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In the famous equation $E=mc^2$, the variables stand for:

$E$ is energy, $m$ is mass, and $c$ is the speed of light (in vacuum).

And I understand the equation fairly but limited in knowing in its effects/outputs in reality, honestly.

  • Some example? Could anybody please give me some sort of examples in real world which are done by this equation?
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closed as too broad by Alan Rominger, Waffle's Crazy Peanut, Manishearth Jul 3 '13 at 6:32

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

Nuclear fission, and fusion. – user7757 Jan 9 '13 at 2:46
-1 vague question. ! – centralcharge Jun 29 '13 at 13:27

This equation is incredibly generic and describes many phenomena outside of nuclear phenomena.

For example: place the the following setup in a box (a spring and some bars) and weigh them:

enter image description here.

Now, loosen the spring and repeat. You should measure a smaller mass because you've removed some of the energy.

In reality you couldn't possibly measure the difference in this experiment. The energy content of a typical spring is in the millijoules, so the mass difference is $E/c^2 = 10^{-20}\;\textrm{kg}$. That's the mass of a virus. However, in atomic system, such a difference is (barely) measurable.

Mass of vacuum

Now, for a much weirder example: the vacuum state. It's well known from quantum mechanics that the energy of a standing wave, such as a spring or photon in a box, has a minimum zero-point energy. An empty box is filled with many possible photon modes, even if it has no photons in it. Each of these modes has some energy. If that doesn't make any sense, that's okay, here's my point: quantum mechanics predicts that an empty box has energy.

That means an empty box has mass. Is it a lot? Well, let's consider only photons. I want to calculate the total energy of all the modes in a box, so I integrate the number of standing waves per energy times the energy.

$$ E_\textrm{zero} = \int \rho(E) E \;dE = \frac{8\pi V}{(hc)^3} \int E^3\;dE $$

Now, this is the kind of integral you never want to see because it's badly divergent. Physicists usually try to bluff their way out of these problems by saying: "I know how physics works at low energy, but I have no idea what happens at very high energy. So, I'll just integrate until I get to the part where something else happens." So, we integrate up to $E_\textrm{cutoff}$ and find $E_\textrm{zero} = (2\pi V E_\textrm{cutoff}^4)/(hc)^3$. What do we use as a cutoff? An optimist says that we understand everything up to the Planck energy of $10^{28}\;\textrm{eV}$. A pessimist says we've got everything up to the LHC, which is $10^{14}\;\textrm{eV}$.

$$ E_\textrm{zero} = \left\{ \begin{array}{ll} 10^{111}\;\textrm{J/m}^3 & \textrm{optimist} \\ 10^{55}\;\textrm{J/m}^3 & \textrm{pessimist} \end{array}\right.$$

These are some very large numbers. Note that the optimist and pessimist disagree by 56 orders of magnitude! Do they have meaning? Well, yes and no.

  1. Yes, when we look at differences. If there is an energy density, as shown here, there is also a force. You would expect that a box would feel a force inward. This has been observe and is called the Casimir Force. It turns out that the cutoff energy we used does not matter.
  2. No, when we consider the actual value. If that's the energy density of the box, then the mass density is $10^{38} - 10^{94} \;\textrm{kg/m}^3$. (We're finally back to your question!) We know from astronomical observations of the universe's expansion that the actual mass density is something around $10^{-26} \;\textrm{kg/m}^3$. So, we're off by something around 64 to 120 orders of magnitude. This is an unbelievably large error (and error bar!).

So, instead of telling you why $E=mc^2$ is useful, I've shown you an example where it's 120 orders of magnitude off, which happens when you compare a prediction from quantum mechanics with observations from cosmology. I hope that's more interesting.

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The more general equation is: $$E=\sqrt {(mc^2)^2+(pc)^2}$$

where p is momentum

$E=mc^2$ is a special case when momentum of system is zero. This popular equation says following things:

  • Energy has all properties of mass. It means, energy also posses momentum. And, it can be influenced by gravity. It can create gravity, too.

  • Mass has all properties of energy. Consequently, it can be converted to another form of energy. In fact, mass is most concentrated form of energy.

Its applications:

  • New particles can be created with enormous amount of energy in other forms. Currently, its useful for research purpose only. But, who knows, one day we'd utilize it to synthesize objects using matter patterns stored in computer memory.

  • Matter can be converted to other forms of energy. Nuclear bombs, Sun etc do the same thing, but tapping out energy isn't easy thing. Nuclear bombs, Sun etc destroy less than 1% of their fuel to make other forms of energy. In acceretion disk of black holes, the number is 43% thanks to enormous gravity. But, there's one process in which the number is 100%: Matter-Antimatter annihilation. Matter-Antimatter annihilation can be used in energy sources of future mankind or in ultimate weapons of future. The problem is: We don't have access to antimatter repo. To create antimatter, we need more energy than annihilation output. Currently, antimatter is created for research purposes only.

