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prove-that-E-mc-2-




Question Number 98003 by  M±th+et+s last updated on 11/Jun/20
prove that   E=mc^2
$${prove}\:{that}\: \\ $$$${E}={mc}^{\mathrm{2}} \:\: \\ $$
Answered by Rio Michael last updated on 11/Jun/20
Let me begin with Plank′s equation   E = hf   ⇒ E = h((c/λ))                c = fλ   ⇒ E = ((hc)/λ)  also E = pc as we know momentum = (h/λ)  ⇒ E = mc(c) , momentum = mv in this case v = c  ⇒ E = mc^2   let me use other methods to prove this
$$\mathrm{Let}\:\mathrm{me}\:\mathrm{begin}\:\mathrm{with}\:\mathrm{Plank}'\mathrm{s}\:\mathrm{equation} \\ $$$$\:{E}\:=\:{hf}\: \\ $$$$\Rightarrow\:{E}\:=\:{h}\left(\frac{{c}}{\lambda}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}\:=\:{f}\lambda \\ $$$$\:\Rightarrow\:{E}\:=\:\frac{{hc}}{\lambda} \\ $$$$\mathrm{also}\:{E}\:=\:{pc}\:\mathrm{as}\:\mathrm{we}\:\mathrm{know}\:\mathrm{momentum}\:=\:\frac{{h}}{\lambda} \\ $$$$\Rightarrow\:{E}\:=\:{mc}\left({c}\right)\:,\:\mathrm{momentum}\:=\:{mv}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{v}\:=\:{c} \\ $$$$\Rightarrow\:{E}\:=\:{mc}^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{me}\:\mathrm{use}\:\mathrm{other}\:\mathrm{methods}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{this} \\ $$
Answered by Rio Michael last updated on 11/Jun/20
 P_(photon) = (E/c) .....(i)  P = mv .....(ii)   v = ((Δx)/(Δt)) .....(iii)  by conservation of momentum,   M((Δx)/(Δt)) = (E/c)   Δt = (L/c) ......(iv)  ⇒ MΔx = ((EL)/c^2 )  also x^�  = ((Mx_1  + mx_2 )/(M + m))  ⇒ ((Mx_1  + mx_2 )/(M + m)) = ((M(x_1  −Δx) + mL)/(M+ m))  ⇒ ML = MΔx  ⇒ ML = ((EL)/c^2 )  E = Mc^2
$$\:{P}_{{photon}} =\:\frac{{E}}{{c}}\:…..\left({i}\right) \\ $$$${P}\:=\:{mv}\:…..\left({ii}\right) \\ $$$$\:{v}\:=\:\frac{\Delta{x}}{\Delta{t}}\:…..\left({iii}\right) \\ $$$$\mathrm{by}\:\mathrm{conservation}\:\mathrm{of}\:\mathrm{momentum}, \\ $$$$\:{M}\frac{\Delta{x}}{\Delta{t}}\:=\:\frac{{E}}{{c}} \\ $$$$\:\Delta{t}\:=\:\frac{{L}}{{c}}\:……\left({iv}\right) \\ $$$$\Rightarrow\:{M}\Delta{x}\:=\:\frac{{EL}}{{c}^{\mathrm{2}} } \\ $$$$\mathrm{also}\:\bar {{x}}\:=\:\frac{{Mx}_{\mathrm{1}} \:+\:{mx}_{\mathrm{2}} }{{M}\:+\:{m}} \\ $$$$\Rightarrow\:\frac{{Mx}_{\mathrm{1}} \:+\:{mx}_{\mathrm{2}} }{{M}\:+\:{m}}\:=\:\frac{{M}\left({x}_{\mathrm{1}} \:−\Delta{x}\right)\:+\:{mL}}{{M}+\:{m}} \\ $$$$\Rightarrow\:{ML}\:=\:{M}\Delta{x} \\ $$$$\Rightarrow\:{ML}\:=\:\frac{{EL}}{{c}^{\mathrm{2}} } \\ $$$${E}\:=\:{Mc}^{\mathrm{2}} \\ $$
Answered by Rio Michael last updated on 11/Jun/20
 KE = ∫_0 ^s F ds   where F = ((d(mv))/dt)  ⇒ KE = ∫_0 ^s ((d(mv))/dt) ds = ∫_0 ^(mv) v d(mv) = ∫_0 ^v v d[((m_0 v)/( (√(1 −v^2 /c^2 ))))]  integrating by parts  KE = ((m_0 v^2 )/( (√(1−v^2 /c^2 )))) −m_0  ∫_0 ^v ((v dv)/( (√(1−v^2 /c^2 ))))          = ((m_0 v^2 )/( (√(1−v^2 /c^2 )))) + ∣m_0 v^2           = ((m_0 c^2 )/( (√(1−v^2 /c^2 )))) + m_0 c^2          = mc^2  − m_0 c^2   ⇒ mc^2  = m_0 c^2  +KE  ⇒ E = E_0  + KE  where E_0  = m_0 c^2    also see that E = mc^2  = ((m_0 c^2 )/( (√(1−v^2 /c^2 )))) .
