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A question about E=mc^2

  1. Mar 24, 2008 #1
    If we want to know the totality of the energy of an object/particle do we only have to look at the mass, since mass changes with speed, or do we have to also add the classical kinetic energy of the object to the changed mass energy?

    I think it is is former but it is not obvious to me right now why.
     
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  3. Mar 24, 2008 #2

    CompuChip

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    That depends on how you look at it. Personally, I prefer the view where mass is constant (the "rest" mass) and the energy is given by
    [tex]E = \gamma m c^2[/tex]
    and all the change is in the gamma factor.
    For small velocities, one can expand around [itex]\gamma = 1[/itex] to get
    [tex]E = m c^2 + \frac12 m v^2 + \mathcal O(v^4)[/tex]
    which contains the rest energy and the classical kinetic energy terms already.
     
  4. Mar 24, 2008 #3

    jtbell

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    If you are thinking in terms of relativistic mass, that's all you need:

    [tex]E = m_{rel} c^2[/tex]

    If you are thinking in terms of "rest mass" (more properly called "invariant mass"), then you need to include kinetic energy:

    [tex]E = m_0 c^2 + K[/tex]

    In other words, the difference between "relativistic mass" and "rest mass" corresponds to the kinetic energy:

    [tex]K = m_{rel} c^2 - m_0 c^2[/tex]
     
    Last edited: Mar 24, 2008
  5. Mar 24, 2008 #4

    rbj

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    it seems presently out of vogue to talk of relativistic mass and rest mass:

    [tex] m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]

    most folks here like to refer only to the rest mass calling it "invariant mass", which is not a bad name for it. the same people will say that photons are "massless" without qualification (whereas i would say that photons have no rest mass, but do have relativistic mass).

    anyway, if the mass you mean is the above m, then the ubiquitous equation

    [tex]E = m c^2[/tex]

    means the total energy, rest energy plus kinetic energy:

    [tex]E = m c^2 = E_0 + T[/tex]

    so that the kinetic energy is

    [tex] T = E - E_0 = (m - m_0) c^2 [/tex]

    if you plug the m at top into the equation just above, let v<<c, and solve, you will get an expression that, in the limit, becomes the same as the classical kinetic energy

    [tex] T \approx \frac{1}{2} m v^2 [/tex]


    now, on the other hand, if what is meant by the mass m is really only the invariant mass (what i call the "rest mass" m0), then what is mean by E=mc2 is what i called the "rest energy" E0 = m0c2.
     
    Last edited: Mar 24, 2008
  6. Mar 24, 2008 #5
    Probably a stupid question, but if one uses E=ymc^2, does that include all potential energy as well? So, say we had an electron experiencing an electric potential V. Does E=ymc^2 account for V as well?
     
  7. Mar 24, 2008 #6

    jtbell

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    Potential energy is a property of a system, not of individual particles. The potential energy of a system is included in the invariant mass of the system. For example, the invariant mass of a nucleus is less than the sum of the invariant masses of the protons and neutrons that comprise it, because the potential energy is less when the p's and n's are together than when they are separated.
     
  8. Mar 24, 2008 #7
    So if you take the potential energy from an electron interacting with itself at the "classical electron radius" you get, not surprisingly, the known mass of the electron. So is mass just a form of potential energy?
     
  9. Mar 24, 2008 #8

    jtbell

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    Indeed it's not surprising, because the "classical electron radius" is defined as the (hypothetical) radius that a uniform spherical charge distribution would have to have in order for its electrostatic potential energy to correspond to the mass of the electron.
     
  10. Mar 24, 2008 #9
    Right. So does that imply that all mass can be interpreted as some kind of potential energy confined to a small point in space? Or has the "classical electron radius" idea, and everything related to it, been rendered obsolete by QM? Put another way, does the classical electron radius actually mean anything or was it a pointless exercise?
     
  11. Mar 24, 2008 #10

    jtbell

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    If an electron actually had a definite radius in the range given by the "classical electron radius" (about [itex]10^{-15}[/itex] m) its effects would have been seen in electron-electron or electron-positron scattering experiments by now. For example, see http://arxiv.org/abs/hep-ph/0002172 which uses an experiment at LEP to set an upper limit of about [itex]10^{-19}[/itex] m for the electron radius. (That doesn't mean that they actually found such a size, it just means that if the electron has a size, it must be smaller than that or else they would have "seen" it.)
     
  12. Mar 24, 2008 #11
    Ok cool. So I guess the short answer is, yes, the "classical electron radius" is more or less meaningless at this point?
     
