{x})=E(x)+\frac{dE}{dx}\Delta{x} \rangle\\ &=\left[E(x)+\frac{dE}{dx}\Delta{x}\right]h-E(x,t)h;\\ &=\underline{\frac{dE}{dx}\Delta{x}\:h;} \end{align*} ∮ E ⋅ d S = E ( x + Δ x , t ) h + 0 − E ( x , t ) h + 0 ; = [ E ( x ) + d x d E Δ x ] h − E ( x , t ) h ; = d x d E Δ x h ; ⟨ E ( x + Δ x ) = E ( x ) + d x d E Δ x ⟩
∮ E ⃗ ⋅ d S ⃗ = − d Φ B d t = − d d t ( ∮ B ⃗ ⋅ d A ⃗ ) = − d d t ( B A cos θ ) = − d d t ( B Δ x h ) = − d B d t Δ x h ; ‾ \begin{align*} \oint{\vec{E}\cdot{d}\vec{S}} &=-\frac{d\Phi_B}{dt}=-\frac{d}{dt}\left(\oint{\vec{B}\cdot{d}\vec{A}}\right)=-\frac{d}{dt}\left(B\:A\:\cos{\theta}\right)\\ &=-\frac{d}{dt}\left(B\:\Delta{x}\:h\right)=\underline{-\frac{dB}{dt}\Delta{x}\:h;}\\\\ \end{align*} ∮ E ⋅ d S = − d t d Φ B = − d t d ( ∮ B ⋅ d A ) = − d t d ( B A cos θ ) = − d t d ( B Δ x h ) = − d t d B Δ x h ;
d E d x Δ x h = − d B d t Δ x h ∴ d E d x = − d B d t \begin{align*} \frac{dE}{dx}\Delta{x}\:h&=-\frac{dB}{dt}\Delta{x}\:h\\ \therefore\frac{dE}{dx}&=-\frac{dB}{dt} \end{align*} d x d E Δ x h ∴ d x d E = − d t d B Δ x h = − d t d B
∮ B ⃗ ⋅ d S ⃗ = − B ( x + Δ x ) h + 0 + B ( x , t ) h + 0 , ⟨ B ( x + Δ x ) = B ( x ) + d B d x Δ x ⟩ = − [ B ( x , t ) + d B d x Δ x ] h + B ( x , t ) h = − d B d x Δ x h ; ‾ \begin{align*} \oint{\vec{B}\cdot{d}\vec{S}}&=-B(x+\Delta{x})h+0+B(x,t)h+0,&&\langle B(x+\Delta{x})=B(x)+\frac{dB}{dx}\Delta{x} \rangle\\ &=-\left[B(x,t)+\frac{dB}{dx}\Delta{x}\right]h+B(x,t)h\\ &=\underline{-\frac{dB}{dx}\Delta{x}\:h;} \end{align*} ∮ B ⋅ d S = − B ( x + Δ x ) h + 0 + B ( x , t ) h + 0 , = − [ B ( x , t ) + d x d B Δ x ] h + B ( x , t ) h = − d x d B Δ x h ; ⟨ B ( x + Δ x ) = B ( x ) + d x d B Δ x ⟩
∮ B ⃗ ⋅ d S ⃗ = μ 0 [ i through + ϵ 0 d Φ E d t ] , ⟨ i through = 0 ⟩ = ϵ 0 μ 0 d Φ E d t = ϵ 0 μ 0 d d t ∮ E ⃗ ⋅ d S ⃗ = ϵ 0 μ 0 d E d t Δ x h ‾ ; \begin{align*} \oint{\vec{B}\cdot{d}\vec{S}}&=\mu_0\left[i_{\text{through}}+\epsilon_0\frac{d\Phi_E}{dt}\right],&&\langle i_{\text{through}}=0 \rangle\\ &=\epsilon_0\mu_0\frac{d\Phi_{E}}{dt}=\epsilon_0\mu_0\frac{d}{dt}\oint{\vec{E}\cdot{d}\vec{S}}\\&=\underline{\epsilon_0\mu_0\frac{dE}{dt}\Delta{x}\:h};\\\\ \end{align*} ∮ B ⋅ d S = μ 0 [ i through + ϵ 0 d t d Φ E ] , = ϵ 0 μ 0 d t d Φ E = ϵ 0 μ 0 d t d ∮ E ⋅ d S = ϵ 0 μ 0 d t d E Δ x h ; ⟨ i through = 0 ⟩
