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Rest Energy

  1. Jun 14, 2005 #1
    E = mc^2

    Please explain to me why the speed of light is squared.
  2. jcsd
  3. Jun 14, 2005 #2


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    Write it

    [tex] E=m^{a}c^{b} [/tex]

    and from the units get the values of "a" & "b".

  4. Jun 14, 2005 #3
    I don't understand what you mean. What logic was involved that determined whether the energy-mass relationship was proportional to the speed of light squared?
  5. Jun 14, 2005 #4


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    It follows from the unit analysis that E~m_{0}c^{2} and equal to it from the postulates.

  6. Jun 14, 2005 #5
    dextercioby is trying to prove you the formula through the dimension analysis

    For E, the dimensions are [itex]ML^2 T^-2[/itex]

    As E=mc^2 , keeping aside M , we are left with [itex]L^2 T^-2[/itex] which is a dimension for [itex]c^2[/itex] ... as dimension for a velocity is [itex]LT^-1[/itex]
  7. Jun 14, 2005 #6
    I'm sorry. I wasn't aware if there was any specific mathematics involved.

    Can you specify which courses are?
  8. Jun 14, 2005 #7


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    The topic is called "dimensional analysis" as others have written. It's usually taught as a part of standard physics courses (High school physics, I think, I'm not sure anymore).

    Google for the term "dimensional analysis" or try

  9. Jun 14, 2005 #8
    Oh man. What was I talking about?!

    haha damn

    kg * (m/s)^2 = J
  10. Jun 14, 2005 #9
    Yes, it is taught in high school.
  11. Jun 14, 2005 #10


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    but dimensional analysis isn't enough to show why [itex] E = m c^2 [/itex]. perhaps it's

    [tex] E = \frac{1}{2} m c^2 [/tex]


    [tex] E = m c v [/tex]


    [tex] E = \frac{m c^3}{v} [/tex]

    where [tex] v [/tex] is the velocity of my daughter's bicycle at full clip. those are both dimensionally correct. or maybe it's

    [tex] E = \frac{(m c)^2}{m_e} [/tex]

    where [tex] m_e [/tex] is the mass of the electron.

    all of these are dimensionally correct, but they ain't correct.

    r b-j
  12. Jun 14, 2005 #11
    Firstly, in your counter-argument for the dimensional analysis, you're ignoring the fact that the 'full' equation really reads


    So in the case where the particle is at rest, this simplifies to [itex]E=mc^2[/itex]. Anywho, ignoring all that, the equation is pretty simply derived by applying the relativistic Lorentz transforms to energy and momentum. The transforms themselves are derived from the postulates of special relativity (Physical laws hold in all inertial reference frames and the speed of light is constant in all intertial reference frames).
  13. Jun 14, 2005 #12


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    something happened to the textbook notation since i was in school. i was taught (ca. 1975) that the famous equation

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

    resulted from an interpretation of the derivation of the relativistic kinetic energy of a particle taking into consideration of the change in mass due to relativistic considerations.

    Kinetic energy: [tex] T = m c^2 - m_0 c^2 [/tex]

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

    and [tex] m_0 [/tex] is the rest mass. then we rearraged and interpreted that equation as

    [tex] m c^2 = m_0 c^2 + T [/tex]

    or [tex] E = E_0 + T [/tex]

    or "total energy equals rest energy plus kinetic energy."

    if [tex] m [/tex] is the relativistic mass,

    [tex] E = m c^2 [/tex] is the full equation.

    if i plug everything in

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

    [tex] E^2 = \frac{m_0^2 c^4}{1-\frac{v^2}{c^2}} [/tex]

    i s'pose it comes out to what you're saying is the full equation when [tex] |v| << c [/tex]. i don't assume [tex] m [/tex] is the rest mass.

    (new edit)

    okay, i just worked it out and there is no need for [tex] |v| << c [/tex].

