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E=mc^2 and it's derivation

  1. Jan 1, 2007 #1
    I'm just a highschool student and thus not particularly knowledgeble about special relativity. However, I've always wanted to find a derivation of the famous equation E=MC^2. I could never find one in textbooks or on the web that didnt rely on oversimplifications or unexplained steps, so I tried to formulate one myself. I already knew the simple stuff, time dialation equations/relativistic mass stuff. That's easy to find. I figured it should be possible to derive E=MC^2 from that + classical mechanics stuff, so I went ahead and tried. To my astonishment, I think I may (emphasize may) have succeded. And so I was wondering if someone could take a look at my logic and tell me if it's valid:

    I started with two equations, M2=M1[1/ √(1-V^2/C^2)] (relativistic mass equation) and KE=1/2(MV^2) (classical kinetic energy equation). I realized one could, by solving for V^2 in the first equation and plugging that into the latter (and recognizing that M2= M1+ΔM), obtain an equation for kinetic energy based on initial mass and the change in the mass due to the effects of relativity. This comes out to be 2KE= ΔMC^2(2+ ΔM/M1) after some algebraic manipulation. So what I then realized is that, because V^2=[1-(M1/{M1+ ΔM})^2] from the original equation, as M1 approaches infinity the energy imparted into an object results more and more solely in a relativistic mass increase because V approaches 0. Thus kinetic energy imparted into an object with "infinite" rest mass results in 0 change in velocity, meaning all that imparted energy results in a mass increase. This is what we want, a relation between mass and energy alone, regardless of velocity. So, if we take the limit of the equation formulated for KE based on M1 and ΔM as M1 approaches infinity, we get 2KE= ΔMC^2(2+0). This simplifies to KE= ΔMC^2, or more properly E=MC^2.

    (Hopefully that made sense. I can elaborate on it a little more if isn't clear enough)

    So, is that a valid "proof" or not? Thanks in advance for any replies.
     
  2. jcsd
  3. Jan 1, 2007 #2
    Probably not - I didn't actually read what you did because it was a bit messy. E=mc^2 is actually just a generalization for a particle with no kinetic energy.

    For the derivation you have to use the work energy theorem.

    [tex]W = \int_{x_1}^{x_2} F dx = \int_{x_1}^{x_2} \frac{dp}{dt} dx[/tex]

    If you have done Calculus it becomes an easy u-sub problem.

    Eventually you will get that

    [tex]K = \gamma mc^2 - mc^2[/tex]
    where
    [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]
     
  4. Jan 1, 2007 #3
    Ah, thank you. I figured it was some simple calc method, but I had no idea how to go about that.

    I would still like to know if my argument is valid. Basically what I'm saying is that you can formulate an equation for kinetic energy based on mass and relativistic mass. Thats step one. The second is to figure the relationship for energy and mass, ignoring rest mass and kinetic energy. Taking that equation as a limit as rest mass approaches infinity (and thus velocity to zero) essentially rids us of kinetic energy because there is none. An infinite mass cannot gain kinetic energy. It can only gain mass energy. If we take that limit, the equation pans out to be E=MC^2...which seems so perfect.

    I would imagine that, if there is a problem, it involves one of these steps: Applicability of classical Kinetic energy equation, the limit process, or my understanding of the concept. It just seems too weird to have worked out that way, which makes me think its true enough . It's admittedly an ugly work around at best, and not terribly rigorous, but is it logically consistent?
     
  5. Jan 1, 2007 #4
    Looking at your start, you might have a misunderstanding with relativistic mass. It looks like you thought that a mass has an initial mass, m_0 or m1 as you have it, and that the relativistic mass is additive, where relativistic mass varies by the factor of gamma. I don't follow where you get the +∆M from, the mass-energy relationship you are trying to make. Another problem I see is that there isn't a physical reason for your rest mass to approach infinity.