$$\:{KE}\:=\:\int_{\mathrm{0}} ^{{s}} {F}\:{ds}\: \\ $$$$\mathrm{where}\:{F}\:=\:\frac{{d}\left({mv}\right)}{{dt}} \\ $$$$\Rightarrow\:{KE}\:=\:\int_{\mathrm{0}} ^{{s}} \frac{{d}\left({mv}\right)}{{dt}}\:{ds}\:=\:\int_{\mathrm{0}} ^{{mv}} {v}\:{d}\left({mv}\right)\:=\:\int_{\mathrm{0}} ^{{v}} {v}\:{d}\left[\frac{{m}_{\mathrm{0}} {v}}{\:\sqrt{\mathrm{1}\:−{v}^{\mathrm{2}} /{c}^{\mathrm{2}} }}\right] \\ $$$$\mathrm{integrating}\:\mathrm{by}\:\mathrm{parts} \\ $$$${KE}\:=\:\frac{{m}_{\mathrm{0}} {v}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{v}^{\mathrm{2}} /{c}^{\mathrm{2}} }}\:−{m}_{\mathrm{0}} \:\int_{\mathrm{0}} ^{{v}} \frac{{v}\:{dv}}{\:\sqrt{\mathrm{1}−{v}^{\mathrm{2}} /{c}^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{{m}_{\mathrm{0}} {v}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{v}^{\mathrm{2}} /{c}^{\mathrm{2}} }}\:+\:\mid{m}_{\mathrm{0}} {v}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{{m}_{\mathrm{0}} {c}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{v}^{\mathrm{2}} /{c}^{\mathrm{2}} }}\:+\:{m}_{\mathrm{0}} {c}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\:{mc}^{\mathrm{2}} \:−\:{m}_{\mathrm{0}} {c}^{\mathrm{2}} \\ $$$$\Rightarrow\:{mc}^{\mathrm{2}} \:=\:{m}_{\mathrm{0}} {c}^{\mathrm{2}} \:+{KE} \\ $$$$\Rightarrow\:{E}\:=\:{E}_{\mathrm{0}} \:+\:{KE} \\ $$$$\mathrm{where}\:{E}_{\mathrm{0}} \:=\:{m}_{\mathrm{0}} {c}^{\mathrm{2}} \: \\ $$$$\mathrm{also}\:\mathrm{see}\:\mathrm{that}\:{E}\:=\:{mc}^{\mathrm{2}} \:=\:\frac{{m}_{\mathrm{0}} {c}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{v}^{\mathrm{2}} /{c}^{\mathrm{2}} }}\:.\: \\ $$
Commented by  M±th+et+s last updated on 11/Jun/20
well done thank you
$${well}\:{done}\:{thank}\:{you} \\ $$
Commented by Rio Michael last updated on 11/Jun/20
welcome sir.
