Energy in E=mc2

1. Jun 20, 2012

karanbir

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?

2. Jun 20, 2012

Pengwuino

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.

3. Jun 20, 2012

karanbir

4. Jun 20, 2012

karanbir

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?

5. Jun 21, 2012

jartsa

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.

6. Jun 21, 2012

supernova1387

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.

7. Jun 21, 2012

karanbir

thank you very much supernova1387. Your reply was really helpful.

8. Jun 21, 2012

clamtrox

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.

9. Jun 21, 2012

the_emi_guy

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. Jun 22, 2012

clamtrox

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. Jun 22, 2012

ZapperZ

Staff Emeritus
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. Jun 22, 2012

harrylin

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.

Last edited: Jun 22, 2012
13. Jun 22, 2012

DrStupid

And so is the total rest mass.

14. Jun 22, 2012

harrylin

That's wrong, see #11.

15. Jun 22, 2012

Naty1

While posts #2 and #6 are correct, there is an additional perspective that I found very helpful to understand.

Karanbir:

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. Jun 23, 2012

TrickyDicky

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. Jun 23, 2012

harrylin

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. Jun 23, 2012

Staff: Mentor

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.

Last edited: Jun 23, 2012
19. Jun 23, 2012

TrickyDicky

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. Jun 23, 2012

harrylin

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.