    From the equation for kinetic energy you can get to Einstein's famous eq:

    [tex]K+mc^2=\gamma m c^2[/tex]

    The left side is said to be the total energy.

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

    So, for a particle at rest you find that

    [tex]E=m_0 c^2[/tex]
     
  6. Jan 1, 2007 #5
    Yeah I know it varies by gamma, I used delta M to represent the difference between an object's relativistic mass due to some velocity and it's rest mass. I am simply using it because I want the equation to come out in terms of delta M and the rest mass. Maybe I should re-explain this, step by step.
    KE=(1/2)mV^2
    m2=m1[1/(1-{V/C}^2)]^(1/2)
    From this we can get V^2 in terms of M2 and M1:
    (V^2)=(C^2)[1-{m1/m2}^2] Fair enough so far, right?
    We could then say that an object moving has KE based on its velocity and relativistic mass (right?):
    KE=(1/2)(m2)V^2 Becomes (with V^2 substituition):
    2KE=m2[1-{m1/m2}^2]C^2
    So here we have Kinetic energy of an object as a function of rest mass and relativistic mass. For my purposes I don't want relativistic mass in the equation, I want delta M, the change in the mass. Substituting that fact in to get rid of m2, and doing some algebra gives:
    2KE= ΔMC^2(2+ ΔM/m1)
    This equation would tell you the kinetic energy of an object as a function of rest mass and change in the mass (defined as relativistic mass minus rest mass)
    *Just at this point, the equation looks suspiciously close to E=MC^2.

    Assuming that is valid, we just need to take one more step. From previous equations it is shown that as rest mass m1 approaches infinity, velocity approaches 0. Let me put it this way: When a tiny object is imparted with energy, most of the energy goes into velocity and a tiny bit goes into a relativistic mass increase. When a large object is imparted with energy, its velocity increases less and its mass increases more. Taken to the limit then, the above equation simplifies to a direct relationship between mass and energy because at the so called point at infinity all energy imparted into an object would do is cause a relativistic mass increase, no velocity increase. (Yes, I understand that this point at infinity does not exist. That's why I used a limit). My point is that the equation should become increasingly just a relationship between mass and energy as m1 approaches infinity. And it does. The equation becomes:
    2KE=ΔMC^2(2+0) More properly written:
    2E=2ΔMC^2 Simplifies to:
    E=ΔMC^2 Which basically states:
    E=MC^2

    Again, what I'm saying is the KE equation in terms of delta M and m1 tends asypmtotically towards a relationship between Energy and mass change. That relationship comes out to be, unsurprisingly, E=mc^2. So, it would seem that this is a reach around way to derive the equation.

    Hopefully that was clearer. Again, I realize its probably a rather idiotic way of obtaining it, but I still don't see whats wrong with it. It seems to work out so well. But of course I would hardly be surprised if I made a grave error. I just want to know specifically what that error was.

    Thanks again.
     
  7. Jan 1, 2007 #6
    It could be right, having learned the work-energy derivation I might forever be partial to it. A good deal of physics problems can be done with more than one approach but I'm still hung up on the masses.

    [tex]m'=\gamma m_0 c^2[/tex]

    [tex]\Delta m = m' - m_0[/tex]

    [tex]\Delta m = \gamma m_0 c^2 - m_0[/tex]

    and from here you solved for v and put it into kinetic energy?
     
  8. Jan 2, 2007 #7
    I did that yeah, but in a different order. I got rid of m' (or m2)*at the end, but same thing. Probably should have done it your way, would have been clearer. Sorry.

    And yes, the work energy theorem derivation you showed is obviously vastly superior. Thank you for showing it to me. As I said, I'm just curious if this method (while perhaps uglier) is still acceptable in some manner.

    *Sorry about the notation as well. I'm not as well versed in proper notation as I should be. Also, I am constrained by the limits of my keyboard and MS word level symbols.
     