$$\mathrm{welcome}\:\mathrm{sir}. \\ $$
Answered by smridha last updated on 11/Jun/20
let me show how I usually like to  do this...this is not the method   of the god of physics[EINSTEIN].  with due respect to him let′s start.    ★let′s do some thought expt:★  consider two inertial frame of reference  s and s′.now an atom is placed   into the s′which is moving very   very fast at a speed u with respect  to s.now consider two observers  A_s and A_(s′) .now suppose the atom  emits two photon of frequency  𝛎,one is going towards both the  observers and another is going  farther away from them.  now let′s see how the energy and  momentum changed with respect  to two observers.  for the observer A_(s^′ ) the change of  momentum is (𝚫p)_s^′  =((h𝛎)/c)−((h𝛎)/c)=0  because it′s remain standing   with rest with respect to A_s^′  (acc  to 1st postulate of Relativity)  now change of of energy  (𝚫E)_s^′  =2h𝛎.  now let′s see what is going on with  A_s .we fast consider the atom emits  two photon from the moving fram  of reference s^′ (respect to s)..  and the s^′ frame moving very very  fast so now Relativistic doppler  effect comes with a magical  spell on observer A_s .  the change frequency of photon  which comes towards A_s   𝛎_1 =𝛎(√((1+(u/c))/(1−(u/c)))) (blueshift)  and which is moving away from   A_(s ) ,ν_2 =𝛎(√((1−(u/c))/(1+(u/c)))) (red shift)  so the change energy in  frame s with respect to s^′  is      (𝚫E)_s =h𝛎_1 +h𝛎_2 =((2h𝛎)/( (√(1−(u^2 /c^2 )))))=𝛄(𝚫E)_s^′    where 𝛄=[1−(u^2 /c^2 )]^(−(1/2))   now change in momentum  (𝚫p)_s =((h𝛎_1 )/c)−((h𝛎_2 )/c)=((𝛄(𝚫E)_s^′  )/c^2 )u=(((𝚫E)_s )/c^2 )u....(a)  now consider the mass of the   atom is (𝚫m) very small.and  for ovserver A_s  the mass is moving  so it has momentum so now from  eq^n ( a)we get  (𝚫m)_s u=(((𝚫E)_s )/c^2 )u[c speed of light]  now our worlds famous eq^n come  which really changed the world  is...                   (𝚫E)_s =(𝚫m)_s c^2   but acctually most logical  eq^n  is E=(√(p^2 c^2 +m^2 c^4 ))  in  original paper of  Einstein  the equantion is like that           m=(L/v^2 )   ′′ does the inertia of body depend  upon it′s energy content′′  the titel of this paper.
$$\boldsymbol{{let}}\:\boldsymbol{{me}}\:\boldsymbol{{show}}\:\boldsymbol{{how}}\:\boldsymbol{{I}}\:\boldsymbol{{usually}}\:\boldsymbol{{like}}\:\boldsymbol{{to}} \\ $$$$\boldsymbol{{do}}\:\boldsymbol{{this}}…\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{not}}\:\boldsymbol{{the}}\:\boldsymbol{{method}}\: \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{god}}\:\boldsymbol{{of}}\:\boldsymbol{{physics}}\left[\mathbb{EINSTEIN}\right]. \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{due}}\:\boldsymbol{{respect}}\:\boldsymbol{{to}}\:\boldsymbol{{him}}\:\boldsymbol{{let}}'\boldsymbol{{s}}\:\boldsymbol{{start}}. \\ $$$$\:\:\bigstar\boldsymbol{{let}}'{s}\:\boldsymbol{{do}}\:\boldsymbol{{some}}\:\boldsymbol{{thought}}\:\boldsymbol{{expt}}:\bigstar \\ $$$$\boldsymbol{{consider}}\:\boldsymbol{{two}}\:\boldsymbol{{inertial}}\:\boldsymbol{{frame}}\:\boldsymbol{{of}}\:\boldsymbol{{reference}} \\ $$$$\boldsymbol{{s}}\:\boldsymbol{{and}}\:\boldsymbol{{s}}'.