− d B d x Δ x h = ϵ 0 μ 0 d E d t Δ x h ∴ − d B d x = ϵ 0 μ 0 d E d t \begin{align*} -\frac{dB}{dx}\Delta{x}\:h&=\epsilon_0\mu_0\frac{dE}{dt}\Delta{x}\:h\\ \therefore-\frac{dB}{dx}&=\epsilon_0\mu_0\frac{dE}{dt} \end{align*} − d x d B Δ x h ∴ − d x d B = ϵ 0 μ 0 d t d E Δ x h = ϵ 0 μ 0 d t d E
Use these two equations to come up with something new
{ d E d x = − d B d t − d B d x = ϵ 0 μ 0 d E d t \left\{ \begin{align*} \frac{dE}{dx}&=-\frac{dB}{dt}\\\\ -\frac{dB}{dx}&=\epsilon_0\mu_0\frac{dE}{dt} \end{align*} \right. ⎩ ⎨ ⎧ d x d E − d x d B = − d t d B = ϵ 0 μ 0 d t d E
( d d x ) d E d x = − ( d d x ) d B d t d 2 E d x 2 = d d t ( − d B d x ) , ⟨ − d B d x = ϵ 0 μ 0 d E d t ⟩ = d d t ( ϵ 0 μ 0 d E d t ) = ϵ 0 μ 0 d 2 E d t 2 \begin{align*} \left(\frac{d}{dx}\right)\frac{dE}{dx}&=-\left(\frac{d}{dx}\right)\frac{dB}{dt}\\ \frac{d^2E}{dx^2}&=\frac{d}{dt}\left(-\frac{dB}{dx}\right),&&\langle -\frac{dB}{dx}=\epsilon_0\mu_0\frac{dE}{dt} \rangle\\ &=\frac{d}{dt}\left(\epsilon_0\mu_0\frac{dE}{dt}\right)\\ &=\epsilon_0\mu_0\frac{d^2E}{dt^2} \end{align*} ( d x d ) d x d E d x 2 d 2 E = − ( d x d ) d t d B = d t d ( − d x d B ) , = d t d ( ϵ 0 μ 0 d t d E ) = ϵ 0 μ 0 d t 2 d 2 E ⟨ − d x d B = ϵ 0 μ 0 d t d E ⟩
Reminder : wave equation (PHYS032)
d 2 y d x 2 = 1 v 2 d 2 y d t 2 \begin{align*} \frac{d^2y}{dx^2}&=\frac{1}{v^2}\frac{d^2y}{dt^2}\\ \end{align*} d x 2 d 2 y = v 2 1 d t 2 d 2 y
So we use this relation to make sense of our derived equation
{ d 2 y d x 2 = ( 1 v 2 ) d 2 y d t 2 d 2 E d x 2 = ( ϵ 0 μ 0 ) d 2 E d t 2 \left\{ \begin{align*} \frac{d^2y}{dx^2}&=\left(\frac{1}{v^2}\right)\frac{d^2y}{dt^2}\\\\ \frac{d^2E}{dx^2}&=\left(\epsilon_0\mu_0\right)\frac{d^2E}{dt^2} \end{align*} \right. ⎩ ⎨ ⎧ d x 2 d 2 y d x 2 d 2 E = ( v 2 1 ) d t 2 d 2 y = ( ϵ 0 μ 0 ) d t 2 d 2 E
1 v 2 = ϵ 0 μ 0 ; v = 1 ϵ 0 μ 0 = 1 ( 1 0 − 9 36 π ) ( 4 π × 1 0 7 ) = 3.0 × 1 0 8 \begin{align*} \frac{1}{v^2}&=\epsilon_0\mu_0;\\ v&=\frac{1}{\sqrt{\epsilon_0\mu_0}}=\frac{1}{\sqrt{\left(\frac{10^{-9}}{36\pi}\right)\left(4\pi\times10^{7}\right)}}\\ &=3.0\times10^{8} \end{align*} v 2 1 v = ϵ 0 μ 0 ; = ϵ 0 μ 0 1 = ( 36 π 1 0 − 9 ) ( 4 π × 1 0 7 ) 1 = 3.0 × 1 0 8
Light is no different from EM wave (i.e. microwave, radio wave, x-ray, etc.)
