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Derivation of relativistic energy

  1. Feb 10, 2005 #1
    I am now currently using an introductory modern physics textbook but they did not give me the derivation for relativistic KE for a moving object which is [tex] \gamma mc^2 [/tex]. Is the derivation too difficult to be put in the introductory text?? Anyway, what is its derivation? Thanks alot.
     
  2. jcsd
  3. Feb 10, 2005 #2

    jtbell

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    Staff: Mentor

    Some books at that level do have that derivation, but it takes a bit of fancy footwork with calculus. Basically, you start with an object at rest, integrate the work-energy theorem, apply the form of Newton's Second Law that says F = dp/dt, and use relativistic momentum:

    [tex]K = \int {F dx} = \int {\frac {dp}{dt} dx} = \int {\frac {dx} {dt} \frac {dp}{dv} dv} = \int {v \frac {dp}{dv} dv} = \int {v \frac {d}{dv} ( \gamma mv ) dv } [/tex]

    Integrate from 0 to v. The result is [tex]K = (\gamma - 1) mc^2[/tex], not [tex]\gamma mc^2[/tex].
     
  4. Feb 10, 2005 #3
    Let the following mathematical formula be true, in any inertial reference frame, and only inertial reference frames, regardless of whether or not the theory of relativity is correct.

    [tex] \vec F = \frac{d \vec P}{dt} = \frac{d (M\vec v)}{dt} [/tex]

    P is the momentum, M is inertia, and v is the velocity of the center of inertia in some inertial reference frame.

    Now, regard the following as a postulate of relatativistic dynamics:

    Postulate: The inertial mass of an object is a function of its speed in a frame, and is given by:

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

    Where c is a numerical physical constant, with units of speed, and is equivalent to 299792458 meters per second. Thus, dc/dt =0, no matter what frame the derivative is taken in. The parameter m0, called the rest mass, is the inertial mass of a body, as measured while it isn't moving, and hence cannot depend upon v.

    Now, substitute the relativistic mass formula for M, and differentiate.

    [tex] \vec F = d/dt (\frac{m_0}{\sqrt{1-v^2/c^2}} \vec v)}[/tex]

    Assuming the rest mass m0 cannot change in time for any reason (for example if the rest mass depends on temperature then the rest mass could vary in time), we can pull it outside of the derivative to obtain:

    [tex] \vec F = m_0 d/dt (\frac{1}{\sqrt{1-v^2/c^2}} \vec v)} [/tex]

    The derivative with respect to time of [tex] (1-v^2/c^2)^{-1/2} [/tex] is given by:

    [tex] (-1/2) (1-v^2/c^2)^{-3/2} ( \frac{-2v}{c^2} dv/dt) = (1-v^2/c^2)^{-3/2} ( \frac{v}{c^2} dv/dt) [/tex]

    Now, use the product rule to to obtain the following:

    [tex] \vec F = m_0 [(1-v^2/c^2)^{-3/2} ( \frac{v \vec v}{c^2} dv/dt)
    +(\frac{1}{\sqrt{1-v^2/c^2}} \frac{d\vec v}{dt})} ] [/tex]

    Now, assume that the acceleration is in the same direction as the velocity, to obtain the following formula:

    [tex] \vec F = m_0 [(1-v^2/c^2)^{-3/2} ( \frac{vv}{c^2} d\vec v/dt)
    +(\frac{1}{\sqrt{1-v^2/c^2}} \frac{d\vec v}{dt})} ] [/tex]

    The quantity [tex] \frac{d \vec v}{dt} [/tex] is just the acceleration [tex] \vec a [/tex] so that we have:

    [tex] \vec F = m_0 [(1-v^2/c^2)^{-3/2} ( \frac{vv}{c^2} \vec a)
    +(\frac{1}{\sqrt{1-v^2/c^2}} \vec a)} ] [/tex]

    Now factor out the acceleration, and write vv as [tex] v^2 [/tex] to obtain:

    [tex] \vec F = m_0 \vec a[(1-v^2/c^2)^{-3/2} ( \frac{v^2}{c^2} )
    +(\frac{1}{\sqrt{1-v^2/c^2}} )} ] [/tex]

    Now factor out (1-v^2/c^2)^-1/2, and write [tex] \frac{m_0}{\sqrt{1-v^2/c^2}} [/tex] as M, to obtain:

    [tex] \vec F = M \vec a[(1-v^2/c^2)^{-1} ( \frac{v^2}{c^2} ) + 1} ] [/tex]

    Which is equivalent to:

    [tex] \vec F = M \vec a[\frac{v^2}{c^2-v^2} + 1 } ] [/tex]

    Which is equivalent to:

    [tex] \vec F = M \vec a[\frac{c^2}{c^2-v^2} } ] [/tex]

    Which is equivalent to:

    [tex] \vec F = M \vec a[\frac{1}{1-v^2/c^2} } ] [/tex]

    Which is eqivalent to:

    [tex] \vec F = m_0 \vec a (1-v^2/c^2)^{-3/2} [/tex]

    Now, let the following mathematical formula correctly define the kinetic energy of a body of inertial mass M, regardless of whether or not the relativistic formula is correct:


    [tex] T = \int d \vec P \bullet \vec v [/tex]

    And since v = dr/dt we also have:

    [tex] T = \int d \vec P \bullet \frac{d\vec r}{dt} [/tex]

    Which is equivalent to:

    [tex] T = \int \vec F \bullet d\vec r [/tex]

    Case 1: M is not a function of v.