\boldsymbol{{now}}\:\boldsymbol{{an}}\:\boldsymbol{{atom}}\:\boldsymbol{{is}}\:\boldsymbol{{placed}}\: \\ $$$$\boldsymbol{{into}}\:\boldsymbol{{the}}\:\boldsymbol{{s}}'\boldsymbol{{which}}\:\boldsymbol{{is}}\:\boldsymbol{{moving}}\:\boldsymbol{{very}}\: \\ $$$$\boldsymbol{{very}}\:\boldsymbol{{fast}}\:\boldsymbol{{at}}\:\boldsymbol{{a}}\:\boldsymbol{{speed}}\:\boldsymbol{{u}}\:\boldsymbol{{with}}\:\boldsymbol{{respect}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{s}}.\boldsymbol{{now}}\:\boldsymbol{{consider}}\:\boldsymbol{{two}}\:\boldsymbol{{observers}} \\ $$$$\boldsymbol{{A}}_{\boldsymbol{{s}}} \boldsymbol{{and}}\:\boldsymbol{{A}}_{\boldsymbol{{s}}'} .\boldsymbol{{now}}\:\boldsymbol{{suppose}}\:\boldsymbol{{the}}\:\boldsymbol{{atom}} \\ $$$$\boldsymbol{{emits}}\:\boldsymbol{{two}}\:\boldsymbol{{photon}}\:\boldsymbol{{of}}\:\boldsymbol{{frequency}} \\ $$$$\boldsymbol{\nu},\boldsymbol{{one}}\:\boldsymbol{{is}}\:\boldsymbol{{going}}\:\boldsymbol{{towards}}\:\boldsymbol{{both}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{observers}}\:\boldsymbol{{and}}\:\boldsymbol{{another}}\:\boldsymbol{{is}}\:\boldsymbol{{going}} \\ $$$$\boldsymbol{{farther}}\:\boldsymbol{{away}}\:\boldsymbol{{from}}\:\boldsymbol{{them}}. \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{let}}'\boldsymbol{{s}}\:\boldsymbol{{see}}\:\boldsymbol{{how}}\:\boldsymbol{{the}}\:\boldsymbol{{energy}}\:\boldsymbol{{and}} \\ $$$$\boldsymbol{{momentum}}\:\boldsymbol{{changed}}\:\boldsymbol{{with}}\:\boldsymbol{{respect}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{two}}\:\boldsymbol{{observers}}. \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{the}}\:\boldsymbol{{observer}}\:\boldsymbol{{A}}_{{s}\:^{'} } \boldsymbol{{the}}\:\boldsymbol{{change}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{momentum}}\:\boldsymbol{{is}}\:\left(\boldsymbol{\Delta}{p}\right)_{\boldsymbol{{s}}^{'} } =\frac{\boldsymbol{{h}\nu}}{\boldsymbol{{c}}}−\frac{\boldsymbol{{h}\nu}}{\boldsymbol{{c}}}=\mathrm{0} \\ $$$$\boldsymbol{{because}}\:\boldsymbol{{it}}'{s}\:\boldsymbol{{remain}}\:\boldsymbol{{standing}}\: \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{rest}}\:\boldsymbol{{with}}\:\boldsymbol{{respect}}\:\boldsymbol{{to}}\:\boldsymbol{{A}}_{\boldsymbol{{s}}^{'} } \left(\boldsymbol{{acc}}\right. \\ $$$$\left.\boldsymbol{{to}}\:\mathrm{1}\boldsymbol{{st}}\:\boldsymbol{{postulate}}\:\boldsymbol{{of}}\:\boldsymbol{{R}}{e}\boldsymbol{{lativity}}\right) \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{change}}\:\boldsymbol{{of}}\:\boldsymbol{{of}}\:\boldsymbol{{energy}} \\ $$$$\left(\boldsymbol{\Delta{E}}\right)_{{s}^{'} } =\mathrm{2}\boldsymbol{{h}\nu}. \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{let}}'{s}\:\boldsymbol{{see}}\:\boldsymbol{{what}}\:\boldsymbol{{is}}\:\boldsymbol{{going}}\:\boldsymbol{{on}}\:\boldsymbol{{with}} \\ $$$$\boldsymbol{{A}}_{{s}} .