  13. Mar 25, 2008 #12
    In the most general case you have to use the stress-energy-momentum tensor T to find the energy. The energy of an object equals T00. In general this is a function of the object's speed, proper mass (aka rest mass) and the stress exerted on the body by external sources. In some cases the inertial mass is a function of the pressure inside the object. An example is a gas in a container. The inertial mass of the gas is a function of the pressure of the gas. If you consider the gas+container then this is a closed system and the energy is a function only of the proper mass of the gas+container system and the system's speed.
    I disagree. When one derives the expression for the mass of a charged object one must include the stresses holding the object together. These stresses are known as Poincare stresses. Omitting them in the derivation will result in an error.
    If one is to calculate the expression for the mass of a nucleus, treating it as a classical object with interacting particles, then one has to account for the proper mass of the particles, the energy associated with the field and the stresses in the field.

    Pete
     
  14. Mar 25, 2008 #13
    Um, you can disagree, but that is indeed how the "classical electron radius" is derived.
     
  15. Mar 26, 2008 #14
    Actually I was trying to be polite. I was really saying that you're quite wrong. One simply cannot derive the classical electron radius by simply using the energy of the field. If Poincare stresses are ignored then the answer will be wrong. This is a well-known fact and can be found in any decent physics text such as the Feynman lectures. Would you like me to quote Feynman to you? Its also in a well known article in the American Journal of Physics by Fritz Rohrlich. I can make that paper available to you too.

    However, if you think you can do it then please feel free to provide the derivation. If the derivation is valid then publish it.

    Pete
     
  16. Mar 26, 2008 #15
    Wow, yet another unnecessarily bad attitude just walked in the room! (Why are these forums so plagued by them?)

    Anyway -

    http://en.wikipedia.org/wiki/Classical_electron_radius

    r_e = k * e^2 / (m_e * c^2)
    = 2.8179 x 10^-15 m

    Maybe you and I are talking about two different things, but there's no reason for the attitude either way.
     
  17. Mar 27, 2008 #16
    If you concluded that I have a bad attitude then you are mistaken. I try to be polite as possible. That means that instead of saying someone is wrong when I see a mistake that I instead disagree with them. Unfortunately you understood that to mean something otherwise, as if I wasn't familiar with this derivation and was being a jerk about it. Totally wrong. However I was being kind of sarcastic of suggesting you publish it, but we all have our off days, right? :redface:

    And yes. You're correct. You and I are talking about two different things. I had something else in mind. Sorry for the confusion. There is another way to derive the electomagnetic mass and it is that derivation that is incomplete without taking into account Poincare stress. The reason I was referring to the relation m = p/v is because this is how mass is defined in relativity, not by E = mc^2 which is a derived equality. Therefore it is important that one be able to calculate mass in this way. However if the mass is calculated by m = E0/c2 then one does get the correct result. So please take note of my correction to my earlier comments. Again, sorry for the confusion.

    Let me clarify what I was referring to. Consider The Feynman Lectures on Physics - Volume II, Feynman, Leighton and Sands, page 28-4. In the previous section Feyman et al showed that there are two ways of deriving the electromagnetic mass of an electron. One is found by using E = mc^2 and the other is found by calculating m = p/v and letting v->0. The results are contradictory. This was a source of great confusion. Feynman explains how this contradiction was resolved
    Feynman goes on to explain what was left out, i.e. the Poincare stresses. I can make the rest available if you'd like. Then there is the article I mentoned. That article is

    Self-Energy and Stability of the Classical Electron, F. Rohrlich, Am. J. Phys. 28, 639 (1960)
    I can make this available to those of you who'd like to read it.

    Pete
     
    Last edited: Mar 27, 2008
  18. Mar 27, 2008 #17
    If you calculate the electromagnetic mass of a spherical shell with total charge e then you get the same answer. In fact this is how Feynman does it in his Lectures. Otherwise you'd be able to tell what the structure was from the result and that doesn't seem possible as of yet.

    A good paper on this (which contains the derivation for electromagnetic mass of a spherically charged shell) is

    Mass renormalization in classical electrodynamics, Griffiths and Owen, Am. J. Phys., 51 (12), December 1983.

    Of course I can make this available to those who'd like to read this one too. :smile:

    Note: I recommend caution when you come across the relation E = mc2. There is a tendancy to use m to refer to proper mass and as such this expression would be wrong. Some people can being sloppy about this. If m is the proper mass then the correct expression for mc2 is E0 = mc2.

    Pete
     
    Last edited: Mar 27, 2008
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