    In this case we have:

    F = dP/dt = d(Mv)/dt = Mdv/dt+vdM/dt = Ma+v dM/dt.

    If we presume that the inertial mass of an object cannot change in time for any reason at all, then we have:

    F = Mdv/dt, and if we substitute this into the expression for T above, we have:

    [tex] T = \int M \frac{d\vec v}{dt} \bullet d\vec r [/tex]
    Which is equivalent to:

    [tex] T = M \int v d\vec v [/tex]

    And now, if we switch to an indefinite integral we have:

    [tex] \Delta T = M \int^{v2}_{v1} v d\vec v [/tex]

    Which leads to:

    [tex] \Delta T = \frac{M(v2)^2}{2} - \frac{M(v1)^2}{2} [/tex]

    From which we can infer that:

    [tex] T = \frac{Mv^2}{2} [/tex]

    Noticing that if v=0 then T=0 we can write:

    [tex] T = \int^{V}_0 \vec F \bullet d\vec r [/tex]


    Now, use the relativistic expression for force instead, to obtain:

    [tex] T = \int m_0 \vec a (1-v^2/c^2)^{-3/2} \bullet d\vec r [/tex]

    In the case where the center of mass is accelerated from rest, the direction of its acceleration will be identical to the direction of [tex] d\vec r [/tex], so that we have:

    [tex] \vec a \bullet d\vec r = |\vec a| |dr| cos (0) = adr [/tex]

    Where a is the magnitude of the acceleration.

    Hence we have:

    [tex] T = \int m_0 a (1-v^2/c^2)^{-3/2} dr [/tex]

    And the magnitude of the acceleration is dv/dt, so that we have:

    [tex] T = \int m_0 (dv/dt) (1-v^2/c^2)^{-3/2} dr [/tex]

    From which it follows that:


    [tex] T = \int m_0 (dv) (1-v^2/c^2)^{-3/2} (dr/dt) [/tex]

    And dr/dt is just the speed v of the center of inertial mass, so that we obtain:

    [tex] T = \int m_0 (1-v^2/c^2)^{-3/2} v dv [/tex]

    Now, let us change from an indefinite integral to a definite integral. Thus, we have:

    [tex] \Delta T = \int^{v2}_{v1} m_0 (1-v^2/c^2)^{-3/2} v dv [/tex]

    Stipulate that if the center of mass of an object is at rest in a frame, that T=0. Now let the initial speed v1 be zero, therefore we have:

    [tex] T = \int ^{v2}_0 m_0 (1-v^2/c^2)^{-3/2} v dv [/tex]

    Now, just perform the integral. Let U = (1-v^2/c^2), so that dU = (-2vdv/c^2), from which it follows that vdv = (-c^2/2) dU, and make substitutions. After pulling m0 outside of the integral, you obtain the following:

    [tex] T = -m_0 \int ^{v2}_0 U^{-3/2} \frac{c^2}{2} dU [/tex]

    And we can pull c^2/2 out, to obtain:

    [tex] T = \frac{-m_0 c^2}{2} \int ^{v2}_0 U^{-3/2} dU [/tex]

    Which is simple to integrate. After integrating we have:

    [tex] T = \frac{-m_0 c^2}{2} [ -2U^-1/2]^{v2}_0 [/tex]

    Factoring out the -2 we have:

    [tex] T = m_0 c^2 [U^-1/2]^{v2}_0 [/tex]

    And restoring U to what it originally was we have:

    [tex] T = m_0 c^2 [\frac{1}{\sqrt{1-v^2/c^2}}]^{v2}_0 [/tex]

    Which is equivalent to:

    [tex] T = m_0 c^2 [ \frac{1}{\sqrt{1-(v2)^2/c^2}} - \frac{1}{\sqrt{1-0^2/c^2}} ] [/tex]

    Which is equivalent to:

    [tex] T = m_0 c^2 [ \frac{1}{\sqrt{1-(v2)^2/c^2}} - 1 ] [/tex]

    Now, just replace v2 by V, where V is the final speed of some object which is being accelerated in some inertial reference frame. Thus, we have:

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

    And now, if you distribute m0, you can write the first term inside the brackets as the relativistic inertial mass M, so that you get:

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

    Now, distribute c^2 to obtain:

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

    Now, add m0c^2 to both sides to obtain:

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

    Let [tex] E_0 = m_0 c^2 [/tex]
    Let [tex] E = Mc^2 [/tex]

    Therefore we have:

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

    When the body is at rest, T=0, so that the total energy of the body is equal to E0, which is called the rest energy. When a body is moving at speed v in some frame, the total energy of the body is equal to the sum of its rest energy E0, and its kinetic energy T. Thus, we can interpret the rest energy E0 as potential energy, so that the total energy E is equal to the sum of the object's potential energy and kinetic energy.

    Regards,

    Guru
     
    Last edited: Feb 10, 2005
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