\boldsymbol{{we}}\:\boldsymbol{{fast}}\:\boldsymbol{{consider}}\:\boldsymbol{{the}}\:\boldsymbol{{atom}}\:\boldsymbol{{emits}} \\ $$$$\boldsymbol{{two}}\:\boldsymbol{{photon}}\:\boldsymbol{{from}}\:\boldsymbol{{the}}\:\boldsymbol{{moving}}\:\boldsymbol{{fram}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{reference}}\:\boldsymbol{{s}}^{'} \left(\boldsymbol{{respect}}\:\boldsymbol{{to}}\:\boldsymbol{{s}}\right).. \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{the}}\:\boldsymbol{{s}}^{'} {frame}\:\boldsymbol{{moving}}\:\boldsymbol{{very}}\:\boldsymbol{{very}} \\ $$$$\boldsymbol{{fast}}\:\boldsymbol{{so}}\:\boldsymbol{{now}}\:\boldsymbol{{R}}{e}\boldsymbol{{lativistic}}\:\boldsymbol{{doppler}} \\ $$$$\boldsymbol{{effect}}\:\boldsymbol{{comes}}\:\boldsymbol{{with}}\:\boldsymbol{{a}}\:\boldsymbol{{magical}} \\ $$$$\boldsymbol{{spell}}\:\boldsymbol{{on}}\:\boldsymbol{{observer}}\:\boldsymbol{{A}}_{{s}} . \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{change}}\:\boldsymbol{{frequency}}\:\boldsymbol{{of}}\:\boldsymbol{{photon}} \\ $$$$\boldsymbol{{which}}\:\boldsymbol{{comes}}\:\boldsymbol{{towards}}\:\boldsymbol{{A}}_{{s}} \\ $$$$\boldsymbol{\nu}_{\mathrm{1}} =\boldsymbol{\nu}\sqrt{\frac{\mathrm{1}+\frac{\boldsymbol{{u}}}{\boldsymbol{{c}}}}{\mathrm{1}−\frac{\boldsymbol{{u}}}{\boldsymbol{{c}}}}}\:\left(\boldsymbol{{blueshift}}\right) \\ $$$${and}\:{whi}\boldsymbol{{ch}}\:\boldsymbol{{is}}\:\boldsymbol{{moving}}\:\boldsymbol{{away}}\:\boldsymbol{{from}}\: \\ $$$$\boldsymbol{{A}}_{{s}\:} ,\nu_{\mathrm{2}} =\boldsymbol{\nu}\sqrt{\frac{\mathrm{1}−\frac{\boldsymbol{{u}}}{\boldsymbol{{c}}}}{\mathrm{1}+\frac{\boldsymbol{{u}}}{\boldsymbol{{c}}}}}\:\left(\boldsymbol{{red}}\:\boldsymbol{{shift}}\right) \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{the}}\:\boldsymbol{{change}}\:\boldsymbol{{energy}}\:\boldsymbol{{in}} \\ $$$$\boldsymbol{{frame}}\:\boldsymbol{{s}}\:\boldsymbol{{with}}\:\boldsymbol{{respect}}\:\boldsymbol{{to}}\:\boldsymbol{{s}}^{'} \:{is} \\ $$$$\:\:\:\:\left(\boldsymbol{\Delta{E}}\right)_{\boldsymbol{{s}}} =\boldsymbol{{h}\nu}_{\mathrm{1}} +\boldsymbol{{h}\nu}_{\mathrm{2}} =\frac{\mathrm{2}\boldsymbol{{h}\nu}}{\:\sqrt{\mathrm{1}−\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\boldsymbol{{c}}^{\mathrm{2}} }}}=\boldsymbol{\gamma}\left(\boldsymbol{\Delta{E}}\right)_{\boldsymbol{{s}}^{'} } \\ $$$$\boldsymbol{{where}}\:\boldsymbol{\gamma}=\left[\mathrm{1}−\frac{\boldsymbol{{u}}^{\mathrm{2}} }{\boldsymbol{{c}}^{\mathrm{2}} }\right]^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{change}}\:\boldsymbol{{in}}\:\boldsymbol{{momentum}} \\ $$$$\left(\boldsymbol{\Delta{p}}\right)_{\boldsymbol{{s}}} =\frac{\boldsymbol{{h}\nu}_{\mathrm{1}} }{\boldsymbol{{c}}}−\frac{\boldsymbol{{h}\nu}_{\mathrm{2}} }{{c}}=\frac{\boldsymbol{\gamma}\left(\boldsymbol{\Delta{E}}\right)_{{s}^{'} } }{\boldsymbol{{c}}^{\mathrm{2}} }\boldsymbol{{u}}=\frac{\left(\boldsymbol{\Delta{E}}\right)_{{s}} }{\boldsymbol{{c}}^{\mathrm{2}} }\boldsymbol{{u}}….\left(\boldsymbol{{a}}\right) \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{consider}}\:\boldsymbol{{the}}\:\boldsymbol{{mass}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\: \\ $$$$\boldsymbol{{atom}}\:\boldsymbol{{is}}\:\left(\boldsymbol{\Delta{m}}\right)\:\boldsymbol{{very}}\:\boldsymbol{{small}}.\boldsymbol{{and}} \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{ovserver}}\:\boldsymbol{{A}}_{\boldsymbol{{s}}} \:\boldsymbol{{the}}\:\boldsymbol{{mass}}\:\boldsymbol{{is}}\:\boldsymbol{{moving}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{it}}\:\boldsymbol{{has}}\:\boldsymbol{{momentum}}\:\boldsymbol{{so}}\:\boldsymbol{{now}}\:\boldsymbol{{from}} \\ $$$$\boldsymbol{{eq}}^{\boldsymbol{{n}}} \left(\:\boldsymbol{{a}}\right)\boldsymbol{{we}}\:\boldsymbol{{get}} \\ $$$$\left(\boldsymbol{\Delta{m}}\right)_{\boldsymbol{{s}}} \boldsymbol{{u}}=\frac{\left(\boldsymbol{\Delta{E}}\right)_{{s}} }{\boldsymbol{{c}}^{\mathrm{2}} }\boldsymbol{{u}}\left[\boldsymbol{{c}}\:\boldsymbol{{speed}}\:\boldsymbol{{of}}\:\boldsymbol{{light}}\right] \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{our}}\:\boldsymbol{{worlds}}\:\boldsymbol{{famous}}\:\boldsymbol{{eq}}^{\boldsymbol{{n}}} \boldsymbol{{come}} \\ $$$$\boldsymbol{{which}}\:\boldsymbol{{really}}\:\boldsymbol{{changed}}\:\boldsymbol{{the}}\:\boldsymbol{{world}} \\ $$$$\boldsymbol{{is}}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\Delta{E}}\right)_{{s}} =\left(\boldsymbol{\Delta{m}}\right)_{\boldsymbol{{s}}} \boldsymbol{{c}}^{\mathrm{2}} \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{acctually}}\:\boldsymbol{{most}}\:\boldsymbol{{logical}} \\ $$$$\boldsymbol{{eq}}^{\boldsymbol{{n}}} \:\boldsymbol{{is}}\:\boldsymbol{{E}}=\sqrt{\boldsymbol{{p}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{2}} +\boldsymbol{{m}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{4}} } \\ $$$$\boldsymbol{{in}} \\ $$$$\boldsymbol{{original}}\:\boldsymbol{{paper}}\:\boldsymbol{{of}}\:\:\boldsymbol{{E}}{i}\boldsymbol{{nstein}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{equantion}}\:\boldsymbol{{is}}\:\boldsymbol{{like}}\:\boldsymbol{{that}} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{m}}=\frac{\boldsymbol{{L}}}{\boldsymbol{{v}}^{\mathrm{2}} }\: \\ $$$$''\:\boldsymbol{{does}}\:\boldsymbol{{the}}\:\boldsymbol{{inertia}}\:\boldsymbol{{of}}\:\boldsymbol{{body}}\:\boldsymbol{{depend}} \\ $$$$\boldsymbol{{upon}}\:\boldsymbol{{it}}'\boldsymbol{{s}}\:\boldsymbol{{energy}}\:\boldsymbol{{content}}'' \\ $$$${t}\boldsymbol{{he}}\:\boldsymbol{{titel}}\:\boldsymbol{{of}}\:\boldsymbol{{this}}\:\boldsymbol{{paper}}. \\ $$$$ \\ $$
Commented by  M±th+et+s last updated on 11/Jun/20
nice work thank you sir
$${nice}\:{work}\:{thank}\:{you}\:{sir} \\ $$
Commented by smridha last updated on 11/Jun/20
welcome.
$$\boldsymbol{{welcome}}. \\ $$

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