# Alternate derivation of kinetic mass increase in special relativity, from Maudlin

I've recently come across a derivation, which I've not seen before, of mass increase in special relativity. It seems to make sense, but I get tripped-up on an intermediate step, and I can't seem to find any justification for this step. The derivation is from "Quantum Non-Locality and Relativity (3rd ed.)", by Tim Maudlin (Wiley-Blackwell, ISBN-13: 978-1444331271), and it's covered in pages 60-63. I'll run through the derivation below, and point-out the step I have trouble justifying; but I'll finish the derivation before I revisit that step in more detail.

In the diagram below, we have two objects of equal mass, moving in opposite directions at the same speed, with world lines described by $A: x=-vt$ and $B: x=+vt$. Due to the symmetry of the arrangement, the center of mass is not moving, and coincides with the $ct$ axis.

A simultaneity slice $S$ is chosen from the rest frame of $B$. The points $P$, $Q$, and $R$ indicate the intersections of $S$ with $A$, the $ct$ axis, and $B$ (respectively). Choosing: $$P=(cT, -vT)=cT(1,-\beta)$$

(where $\beta = v/c$), and designating the slope of $S$ as $\sigma$, we find the equation describing the simultaneity slice: $$S: ct=\sigma x + cT(1 + \sigma\beta)$$ and the points: $$Q=cT(1+\sigma\beta, 0)$$

$$R=\rho (cT, +vT)=\rho cT(1, +\beta)$$

(where $\rho = \frac{1+\sigma\beta}{1-\sigma\beta}$).

From this, we can determine the invariant intervals for the segments $PQ$ and $QR$: $$s_{PQ}^2 = c^2\Delta t_{PQ}^2 - \Delta x_{PQ}^2 = c^2 T^2 \beta^2 ( \sigma^2 - 1 )$$

$$s_{QR}^2 = c^2\Delta t_{QR}^2 - \Delta x_{QR}^2 = c^2 T^2 \rho^2 \beta^2 ( \sigma^2 - 1 )$$

By definition, each of these intervals should have a constant value regardless of reference frame. Their ratio (which is also invariant) is $s_{PQ}:s_{QR}=1:\rho$. In the rest frame of $B$, the lengths of $PQ$ and $QR$ (at rest) are equal to their intervals (since the time coordinate is difference is zero).

Now in order for the center of mass to remain on an inertial trajectory in the rest frame of $B$ (or so the argument goes), the masses must be different by a ratio which is the inverse of that between the segment intervals. That is, the masses must be in the ratio $m_A:m_B=\rho:1$. (This is the step I have trouble justifying - more about that later.)

Because the slope of the world-line of $B$ is $1/\beta$, then the slope of its simultaneity slice is $\beta$, and $\sigma=\beta$. This lets us express $\rho$ entirely in terms of $\beta$: $$\rho=\frac{1+\beta^2}{1-\beta^2}$$

We're interested in the ratio of the masses in the rest frame of $B$ in terms of $w$, the speed of $A$ in the same reference frame. However, the above expression for the ratio is in terms of the speed of each mass in the center-of-mass rest frame. So we use the velocity transformation, $$\beta_u' = \frac{ \beta_u - \beta }{ 1 - \beta_u\beta}$$ (Substitute $\beta_u = u / c$, $\beta_u' = u' / c$, and $\beta = v / c$ to recover the more familiar $u' = \frac{ u - v }{ 1 - uv/c^2 }$.) Except $u'=w$ and $u=-v$, so it reduces to: $$\beta_w = \frac{ -2\beta }{ 1 + \beta^2 }$$

Using this, we can find the ratio of the masses to be: $$\frac{m_A}{m_B} = \rho = \frac{ 1 + \beta^2 }{ 1 - \beta^2 } = \left( \frac{ 1 - \beta^2 }{ 1 + \beta^2 } \right)^{-1} = \left( \frac{ ( 1 - \beta^2 )^2 }{ ( 1 + \beta^2 )^2 } \right)^{-1/2} = \left( \frac{ 1 - 2\beta^2 + \beta^4 }{ 1 + 2\beta^2 + \beta^4 } \right)^{-1/2} = \left( 1 - \frac{ 4\beta^2 }{ 1 + 2\beta^2 + \beta^4 } \right)^{-1/2} = \left( 1 - \left( \frac{ -2\beta }{ 1 + \beta^2 } \right)^2 \right)^{-1/2} = ( 1 - \beta_w^2 )^{-1/2} = \gamma_w$$

Or rearranging: $m_A = \gamma_w m_B$ If we make use of the symmetry of the situation to note that the mass of either object in its own rest frame will be the same value, then we can use $m_0$ to indicate the mass at rest, and $m$ to indicate the other mass. In this case, $m_B = m_0$ and $m_A = m$, but the roles would be reversed for the case of the rest frame of $A$. In either case, we have $m = \gamma_w m_0$, which is the usual mass increase relation.

For at-least two distinct reasons, I have trouble understanding the pivotal step in the middle of this derivation.

The first reason is that $m_A:m_B=\rho:1$ can't possibly apply in all frames, as the argument seems to imply - it seems to only apply to the rest frame of $B$. It certainly can't apply to the rest frame of $A$, due to the symmetry just described. And it conflicts with the original premise, that the masses are equal in the rest frame of the center-of-mass ($m_A = m_B = \mu$). In the center-of-mass frame, we have: $$m_A x_A + m_B x_B = \mu x - \mu x = 0$$ (which is also true by symmetry anyway). This wouldn't be true if the masses were not equal (never mind them being in the ratio of $\rho$ to each-other).

The other reason I have is that I can't find a way to use center-of-mass-based arguments to fix the ratio of the masses to $\rho$ even in the rest frame of $B$. It seems that any fixed ratio (independent of time) would ensure the center of mass is on an inertial trajectory.

Is there some detail or set of details that I'm missing that would make more sense of this step in the derivation? Or is it just a lucky coincidence that the originator stumbled upon?

My apologies for the sloppiness of the diagram. I use Inkscape, though I'm not very good at it. And in case you consult the cited work, just be aware that I've also used slightly different labels and notation than the author (I've never been a fan of setting $c=1$ to make the math simpler, so I've used other means of arriving at dimensionless quantities to work with.)

Update for Diagram:

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