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I Difficulty in understanding the proof of mass-speed relation

  1. Jul 10, 2016 #1
    I have a hard time understanding the variation of mass with velocity, more precisely the proof. In almost every material I've found, the author analyses 2 bodies colliding. The idea of looking at the collision is not hard to grasp and by considering one of the velocities equal zero, you get a nice relation between the masses of the two bodies. You cand thus take the resting mass to be m0 and the other m and congrats, you've done it!
    Here is a link to such a proof: http://www.physics-assignment.com/variation-of-mass-with-velocity
    Here are the exact points I don't understand:
    1) The example starts with two bodies and finishes with a relation describing only one body.
    2) We make a mass be m0 and say that the other varies with respect to it. But m0 is supposed to be the mass of the moving body if it would be at rest. So, does this mean that the bodies used have the same mass, no matter what?


    I've also found something different and very easy to understand. From what I heard, this is a proof designed by Einstein himself:
    https://proofwiki.org/wiki/Einstein's_Mass-Velocity_Equation
    I like this one and I have no problem with its validity. But it seems a bit like a particular case. What I mean is that it starts from a restrictive scenario: "The space ship's trajectory be perpendicular to the comet's trajectory towards the planet, so there is no length contraction parallel to the trajectory of the comet."
    Is this second proof really general or not.


    A third one I've found in a book that is not printed in English, from what I know. It uses space-time vectors and is very short. But is not well enough explained in the book at one particular point. Maybe two.
    I follows as this:

    xμ = (ct, x, y, z)
    uμ = dxμ(τ)/dτ = (c dt/dτ, dx/dτ, dy/dτ, dz/dτ)
    Where τ = Δs/c (the studied body's time)
    *Here I have the first lack of understanding with this proof: why use τ (the body's time). What significance has using the ratio of the coordinates measured by you and the time measured by the moving body. After all, in the body's frame, it is not moving with respect to itself.
    uμ = dt/dτ (c, dx/dt, dy/dt, dz/dt)= dt/dτ (c, vx, vy, vz) = γ(c, vx, vy, vz)
    Where γ = 1/√(1-v2/c2)
    The author says that v = (vx, vy, vz) is the speed measured by us. So if we multiply by the rest mass m0 we get:
    pμ = m0γ(c, vx, vy, vz)
    **Here I really don't get it: why is pμ = m0uμ. If we differentiate xμ by our time t and then multiply by m0 we simply make γ disappear, so the real question here is: what's the difference.
    ***Also, could anyone explain what uμ = γ(c, vx, vy, vz) actually represents?


    Note: I don't say that any of those proofs are wrong, I just want to understand them better. I would also be grateful if anyone has an easier proof or some material that explains one of those better.
    Note: I would also be grateful for a nice source of problems for both special and general relativity. What I have found so far is usually just easy exercises that ask you to apply Lorenz transforms, which aren't actual problems. I'm looking for more challenging stuff, that involves Minkovski space-time representation, space-time interval and things like those.
     
  2. jcsd
  3. Jul 10, 2016 #2

    Nugatory

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    before you dig deeper into relativistic mass increase, you will want to take a look at our Insights article on the subject: https://www.physicsforums.com/insights/what-is-relativistic-mass-and-why-it-is-not-used-much/

    You may find it better to start over again with a more modern and systematic treatment of relativity. For example:
    Try Taylor and Wheeler's "Spacetime Physics".
     
  4. Jul 10, 2016 #3

    vanhees71

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  5. Jul 12, 2016 #4

    stevendaryl

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    The explanation in that link is not very clear, in my opinion. What they appear to be doing is this:
    1. They are assuming that the relativistic momentum of an object has the same form as the classical momentum, [itex]p = m v[/itex], except that they are allowing the mass to vary with velocity (so-called "relativistic mass").
    2. They are assuming that relativistic mass is conserved in a collision.
    The modern way of doing this is to let the mass of an object be a quantity that is independent of rest frame, and that
    1. Momentum is in the same direction as velocity, is proportional to the mass, with a constant of proportionality that depends on velocity: [itex]\vec{p} = m_0 f(v) \vec{v}[/itex]
    2. Energy is also proportional to the mass, with a constant of proportionality that depends on velocity: [itex]E = m_0 g(v)[/itex]
    In the link you gave, they are defining relativistic mass (or they call it "velocity-dependent mass") to be [itex]m = m_0 f(v)[/itex], but they are also implicitly assuming without proof that the energy is proportional to this relativistic mass: [itex]E = K m_0 f(v)[/itex]. They don't say this, but they are assuming it, because they assume that relativistic mass is conserved by collisions, so relativistic mass has to have the same dependence on velocity that energy does. So with those assumptions, the problem is to derive the dependence of relativistic mass on velocity, which means finding the function [itex]f(v)[/itex] such that [itex]m = m_0 f(v)[/itex].

    To this end, they start with a collision problem where we know the answer. Consider a frame [itex]S'[/itex] in which we have two identical masses (in my terms, [itex]m_0[/itex] is the same for the two masses) moving in opposite directions at speed [itex]u'[/itex]. So they have equal and opposite momentum, and the total momentum must be zero. The two masses collide and stick together to form a larger mass, which must be at rest in frame [itex]S'[/itex] (since the momentum is zero). So in [itex]S'[/itex], the collision looks like: ([itex]m'_1[/itex] is the relativistic mass of one object, [itex]m'_2[/itex] is the relativistic mass of the other, [itex]u'_1[/itex] is the velocity of one object, [itex]u'_2[/itex] is the velocity of the other. [itex]M'[/itex] is the relativistic mass of the conjoined object, and [itex]U'[/itex] is its velocity.)