from GALILEAN TRANSFORMATION
x ′ = x − V t ; x'=x-\mathbf{V}t; x ′ = x − V t ;
Lorentz Transformation says that, in special relativity,
x ′ = γ ( x − V t ) x'=\gamma\left(x-\mathbf{V}t\right) x ′ = γ ( x − V t )
Einstein's Postulate says if we consider the distance to be
d = r t d=rt d = r t
{ S ′ : x ′ = c t ′ ; S : x = c t ; \left\{\begin{align*} S'&:x'=ct';\\ S&: x=ct; \end{align*}\right. { S ′ S : x ′ = c t ′ ; : x = c t ;
back to our Lorentz Transformation
x ′ = γ ( x − V t ) ( c t ′ ) = γ ( ( c t ) − V t ) c t ′ = γ ( c − V ) t ‾ Lorentz TF form x = γ ( x ′ + V t ′ ) ( c t ) = γ ( ( c t ′ ) + V t ′ ) c t = γ ( c + V ) t ′ ‾ Inverse Lorentz TF form \begin{matrix} \underset{\text{Lorentz TF form}}{\underline{\begin{align*} x'&=\gamma\left(x-\mathbf{V}t\right)\\ (ct')&=\gamma\left((ct)-\mathbf{V}t\right)\\ ct'&=\gamma\left(c-\mathbf{V}\right)t \end{align*}}}&& \underset{\text{Inverse Lorentz TF form}}{\underline{\begin{align*} x&=\gamma\left(x'+\mathbf{V}t'\right)\\ (ct)&=\gamma\left((ct')+\mathbf{V}t'\right)\\ ct&=\gamma\left(c+\mathbf{V}\right)t' \end{align*}}}\end{matrix} Lorentz TF form x ′ ( c t ′ ) c t ′ = γ ( x − V t ) = γ ( ( c t ) − V t ) = γ ( c − V ) t Inverse Lorentz TF form x ( c t ) c t = γ ( x ′ + V t ′ ) = γ ( ( c t ′ ) + V t ′ ) = γ ( c + V ) t ′
Use this relation to derive Gamma
1 = 1 ; ( c t ′ γ ( c − V ) t ) = ( γ ( c + V ) t ′ c t ) ; c γ ( c − V ) = γ ( c + V ) c c 2 = ( γ ) 2 ( c − V ) ( c + V ) = γ 2 ( c 2 − V 2 ) ; γ 2 = c 2 c 2 − V = 1 c 2 − V c 2 = 1 1 − V 2 c 2 ; γ = ( 1 1 − V 2 c 2 ) = 1 1 − ( V c ) 2 = 1 1 − ( β ) 2 , \begin{align*} 1&=1;\\ \left(\frac{ct'}{\gamma\left(c-\mathbf{V}\right)t}\right)&=\left(\frac{\gamma\left(c+\mathbf{V}\right)t'}{ct}\right);\\ \frac{c}{\gamma\left(c-\mathbf{V}\right)}&=\frac{\gamma\left(c+\mathbf{V}\right)}{c}\\ c^2&=(\gamma)^2(c-\mathbf{V})(c+\mathbf{V})\\ &=\gamma^2\left(c^2-\mathbf{V}^2\right);\\\\ \gamma^2&=\frac{c^2}{c^2-\mathbf{V}}=\frac{1}{\frac{c^2-\mathbf{V}}{c^2}}=\frac{1}{1-\frac{\mathbf{V}^2}{c^2}};\\ \gamma&=\sqrt{\left(\frac{1}{1-\frac{\mathbf{V}^2}{c^2}}\right)}=\frac{1}{\sqrt{1-\left(\frac{\mathbf{V}}{c}\right)^2}}\\ &=\frac{1}{\sqrt{1-\left(\beta\right)^2}}, \end{align*} 1 ( γ ( c − V ) t c t ′ ) γ ( c − V ) c c 2 γ 2 γ = 1 ; = ( c t γ ( c + V ) t ′ ) ; = c γ ( c + V ) = ( γ ) 2 ( c − V ) ( c + V ) = γ 2 ( c 2 − V 2 ) ; = c 2 − V c 2 = c 2 c 2 − V 1 = 1 − c 2 V 2 1 ; = ( 1 − c 2 V 2 1 ) = 1 − ( c V ) 2 1 = 1 − ( β ) 2 1 ,
where
β = V c { β → 0 V ≪ c β → 1 V ≈ c \begin{matrix} \begin{align*} \beta=\frac{\mathbf{V}}{c} \end{align*}& \begin{cases} \beta\to0&\mathbf{V}\ll{c}\\ \beta\to1&\mathbf{V}\approx{c} \end{cases} \end{matrix} β = c V { β → 0 β → 1 V ≪ c V ≈ c