    [tex] E^2 = \frac{m_0^2 c^4}{1-\frac{v^2}{c^2}} [/tex]


    [tex] E^2 \left( 1-\frac{v^2}{c^2} \right) = m_0^2 c^4 [/tex]

    [tex] E^2 -\frac{E^2 v^2}{c^2} = m_0^2 c^4 [/tex]

    [tex] E^2 -\frac{(m c^2)^2 v^2}{c^2} = m_0^2 c^4 [/tex]

    [tex] E^2 - (m v c)^2 = m_0^2 c^4 [/tex]

    [tex] E^2 - p^2 c^2 = m_0^2 c^4 [/tex]

    [tex] E^2 = m_0^2 c^4 + p^2 c^2[/tex]

    still, [tex] E = m c^2 [/tex] is the "full" equation if [tex]m[/tex] is relativistic mass.

    r b-j
    Last edited: Jun 14, 2005
  14. Jun 14, 2005 #13
    In that 'derivation' you're assuming that [itex]E=mc^2[/itex] to begin with. Einstein derived it by looking at momentum carrying photons being emitted and adsorbed.

    Anyway, I can't see how your final equation can possibly become [itex]E^2=m^2c^4+p^2c^2[/itex] (note this is Lorentz invarient - it holds in any frame).
  15. Jun 15, 2005 #14
    Well when u derive E=mc² u begin by integrating the kinetic energy, the actual start point is KE=1/2 mv²...
  16. Jun 15, 2005 #15
    That will suffice
  17. Jun 15, 2005 #16
    He actually starts with the relativistic mass = m.

    [tex]E = m c^2
    = \gamma m_0 c^2[/tex]
  18. Jun 15, 2005 #17


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    no, James. the assumption that i started with was the relativistic expression of kinetic energy:

    [tex] T = m c^2 - m_0 c^2 [/tex]

    where [itex] m [/itex] is the relativistic mass. do you disagree with that? will i have to derive that? (it's a b!tchier job than i want to do.) then we interpret that to be

    [tex] T = E - E_0 [/tex]


    [tex] E = E_0 + T [/tex] .

    are you okay with that? BTW, it ain't hard to show that

    [tex] T = m c^2 - m_0 c^2 [/tex]


    [tex] T = \frac{1}{2} m_0 v^2 [/tex]

    in the limit as [itex] |v| << c [/itex].

    perhaps ... i dunno. i am only reverberating what is in my "modern" physics textbook and it's a treatment that i understand. i dunno how they teach it now.

    remember, my [itex]m[/itex] ain't the same as your [itex]m[/itex]. your [itex]m[/itex] is my [itex]m_0[/itex]. all's i was doing was showing that

    [tex]E^2 = m_0^2 c^4 + p^2 c^2[/tex]

    is consistent with

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

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

    and [tex] p = m v [/tex].

    r b-j
  19. Jun 15, 2005 #18
    "the assumption that i started with was the relativistic expression of kinetic energy"

    You said it yourself. This isn't a derivation of [itex]E=mc^2[/itex] as you started with that!
  20. Jun 15, 2005 #19


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    but it's not what you said. you said that i started with the assumption [itex]E = m c^2[/itex] and i said that i started with [itex]T = m c^2 - m_0 c^2[/itex].

    you said that there was no way you could see how i could get [itex] E^2 = m^2 c^4 + p^2 c^2 [/itex] from [itex]E = m c^2[/itex] and i said that did get [itex] E^2 = m_0^2 c^4 + p^2 c^2 [/itex] from [itex]E = m c^2[/itex] and i showed how.

    no, i started with [itex]T = m c^2 - m_0 c^2[/itex] and i never attempted to prove it. i did not start with [itex]E = m c^2[/itex].

    i dunno if you're intending to prop a straw man up or not, but where can you quote me where i said i was deriving [itex]E = m c^2[/itex]?

    my only intention in adding to this thread was to point out that dimensional analysis was insufficient to point out why it's [itex]E = m c^2[/itex]. there are a zillion incorrect physical equations that are dimensionally correct. if you're gonna assume that it's [itex]E = m^a c^b[/itex], then dimensional analysis is sufficient to show that [itex]a = 1[/itex] and [itex]b = 2[/itex], and, perhaps that is sufficient to answer the OP's question. but it doesn't answer why it's [itex]E = m c^2[/itex].

    dimensional analysis can really only be used to prove that some physical equation is wrong. it is not sufficient to prove that any physical equation is right. if a derived equation comes out to be dimensionally inconsistent across "=" or "+" or "-" signs, then you know something is wrong and you may as well not proceed further with it until you fix the problem.

    if you want a derivation for [itex]E = m c^2[/itex] or for [itex]T = m c^2 - m_0 c^2[/itex], i might suggest getting a textbook. i'm too lazy to.

    perhaps, James, you would be kind enough to post a derivation.

    r b-j
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