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Where is my error?

  1. Jun 6, 2004 #1


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    I was trying to get to the equation E=mc^2, but I can't seem to.

    I started with the example of a moving observer with a stationary observer with a beam of light.. (you know, c^2*t^2 = c^2*t'^2+t^2*v^2) and isolated v^2:

    v^2 = c^2(t^2-t'^2)/t^2 = c^2(1-(t'/t)^2)

    Then took the formula for kinetic energy: E(k) = mv^2/2 and replaced v^2:

    E(k) = m*c^2*(1-(t'/t)^2)/2

    Now.. I got the right equation except that for it to be correct (1-(t'/t)^2)/2 = 1 so 1-(t'/t)^2 = 2, (t'/t)^2 = -1, t'/t = sqr(-1)

    ...I think it's pretty obivous I made a mistake somewhere... but where?
  2. jcsd
  3. Jun 6, 2004 #2
    If you really intend for mv2/2 to equal mc2, then you are in fact saying that v2 = 2c2, which makes v approximately equal to 1.4c. Oops! v is larger than c and outside the range of standard special relativity. Substitute v2 = 2c2 back into your original equation and you will get (t')2 = -t2 and your unfortunate conclusion.

    I believe you are wrong to attempt pulling mc2 out of mv2/2, tempting as it may look. In fact, mv2/2 comes out of mc2 as a single term of an expansion. To be totally correct, the kinetic energy in special relativity is

    mc2(1 - 1/SQRT(1 - v2/c2))

    This expands in powers of v/c as

    mc2(v2/2c2 + 3v4/8c4 + 5v6/16c6 + ...)

    and so on to more terms. Multiply mc2 through this series to yield

    mv2/2 + 3mv4/8c2 + 5mv6/16c4 + ...

    . If v is tiny enough compared to c, then only the first term really counts. Otherwise, it is just one term of a series for the kinetic energy.


    I will (later) try to produce a demonstration of E=mc2 here that is similar to your attempt, but uses another fact of special relativity.
  4. Jun 6, 2004 #3


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    I think your mistake is conceptual. E = mc2 is not a statement of kinetic energy; it is either a statement of inherent energy (if m is the rest mass) or total energy (if m is the relativistic mass).

    AFAIK, the famous relationship can be derived from conservation of 4-momentum, identification of the time-like part of the momentum with energy, and the idea that a contraction of two 4-vectors forms an invariant scalar.
  5. Jun 6, 2004 #4
    Get to E=mc2.

    Suppose we have a particle of mass m0 standing still at time t=0. From that time on it gets accelerated continuously along the positive x line. We will represent its instantaneous speed as a variable v, function of time t.

    Suppose the initial energy of the particle is E0 and the instantaneous energy of the accelerated particle is represented by variable E, function of time t.

    We define a variable m of time t as m0/SQRT(1 - v2/c2). It has the dimension of a mass. It is usually called the relativistic mass. For those who don't like relativistic mass and say that the particle always has an invariant mass m0, I assuage them by calling m just a defined variable. Its purpose is to allow us to represent the magnitude of the momentum of a particle in the way Newton did it: p = mv. This is then just p = m0v/SQRT(1 - v2/c2), and I think all agree that this variable is the instantaneous magnitude of the momentum of our particle at time t.

    In fact I will get a little more extra mileage from having this defined m variable around, as you will presently see.

    Start with the following equation of motion:

    dE/dt = vdp/dt

    . This is close to the work definition of energy gained on the particle and I think it can be obtained by a change of integration variable.

    Now p = mv, so

    dE/dt = 1/m*(pdp/dt)

    . pdp/dt = 1/2*d(p2)/dt

    , so

    2mdE/dt = d(p2)/dt

    . This is our dynamic equation.

    I think this equation is perfectly general for the newtonian as well as the relativistic case. I haven't used the actual definition of the variable m yet. Suppose I replace my definition of m from the earlier paragraph to a simple m = m0. Then the variable m does NOT depend on the speed v of the particle at any time t from 0 on. This means dm/dt = 0.

    newtonian case:
    For m = m0 (at all times), we can divide by nonzero 2m and get

    dE/dt = 1/2m*d(p2)/dt = d(p2/2m0)/dt

    . Then

    E - p2/2m0 = constant-in-time

    . Since at t=0 E=E0 and p = 0,

    E - p2/2m0 = E0 - 02/2m0

    . lastly,

    E = E0 + p2/2m0 = E0 + 1/2*m0v2

    . This case gives us no clue what the initial energy E0 is supposed to be.


