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Unable derive mass as Resnick \& Halliday describe it

  1. Dec 28, 2012 #1
    I am reading Resnick and Halliday's "Basic Concepts in Relativity" (ISBN: 0023993456) and have come to an impass about deriving equation 3-4a
    \begin{equation}m = \gamma m_0\end{equation}
    in the text. The authors wrote that the equation can be derived from the following equations
    \begin{array}{cr}
    M = m^{\prime} + m_0 & \\
    Mu = m^{\prime} u^{\prime} & \\
    u^{\prime} = 2u / (1+u^2/c^2)& \\
    \end{array}
    based on the attached figure. The derivation is left as an exercise (problem 18 pp 133). The problem is to derive m in the first equation above from the three equations below it, but I only keep coming up with
    \begin{equation} \frac{m^{\prime}}{m_0} = \frac{1+u^2/c^2}{1-u^2/c^2} \end{equation}
    which is an intermediate equation in Tolman's text that leads to the desired equation after substituting a transformation equation for γ into the rhs ratio (e.g. see the sometimes "colorful" thread at https://www.physicsforums.com/forumdisplay.php?f=70\&order=desc\&page=23). But this approach seems to deviate far afield of the three equations given, and it leaves me feeling like I am doing something wrong. What I mean is that I cannot complete the derivation without including the transformation for γ from Tolman, and it seems to me that the authors think I should. Can anyone tell me what I am doing wrong? Hints, recommendations, anything? I am stuck. Apparently the book is out of print, so I can provide more info if need be. Thnx for any help.
     

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    Last edited: Dec 28, 2012
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  3. Dec 28, 2012 #2

    PeterDonis

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    [Edit: Deleted, I was misinterpreting the m's as rest masses instead of relativistic masses. Will follow up with a corrected re-do.]
     
    Last edited: Dec 28, 2012
  4. Dec 28, 2012 #3

    PeterDonis

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    Okay, after fixing my interpretation of the m's, here's a re-do of my previous (now deleted) post:

    I don't agree with the first two of the three equations you wrote down that you say you're supposed to derive the result from. When I write down energy and momentum conservation in the two frames, here's what I get (I'm using units where c = 1, and the first and second values given for E and p in each frame are before and after the collision):

    Unprimed Frame

    [tex]E = 2 m = M[/tex]

    [tex]p = 0 = 0[/tex]

    Primed Frame

    [tex]E' = m' + m_0 = \gamma M[/tex]

    [tex]p' = m' u' = \gamma M u[/tex]

    where [itex]\gamma = 1 / \sqrt{1 - u^2}[/itex] has to be included on the RHS of the primed frame formulas because the single object with rest mass [itex]M[/itex] is *moving* in that frame.

    If you start with the above equations (and your third equation for [itex]u'[/itex]) and eliminate [itex]m'[/itex] and [itex]u'[/itex], you should be able to easily derive [itex]m = \gamma m_0[/itex]. I would suggest checking the textbook to make sure you haven't left out a gamma factor in the two equations involving [itex]M[/itex].
     
  5. Dec 28, 2012 #4

    bcrowell

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    Halliday and Resnick's treatment of relativity is truly awful. You'd do better to read almost any other treatment.
     
  6. Dec 28, 2012 #5

    PeterDonis

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    Awful would be bad enough, but if the formulas in the OP actually appear in the book, it would seem that it goes beyond awful and into outright wrong.
     
  7. Dec 28, 2012 #6

    TSny

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    As PeterDonis noted, the figure from the text has a misprint: the M in the upper right of the figure should have a supscript o because the mass is at rest: ##M_o##.

    You can use the figure to derive the explicit expression for ##\gamma## if you assume that relativistic mass and rest mass are related by ##m = \gamma m_o## where the explicit form of ##\gamma## is yet to be determined.

    The upper part of the figure gives:

    ##2m = M_o## [itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;[/itex] (1)

    And the lower part of the figure gives:

    ##m’ + m_o = M## ##\;\;\;\;\;\;\;\;## (2)


    ##m'u' = Mu## ##\;\;\;\;\;\;\;\;\;\;\;## (3)


    Using ##u’ = 2u/(1+u^2/c^2)## and eliminating ##m'## between (2) and (3) yields

    ##(1-u^2/c^2)M = 2m_o\;\;\;\;## or ##\;\;\;\;(1-u^2/c^2)\gamma M_0 = 2m_o##

    Using (1) to replace ##M_o## by ##2m = 2\gamma m_o## yields

    ##(1-u^2/c^2)\gamma ^2 2m_o = 2m_o## .

