Pls help me dissect e=mc^2

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Now i know it will be impossible to explain to me in depth on a forum the full explaination of the equation. But maybe you guys could give me some pointers as to why in so many of einsteins equations the speed of light is squared. Is it because of the fact that if u had a flashlight traveling at the speed of light, and then u turn the flashlight on, the light from the flashlight would fly from the flashlight at the speed of light? Thx :)
 

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  • #2
malawi_glenn
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The c^2 is there just for unit conversions, from mass to energy. One can work in units where c = 1 (unitless) which makes the formula become E = m.

Many of einsteins equations are not derived, but postulated, https://www.physicsforums.com/showthread.php?t=76010
 
  • #3
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The c^2 is there just for unit conversions, from mass to energy. One can work in units where c = 1 (unitless) which makes the formula become E = m.

Many of einsteins equations are not derived, but postulated, https://www.physicsforums.com/showthread.php?t=76010
thx for the reply!

could pls expand on what you said about unit conversions? maybe give me an example. :)
 
  • #4
Fredrik
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Two posts that I made in the Science Advisor subforum are relevant here. I'll just repost the text here. See also #15 in this thread.

Post 1

Regarding the "why c?" question, here's an outline of an answer: We define (four-)momentum as mass times the (four-)velocity, so the square of momentum is (in units such that c=1): [itex]p^2=m^2u^2=-m^2[/itex], but also [itex]p^2=-(p^0)^2+\vec p^2[/itex], so [itex]E^2=\vec p^2+m^2[/itex]. When we restore factors of c, we get [itex]E^2=\vec p^2c^2+m^2c^4[/itex]. There are two ways to interpret E=mc2. One is to set [itex]\vec p=0[/itex] in the previous equation, and take the square root. The other is to write the mass as m0 instead, and define the "relativistic mass" by E=mc2. I'm one of those who feel that the latter is a really pointless thing to do.

So where did the c "come from"? If we include explicit factors of c in every step, one must be included in the definition of the four-velocity. See this post for more about four-velocity. The place where c enters is where we note that the slope of the world line is [itex]1/v=u^0/(c|\vec u|)[/itex].

Post 2

Yes, what I said above doesn't really explain why the 0th component of the four-momentum must be interpreted as energy, or why four-momentum is defined as mass times four-velocity. The post I linked to has a partial answer to the latter. (But I think a full answer should explain why this definition makes four-momentum a conserved quantity in particle interactions).

How are you going to derive E=mc2? The only other way I know to obtain that result as a consequence of something else is to calculate the work performed when accelerating a massive particle from 0 to v. (The result is [itex]\gamma m-m[/itex]). But to obtain that result, we "just assumed" that the relativistic expression for work is obtained from the non-relativistic

[tex]W=\int F\ dx=\int m\ddot x\dot x\ dt=\int m\frac{d\dot x}{dt}\dot x\ dt[/tex]

by substituting the first of the two velocities in the last step (but not the second) for the spatial components of the four-velocity. I know that we can make this look like a "natural" thing to do by expressing the non-relativistic work as

[tex]W=\int\dot\vec p\cdot d\vec x[/tex]

but that doesn't change the fact that we have made an assumption about the relationship between energy and the spatial components of four-momentum, and got E=mc2 from that. This justification of E=mc2 isn't better just because the assumption is a few algebraic steps away from the conclusion.

I think that to really answer the question, we'd have to show that the components of four-momentum behave like energy and momentum in non-relativistic mechanics. In particular, we should show that they are all conserved quantities in particle interactions.
 
  • #5
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You can see a high school level derviation of the Lorentz transformation from Einstein himself in RELATIVITY, Chapter XI and Appendix 1....at http://www.bartleby.com/173/....

and note the role c2 plays.

Note the entire book is available free

Einstein explains why the "older mechanics" based on moving coordinate frames of the form x' = x - vt (Galilei transformation) doesn't work...but Lorentz transformations (involving c2) does work.

You can also try Wikipedia at
http://en.wikipedia.org/wiki/E=mc_squared
which has lots of practical discussion.
 
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  • #6
chroot
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Break the units down into fundamental SI units.

Energy is measured in joules, and 1 joule is 1 kg * m^2/s^2.
Mass is measured in kg.

Let's say you wanted to relate mass and energy, and tried to do so with an equation of the form E = m k^n, where k is some speed and n is some integer. The only such integer that makes the units equivalent on both sides is n = 2.

What's really amazing is not the fact that k (a speed) must be squared to make the units work -- what's amazing is that k turns out to be the speed of light, and we can write the equation E = m c^2. The speed of light plays an interesting role in this universe: it's not only universal "speed limit," it's also the only speed that everyone in the universe will agree upon, no matter how they're moving with respect to one another.

- Warren
 
  • #7
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The c^2 is there just for unit conversions, from mass to energy. One can work in units where c = 1 (unitless) which makes the formula become E = m.

Many of einsteins equations are not derived, but postulated, https://www.physicsforums.com/showthread.php?t=76010

Why are mass and energy related by c^2? What connects mass and energy? Why is c^2 the constant between them?
 
  • #8
Fredrik
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Why are mass and energy related by c^2? What connects mass and energy? Why is c^2 the constant between them?
See post #4, specifically the stuff under the words "Post 1". You may also want to have a look at the post about 4-velocity (for which you can find a link in #4). The "c" that's used there (and actually set to 1 by a choice of units) is the invariant speed. A measurement of the velocity of something moving at that speed yields the same result regardless of the velocity of the measuring device. It's an experimental fact that this "c" isn't infinite as you might expect, but instead equal to the speed of light.
 

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