What Does the 'E' in E=mc² Really Represent: Total Energy or Mass Energy?

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SUMMARY

The discussion centers on the interpretation of the equation E=mc², specifically whether 'E' represents total energy or only mass energy. Participants clarify that E=mc² applies strictly when momentum is zero, and the complete equation is E=√[(mc²)²+(pc)²], where 'p' is momentum. They emphasize that kinetic and potential energies are not included in the mass-energy equivalence unless considered in a broader context. The conversation also touches on the implications of energy transformations in systems like nuclear fission and power plants.

PREREQUISITES
  • Understanding of Einstein's mass-energy equivalence principle
  • Familiarity with special relativity concepts, including momentum
  • Basic knowledge of kinetic and potential energy
  • Awareness of the implications of energy transformations in physical systems
NEXT STEPS
  • Study the complete energy-momentum relation E=√[(mc²)²+(pc)²]
  • Explore the implications of mass-energy equivalence in nuclear physics
  • Investigate the role of potential energy in energy transformations
  • Read "Fundamentals of Physics" by David Halliday, focusing on the special relativity chapter
USEFUL FOR

Students of physics, educators, and anyone interested in understanding the nuances of energy transformations and mass-energy equivalence in the context of special relativity.

karanbir
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i'm really a beginner in this topic. So, I'm really confused even about this simple thing

an object has so many types of energies like energy due to mass in it, kinetic energy, pot. energy etc. So does the 'E' in E=mc2 give sum of all these energies or only the energy due to mass?

also when we consider the wave motion of any particle, is the energy of the wave equal to mc2?
 
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The energy is strictly the energy due to the mass or the rest energy. In that formulation, there is no kinetic energy or potential energy.
 
please answer the second part also
 
if we say E in E=mc2 is energy due to mass only, then where does the kinetic energy or pot. energy that the object had (before its mass was converted to energy) go?
 
karanbir said:
if we say E in E=mc2 is energy due to mass only, then where does the kinetic energy or pot. energy that the object had (before its mass was converted to energy) go?

Like a power plant on a hill, where does the potentential energy of its fuel go?

If it's a water boiling power plant, potential energy of the fuel turns into potential energy of heat energy.

It may be so that massless potential energy of the fuel turned into massless potential energy of the heat.

Or it may be so that potential energy of the fuel had mass, and potential energy of the heat energy has mass.

I leave the question whether potential energy has mass to wiser people.
 
karanbir said:
i'm really a beginner in this topic. So, I'm really confused even about this simple thing

an object has so many types of energies like energy due to mass in it, kinetic energy, pot. energy etc. So does the 'E' in E=mc2 give sum of all these energies or only the energy due to mass?

also when we consider the wave motion of any particle, is the energy of the wave equal to mc2?

I like to point out that E=mc^2 is a very misleading formula and unfortunately even in many scientific documentaries , they just use it frequently without knowing what they say. The actual formula derived by Einstein was:

E=√[(mc^2)^2+(pc)^2] (1)

where p is the momentum of the particle relative to the base frame. If the 2 bodies have zero velocity with respect to each other, then p=0 and in this SPECIAL case you get E=mc^2. The kinetic energy is thus included in the second term. The potential energy is normally zero in special relativity because it is very small compare with the rest of the components.

Also I should point out that mass-energy equivalence is valid ONLY if the 2 bodies have zero velocity with respect to each other.

For your second question, if by wave you mean electromagnetic waves which are made of photons, then no. Because in case of a photon, m=0 and you get E=pc. But if the particle has some kind of mass, then you need to consider equation (1) in its general form.

For more info have a look at Fundamentals of Physics by David Halliday , special relativity chapter.
 
thank you very much supernova1387. Your reply was really helpful.
 
jartsa said:
I leave the question whether potential energy has mass to wiser people.

An uranium nucleus is heavier than the fission products created when it splits. Likewise, carbon and oxygen molecules weigh more than the carbon mono/dioxides produced in coal burning. Same holds true for a power plant on a hill, but then the thought experiments gets a bit involved, because it's the total mass of earth+power plant which is different.
 
clamtrox said:
An uranium nucleus is heavier than the fission products created when it splits.