    Before:

    [itex]m'_1 = m'_2[/itex]
    [itex]u'_1 = u'[/itex]
    [itex]u'_2 = - u'[/itex]
    [itex]p_{total} = m'_1 u'_1 + m'_2 u'_2 = 0[/itex]

    After:

    [itex]M' = m'_1 + m'_2[/itex]
    [itex]U' = 0[/itex]

    Now, they switch to a new frame, [itex]S[/itex], whose spatial origin is moving at velocity [itex]v[/itex] relative to [itex]S'[/itex]. We use the relativistic velocity addition formula to compute velocities in this frame, ([itex]m_1[/itex] is the relativistic mass of one object in this frame, [itex]m_2[/itex] is the relativistic mass of the other, [itex]u_1[/itex] is the velocity of one object, [itex]u_2[/itex] is the velocity of the other. [itex]M[/itex] is the relativistic mass of the conjoined object, and [itex]U[/itex] is its velocity.)

    [itex]u_1 = \frac{v + u'}{1 + \frac{u' v}{c^2}}[/itex]
    [itex]u_2 = \frac{v - u'}{1 - \frac{u' v}{c^2}}[/itex]
    [itex]U = v[/itex]

    In this frame, the conjoined mass, [itex]M[/itex] moves at speed [itex]v[/itex]. Conservation of relativistic mass (I would call it "energy") in this frame tells us:

    [itex]m_1 + m_2 = M[/itex]

    Conservation of momentum in this frame tells us:

    [itex]m_1 u_1 + m_2 u_2 = M U[/itex]

    or using what we know about [itex]u_1, u_2, M, U[/itex]:

    [itex]m_1 \frac{v+u'}{1 + \frac{u' v}{c^2}} + m_2 \frac{v - u'}{1 - \frac{u' v}{c^2}} = (m_1 + m_2) v[/itex]

    Skipping a lot of messy algebraic manipulation, this implies the following about [itex]m_1[/itex] and [itex]m_2[/itex]:

    [itex]\frac{m_1}{m_2} = \frac{\frac{1}{\sqrt{1 - (\frac{u_1}{c})^2}}}{\frac{1}{\sqrt{1 - (\frac{u_2}{c})^2}}}[/itex]

    This implies that [itex]m_1 = m_0 \frac{1}{\sqrt{1 - (\frac{u_1}{c})^2}}[/itex] and [itex]m_2 = m_0 \frac{1}{\sqrt{1 - (\frac{u_1}{c})^2}}[/itex]
     
  6. Jul 12, 2016 #5

    vanhees71

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    It's highly misleading to call an energy a mass. Today, we don't use the idea of "relativistic speed dependent mass" anymore but characterize the mass of a particle or body by the invariant mass. Defining the energy of a body such that together with momentum it builds a four-vector the energy of the body at rest is ##E_0=m c^2##, where ##m## is the invariant mass. For a moving particle the energy is ##E=\sqrt{\vec{p}^2c^2+m^2 c^4}##. The covariant definition of mass is ##p_{\mu} p^{\mu}=m^2 c^2##, where the energy-momentum four-vector is defined by ##(p^{\mu})=(E/c,\vec{p})##. For details see

    http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf
     
  7. Jul 12, 2016 #6

    stevendaryl

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    This is a very important concept that will be even more important in General Relativity: the parametrized path through spacetime. From the point of view of relativity, time is put on the same footing as space. So objects move through 4-dimensional spacetime, rather than just 3-dimensional space. An important thing to understand is the distinction between a "location" in spacetime (usually called an "event") and the coordinates that are used to describe it. To describe a path through spacetime, you need a path parameter, call it [itex]s[/itex], which is something that increases as you move along the path. For slower-than-light travel, proper time is a convenient choice for the parameter [itex]s[/itex]. Then the path itself is a continuous function [itex]\mathcal{P}(s)[/itex] which returns a location in spacetime (that is, an event) for each possible value of [itex]s[/itex]. If you pick a particular coordinate system, then this path can be represented by 4 functions:

    [itex]x(s), y(s), z(s), t(s)[/itex]

    It's the same spacetime path, regardless of what coordinate system you use. Associated with a spacetime path is a corresponding 4-velocity:

    [itex]U^x = \frac{dx}{ds}, U^y = \frac{dy}{ds}, U^z = \frac{dz}{ds}, U^t = \frac{dt}{ds}[/itex]

    This 4-velocity is a vector quantity that has different components in different frames. Under a Lorentz transformation in the x-direction, these components transform in the following way:

    [itex](U^x)' = \frac{dx'}{ds} = \gamma (U^x - v U^t)[/itex]
    [itex](U^t)' = \frac{dt'}{ds} = \gamma (U^t - \frac{v}{c^2} U^x)[/itex]
    [itex](U^y)' = \frac{dy'}{ds} = U^y[/itex]
    [itex](U^z)' = \frac{dz'}{ds} = U^z[/itex]

    Note: the parameter [itex]s[/itex] is NOT changed when you change frames. That's important. The parameter is called a "scalar", which means that it is a real number that has the same value in every coordinate system.

    You're right, that if the object is moving inertially, then you can find a coordinate system in which:

    [itex]x(s) = y(s) = z(s) = 0[/itex]
    [itex]t(s) = s[/itex]

    So in this coordinate system, [itex]U^x = U^y = U^z = 0[/itex], and [itex]U^t = 1[/itex]
     
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