# I Real life example of the energy contained at E=γmc^2 (1 Viewer)

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#### Foruer

We've just begun studying about relativity, and I find it amazing that bodies have the energy of E=γmc^2. Even at rest they have E=mc^2.
But where exactly is this energy present in real life? For example the keyboard I am currently typing this post with has a huge amount of energy, according to this equation, but how is it usable?

#### Orodruin

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But where exactly is this energy present in real life? For example the keyboard I am currently typing this post with has a huge amount of energy,
Most of that energy is due to the rest mass of the constituents.

but how is it usable?
Why does it have to be usable for anything?

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#### Foruer

Because it's a big amount of energy in small mass bodies, so you'd think that if it was really usable somehow, you would could provide energy to the entire world by using small amount of matter. I just wonder if it can be expressed somehow in real life?

#### jbriggs444

Because it's a big amount of energy in small mass bodies, so you'd think that if it was really usable somehow, you would could provide energy to the entire world by using small amount of matter. I just wonder if it can be expressed somehow in real life?
If you throw a baseball toward home plate, the difference, $\gamma mc^2 - mc^2$ gives the kinetic energy of the baseball.

• Foruer

#### russ_watters

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For example the keyboard I am currently typing this post with has a huge amount of energy, according to this equation, but how is it usable?
Well, you could collide it with an anti-keyboard....

In reality, this isn't usable energy, but rather an energy equivalence. However, if you've studied nuclear energy yet, you'll see that there are small differences in the starting and ending mass that you can evaluate.

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#### pervect

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Aside from colliding your keyboard with an anti-keyboard, you could extract energy from it by lowering it into a black hole. You could extract this energy in principle the same way as you could extract energy from a dropping weight, by having it turn a gearshaft that cranks a generator or whatever.

If you had a strong enough cable, you could in theory extract the entire amount of energy from it, but any sort of realistic cable would break.

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#### PAllen

A quarter teaspoon of water plus a quarter teaspoon of anti water would yield an explosion approximately twice the energy of the atomic bomb dropped on Nagasaki in WW II.

#### Nugatory

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But where exactly is this energy present in real life?
If you look closely at any reaction that releases energy, you will find that the mass of the material coming out is very slightly less than the mass of the material that went in (and the other way around for a reaction that consumes energy, like charging a battery). For example, if I burn a chunk of wood... the weight of the wood plus the weight of the oxygen from the air will be very slightly more than the weight of the ashes and the gases released by combustion. The difference is exactly what you'd calculate from the energy change using the $E=\gamma{m}c^2$ formula (easiest if everything is at rest so $\gamma=1$, of course).

So the best answer for where the energy is will be something along the lines of "the mass is the energy, just in a different form".

As a practical matter, only nuclear reactions involve enough energy for the effect to be detectable; $c^2$ is a very big number. For example, a lead-acid automobile battery weighs only a few nanograms more when charged than discharged.

#### Mister T

Gold Member
Because it's a big amount of energy in small mass bodies,
No, it's not. The rest energy is equivalent to the mass, so it doesn't make sense to say that one is larger than the other.

#### nitsuj

No, it's not. The rest energy is equivalent to the mass, so it doesn't make sense to say that one is larger than the other.
I found it clear that they were referring to energy density, not a stress energy tensor

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#### nitsuj

But where exactly is this energy present in real life? For example the keyboard I am currently typing this post with has a huge amount of energy, according to this equation, but how is it usable?
Read about the constituents of an atom. It's really neat, allot of it (mass in "real life") is kinetic energy of quarks (things that form protons neutrons) bound by gluons; this is the strong nuclear force.

there's also such a thing as a nuclear battery, weak nuclear not strong...using tritium beta decay! That's hydrogen turning into helium "exactly" over time.

with both fission and fusion we can (potentially) do work.

there was allot of added info in the replies.

Yes to "power the world"; interesting political topic regarding nuclear power...far to complicated for discussion in a physics thread :D

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#### Sorcerer

I found it clear that they were referring to energy density, not a stress energy tensor
May I ask how that matters with respect to the equation in question? If the rest energy is equivalent to the mass, then they are one and the same. It's not like the left side of the equation takes up more volume than the right side, or something like that. I mean, does it make sense to ask if mass*acceleration is larger than force? So yeah, your response has confused me somewhat. Would you care to elaborate why Mr. T is wrong?

#### stevendaryl

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A pet peeve of mine about $E = mc^2$ is that people often say that nuclear reactions are demonstrations of that formula, that the reason that so much energy is produced in an atomic explosion is because matter is being turned into energy. I consider that both true and misleading.

The facts for nuclear fusion is that there is a transformation that looks like this (yes, I know, that is not the way that fusion happens, directly, but I'm just simplifying for illustration)

$4H \Rightarrow He + 2 \bar{\nu_e}$

The fact that this reaction is exothermic (the total energy of the right-hand side is less than the total energy of the left hand side) is what makes it possible for fusion to produce energy. I don't see that it essentially involves $E = mc^2$ any more than the chemical reaction

$2H + O \Rightarrow H_2 O$

The relevance of $E = mc^2$ is that if you measure the total mass of the products on the right-hand side, it will be slightly less than the total mass of the constituents on the left-hand side. But that's as true for the chemical reaction as it is for the nuclear reaction: Water has slightly less mass than the hydrogen and oxygen atoms it was made of. I don't see why the nuclear process is an example of matter being turned into energy any more than the chemical process is.

