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Everyone knows Einstein’s $E=mc^2$ equation: it is the beautiful relation that links energy with mass. My high school (tenth grade) book says that the equation can be used to find the energy released from nuclear fission and gives several examples where mass is mindlessly multiplied by 9x10^16. I, however, don’t believe in what the book says - which is by the way written by the school for the school. What energy does the equation give? Is it the energy from nuclear fission or nuclear fusion or matter-antimatter annihilation, because each one of these is orders less powerful than the other? Wikipedia (third paragraph, second line) says that the way to release the energy expressed in the Einstein equation is by matter-antimatter annihilation. If this is true, how did the equation mark the beginning of many great inventions that have nothing to do with antimatter?

My questions are

  1. What energy does the equation $E=mc^2$ give?

  2. If it is the absolute energy contained in matter that can only by released by matter-antimatter destruction, how was the equation useful at all in our limited technology that cannot utilize or even find antimatter?

Please give me a complete answer that explains things in a simple but non-dumbed down way and please excuse my naivety, lack of understanding, lack of knowledge; I am fifteen years old.

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    $\begingroup$ By the way the complete equation is $E^2=p^2c^2 +m^2c^4$ $\endgroup$ – Paul Feb 25 '17 at 9:11
  • $\begingroup$ Should be potential because it is the energy locked up in the masses. The energy locked up in the momentums would be kinetic energy. $\endgroup$ – Roghan Arun May 19 at 21:44
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If this is true, how did the equation mark the beginning of many great inventions like the steam engine?

It did not mark the beginning of the steam engine and has nothing to do with steam power.

The final paper that Einstein wrote about his then new theory of Special Relatively was published in September 1905, and introduced $m = E/c^2$. This is according to Einstein's Miracle Year

Thanks to anna v for this correction

A particle at rest has its rest mass equal to energy but the m in the formula is the relativistic mass, en.wikipedia.org/wiki/… the rest mass is the $m_0$in the formula for relativistic mass.

The energy released on total annihilation is exactly what the equation says, the product of the mass of the object and the speed of light squared, I will leave it to you to ensure this is dimensionally correct.

If it is the absolute energy contained in matter that can only by released by matter-antimatter destruction, how was the equation useful at all in our limited technology that cannot utilize or even find antimatter?

True, we do not utilise antimatter, but we can make tiny amounts of it, amounts so small in mass that even their total energy on annihilation is tiny.

Antimatter is also produced naturally, from Cosmic Rays, for example:

Of primary cosmic rays, which originate outside of Earth's atmosphere, about 99% are the nuclei (stripped of their electron shells) of well-known atoms, and about 1% are solitary electrons (similar to beta particles). Of the nuclei, about 90% are simple protons, i. e. hydrogen nuclei; 9% are alpha particles, identical to helium nuclei, and 1% are the nuclei of heavier elements, called HZE ions. A very small fraction are stable particles of antimatter, such as positrons or antiprotons. The precise nature of this remaining fraction is an area of active research. An active search from Earth orbit for anti-alpha particles has failed to detect them.

You have lots of things correctly set out in your post, except you have the dates mixed up.

The equation is far more useful (actually essential) in many theories of modern physics than in any practical application, excluding nuclear bombs, in which a small amount of matter is converted to energy.

You need to look up the definition of energy in an article such as Wikipedia Energy, as this explains the way energy is viewed today, and how difficult it is to define it, as it is a fundamental concept.

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  • $\begingroup$ Excuse me, but what does the phrase 'dimensionally correct' mean? $\endgroup$ – InterestedLearner Feb 25 '17 at 11:13
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    $\begingroup$ Sorry, this it saves a lot of time and you can do it before starting any problem. An equation has an equals sign, in numbers that obviously means 5 = 2 +3. No problem. But in physics, as energy is in joules on one side, it should be in joules on the other side of the equation as well. You might be doing it already, I just used my version of it (Dimensional Analysis)[en.wikipedia.org/wiki/Dimensional_analysis] describes it. It's just a fancy name for a simple check on what you are doing makes sense, so you don't end with up with 1 hour = 2 miles, $\endgroup$ – user146020 Feb 25 '17 at 11:29
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    $\begingroup$ alysion.org/dimensional/fun.htm also describes it. $\endgroup$ – user146020 Feb 25 '17 at 11:36
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To get to the inside working of $E=mc^2$, we first need to understand what it actually means. Einstein wrote it in his paper like "$m=E/c^2$",why?

Because according to me when you write it as $E=mc^2$, you sort of think, 'hmm I think this equation says that mass is a kind of energy or that matter can be converted into energy", and that is really really misleading, in fact what this equation says is that mass is an intrinsic property of all kinds of energy, the inertia content of a body depends on the energy content of the body, the more energy you have, harder it is to push or pull. So if you have a ball having a mass $10 kg$( I will explain this $mass$ later ) and you kick it so that it gains some kinetic energy , now Einstein says that the moving ball has a high inertia as compared to the stationary ball, and why is that? Because it simply has more energy content, now all that kinetic energy is not being converted into matter, but it is $mimicking$ the original behaviour of matter, and it appears to us that mass has increased. Now you may ask that where does that $10 kg$ mass come from, is it $fundamental?$ The answer is still- "NO" , what's fundamental is energy, and that $10 kg$ mass is just the outcome of "atoms interacting with $Higgs field$".

Now your questions :- 1. What energy does $E=mc^2$ give?

It gives energy much like all the other Energies, for example in nuclear fusion, protons and neutrons come together and form a nucleus, right ? And there is a huge energy release, now some people say that this happens because mass is being converted to energy, but that's not correct, see when you bring a proton and neutrons together, their potential energy decreases due to the "strong nuclear force attraction", see energy content is getting low, so due to this it appears to us that mass has been decrased, but that decrease occurred due to the lowering of energy content, and that lowering in kinetic and potential energy was released as heat and light, see it's energy that's being converted not mass. Mass itself is a measure of energy content, in other words you are not getting energy from MASS, but you are getting it from another kind of energy.

Question 2 :-

To release the energy contained in matter, you just need to distort its stable kinetic and potential Energies, for example when electrons jumps from higher state to lower state, its potential energy decreases and is converted to light energy, now you can also visualize it like this, when it jumps from higher shell to lower shell, its mass decreases ( due to lowering energy) and that loss of mass is equally compensated by energy release but this picture is certainly not correct, all the conversions are going from energy to energy, you dont need any mass to energy conversion.

When you annihilate matter and antimatter, all their potential Energies and kinetic Energies are converted into light and heat, and mass becomes 0 because photons don't interact with Higgs field, so no Higgs potential means 0 inertia content. Hope that answers your questions.

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    $\begingroup$ If you put your formulas inside dollar signs, you get them to look nice. E.g. $mc^2$ vs mc^2. $\endgroup$ – Magicsowon Feb 25 '17 at 10:10
  • $\begingroup$ @Magicsowon Thanks for the suggestion, I will surely do it. $\endgroup$ – Sarthak Sharma Feb 25 '17 at 10:19

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