The *rest mass* of a uranium nucleus is heavier than the *rest masses* of the fission products created when it splits.

The total relativistic mass is conserved.
 
  • #10
the_emi_guy said:
The *rest mass* of a uranium nucleus is heavier than the *rest masses* of the fission products created when it splits.

The total relativistic mass is conserved.

If you insist calling a particles energy it's mass, then sure :D I usually call mass mass and energy energy, so there's no confusion.
 
  • #11
the_emi_guy said:
The *rest mass* of a uranium nucleus is heavier than the *rest masses* of the fission products created when it splits.

The total relativistic mass is conserved.

What "relativistic mass"? The "missing mass" is in the binding energy!

I definitely agree with clamtrox here. Mixing up those two, especially when someone isn't familiar with the concept, is causing nothing but confusion.

Zz.
 
  • #12
supernova1387 said:
[..]The actual formula derived by Einstein was:

E=√[(mc^2)^2+(pc)^2] (1) [..]
Please show me, I can't find it:
http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

However, I do think that your reply was very useful, as you provided a good equation for m= rest mass.

Moreover, indeed the formula as first derived was slightly different (and it was even just put in words):

" If a body gives off the energy L in the form of radiation, its mass diminishes by L/c² "

Δm = L/c2 with L=emitted radiation energy

And that formula was derived for an atom at rest ("a stationary body").
The derivation does not explicitly exclude potential energy; however it only discusses kinetic energy, as it was directed at answering a specific question which was correctly answered as cited.
 
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  • #13
the_emi_guy said:
The total relativistic mass is conserved.

And so is the total rest mass.
 
  • #14
DrStupid said:
And so is the total rest mass.
That's wrong, see #11.
 
  • #15
While posts #2 and #6 are correct, there is an additional perspective that I found very helpful to understand.

Karanbir:
..an object has so many types of energies like energy due to mass in it, kinetic energy, pot. energy etc. So does the 'E' in E=mc2 give sum of all these energies or only the energy due to mass?...if we say E in E=mc2 is energy due to mass only, then where does the kinetic energy or pot. energy that the object had
As #6, explains, E = mc2 applies when the momentum [p] [lateral movement] is zero. An equivalent way of looking at this is that the first part [mc2 is the total energy in the frame of the matter particle while the second part is it's additional energy in the frame of the observer.

However, there are different types of energy in the frame of the particle. So for example, if we are considering a particle with structure, like an atom with orbital electrons, heating the atom increases the energy in the frame of the particle: the electrons move to more energetic orbitals, the structure now exhibits additional 'mass' due to this random kinetic energy. It has more 'rest mass' than a cold atom.

You can think of this as analogous to a coiled spring: In that situation energy is also contained within the structure, but there it is mostly potential energy; a coiled spring has infinitisimally more 'rest mass' than an uncoiled spring! [It also heats a bit when compressed.]

But you cannot 'heat' a fundamental matter particle, say an electron, because it has no 'structure' with which to absorb energy. No degrees of freedom like an atom. In an atom, it is the structure that absorbs heat energy, not the individual electrons themselves...they just bump up or down in orbitals of different energy levels, say as when reacting with chemical reactions. [Neutrons and protons do have structure...quarks.]

The binding energy mentioned in previous posts is also an analogous energy component, but in that case is an energy component of the nucleus rather the the orbital electrons.
 
  • #16
I find some ambiguity in the term rest(invariant) mass, because if rest is considered to be relative in SR, how can it be invariant? I mean if rest mass is the same in all reference frames, then all observers are agreeing about the rest frame, aren't they?

Unfortunately mass (like energy) is not a well defined quantity in GR (momentum-energy is), but it is well defined in SR so it should be possible to clarify the above within SR.
 
  • #17
TrickyDicky said:
I find some ambiguity in the term rest(invariant) mass, because if rest is considered to be relative in SR, how can it be invariant? I mean if rest mass is the same in all reference frames, then all observers are agreeing about the rest frame, aren't they?