    Sorry for the derail to the newtonian m = m0 case! I just felt compelled to show how the familiar kinetic energy term comes out of the dynamic equation.

    relativistic case:
    For m = m0/SQRT(1 - v2/c2) at all times,

    Square this.

    m2 = m02/(1 - v2/c2)

    Multiply by that denominator and then by c2 to get

    m2c2 - m2v2 = m02c2

    p2 = m2v2

    from the p = mv definition, so

    m2c2 - p2 = m02c2

    , and then

    p2 = m2c2 - m02c2



    d(p2)/dt = d(m2c2)/dt - d(m02c2)/dt

    d(p2)/dt = d(m2c2)/dt

    , since dm0/dt and dc/dt equal 0.

    d(p2)/dt = d(m2)/dt*c2 = 2m*dm/dt*c2

    , since dc/dt is 0.

    Remember the dynamic equation:

    2mdE/dt = d(p2)/dt

    . Substitute for d(p2)/dt:

    2mdE/dt = 2m*dm/dt*c2

    Divide by nonzero 2m

    dE/dt = dm/dt*c2

    dE/dt = d(mc2)/dt

    , since dc/dt = 0. Then

    E - mc2 = constant-in-time

    At t=0,

    E0 - m0c2 = constant-in-time

    E - mc2 = E0 - m0c2

    E = E0 + (m - m0)c2


    NOTE: If E0 were to equal m0c2, then

    E would equal mc2 at each time t. That is a hint and NOT a proof, but very seductive.

    In that particular case, the particle has an initial resting energy E0 and hence a kinetic energy of

    E - E0 = m0(1 - 1/SQRT(1 - v2/c2)).
    Last edited: Jun 6, 2004
  6. Jun 7, 2004 #5
    Just a bit of lily-gilding for my previous reply

    I had earlier promised
    It may not be obvious that I did this in my last post. Let me try to show that now.

    In the last post, I derived (for the relativistic case) this equation from the assumption m = m0/SQRT(1 - v2/c2) for all times t:

    p2 = m2c2 - m02c2

    and went on to take the derivative to use in the dynamic equation. Now I will derive something straight away from this equation.

    p2 = m2v2 from the definition of p, so

    m2v2 = m2c2 - m02c2

    . Rearrange,

    m2c2 = m02c2 + m2v2


    c2m2 = c2 m02+ v2m2

    . Let m' be just another way to write m0 (particle rest frame value for particle mass).

    c2m2 = c2 m'2+ v2m2
    This looks just like Alkatran's

    c2t2 = c2 t'2+ v2t2

    , except m's (mass variables) occur where t's (time variables) occurred.



    I read somewhere that m/t = m'/t' is an invariant of interest that might be used to prove the dynamical relation

    m = m0/SQRT(1 - v2/c2)

    from the kinematical relation

    t = t'/SQRT(1 - v2/c2)

    . But I don't know what that mass/time is supposed to signify.


    continuing ...

    m2v2 = m2c2 - m02c2

    m2v2 = (m2- m02)c2

    Factor difference of squares:

    m2- m02 = (m + m0)(m - m0)


    m2v2 = (m + m0)(m - m0)c2

    Divide by 2:

    m2v2/2 = (m + m0)/2*(m - m0)c2

    Divide by nonzero (m + m0)/2:

    (m2/((m + m0)/2)* v2/2 = (m - m0)c2

    . The relativistic kinetic energy is

    K = (m - m0)c2

    . So,

    K = (m2/((m + m0)/2)* v2/2

    . Define a new mass variable M by m2/((m + m0)/2.


    K = Mv2/2

    . This looks like the classic formula but has a different mass element. Note that (m + m0)/2 means just the average value of m between time 0 and the time t of interest.


    M = m*(m/average-m-value)

    . When m0 < m, then

    m0 < average-m-value < m

    . So

    1 < m/average-m-value < m/m0

    , all of these mass elements having strict positive values.

    So, I have at long last explained Alkatran's puzzle by showing that the kinetic energy of a particle is

    K = Mv2/2 = (m2/((m + m0)/2)* v2/2

    , and that is NOT the same thing as




    If we let let v -> 0, then

    m -> m0


    M -> m02/((m0 + m0)/2) = m0

    , which yields the classical kinetic energy as a limit value:

    K -> m0v2/2

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