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#### Orodruin

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I don't see why the nuclear process is an example of matter being turned into energy any more than the chemical process is.
A pet peeve of mine is that matter is not being turned into energy. It is converted from one form of energy (mass energy) to another (kinetic energy in the products that in the end goes to heat).

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#### stevendaryl

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A pet peeve of mine is that matter is not being turned into energy. It is converted from one form of energy (mass energy) to another (kinetic energy in the products that in the end goes to heat).
But I would say that it's not really doing that, either. I would say that in both nuclear and chemical processes, it's a matter of binding energy, rather than mass energy. If you have a nucleus with N protons and M neutrons, in some cases a neutron can transform to a proton + electron + anti-neutrino + kinetic energy, and sometimes a proton can transform to a neutron + positron + neutrino + kinetic energy. The reason one or the other is favored is because of the binding energy for nucleons (which involves both strong forces and electromagnetic forces). So it's really a matter of converting potential energy into kinetic energy, just like chemical processes.

#### nitsuj

May I ask how that matters with respect to the equation in question? If the rest energy is equivalent to the mass, then they are one and the same. It's not like the left side of the equation takes up more volume than the right side, or something like that. I mean, does it make sense to ask if mass*acceleration is larger than force? So yeah, your response has confused me somewhat. Would you care to elaborate why Mr. T is wrong?
their statement wasn't wrong at all, the interpretation of the question was. Mister T has allot of great replies to questions and has enlightened me a number of times.

To be even more clear, surely the OP is, like the vast majority of people, used to "chemical energy", and in turn the concept of energy density.

They are not "one and the same", they are equivalent...clearly on opposite sides of the equation.

#### nitsuj

But I would say that it's not really doing that, either. I would say that in both nuclear and chemical processes, it's a matter of binding energy, rather than mass energy. If you have a nucleus with N protons and M neutrons, in some cases a neutron can transform to a proton + electron + anti-neutrino + kinetic energy, and sometimes a proton can transform to a neutron + positron + neutrino + kinetic energy. The reason one or the other is favored is because of the binding energy for nucleons (which involves both strong forces and electromagnetic forces). So it's really a matter of converting potential energy into kinetic energy, just like chemical processes.
So the distinction between one being a nuclear force and the other being em force is moot, because with either we use the work in the same way ?

#### stevendaryl

Staff Emeritus
So the distinction between one being a nuclear force and the other being em force is moot, because with either we use the work in the same way ?
The big distinction that I see between common nuclear reactions and chemical reactions is that in the case of chemical reactions, you can think of the transformation as a rearrangement of constituent particles. In contrast, when a neutron transforms into a proton + electron + anti-neutrino, it's not literally the case that it's just a rearrangement of constituent particles. (There is no proton or electron or neutrino inside a neutron).

That's something new with nuclear physics: transformations that are not just rearrangements of constituents. But I would still say that in nuclear reactions, it's the binding energy that makes the reaction go, rather than converting rest mass energy into kinetic energy.

Let's take two different nuclear reactions:
1. $^{14}C \Rightarrow ^{14}N + e + \bar{\nu_e}$: Carbon 14 decays into Nitrogen
2. $^{23}Mg \Rightarrow ^{23}Na + e^+ + \nu_e$. Magnesium 23 decays into Sodium
Now, let's combine the two reactions into one:

$^{14}C +\ ^{23}Mg \Rightarrow\ ^{14}N +\ ^{23}Na$

That's a nuclear reaction which is possible (though extremely unlikely) that is simply a rearrangement of protons and neutrons. In this case, it's much more like a chemical reaction: A rearrangement of constituent particles results in a lower total energy, so it releases some kinetic energy. But to me, it doesn't make sense to say that the first two reactions involve transmuting rest mass energy into kinetic energy, but the last one does not. But if you say that the latter illustrates converting rest mass energy into kinetic energy, then I don't see how that description wouldn't apply equally well to chemical reactions.

To me, the lesson from $E=mc^2$ is that if you have a reaction that releases kinetic energy, then the rest mass of the products will be less than the rest mass of the original ingredients. That's true of chemical reactions as well as nuclear reactions.

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#### martinbn

Since we are talking about peeves, I have this one. It bothers me that people say things like "matter is turned in this or that" or "matter is equivalent/equal to so and so". The $m$ in the equation stands for mass, matter is not mentioned at all.

#### stevendaryl

Staff Emeritus
Since we are talking about peeves, I have this one. It bothers me that people say things like "matter is turned in this or that" or "matter is equivalent/equal to so and so". The $m$ in the equation stands for mass, matter is not mentioned at all.
Matter is just an informal word meaning stuff that has mass, isn't it?

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