Unfortunately mass (like energy) is not a well defined quantity in GR (momentum-energy is), but it is well defined in SR so it should be possible to clarify the above within SR.
Rest mass simply means that the mass is determined with the particle or object "in rest", which distinguishes it from other definitions. Similar expressions are rest length or proper length, and proper time.
 
  • #18
For a specific object, the quantity ##mc^2 = \sqrt{E^2 - (pc)^2}## has the same value in all inertial reference frames, even though E and p vary from one frame to another. Therefore we say m is invariant across inertial reference frames (a.k.a. Lorentz invariant). In some particular frame the object has p = 0 (unless it is a photon or other "massless" particle which can never have p = 0). We say it is "at rest" in that frame, and in that frame mc2 = E.
 
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  • #19
jtbell said:
For a specific object, the quantity ##mc^2 = \sqrt{E^2 - (pc)^2}## has the same value in all inertial reference frames, even though E and p vary from one frame to another. Therefore we say m is invariant across inertial reference frames (a.k.a. Lorentz invariant). In some particular frame the object has p = 0 (unless it is a photon or other "massless" particle which can never have p = 0). We say it is "at rest" in that frame, and in that frame mc2 = E.
But that frame is agreed by all inertial observers, right? I mean you just said it has the same value in all inertial reference frames. So do you mean we say is "at rest" from any frame, but it is not really "at rest"? Is it some idealized rest?
 
  • #20
TrickyDicky said:
But that frame is agreed by all inertial observers, right? I mean you just said it has the same value in all inertial reference frames. So do you mean we say is "at rest" from any frame, but it is not really "at rest"? Is it some idealized rest?
To use the topic as example: the "rest energy" of a particle is mc2.
Now consider that you are discussing a "high energy" electron (as it's accelerated in a particle accelerator), and you want to know how many times more energy it has got. To answer that, one has to know the energy in the lab frame and the energy that it would have in rest - the rest energy. But of course, it's not at rest.
 
  • #21
harrylin said:
To use the topic as example: the "rest energy" of a particle is mc2.
Now consider that you are discussing a "high energy" electron (as it's accelerated in a particle accelerator), and you want to know how many times more energy it has got. To answer that, one has to know the energy in the lab frame and the energy that it would have in rest - the rest energy. But of course, it's not at rest.

I don't mean that an elementary particle with invariant mass has necessarily to be at rest. That is silly. But at least seems to imply that there's a theoretical rest frame state for the particle. I guess you mean that frame doesn't really exist even if we have an invariant rest mass frame that is agreed by any observer (for instance in the case of the electron 0.51 MeV/c^2).
 
  • #22
TrickyDicky said:
But that frame is agreed by all inertial observers, right? I mean you just said it has the same value in all inertial reference frames. So do you mean we say is "at rest" from any frame, but it is not really "at rest"? Is it some idealized rest?

The equation that always holds, in all inertial frames, is ##mc^2 = \sqrt{E^2 - (pc)^2}##. The values of m and p are different in different frames, but they are always different in ways that make this equality hold in every frame.

There is one frame where p=0, namely the frame of an (imaginary) observer who is sitting on the back of the electron, riding it through space like a cowboy on a horse. We call that frame the "rest frame of the electron" and we call the mass of the electron in that frame the "rest mass". Aside from the fact that it is sometimes easier to do the math in that frame, there's nothing special or unique about it - it's just a frame in which the electron is at rest and everything else is moving at various speeds, as opposed to a frame in which some other object is at rest and everything else including the electron is moving at various speeds.

Often we choose to think about a problem in a whatever frame makes it easiest. For example, if you wanted to answer the question "How much mass does a coil spring gain when it is compressed", you'd certainly want to solve it in a frame in which the momentum p of the spring is zero - its rest frame. On the other hand, it's damn near impossible to make a free electron stand still in a lab, so if we find ourselves sitting in a lab somewhere contemplating an electron, we might find it easiest and most natural to measure its momentum p in our lab frame then proceed with our calculations in that frame.
 