An observer from frame S S S S ′ S' S ′ dilated when the frame S ′ S' S ′ close to speed of light . On that note, we need to know how to get the proper time, t ′ t' t ′ S S S
x = γ ( x ′ + V t ′ ) = γ ( ( x − V t ) + V t ′ ) = γ 2 x − γ 2 V t + γ V t ′ ; γ V t ′ = x − γ 2 x + γ 2 V t ; t ′ = x ( γ V ) − γ 2 x ( γ V ) + γ 2 V t ( γ V ) = x γ V − γ V x + γ t = x V ( 1 γ − γ ) + γ t = x V ( 1 − γ 2 γ ) + γ t = x V ( γ γ ) ( 1 − γ 2 γ ) + γ t = x γ V ( 1 − γ 2 γ 2 ) + γ t , \begin{align*} x&=\gamma(x'+\mathbf{V}t')\\ &=\gamma\left(\left(x-\mathbf{V}t\right)+\mathbf{V}t'\right)\\ &=\gamma^2x-\gamma^2\mathbf{V}t+\gamma\mathbf{V}t';\\ \gamma\mathbf{V}t'&=x-\gamma^2x+\gamma^2\mathbf{V}t;\\ t'&=\frac{x}{(\gamma\mathbf{V})}-\frac{\gamma^2x}{(\gamma\mathbf{V})}+\frac{\gamma^2\mathbf{V}t}{(\gamma\mathbf{V})}=\frac{x}{\gamma\mathbf{V}}-\frac{\gamma}{\mathbf{V}}x+\gamma{t}\\ &=\frac{x}{\mathbf{V}}\left(\frac{1}{\gamma}-\gamma\right)+\gamma{t}=\frac{x}{\mathbf{V}}\left(\frac{1-\gamma^2}{\gamma}\right)+\gamma{t}\\ &=\frac{x}{\mathbf{V}}\left(\frac{\gamma}{\gamma}\right)\left(\frac{1-\gamma^2}{\gamma}\right)+\gamma{t}\\ &=\frac{x\gamma}{\mathbf{V}}\left(\frac{1-\gamma^2}{\gamma^2}\right)+\gamma{t},\\ \end{align*} x γ V t ′ t ′ = γ ( x ′ + V t ′ ) = γ ( ( x − V t ) + V t ′ ) = γ 2 x − γ 2 V t + γ V t ′ ; = x − γ 2 x + γ 2 V t ; = ( γ V ) x − ( γ V ) γ 2 x + ( γ V ) γ 2 V t = γ V x − V γ x + γ t = V x ( γ 1 − γ ) + γ t = V x ( γ 1 − γ 2 ) + γ t = V x ( γ γ ) ( γ 1 − γ 2 ) + γ t = V x γ ( γ 2 1 − γ 2 ) + γ t ,
{ 1 − ( γ 2 ) γ 2 = 1 γ 2 − 1 = ( c 2 − V 2 c 2 ) − ( c 2 c 2 ) = − V 2 c 2 ; } \left\{ \begin{align*} \frac{1-\left(\gamma^2\right)}{\gamma^2}&=\frac{1}{\gamma^2}-1\\ &=\left(\frac{c^2-\mathbf{V}^2}{c^2}\right)-\left(\frac{c^2}{c^2}\right)\\ &=-\frac{\mathbf{V}^2}{c^2}; \end{align*}\right\} ⎩ ⎨ ⎧ γ 2 1 − ( γ 2 ) = γ 2 1 − 1 = ( c 2 c 2 − V 2 ) − ( c 2 c 2 ) = − c 2 V 2 ; ⎭ ⎬ ⎫
t ′ = x γ V ( − V 2 c 2 ) + γ t = γ [ t − V c 2 x ] ; \begin{align*} t'&=\frac{x\gamma}{\mathbf{V}}\left(-\frac{\mathbf{V}^2}{c^2}\right)+\gamma{t}\\ &=\gamma\left[t-\frac{\mathbf{V}}{c^2}x\right]; \end{align*} t ′ = V x γ ( − c 2 V 2 ) + γ t = γ [ t − c 2 V x ] ;
t ′ = γ [ t − V c 2 x ] ‾ Lorentz TF form t = γ [ t ′ + V c 2 x ′ ] ‾ Inverse Lorentz TF form \begin{matrix} \underset{\text{Lorentz TF form}}{\underline{\begin{align*} t'&=\gamma\left[t-\frac{\mathbf{V}}{c^2}x\right] \end{align*}}}&& \underset{\text{Inverse Lorentz TF form}}{\underline{\begin{align*} t&=\gamma\left[t'+\frac{\mathbf{V}}{c^2}x'\right] \end{align*}}}\end{matrix} Lorentz TF form t ′ = γ [ t − c 2 V x ] Inverse Lorentz TF form t = γ [ t ′ + c 2 V x ′ ]
