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Relativistic mass

  1. May 3, 2014 #1
    I have spent some time reading other threads and other source. However I could not find any clear derivation of relativistic mass and momentum

    In this case for example
    http://www.phys.ufl.edu/~acosta/phy2061/lectures/Relativity4.pdf

    On the first page, how come you can give such a new definition of momentum in the moving frame as:

    p = dX / dTo

    I understand if it would be p = dXo / dTo

    Could anyone please recommend/write derivations for relativistic momentum and relativistic mass
     
  2. jcsd
  3. May 3, 2014 #2

    mfb

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    Don't use relativistic mass. Science abandoned it decades ago and it just leads to all sorts of problems.

    What is x0?

    You can define momentum every way you want. You just have to do it consistently, and this definition of a relativistic momentum is very useful.
     
  4. May 3, 2014 #3

    Dale

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    As mfb mentioned, avoid relativistic mass. It is a historical artifact that is not used much by modern scientists but lingers in outdated descriptions of relativity.

    When modern physicists talk about "mass" they usually are referring to the "invariant mass" which is given by ##m^2 c^2 = E^2/c^2 - p^2##. In particular, this is the quantity that is intended when scientists talk about the mass of fundamental particles, such as the zero mass of a photon. This quantity is very useful because all reference frames agree on its value, i.e. it is frame-invariant.

    If you absolutely must speak of relativistic mass, then simply use E/c^2 from the above equation. Since we often use units where c=1, relativistic mass is considered to be simply another superfluous name for energy. In contrast, invariant mass is something fundamentally different.
     
    Last edited: May 3, 2014
  5. May 3, 2014 #4
    Relativistic mass is the mass as used in Newton's definition of momentum under relativistic conditions. In order to make sure that it has the same velocity dependence in all frames of reference I start from

    [itex]m\left( v \right) = m_0 \cdot f\left( v \right)[/itex]

    where [itex]f\left( v \right)[/itex] is a function that is identical for all bodies in all frames of reference. [itex]m_0[/itex] is the mass of the body at rest for

    [itex]f\left( 0 \right) = 1[/itex]

    Isotropy leads to

    [itex]f\left( { - v} \right) = f\left( v \right)[/itex]

    With Newton's definition of momentum and the second law of motion I get

    [itex]F = \left( {f + v \cdot f'} \right) \cdot m_0 \cdot a[/itex]

    In order to simplify the following equations I define

    [itex]K\left( v \right): = f\left( v \right) + v \cdot f'\left( v \right)[/itex]

    The third law now says

    [itex]F_1 + F_2 = K\left( {v_1 } \right) \cdot m_1 \cdot a_1 + K\left( {v_2 } \right) \cdot m_2 \cdot a_2 = 0[/itex]

    and that should be valid for all frames of reference:

    [itex]F'_1 + F'_2 = K\left( {v'_1 } \right) \cdot m_1 \cdot a'_1 + K\left( {v'_2 } \right) \cdot m_2 \cdot a'_2 = 0[/itex]

    Wit Galilei transformation this results in [itex]f\left( v \right) = 1[/itex]. But now I leave classical mechanics and turn to Lorentz transformation. With [itex]c = 1[/itex] that means

    [itex]v' = \frac{{v - u}}{{1 - u \cdot v}}[/itex]

    and

    [itex]a' = a \cdot \left( {\frac{{\sqrt {1 - u^2 } }}{{1 - u \cdot v}}} \right)^3[/itex]

    and therefore

    [itex]\frac{{K\left( {\frac{{v_1 - u}}{{1 - u \cdot v_1 }}} \right)}}{{K\left( {v_1 } \right) \cdot \left( {1 - u \cdot v_1 } \right)^3 }} = \frac{{K\left( {\frac{{v_2 - u}}{{1 - u \cdot v_2 }}} \right)}}{{K\left( {v_2 } \right) \cdot \left( {1 - u \cdot v_2 } \right)^3 }}[/itex]

    This applies for all combination of velocities including [itex]u = v_1 = v[/itex] and [itex]v_2 = 0[/itex]:

    [itex]K\left( v \right) \cdot K\left( { - v} \right) = \left( {1 - v^2 } \right)^{ - 3}[/itex]

    This is given for

    [itex]K\left( v \right) = \sqrt {1 - v^2 } ^{ - 3}[/itex]

    After checking that this is not only a solution for the special case but also for the general equation I just need to solve the resulting differential equation

    [itex]f' = \frac{1}{v}\left( {\sqrt {1 - v^2 } ^{ - 3} - f} \right)[/itex]

    to get

    [itex]f = \sqrt {1 - v^2 } ^{ - 1}[/itex]

    and therefore

    [itex]m\left( v \right) = \frac{{m_0 }}{{\sqrt {1 - v^2 } }}[/itex]

    and

    [itex]p = m\left( v \right) \cdot v = \frac{{m_0 \cdot v}}{{\sqrt {1 - v^2 } }} [/itex]

    With this result you can avoid relativistic mass at least for momentum. To replace it by E/c² you first need to derive E=m·c² but that's another topic.
     
  6. May 3, 2014 #5
    If relativistic mass is simply another superfluous name for energy than invariant mass is simply another superfluous name for rest energy
     
  7. May 4, 2014 #6

    vanhees71

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    Invariant mass is conceptually different from energy, as is clearly seen from its fundamental meaning as a Casimir operator of the Poincare group while energy is the generator of time translations.

    The most simple argument for defining the (special) relativistic momentum within point mechanics is to formulate this theory in a manifestly covariant way. The idea is to describe the motion of a particle in terms of worldlines in Minkowski space rather than the usual trajectories in 3D space. This makes sense, because then one can use the elegant formalism of four-vectors. The worldline is described as an arbitrary parametrization [itex]x^{\mu}=x^{\mu}(\lambda)[/itex], where [itex]\lambda[/itex] is a scalar parameter. As a trajectory describing a causal chain of events we should have [itex]\mathrm{d} x^0/\mathrm{d} \lambda>0[/itex], [itex]\mathrm{d} x^{\mu}/\mathrm{d} \lambda \mathrm{d} x_{\mu}/\mathrm{d} \lambda >0[/itex] . The next step is to choose a simple physically interpretable parameter [itex]\lambda[/itex].

    For a single particle there exists a preferred (local) reference frame, namely the one where the particle is at rest at a certain moment in time. This can be defined for any instant in time by a rotation free proper orthochronous Lorentz transform. From the world line [itex]x^{\mu}[/itex] we can form the incremental invariant
    [tex]\mathrm{d} \tau = \frac{1}{c} \sqrt{\frac{\mathrm{d} x_{\mu}}{\mathrm{d} \lambda} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \lambda}} \mathrm{d} \lambda.[/tex]
    Thus the proper time of the particle [itex]\tau[/itex] is a natural choice for a parameter [itex]\lambda[/itex]. This is the time an observer comoving with the particle would measure with his clock.

    Then it's easy to "guess" the manifestly covariant analogy of the kinematics known from non-relativistic physics. Usually we have equations of motion of 2nd order in time, but this we like to express in terms of the scalar proper-time quantity [itex]\tau[/itex] introduced above. This first leads us to the definition of four-velocity, which is the four-vector
    [tex]u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.[/tex]
    By definition the four components are not independent but subject to the constraint
    [tex]u_{\mu} u^{\mu}=c^2=\text{const},[/tex]
    which directly follows from the definition of proper time above.

    Now in analogy to Newton's 2nd Law we define the idea of forces in a covariant way and write the equation of motion as
    [tex]m \frac{\mathrm{d} u ^{\mu}}{\mathrm{d} \tau}=K^{\mu}.[/tex]
    Contrary to the non-relativistic case, usually in relativistic physics the Minkowski force [itex]K^{\mu}[/itex] should depend not only on the space-time coordinates but also on four-velocity, because the above given constraint implies
    [tex]K^{\mu} u_{\mu}=0.[/tex]
    Obviously [itex]m[/itex] is a scalar parameter, the invariant mass, of the particle. It is very convenient to define intrinsic properties (such as, e.g., mass and electric charge of a particle) as Minkowski scalars, and that's why this is done in modern physics. It reaches back to the great work of Minkowski on the mathematical/geometrical structure of special relativity (1907).