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  • #23
Nugatory said:
There is one frame where p=0, namely the frame of an (imaginary) observer who is sitting on the back of the electron, riding it through space like a cowboy on a horse. We call that frame the "rest frame of the electron" and we call the mass of the electron in that frame the "rest mass". Aside from the fact that it is sometimes easier to do the math in that frame, there's nothing special or unique about it - it's just a frame in which the electron is at rest and everything else is moving at various speeds, as opposed to a frame in which some other object is at rest and everything else including the electron is moving at various speeds.
Well, you keep saying that there is really something special about that frame, namely that it is agreed by all observers that it is at rest (not only by the one sitting on its back, every state of motion considers itself to be at rest wrt itself but the rest of observers not necessarily agree), no other frame has that feature in SR. any other state of motion is not agreed by all observers regardless there own state of motion.
 
  • #24
TrickyDicky said:
Well, you keep saying that there is really something special about that frame, namely that it is agreed by all observers that it is at rest (not only by the one sitting on its back, every state of motion considers itself to be at rest wrt itself but the rest of observers not necessarily agree), no other frame has that feature in SR. any other state of motion is not agreed by all observers regardless there own state of motion.

No other frame in SR has the property of being the rest frame for that electron... But every frame in SR has the property of being the one and only rest frame, agreed upon by all observers regardless of their own state of motion, for *something*... so that doesn't make this particular electron's rest frame especially "special" unless there is something special about this particular electron.
 
  • #25
Nugatory said:
No other frame in SR has the property of being the rest frame for that electron... But every frame in SR has the property of being the one and only rest frame, agreed upon by all observers regardless of their own state of motion, for *something*... so that doesn't make this particular electron's rest frame especially "special" unless there is something special about this particular electron.
True, I'm not asking about the frame of a certain particle properties.
It is hard to explain, I'm referring to the concept that the quantity is invariant regardless the state of motion.
Take momentum instead of energy, there is no invariant momentum. So it might sound somewhat confusing that there is an energy invariant when in SR what is conserved is the momentum-energy.
 
  • #26
TrickyDicky said:
True, I'm not asking about the frame of a certain particle properties.
It is hard to explain, I'm referring to the concept that the quantity is invariant regardless the state of motion.
Take momentum instead of energy, there is no invariant momentum. So it might sound somewhat confusing that there is an energy invariant when in SR what is conserved is the momentum-energy.
If we define invariant momentum like invariant mass at v=0 then we get that invariant momentum - the momentum at zero speed - is 0. I suppose that that is why there is no invariant momentum...
 
  • #27
harrylin said:
If we define invariant momentum like invariant mass at v=0 then we get that invariant momentum - the momentum at zero speed - is 0. I suppose that that is why there is no invariant momentum...

So you are saying v=0 (therefore p=0) can be defined and agreed by all observers?
 
  • #28
TrickyDicky said:
So you are saying v=0 (therefore p=0) can be defined and agreed by all observers?
The properties of an object as determined with a reference system relative to which it is in rest are agreed by all observers - and yes, to be precise: zero speed relative to a certain object is agreed by all inertial reference systems.

Thus the group of reference systems relative to which the object has zero speed is called its "rest frame", and its energy relative to that "frame" is called its "rest energy" m0c2

As a matter of fact, an "inertial frame" is an imaginary or real system with rulers and clocks that are in rest relative to each other - according to that system as well as according to systems in uniform motion relative to it. If this was not agreed by all inertial systems, then no consistent transformations would be possible I'm afraid.
 
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  • #29
harrylin said:
That's wrong

Only if E and p are not conserved or E²/c² = m²·c² + p² is wrong.
 
  • #30
DrStupid said:
Only if E and p are not conserved or E²/c² = m²·c² + p² is wrong.
I wrote "see #11": there is "missing" rest mass
- http://en.wikipedia.org/wiki/Binding_energy

As a matter of fact, rest mass diminishes with L/c2 as the first derivation already showed which I cited (post #12, just above your post!).

Nevertheless, while the extended equation includes radiation, it doesn't include potential energy - it's incomplete.
 
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