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A usage of kinetic energy in proof for E=mc^2

  1. Jul 2, 2017 #1

    georg gill

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    aWc4dW95.png
    http://bildr.no/view/aWc4dW95

    Above is a question that I posted on a school help site. This is the answer I got:

    https://www.scribd.com/document/352...usage-of-E-mc-2-for-v-much-smaller-then-c-ans

    Can you use this answer to show that the type of calculation that I tried in the attachement above where I got 26.5 and 27 would hold? Or if not. Why is the document an answer to my question. Can you put some of the formulas from the answer in use practically to the problem I posed where I ended up with 26.5 and 27?
     

    Attached Files:

  2. jcsd
  3. Jul 2, 2017 #2

    mfb

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    Masses and velocities should have units. Unless you use v/c, but then the velocity cannot be 1 or 2.

    The derivation is a bad piece of mathematics and physics at the same time. It uses "mass" in a way that fell out of use more decades ago, it gives mathematicians headaches in the way the derivatives are treated, and the limit-taking procedure is simply wrong. "One factor of this product goes to zero" is not sufficient to show that the product goes to zero - especially as the other factor diverges.

    Use ##E=\gamma m v^2##. That is the correct equation. m is the invariant mass.
     
  4. Jul 2, 2017 #3
    You have ##m_1<m_0## when ##v_1>v_0##. I don't understand how that's possible if ##m## is the relativistic mass.

    I have a hard time following what you're doing because your attachments are too small for me to read and your second link takes me to a textbook section, not your attempt at a solution. I suggest you state the problem using LaTex so we can read it. As @mfb states, the outdated concept of relativistic mass is used. Are you interested in solving the problem as stated or are you trying to understand it in a more general sense? If the former then I guess you're forced to use the outdated concept of relativistic mass because the author is using it.
     
  5. Jul 3, 2017 #4

    georg gill

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    I will try to rewrite it here:

    First this is the derivation that I found online:

    Consider a body of mass m that moves at a speed v very close to the speed of light. A force F acts on it and, as a result, the energy E of the body increases. The speed of the body v cannot exceed c and, as the force continues to act, the speed v approaches c asymptotically. We have:

    $$\frac{dE}{dt}=Fv=\frac{d(mv)}{dt}v=v^2\frac{dm}{dt}+mv\frac{dv}{dt}$$

    In this last equation, t is the independent variable. So the equation tracks how energy E(t), mass m(t) and speed v(t) grown as a function of time t.

    Our concern, however, is to see how mass grows as we increase the energy of the body. To track that, we need to make energy E the independent variable; and to take time t(E), mass m(E) and speed v(E) all to be functions of E. Multiplying the last equation by

    $$\frac{dt(E)}{dE}$$


    and using the chain rule, we recover:

    $$1=v^2 \frac{dm}{dt} \frac{dt}{dE}+mv\frac{dv}{dt}\frac{dt}{dE}=v^2 \frac{dm}{dE}+mv\frac{dv}{dE}$$

    Rearranging:

    $$\frac{dm}{dE}=\frac{1}{v^2}-\frac{m}{v}\frac{dv}{dE}, (a)$$

    We now take the limit in which the energy E grows large. In that limit, v approaches c asymptotically and

    $$\dfrac{dv}{dE}$$
    approaches zero. In this limit we have:

    $$\dfrac{dm}{dE}= \dfrac{1}{c^2}$$

    My construct:

    I start at (a)

    $$\dfrac{dm}{dE}= \dfrac{1}{v^2}- \dfrac{m}{v}\dfrac{dv}{dE}$$

    rearranging:

    $$v^2dm+mvdv=dE $$

    $$E=[m]_{m_a}^{m_b}v_a^2+\dfrac{1}{2}[v^2]_{v_a}^{v_b}=m_b v_a^2+\dfrac{1}{2}m_av_b^2-m_a\dfrac{3}{2}v_a^2$$
    If $$m_b=10, m_a=9, v_b=3, v_a=2$$

    $$m_b\dfrac{1}{2}v_b^2-m_a\dfrac{1}{2}v_a^2=27$$

    $$m_b v_a^2+\dfrac{1}{2}m_av_b^2-m_a\dfrac{3}{2}v_a^2=26.5$$

    So I can not get this energy construct that I have made. When I asked a phd on a homework site I got this answer:

    https://www.scribd.com/document/352...ft500noi&source=impactradius&medium=affiliate

    Can anyone help me explain my problem about getting 26.5 and 27 for the same mass and velocity changes. Or clarify why this answer from the homework site was proving it wrong or right?
     
  6. Jul 3, 2017 #5

    PeterDonis

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    You are trying to integrate the LHS term by term, but you are doing it incorrectly. For example: you are using ##\int v^2 dm = v^2 m## (evaluated at the limits), but ##v## is a function of ##m## (because ##v## increases as ##m## increases), so you need the function ##v(m)## in order to do the integration. Similarly, in the second term, ##m## is a function of ##v##, so you need the function ##m(v)## to do the integration.

    Also, when you calculate the 27 answer, you are using the formula ##(1/2) m v^2## for kinetic energy; but that formula is not correct in relativity, even if you interpret ##m## as the relativistic mass. The correct formula, if you insist on using relativistic mass, is

    $$
    E_\text{kinetic} = m \left( 1 - \frac{1}{\gamma} \right) = m \left( 1 - \sqrt{1 - \frac{v^2}{c^2}} \right)
    $$

    Your real misconception, though, comes much earlier:

    You are using "mass" to mean "relativistic mass"--but relativistic mass is the total energy of the body. If you insist on just looking at kinetic energy, it is relativistic mass minus rest mass (that is what the formula I wrote above says), but the change in kinetic energy is the same as the change in total energy, so it is the same as the change in relativistic mass. So you are trying to go on a long roundabout path (and making missteps along the way) to get an answer that is just an obvious consequence of the definition of relativistic mass.
     
  7. Jul 3, 2017 #6
  8. Jul 3, 2017 #7
    As I do not see any justification for these assumptions this seems to be a postulate just to get the correct result. If this is allowed why not starting with

    [itex]F \cdot ds = dE: = k \cdot m[/itex]

    (where k is a constant that needs to be derived) or even E = m·c²?
     
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