  9. Jan 2, 2007 #8
    Wait. I thought the equation was [tex]m'=\gamma m_0[/tex] Where did the c^2 come from? Isn't the relativistic mass proportional only to gamma and the invarient/rest mass?
     
  10. Jan 2, 2007 #9
    Oh, you are right that there shouldn't be a c^2, I obviously don't remember special relativity all that well.

    I dunno, could be valid. Anyone else have an input?
     
  11. Jan 3, 2007 #10
    I thought you can use the energy-momentum 4 vector?
     
  12. Jan 3, 2007 #11
    That's how I'd do it.

    Define the four-momentum as (with [itex]u^\mu[/itex] understood to be the four-velocity)

    [tex]p^\mu \equiv [E/c, p_x, p_y, p_z] \equiv m_0 u^\mu = \gamma(v) m_0 [c, u_x, u_y, u_z].[/tex]
    The length of this four-vector is undeniably an invariant:

    [tex]p^\mu p_\mu = g_{\mu\nu}p^\mu p^\nu = \frac{E^2}{c^2}-p_x^2 - p_y^2 - p_z^2.[/tex]​

    Now transform to the rest frame of the object: All the components of velocity and momentum vanish. So we have (from the definition of four-momentum as the rest mass multiplied by the four-velocity):

    [tex]
    \begin{align*}
    \frac{E^2_\begin{large}\mbox{rest}\end{large}}{c^2} &= \gamma(0)^2 m_0^2 c^2 \\
    E^2_\begin{large}\mbox{rest}\end{large} &= m_0^2 c^4 \\
    E_\mbox{\begin{large}rest\end{large}} &= m_0 c^2.
    \end{align*}
    [/tex]​

    Incidentally, once we've shown this, we argue that since it's an invariant, its value is the same in any frame; i.e. this is the length of four-momentum in any frame. Therefore we have the oft-used relation:

    [tex]
    \frac{E^2}{c^2}-p_x^2 - p_y^2 - p_z^2 = m_0^2 c^4.
    [/tex]​

    Incidentally (again), this can also be derived by using the definitions of energy and momentum I have implied above, but not implicitly stated: namely that [itex]p \equiv \gamma m_0 v[/itex] and that [itex]E \equiv \gamma m_0 c^2[/itex], and being careful to factor out a [itex]\gamma^2.[/itex]
     
    Last edited: Jan 3, 2007
  13. Aug 19, 2008 #12
    Re: E=mc^2

    I'm deriving the relation e=mc^2
    Is my working out here valid:

    Firstly I start with the formulas KE = mv^2/2 and p=mv

    Then I assume that in separate frames of reference, what may look like a static object to one observer may appear to move with a certain velocity to another. This raises the question 'what
    absolute measure for m can we formulate if not a single frame of reference allows us to measure it?'

    I then assume that:
    1. A mass has a fixed inertial energy regardless of its velocity.
    2. The total energy in a system where mass is present involves the addition of inertial mass and kinetic energy.

    From here I transpose the relativistic mass/velocity relation m=m0/(1-(v/c)^2)^0.5 to give me m0 (inertial mass/energy) = m*(1-(v/c)^2)^-0.5

    I then expand the above right-hand side (using binomial expansion) to obtain the first two factors to give me:

    m0=m*(1 + 1/2*(v/c)^2.....

    Multiplying both sides by c^2, and then multiplying out the right-hand side brackets by m gives me:

    m0c^2 = mc^2 + mv^2/2...

    Here I see an equation that appears to provide the total energy system possessed by a system involving inertial mass and kinetic energy. When the velocity is allowed to go to zero we are left with:

    m0c^2=mc^2

    And I infer that we are left with the inertial energy of mass whereby m0=m, and finally
    E=mc^2

    Is this reasonable?
     
  14. Aug 19, 2008 #13

    Fredrik

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    Gold Member

    Re: E=mc^2

    Check out post #15 in this thread for a derivation of E=mc2.
     
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