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Rest mass

  1. Oct 6, 2007 #1
    If we cant find a frame of total rest, how can we calculate a rest mass?
     
  2. jcsd
  3. Oct 6, 2007 #2
    Of course, there is no absolute frame of total rest. However, one can always find an inertial frame of reference, such that a given body is at rest there (i.e., its velocity is equal to zero). In this reference frame the total energy E of the body and its "rest mass" m are related by Einstein's formula [itex] E = mc^2 [/itex]

    Eugene.
     
  4. Oct 6, 2007 #3

    rbj

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    and that total energy is the "rest energy" of that body since it is at rest in that inertial frame of reference. the "rest energy", E0 = m0 c2, is the energy a body of mass m0 has in it's own frame of reference.

    Einstein's E = m c2 also works in general for the total energy, E, of some body not at rest with respect to the given inertial frame of reference and moving relative to that frame of reference with velocity v. then the total energy, E, of that moving body, as perceived by an observer in that given inertial frame of reference is

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

    where m is the "relativistic mass" (sometimes called "inertial mass" or perhaps "effective mass") and is a term deprecated by many physicists including some here (but not by me). knowing the velocity v, relative to any inertial frame of reference, and rest mass m0 of the body (the mass as measured in the body's own frame of reference), the relativistic mass is

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

    those who deny or deprecate the concept of "relativistic mass" or contrasting this other notion of mass with rest mass (the only mass worthy of note to those deprecating "relativistic mass" and so is labeled "invariant mass", since ,if nothing intrinsically changes about a body, the rest mass or the perceived mass of the body in the body's own frame of reference is always the same, even if the relativistic mass might be different for different observers traveling different velocities relative to that body) agree with the following relating total energy E to rest mass (what they just call "mass) m0 and inertial momentum p.

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

    they view that formula as fundamental whereas those (like me) who do not deprecate "relativistic mass" see it as simply following the relativistic mass expression above and the previously existing definition for momentum:

    [tex] p = m v [/tex]

    so we actually define the effective mass, m as the magnitude of momentum of the body p (that no one disagrees about) as perceived by an observer in an inertial frame of reference, divided by the magnitude of the velocity of the body (that no one disagrees about), again as perceived by an observer in an inertial frame of reference.

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

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

    the difference between the body's total energy E = m c2 and the rest energy E0 = m0 c2 is the kinetic energy of the body:

    [tex] T = E - E_0 = m c^2 - m_0 c^2 [/tex]

    or

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

    or, more fundamentally, the total energy is the rest energy (whatever a body has just for being there) added to the kinetic energy (what the body gets additionally for being in motion):

    [tex] E \equiv E_0 + T [/tex]

    as |v| << |c|, the above exact expression for kinetic energy degenerates to the familiar expression for kinetic energy seen in classical mechanics:

    [tex] T \approx \frac{1}{2} m_0 v^2 = \lim_{v \rightarrow c} m_0 c^2 \left(1 - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right) [/tex]

    that's how i look at this whole thing regarding rest mass vs. mass, etc.
     
    Last edited: Oct 7, 2007
  5. Oct 7, 2007 #4
    relativistic mass (horibile dictu?)

    I am one who does not deprecate the concept of relativistic mass. Happy to see that you are of the same opinion (simils simile gaudet).
    I used to teach the subject as follows. Start with
    p=mu (1)
    in I and with
    p'=m'u' (2)
    in I'. As long as
    u=u'+V (3)
    m and m' represent the absolute Newtonian mass m=m'. The problem is to find out the physics behind m and m' if
    u=(u'+V)/(1+Vu'/cc) (4)
    From (1), (2) and (4) results
    p/m=(p'/m')[(1+V/u')/(1+Vu'/cc)] (5)
    which suggests to consider that
    p=f(V)p'(1+V/u') (6)
    m=f(V)m'(1+Vu'/cc) (7)
    The relativistic arsenal (linerity, reciprocity, symmetry...) leads to
    f(V)=1/sqrt(1-VV/cc).
    If the considered tardyon is at rest in I' (u'=0, p'=0) observers from that frame measure its rest mass m(0) observers from I relative to which it moves with V measure its relativistic mass m(V) related by (7)
    m(V)=m(0)/sqrt(1-VV/cc) (8)
    For those who ban the concept of relativistic mass multiply both sides of (8) with cc in order to obtain
    E(V)=E(0)/sqrt(1-VV/cc) simply taking into account the physical dimensions of m(0)cc and m(V)cc.
    We can express (6) and (7) as a function of E and E' avoiding m and m'.
    Those interested in an elaborate version of the thoughts presented above could receive an extended version of it.
     
  6. Oct 8, 2007 #5

    rbj

    User Avatar

    of course, there's always a mistake somewhere. (and this is old enough that i can't edit my original post.)

    should be

    [tex] T \approx \frac{1}{2} m_0 v^2 = \lim_{v/c \rightarrow 0} m_0 c^2 \left(1 - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \right) [/tex]
     
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