    Thus, get the desired result: ##\gamma = 1/\sqrt{1-u^2/c^2}##.

    [I believe this is essentially the derivation that PeterDonis outlined]
     
  8. Dec 29, 2012 #7

    PeterDonis

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    I did it the other way, deriving [itex]m = \gamma m_0[/itex] from the other givens (including the formula for [itex]\gamma[/itex]). Also, what I wrote as [itex]M[/itex] is what you are calling [itex]M_0[/itex], the rest mass of the single object after the collision.
     
  9. Dec 29, 2012 #8

    stevendaryl

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    I don't quite like this derivation because it uses a velocity-dependent mass, but otherwise assumes that momentum works like Newtonian momentum. It assumes that this relativistic mass is conserved. Why? I think that's a little strange. The way I would do it, which is essentially the same derivation, but with slightly different assumptions, is this:

    In Newtonian, physics, the following two quantities are conserved in any collision:

    1. [itex]M_{total} = \sum m[/itex]
    2. [itex]\vec{P}_{total} = \sum m \vec{v}[/itex]

    Then for SR, we generalize these by allowing a more complex dependency on velocity:

    1. [itex]M_{total} = \sum F(v) m[/itex]
    2. [itex]\vec{P}_{total} = \sum F(v) m \vec{v}[/itex]

    where [itex]F(v)[/itex] is an unknown scalar function of velocity (assumed to be dependent only on the magnitude of velocity, rather than its direction).

    That's plausible, but I'm not sure what motivates using the same velocity-dependence [itex]F(v)[/itex] in generalizing both total mass and total momentum.
     
  10. Dec 29, 2012 #9

    PeterDonis

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    Not when you recognize that "relativistic mass" is just another name for "total energy".

    Because the invariant mass (or "rest mass") of an object has to be, well, invariant. This leads to a relationship between total energy and total momentum that requires them both to have the same velocity dependence [itex]F(v)[/itex] (which is usually referred to as [itex]\gamma(v)[/itex]).
     
  11. Dec 29, 2012 #10
    Thnx All

    PeterDonis: Your comment that you don't agree with the given equations caught my attention because it crossed my mind if there was a typo among them. But I have double checked and triple checked the text (first thing I do when I am stumped) and the figure and equations are as they appear in the text.

    bcrowell: I cut my teeth on the H & R "Fundamentals of Physics" text and liked another text they wrote (a sophmore/junior level QM text), but this relativity text really is truly awful; I came to it seeking a more elegeant derivation than that in Tolman's text.

    TSny: Your statement "... where the explicit form of γ is yet to be determined." was my "aha" moment as that is how I derived γ for the LT's earlier in my studies.

    Yes then, the typo is certainly in the figure in which M is ambiguous and left as such in the narrative (and caption). Thnx loads for your help folks; I understand this now.

    BTW, where can I learn more about editing posts? I was not able to do inline equations like TSny did. Thnx again.
     
  12. Dec 29, 2012 #11

    stevendaryl

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    I know, I'm trying to get in the spirit of the Halliday and Resnick derivation.

    IF you assume that energy and momentum are part of a 4-vector, then everything else follows. But I thought the point of this derivation was to motivate the relativistic forms of energy and momentum before the concept of 4-vectors was introduced.

    An alternative derivation along the same lines first derives the relativistic form for the momentum, and then uses that result to derive the relativistic form for the energy. The Halliday and Resnick derivation seems more compact, because it does it all in one fell swoop, but it seems a little unmotivated to me.
     
  13. Dec 29, 2012 #12

    PeterDonis

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    I'm not sure that's healthy, but ok. :wink:

    Can you post or link to an example?
     
  14. Dec 29, 2012 #13

    stevendaryl

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    I left out the derivation, because it's very long-winded, but since you asked...

    Let's assume that momentum has the form:

    [itex]\vec{p} = F(v) m \vec{v}[/itex]​

    where [itex]F(v)[/itex] depends on the magnitude of [itex]v[/itex], not its direction.

    To have the right nonrelativistic limit, it must be that [itex]F(0) = 1[/itex]

    Now, consider a collision in which in frame [itex]F[/itex] a very light particle of mass [itex]m[/itex] is moving relativistically with a horizontal component of velocity of [itex]v[/itex] and a very small vertical component of velocity [itex]u[/itex]. The magnitude of the velocity is [itex]\sqrt{v^2 + u^2} \approx v[/itex].

