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A fellow engineering student told me many years ago, that $E = mc^2$ means is that as an object of mass $m$ approaches $c$, the speed of light, it's mass increases and, at the speed of light, becomes infinite. If I solve this equation for $m$,

$m = E/c^2$, how in the world can its mass increase? - the denominator is extremely large(186,000 mi/sec)2, making m a very SMALL number, which makes sense, since the smaller the mass of an object becomes as c is approached, the more it would tend to change it's form into pure energy, E. That makes sense to me. But an object increasing in mass as it speeds up does not make sense to me...

Thanks for anyone's explanation(as simple as possible, please)

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  • $\begingroup$ "can an event exist before it occurs?": no, it is an oxymoron $\endgroup$ – user83548 Nov 9 '15 at 21:43
  • $\begingroup$ The equation doesn't mean that at all. The equation simply describes the rest mass energy of a mass. $E$ is the rest energy of a mass $m$. $\endgroup$ – Prahar Nov 9 '15 at 21:43
  • $\begingroup$ Let me add that it is true that what is known as the relativistic mass $m$ does in fact increase and go to infinity as the speed of the object goes to $c$, the speed of light. However, that is not apparent from the equation you have written here. That is apparent from the equation $m(v) = \frac{ m }{ \sqrt{1-v^2/c^2} }$ $\endgroup$ – Prahar Nov 9 '15 at 21:44
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    $\begingroup$ To correct Prahar's comment, mass does not increase with velocity (cf. this post among others), the momentum increases. However, the $v=c$ reference frame does not exist (because of a divide by zero error). $\endgroup$ – Kyle Kanos Nov 9 '15 at 21:58
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    $\begingroup$ This question (v2) seems spurred by a confusion between rest/invariant mass and relativistic mass. See e.g. this Phys.SE post, and a couple of paragraphs down on this Wikipedia page. $\endgroup$ – Qmechanic Nov 9 '15 at 21:59
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First of all, the speed of the object (relative to you, or relative to anything else) does not appear in the equation $E=mc^2$, so the equation cannot possibly say anything about what happens as the speed of the object increases.

Second, the equation in fact applies only to an object at rest. The correct equation for a moving object is $E=mc^2+(mv^2/2)+hot$, where $v$ is the particle's velocity and $hot$ are higher order terms that can be ignored except at extremely high velocities. Call it $E=mc^2+KE$. ($KE$, the kinetic energy, depends on the velocity of the particle and is therefore observer-dependent).

The significance of this equation is that $E$ is conserved in particle interactions (for any observer!). Therefore any loss in kinetic energy must be compensated for by an increase in rest mass, and vice versa.

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  • $\begingroup$ In $E=mc^2$, you have to consider the relativistic mass and also that E is part of the four-momentum vector. Using the rest mass (invariant mass), you will obtain the equation from John Rennie. $\endgroup$ – Timeless Nov 10 '15 at 18:05
  • $\begingroup$ @JohnRennie: I believe both that my equation is correct and that it follows from your (equally correct) equation. Since mine follows from yours, it's hard to imagine how you could endorse your own and still "disagree vehemently" with mine. $\endgroup$ – WillO Nov 11 '15 at 4:40
  • $\begingroup$ My apologies, I misread your answer and thought you were advocating the use of relativistic mass to calculate the kinetic energy. On more careful reading I see that's not the case. $\endgroup$ – John Rennie Nov 11 '15 at 7:04

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