    One of the most simple examples is the electromagnetic force of a particle moving in an electromagnetic field:
    [tex]K^{\mu}=\frac{q}{c} F^{\mu \nu}(x) u_{\nu}.[/tex]
    Here [itex]F^{\mu \nu}[/itex] is the antisymmetric Faraday tensor, which is the covariant representative for the electromagnetic field. It's components are given in terms of the usual 3D quantities as
    [tex]F^{0j}=E^{j}, \quad F^{jk}=\epsilon^{jkl} B^l.[/tex]
    To see that the four-vector
    [tex]p^{\mu}=(E/c,\vec{p})=m \frac{\mathrm d x^{\mu}}{\mathrm{d} \tau}[/tex]
    indeed has the meaning as energy and momentum as suggested by the labelling of the temporal component as [itex]E/c[/itex], we rewrite it in terms of time derivatives, giving up the manifestly covariant formalism (which has also its advantages when it comes to solving concrete problems, particularly when more than 1 particle is involved, where the choice of a fixed coordinate time [itex]t[/itex] is a simpler concept than using proper times for each of the particles). To that end we note that
    [tex]\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \left (\frac{\mathrm{d} \tau}{\mathrm{d} t} \right )=\frac{1}{\sqrt{1-\vec{v}^2/c^2}} (c,\vec{v}),[/tex]
    where [itex]\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t[/itex] is the usual velocity vector in a fixed inertial reference frame. Note that it is NOT a spacial part of a four-vector!

    Then we get
    [tex]E=\gamma m c^2, \quad \vec{p}=\gamma \vec{v} \quad \text{with} \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.[/tex]
    Now, in the non-relativistic limit we have [itex]|\vec{v}|/c \ll 1[/itex], and we can expand the quantities in powers of this small quantity. In leading order we get
    [tex]E=m c^2 + \frac{m}{2} \vec{v}^2+\cdots, \quad \vec{p}=m \vec{v} + \cdots.[/tex]
    This shows that [itex]E[/itex] is basically the non-relativistic kinetic energy with the remarkable addition of a the "rest energy" [itex]E_0=m c^2[/itex], while [itex]\vec{p}[/itex] has the usual non-relativistic limit [itex]\vec{p} \approx m \vec{v}[/itex]. This shows that a relativistic covariant formulation leads to the inclusion of this "rest energy" in order to get a covariant definition of energy and momentum.

    Physically it's of course far reaching. Thinking a bit further along these lines of covariance arguments, one realizes that mass is no longer an additive quantity as in non-relavistic physics. E.g., if you have a many-body system of interacting particles, usually the mass of the total system is not the sum of the masses of its consituents, but the total mass (defined as the total energy of the system in the restframe of its center of energy divided by [itex]c^2[/itex]) contains contributions from the binding energy and the intrinsic motion of the consituents. E.g., usually atomic nuclei have a mass that is smaller than the sum of the masses of the contained protons and neutrons ("mass defect"). The hadrons (like protons or neutrons) usually have a much larger mass than the proper masses of the quarks as their consituents in the naive parton model. Only about 2% of the mass of the matter around us is in fact caused by the Higgs mechanism, which gives the quarks and leptons their fundamental mass, but the great reminder is due to the binding to hadrons of the strong interaction, including a large amount of intrinsic energy, which is not fully understood on a fundamental level.
     
  8. May 4, 2014 #7

    mfb

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    Sure, but we have two concepts (total energy and rest energy // sum of relativistic masses and rest mass) and two words ("energy" and "mass"). It is convenient to use "energy" for one concept and "mass" for the other, instead of "energy" (or even worse "mass") for both.
     
  9. May 4, 2014 #8
    Later you show [itex]E=\gamma m c^2[/itex]. So why not use Eo/c² as Casimir operator of the Poincare group? Why do you need a conceptually different property for this purpose?
     