    This lighter particle "bounces" off a much more massive particle of mass [itex]M[/itex] initially at rest in frame [itex]F[/itex] imparting a small vertical component of velocity [itex]U[/itex] to the larger mass. The "Before" and "After" pictures are shown in Figure 1:

    figure1.jpg

    I'm going to leave off the detailed argument, but approximately what will happen is that for the light particle, its vertical component of velocity will just flip signs, and its horizontal component of velocity will remain unchanged.

    Conservation of momentum in frame [itex]F[/itex] implies:
    [itex]- F(v) m u = +F(v) m u - MU[/itex]​

    So
    [itex]MU = (2 m u)F(v)[/itex]​

    where I used the nonrelativistic form of momentum for the massive particle. Now, look at the same collision in a frame [itex]F'[/itex] in which the lighter particle initially has no horizontal component of its velocity. The collision in this frame looks like Figure 2:

    figure2.jpg

    I'm going to skip the derivation, but the Lorentz transformations can be used to show that in frame [itex]F'[/itex] the vertical components of velocity for the two particles are related to those in frame [itex]F[/itex] by:
    [itex]u' = \gamma u[/itex]​
    [itex]U' = U/\gamma[/itex]​
    where [itex]\gamma = 1/\sqrt{1-v^2/c^2}[/itex] as usual.

    Conservation of vertical momentum in frame [itex]F'[/itex] gives us:
    [itex]-m u' = + m u' - F(v) M U'[/itex]​

    So

    [itex]2m u' = F(v) M U'[/itex]​

    Rewriting in terms of [itex]u[/itex] and [itex]U[/itex] gives:

    [itex]2m u \gamma = F(v) M U/\gamma[/itex]​

    Substituting the previously derived value for [itex]MU[/itex] gives:

    [itex]2m u \gamma = F(v)^2 (2mu)/\gamma[/itex]​

    So

    [itex]F(v) = \gamma[/itex]​

    So we have the relativistic form of momentum:
    [itex]\vec{p} = \gamma m \vec{v}[/itex]​

    The derivation of the form of energy then proceeds much like the Halliday and Resnick case: Imagine in frame [itex]F[/itex] you have two particles of equal mass [itex]m[/itex] and opposite velocities [itex]v[/itex], which collide head-on to produce a mass [itex]M[/itex] at rest.

    Now, switch to a frame [itex]F'[/itex] in which one of the particles is initially at rest. Then the other particle will have (using Lorentz transforms) velocity
    [itex]v' = 2v/(1+v^2/c^2)[/itex]​
    which gives a "gamma" factor of
    [itex]\gamma' = \dfrac{1+(v/c)^2}{1-(v/c)^2}[/itex]​

    In this frame, after the collision, the new composite mass is not at rest, but is moving at speed v. Conservation of momentum in this frame gives:
    [itex]\gamma' m v' = \gamma M v[/itex]​
    Substituting for [itex]\gamma'[/itex] and [itex]v'[/itex] gives:
    [itex]\dfrac{1+(v/c)^2}{1-(v/c)^2} m \dfrac{2v}{1+(v/c)^2}= \gamma M v[/itex]​

    which simplifies to
    [itex]\dfrac{1}{1-(v/c)^2} 2 m = \gamma M[/itex]​

    So, using [itex]\dfrac{1}{1-(v/c)^2} = \gamma^2[/itex] gives us:
    [itex]M = 2 \gamma m[/itex]​

    Now, assume that relativistic energy has the form:
    [itex]E = G(v) m[/itex]​
    So conservation of energy in the center-of-mass frame gives:
    [itex]2 G(v) m = G(0) M[/itex]​

    Substituting for [itex]M[/itex] gives:
    [itex]2 G(v) m = G(0) 2 \gamma m[/itex]​
    So
    [itex]G(v) = G(0) \gamma[/itex]​

    To find out the constant [itex]G(0)[/itex], we expand the expression for energy for low velocity for comparison with the nonrelativistic limit:
    [itex]E = G(0) \gamma m = G(0) m (1 + 1/2 (v/c)^2 + ...)[/itex]​

    If we identify [itex]G(0) m (1/2 (v/c)^2)[/itex] as the nonrelativistic kinetic energy, [itex]1/2 m v^2[/itex] then we find that [itex]G(0) = c^2[/itex]. So the full expression for relativistic energy is:

    [itex]E = \gamma mc^2[/itex]​
     
  15. Dec 29, 2012 #14

    PeterDonis

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    Thanks! I hadn't seen it done this way before.
     
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