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How do we get E=mc2?by fawk3s
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#1
Feb1210, 12:39 PM

P: 342

Hi
Somewhat usual question for you guys I guess. How do we get the formula E=mc2? I saw in a video where it said Einstein combined these equations Mv=E/c t=L/c x=vt Mx=mL and got EL/c2=mL => E/c2=m But I dont quite follow. What is Mv? What is Mx? And what time does L/c represent? How exactly are they combined? Always interested me. Thanks in advance, fawk3s 


#2
Feb1210, 02:06 PM

P: 1,060



#3
Feb1210, 02:12 PM

P: 342




#4
Feb1210, 02:26 PM

P: 1,345

How do we get E=mc2?
It looks like they assumed some of the results in their derivation which is circular logic.
Perhaps a better derivation is: http://en.wikipedia.org/wiki/Mass%E2...nce#Background 


#5
Feb1210, 02:33 PM

P: 1,060

In fact after assuming [tex]p=\frac{E}{c}[/tex] one could directly use [itex]p=mc[/itex] to get the result. However, I don't see a derivation in Wikipedia either?! 


#6
Feb1210, 03:20 PM

HW Helper
PF Gold
P: 1,915

I've written up a derivation that I think is pretty concise.
go to www.shadycrypt.com Click on the E=mc^{2} link at the top. 


#7
Feb1210, 03:30 PM

Emeritus
Sci Advisor
PF Gold
P: 5,577

Here is the argument that Einstein originally published: http://fourmilab.ch/etexts/einstein/E_mc2/www/
Here is a different argument: http://www.lightandmatter.com/html_b...tml#Section1.3 


#8
Feb1210, 07:13 PM

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#9
Feb1210, 07:30 PM

Emeritus
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PF Gold
P: 5,577

Considering that the argument given at fotonowy.pl is due to Einstein, I really don't think you're going to find obvious logical fallacies in it. 


#10
Feb1210, 07:43 PM

P: 1,060

I need to check the links you proposed.
Can you point me to a link where they show the prove how to derive p=mv from Maxwell? (I'm not surprised it works, since Maxwell is already relativistic?!) However: 


#11
Feb1210, 08:10 PM

Emeritus
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PF Gold
P: 5,577

[EDIT] Actually the link I gave above only proves that p is nonzero for an electromagnetic wave (which is inconsistent with the classical relation p=mv, since m=0 for light). By linearity and units we must have [itex]p=kE/c[/itex], where k is a unitless constant. For the proof that k=1, see this link: http://www.lightandmatter.com/html_b...ml#Section11.6 (subsection 11.6.2). 


#12
Feb1210, 08:35 PM

P: 1,060

Oh, but if you mean the link you posted later... I still have to go through it... 


#13
Feb1210, 08:38 PM

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#14
Mar110, 09:14 PM

P: 72

The normal particles have nonzero rest masses, so their energy is proportional to [tex]p^2[/tex] rather then to [tex]p[/tex]. 


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