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I am answering to the comment in the answer of @Bobrebock

So is there nothing it can do rather then bombs, mostly? – 4lvin

  1. nuclear radiation is used for medical purposes, from x-rays to cancer treatment. It depends on this conversion

  2. It is used to sterilize objects in large quantities.

  3. It is a diagnostic method using the table of isotopes for innumerable needs, from paleontology ( + paleoclimate) to astrophysics, to medical tests, to checking underground sources etc

And of course when fusion becomes commercial once ITER works, or some other competing method, it will supply all the energy needs of the world at low cost.

So no, it is not only good for bombs.

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Equations don't do, they describe. This one describes the equivalence between a mass at rest and energy in a system. Some measurable mass gets converted to energy when some unstable atomic nuclei break apart into others. In other instances putting together of nuclei yields combinations which release excess energy by together having less rest mass. Those are fission reactions and fusion reactions, ie atom bombs and reactors and thermonuclear bombs and the sun respectively Even the much weaker breaking and forming of atomic bonds by electrons stores in mass and releases as energy from and to the surrounding but the changes in mass are not easily directly measured due the smallness of the mass changes.

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So is there nothing it can do rather then bombs, mostly? – 夏期劇場 Jan 9 '13 at 3:03

I find particularly simple to understand the example of binding energy and in particular the so-called mass deficit of atoms.

Mass change (decrease) in bound systems, particularly atomic nuclei, has also been termed mass defect, mass deficit, or mass packing fraction. The difference between the unbound system calculated mass and experimentally measured mass of nucleus (mass change) is denoted by Δm. It can be calculated as follows:

$$\Delta m = \sum m_{unbound} - m_{measured}$$ i.e., (sum of masses of protons and neutrons) - (measured mass of nucleus)

The difference in mass is actually due to a difference in energy.

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That's not really a "real life" example, but something to give you an idea of how energy is related to speed.

It pretty much describes the ultimate energy output from a perfect energy source. Let's assume, that unlike in fusion and fission where only a minor part of the energy is used up, we produce energy through annihilation of matter and antimatter where it all gets converted to energy.

The old good "kinetic energy" is $$ E={m_{payload}v^2 \over 2} \\ E=m_{fuel}c^2 $$

Now to propel a mass of $m=1kg$ to "newtonian equivalent of speed of light" (subjectively speed of light from the point of view of the crew/payload), $v=c$ $$ {m_{payload} c^2 \over 2} = m_{fuel}c^2 \\ m_{payload} = 2 m_{fuel} $$

So to propel a ship of 10 tons to speed that would seem like speed of light from viewpoint of the crew, you'd need to "burn" 2.5 tons of matter and 2.5 tons of antimatter. (and that's not counting energy required to propel that fuel...)

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"(subjectively speed of light from the point of view of the crew/payload)" What? From the point of view of the crew/payload they are at rest. Additionally, the speed of light as measured by the crew/payload is same the speed of light regardless of whether they are moving or not. You're also using the non-relativistic kinetic energy in what seems like a manifestly relativistic situation (I think - it's hard to tell from your description). I'm afraid this will just confuse the OP unnecessarily. – Michael Brown Jan 9 '13 at 9:53
@MichaelBrown: No, if there's 4 light years from here to Proxima Centauri, and we put enough energy into accelerating the ship that it would travel at speed of light according to Newtonian physics (its kinetic energy is m*c^2/2), then exactly 4 years will pass on the crew's clocks before they reach the destination. And no, from their point of view they are definitely not at rest, from their point of view they are traveling the universe exactly at speed of light (or if you really love to nitpick, the y are at rest but the universe moves past them at that speed.) – SF. Jan 9 '13 at 10:22
You're ignoring relativity which changes everything (and is required for the OP's question to even make sense), but I'll grant you that for the sake of argument. Your first sentence is okay. Your second is not. Even Newtonian mechanics is invariant under Galilean boosts. Their rest frame is their rest frame. They see the universe moving past them at the speed of light. In any case I don't see how this at all relevant to the OP's question. If you simply want to illustrate how much energy $mc^2$ is there are better ways to do that. – Michael Brown Jan 9 '13 at 10:42
@MichaelBrown: You are nitpicking. If you're walking down the street, and a friend calls you asking you what are you doing, will you insist you just move your legs in place and the street is moving under you? It's the ages old conflict of strict definitions versus common language. Common language is not wrong, it's merely imprecise, but it's far easier to understand for layman, and explaining the problem in layman's terms is in place here. – SF. Jan 9 '13 at 11:21
We're not talking about walking here. We're talking about interstellar space travel at near the speed of light. If you're not looking out the window you have no motivation whatsoever to say anything about your motion! Anyway, even if you forget about all of that you are still using non-relativistic mechanics to talk about a relativistic situation in the context of a relativistic question for reasons I can't understand. Relativistic math is not that much more complicated and has the advantage of being correct. I have nothing more to add to this conversation at this point. – Michael Brown Jan 9 '13 at 11:30

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