  10. May 4, 2014 #9
    Convenience is a matter of opinion. As you see in the first replies most physicists are not a quarter as tolerant if the term relativistic mass is used. Strictly avoiding relativistic mass for total energy but using rest mass, invariant mass or simply mass for rest energy is inconsequential. The term mass can be completely omitted. Many particle physicists live without it.
     
  11. May 4, 2014 #10

    mfb

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    It is not inconsequential. It is just natural to talk about the mass of an object.
    I am a particle physist (may I ask for your background?), and I don't remember any physicist that did not call the mass of a particle its "mass". It is usually given in eV as we work with c=1, but it is still called mass.
     
  12. May 4, 2014 #11

    ZapperZ

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    Does that mean that you have, for example, read this thread AND the references given and STILL do not understand where the relativistic momentum and mass came from?

    https://www.physicsforums.com/showthread.php?t=642188

    Zz.
     
  13. May 4, 2014 #12

    Dale

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    I agree, which is why I would usually avoid the term "rest energy"; it is both superfluous and cumbersome.

    The important quantity is the four-momentum. In units where c=1 energy is its timelike component, momentum is its spacelike component, and mass is its norm. The norm of a vector is conceptually different from its components, even though you can obviously always choose a basis where there is only one non-zero component.
     
  14. May 4, 2014 #13
    Isn't that just a philosophical debate? In natural science it makes no much sense to assume two entities to be different if they cannot be distinguished experimentally. Resulting from the basics of relativity you will always measure the same for m and Eo/c². They might be different in the mathematical models but these models are not reality but only tools to describe reality (as we see it).
     
  15. May 4, 2014 #14

    DrGreg

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    The length of a 12-inch ruler is always 12 inches, no matter what you do with it. The height of the ruler could be 12 inches if you hold it vertically; it could be, say, 0.1 inches lying flat on a table; or anything between those two extremes in other orientations. So the "length" of a ruler is conceptually different from its "height", even though in some circumstances they could have the same value. (And its easier to say "height" than "vertical length".)
     
  16. May 4, 2014 #15
    Eo/c² and m have always the same value - not only in some circumstances.
     
  17. May 4, 2014 #16

    DrGreg

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    Of course they do. I thought DaleSpam was comparing E with m, not E0 with m. E0 and m are, in effect, two names for the same thing, "mass".
     
  18. May 5, 2014 #17
    Thanks ZapperZ!

    Basically, relativistic mass is a heuristic way to explain that kinetic energy does not obey classical equations when observed in a moving frame? Therefore mass is actually invariant, but kinetic energy "produced" by such mass different depending on a moving or stationary frame?

    If we consider mass to be invariant, I am still confused then, how energy momentum relation was derived, which is based on relativistic force, which is based on relativistic momentum which WAS based on "relativistic mass".

    Derivations with relativistic mass are quite easy to follow, but if you omit term "relativistic mass" these derivations are not valid anymore.

    Because as far as I can see the relation for invariant mass
    mo^2 = E^2 - p^2
    comes from energy-momentum relation

    Thanks to all.
     
  19. May 5, 2014 #18

    Dale

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    I agree, which is why I already said that "rest energy" is another superfluous name for mass which I therefore avoid (in units where c=1).

    As DrGreg already mentioned, I think I was very clear to state that the distinction I was making was between the components of the four-momentum (energy and momentum) and its norm (mass). The fact that you can choose a basis where a component is equal to the norm is not a justification for confounding the components and the norm. Thus, even though "rest energy" is the same as "mass" doesn't mean that "energy" is the same as "mass" (again in units where c=1).
     
  20. May 5, 2014 #19

    atyy

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  21. May 5, 2014 #20
    OK, than we were talking at crossed purposes. Of course there is a difference between energy and rest mass. I was talking about rest energy and rest mass. One of them can be omitted, no matter which one. Traditionally rest mass is used. But using rest energy would work just as well (e.g. with E²=Eo²+p² instead of E²=mo²+p²). It would be even possible to replace energy by relativistic mass mr and write mr²=mo²+p². There are no physical reasons to use energy and rest mass only and to avoid rest energy or relativistic mass. Keeping such conventions just improves communication.
     
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