prove that E=mc^2


Question and Answers Forum

All Questions Topic List
Matrices and Determinants Questions
Previous in All Question Next in All Question
Previous in Matrices and Determinants Next in Matrices and Determinants
Question Number 98003 by M±th+et+s last updated on 11/Jun/20

$p r o v e t h a t$ $E = {m c}^{2}$
	Answered by Rio Michael last updated on 11/Jun/20

	$Let me begin with {Plank}^{'} s equation$ $E = h f$ $\Rightarrow E = h (\frac{c}{λ}) c = f λ$ $\Rightarrow E = \frac{h c}{λ}$ $also E = p c as we know momentum = \frac{h}{λ}$ $\Rightarrow E = m c (c), momentum = m v in this case v = c$ $\Rightarrow E = {m c}^{2}$ $let me use other methods to prove this$
	Answered by Rio Michael last updated on 11/Jun/20

	$P_{p h o t o n} = \frac{E}{c} . . . . . (i)$ $P = m v . . . . . (i i)$ $v = \frac{Δ x}{Δ t} . . . . . (i i i)$ $by conservation of momentum,$ $M \frac{Δ x}{Δ t} = \frac{E}{c}$ $Δ t = \frac{L}{c} . . . . . . (i v)$ $\Rightarrow M Δ x = \frac{E L}{c^{2}}$ $also \bar{x} = \frac{{M x}_{1} + {m x}_{2}}{M + m}$ $\Rightarrow \frac{{M x}_{1} + {m x}_{2}}{M + m} = \frac{M (x_{1} - Δ x) + m L}{M + m}$ $\Rightarrow M L = M Δ x$ $\Rightarrow M L = \frac{E L}{c^{2}}$ $E = {M c}^{2}$
	Answered by Rio Michael last updated on 11/Jun/20

	$K E = \int_{0}^{s} F d s$ $where F = \frac{d (m v)}{d t}$ $\Rightarrow K E = \int_{0}^{s} \frac{d (m v)}{d t} d s = \int_{0}^{m v} v d (m v) = \int_{0}^{v} v d [\frac{m_{0} v}{\sqrt{1 - v^{2} / c^{2}}}]$ $integrating by parts$ $K E = \frac{m_{0} v^{2}}{\sqrt{1 - v^{2} / c^{2}}} - m_{0} \int_{0}^{v} \frac{v d v}{\sqrt{1 - v^{2} / c^{2}}}$ $= \frac{m_{0} v^{2}}{\sqrt{1 - v^{2} / c^{2}}} + ∣ m_{0} v^{2}$ $= \frac{m_{0} c^{2}}{\sqrt{1 - v^{2} / c^{2}}} + m_{0} c^{2}$ $= {m c}^{2} - m_{0} c^{2}$ $\Rightarrow {m c}^{2} = m_{0} c^{2} + K E$ $\Rightarrow E = E_{0} + K E$ $where E_{0} = m_{0} c^{2}$ $also see that E = {m c}^{2} = \frac{m_{0} c^{2}}{\sqrt{1 - v^{2} / c^{2}}} .$
	Commented by M±th+et+s last updated on 11/Jun/20

	$w e l l d o n e t h a n k y o u$
	Commented by Rio Michael last updated on 11/Jun/20

	$welcome sir .$
	Answered by smridha last updated on 11/Jun/20

	$l e t m e s h o w h o w I u s u a l l y l i k e t o$ $d o t h i s . . . t h i s i s n o t t h e m e t h o d$ $o f t h e g o d o f p h y s i c s [EINSTEIN] .$ $w i t h d u e r e s p e c t t o h i m {l e t}^{'} s s t a r t .$ $★ {l e t}^{'} s d o s o m e t h o u g h t e x p t : ★$ $c o n s i d e r t w o i n e r t i a l f r a m e o f r e f e r e n c e$ $s a n d s^{'} . n o w a n a t o m i s p l a c e d$ $i n t o t h e s^{'} w h i c h i s m o v i n g v e r y$ $v e r y f a s t a t a s p e e d u w i t h r e s p e c t$ $t o s . n o w c o n s i d e r t w o o b s e r v e r s$ $A_{s} a n d A_{s^{'}} . n o w s u p p o s e t h e a t o m$ $e m i t s t w o p h o t o n o f f r e q u e n c y$ $ν, o n e i s g o i n g t o w a r d s b o t h t h e$ $o b s e r v e r s a n d a n o t h e r i s g o i n g$ $f a r t h e r a w a y f r o m t h e m .$ $n o w {l e t}^{'} s s e e h o w t h e e n e r g y a n d$ $m o m e n t u m c h a n g e d w i t h r e s p e c t$ $t o t w o o b s e r v e r s .$ $f o r t h e o b s e r v e r A_{s^{^{'}}} t h e c h a n g e o f$ $m o m e n t u m i s {(Δ p)}_{s^{^{'}}} = \frac{h ν}{c} - \frac{h ν}{c} = 0$ $b e c a u s e {i t}^{'} s r e m a i n s t a n d i n g$ $w i t h r e s t w i t h r e s p e c t t o A_{s^{^{'}}} (a c c$ $t o 1 s t p o s t u l a t e o f R e l a t i v i t y)$ $n o w c h a n g e o f o f e n e r g y$ ${(Δ E)}_{s^{^{'}}} = 2 h ν .$ $n o w {l e t}^{'} s s e e w h a t i s g o i n g o n w i t h$ $A_{s} . w e f a s t c o n s i d e r t h e a t o m e m i t s$ $t w o p h o t o n f r o m t h e m o v i n g f r a m$ $o f r e f e r e n c e s^{^{'}} (r e s p e c t t o s) . .$ $a n d t h e s^{^{'}} f r a m e m o v i n g v e r y v e r y$ $f a s t s o n o w R e l a t i v i s t i c d o p p l e r$ $e f f e c t c o m e s w i t h a m a g i c a l$ $s p e l l o n o b s e r v e r A_{s} .$ $t h e c h a n g e f r e q u e n c y o f p h o t o n$ $w h i c h c o m e s t o w a r d s A_{s}$ $ν_{1} = ν \sqrt{\frac{1 + \frac{u}{c}}{1 - \frac{u}{c}}} (b l u e s h i f t)$ $a n d w h i c h i s m o v i n g a w a y f r o m$ $A_{s}, ν_{2} = ν \sqrt{\frac{1 - \frac{u}{c}}{1 + \frac{u}{c}}} (r e d s h i f t)$ $s o t h e c h a n g e e n e r g y i n$ $f r a m e s w i t h r e s p e c t t o s^{^{'}} i s$ ${(Δ E)}_{s} = h ν_{1} + h ν_{2} = \frac{2 h ν}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} = γ {(Δ E)}_{s^{^{'}}}$ $w h e r e γ = {[1 - \frac{u^{2}}{c^{2}}]}^{- \frac{1}{2}}$ $n o w c h a n g e i n m o m e n t u m$ ${(Δ p)}_{s} = \frac{h ν_{1}}{c} - \frac{h ν_{2}}{c} = \frac{γ {(Δ E)}_{s^{^{'}}}}{c^{2}} u = \frac{{(Δ E)}_{s}}{c^{2}} u . . . . (a)$ $n o w c o n s i d e r t h e m a s s o f t h e$ $a t o m i s (Δ m) v e r y s m a l l . a n d$ $f o r o v s e r v e r A_{s} t h e m a s s i s m o v i n g$ $s o i t h a s m o m e n t u m s o n o w f r o m$ ${e q}^{n} (a) w e g e t$ ${(Δ m)}_{s} u = \frac{{(Δ E)}_{s}}{c^{2}} u [c s p e e d o f l i g h t]$ $n o w o u r w o r l d s f a m o u s {e q}^{n} c o m e$ $w h i c h r e a l l y c h a n g e d t h e w o r l d$ $i s . . .$ ${(Δ E)}_{s} = {(Δ m)}_{s} c^{2}$ $b u t a c c t u a l l y m o s t l o g i c a l$ ${e q}^{n} i s E = \sqrt{p^{2} c^{2} + m^{2} c^{4}}$ $i n$ $o r i g i n a l p a p e r o f E i n s t e i n$ $t h e e q u a n t i o n i s l i k e t h a t$ $m = \frac{L}{v^{2}}$ $^{″} d o e s t h e i n e r t i a o f b o d y d e p e n d$ $u p o n {i t}^{'} s e n e r g y {c o n t e n t}^{″}$ $t h e t i t e l o f t h i s p a p e r .$
	Commented by M±th+et+s last updated on 11/Jun/20

	$n i c e w o r k t h a n k y o u s i r$
	Commented by smridha last updated on 11/Jun/20

	$w e l c o m e .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum