Relativistic frameworks and the case for (or against) incommensurability

Abstract

The aim of this paper is to address, from a fresh perspective, the question of whether Newtonian mechanics can legitimately be regarded as a limiting case of the special theory of relativity (STR), or whether the two theories should be deemed so radically different as to be incommensurable in the sense of Feyerabend and Kuhn. Firstly, it is argued that focusing on the concept of mass and its transformation across the two varieties of mechanics is bound to leave the issue unsettled. On the one hand, the idea of a speed-dependent ‘relativistic mass’, which has been invoked in support of incommensurability claims, results from a particular, often innocuous but unnecessary and inappropriate reading of certain basic formulae. On the other hand, the existence of an invariant rest mass in STR does not warrant its identification with the Newtonian mass, be it in a suitable limit. This invariant notwithstanding, those who follow Feyerabend and Kuhn can still uphold their views with regard to the two theories. It is shown, however, that the two mechanics embody relativistic frameworks that are direct consequences of the same set of assumptions. As a result, if Newton’s mechanics cannot simply be regarded as a limiting case of STR, the possibility of ‘recovering’ from the latter some elements of the former can be traced to a common source, belying claims of logical disconnection between the two theories.

Appendix: derivation of relativistic transformations

Given two inertial frames $R\equiv (O;x,y,z)$ and $R^{\prime }\equiv (O^{\prime };x^{\prime }, y^{\prime },z^{\prime })$, we seek a general transformation $T_{R\rightarrow {R}^{\prime }}$ such that the expression P(x, y, z, t) of a physical quantity, relative to frame R, systematically transforms into $Q\left( {x^{\prime },y^{\prime },z^{\prime },t^{\prime }} \right)$ relative to $R^{\prime }$, where t and $t^{\prime }$ refer to clock readings in R and $R^{\prime }$ respectively. Isotropy warrants choosing the direction of the relative rectilinear uniform motion of R and $R^{\prime }$ as that of an arbitrary axis (Ox) with respect to R, and that of ($O^{\prime }x^{\prime })$ with respect to $R^{\prime }$, such that (Ox) and ($O^{\prime }x^{\prime })$ are parallel. Since directions orthogonal to ($Ox)/(O^{\prime }x^{\prime })$ do not contribute anything nontrivial to the change-of-frame transformation $T_{R\rightarrow {R}^{\prime }}$, and assuming that the origins and three pairs of parallel axes ($Ox)/(O^{\prime }x^{\prime })$, ($Oy)/(O^{\prime }y^{\prime })$ and ($Oz)/(O^{\prime }z^{\prime })$ coincide when clocks $C_{R}$ in R and $C_{{R}^{\prime }}$ in $R^{\prime }$ both read 0, then $y=y^{\prime }$ and $z=z^{\prime }$ for all readings of the two clocks. Deriving $T_{R\rightarrow {R}^{\prime }}$ then reduces to working out f and g such that $x^{\prime }=f\left( {x,t} \right)$ and $t^{\prime }=g\left( {x,t} \right)$.

Spatial homogeneity requires that the difference $f\left( {x_2 ,t} \right) -f\left( {x_1 ,t} \right)$ (‘rod length’ in the direction of motion) should not be affected by an arbitrary shift in the x coordinate i.e.

$$\begin{aligned} f\left( {x_2 +\varepsilon ,t} \right) -f\left( {x_1 +\varepsilon ,t} \right) =f\left( {x_2 ,t} \right) -f\left( {x_1 ,t} \right) . \end{aligned}$$

Dividing by $\varepsilon$ and taking the $\varepsilon \rightarrow 0$ limit, this condition amounts to $\left. {\frac{\partial f}{\partial x}} \right| _{x_1 } =\left. {\frac{\partial f}{\partial x}} \right| _{x_2 }$: given the arbitrariness of $x_1$ and $x_2$, f is a linear function of x, and so is g. Similar considerations with respect to the endpoints of any time interval lead to the conclusion that f and g are linear functions of time. Alternatively, since $dx^{\prime } = ~\frac{{\partial f}}{{\partial x}}dx + ~\frac{{\partial f}}{{\partial t}}dt$ and $dt^{\prime } = ~\frac{{\partial g}}{{\partial x}}dx + ~\frac{{\partial g}}{{\partial t}}dt$ , that uniform motion in R appears as uniform in $R^{\prime }$ (UTUM) requires all partial derivatives to be independent of x and t. $T_{R\rightarrow {R}^{\prime }}$ thus takes the form $\left\{ {{\begin{array}{l} {x^{\prime }=\gamma x+\delta t} \\ {t^{\prime }=\alpha x+\beta t} \\ \end{array} }} \right.$, where the coefficients $\alpha , \beta$, $\gamma$, $\delta$ depend only on the relative velocity of the two frames.

As seen from R, the origin $O^{\prime }$ of $R^{\prime }$ ($x^{\prime }=0)$ is such that $\gamma x=-\delta t$: it is observed to be moving with the velocity $v=-\frac{\delta }{\gamma }$. From the viewpoint of $R^{\prime }$, the origin O of R ($x=0)$ moves in the opposite direction with velocity $-v=\frac{\delta }{\beta }$ . Therefore, $\beta =\gamma$ and since $\delta =-\gamma v$, this reduces the task to working out the form of two unknown coefficients  $\alpha$ and $\gamma$, functions of v only, such that $T_v \equiv T_{R\rightarrow {R}^{\prime }} : \left\{ {{\begin{array}{l} {x^{\prime }=\gamma \left( v \right) x-\gamma \left( v \right) vt} \\ {t^{\prime }=\alpha \left( v \right) x+\gamma \left( v \right) t} \\ \end{array} }} \right.$.

$T_v$ should be invariant under a reflection in the parallel planes $(yz)/(y'z')$. If $\tilde{R}$ is the image of R under such reflection, and $\tilde{R^{\prime }}$ that of $R^{\prime }$, then $\tilde{x} = -x, \widetilde{x^{\prime }} =-x^{\prime }, {\tilde{t}} =t, \tilde{t^{\prime }} =t^{\prime }$. As seen from $\tilde{R}, \tilde{R^{\prime }}$ is in uniform rectilinear motion with velocity $-v$ i.e. $T_{\tilde{R}\rightarrow \tilde{R^{\prime }}}\, \equiv \, T_{-v}$ . It follows that $\gamma$ is an even function of $v: \gamma \left( {-v} \right) =\gamma \left( v \right)$, and that $\alpha$ is odd: $\alpha \left( {-v} \right) =-\alpha \left( v \right)$ (the same conclusions obtain upon considering time reversal instead of a mirror image).

A third inertial frame $R^{{\prime }{\prime }}$ is introduced, which is in rectilinear uniform motion relative to $R^{\prime }$, with velocity $v^{\prime }$, in the ($O^{{\prime }{\prime }}x^{{\prime }{\prime }})$ direction parallel to (Ox). A clock $C_{{R}^{{\prime }{\prime }}}$ at rest in $R^{{\prime }{\prime }}$ shows time $t^{{\prime }{\prime }}$. The origins and axes of all three frames are most conveniently chosen so as to coincide when all of their respective clocks read 0. Requiring that $T_{v^{\prime }} T_v =T_{v^{\prime \prime }}$, where $v^{{\prime }{\prime }}$ is the velocity of $R^{{\prime }{\prime }}$ relative to R, is equivalent to the set of simultaneous equations:

$$\begin{aligned} \left\{ {\begin{array}{l} {\gamma ^{\prime }\left( {\gamma - \alpha v^{\prime }} \right) = \gamma ^{{\prime }{\prime }}\qquad \left( a \right) }\\ {\gamma \gamma ^{\prime }\left( {v + v^{\prime }} \right) = \gamma ^{{\prime }{\prime }}v^{{\prime }{\prime }}\qquad \left( b \right) }\\ {\alpha ^{\prime } \gamma + \alpha \gamma ^{\prime } = \alpha ^{{\prime }{\prime }}\qquad \left( c \right) }\\ {\gamma \left( {\gamma ^{\prime } - \alpha ^{\prime } v} \right) = \gamma ^{{\prime }{\prime }}\qquad \left( d \right) } \end{array}} \right. \end{aligned}$$

where $\alpha ^{\prime }\equiv \alpha (v^{\prime }),\gamma ^{{\prime }{\prime }}\equiv \alpha \left( {{v}^{{\prime }{\prime }}} \right) )$ etc.

From $\left( a \right)$ and $\left( d \right) : \frac{\alpha ^{\prime }}{{{\gamma }^{{\prime }}{v^{\prime }}}}=\frac{\alpha }{\gamma v}$ and since $\frac{\alpha }{\gamma v}$ depends on v only, and $\frac{\alpha ^{\prime }}{{{\gamma }^{{\prime }}{v^{\prime }}}}$ on $v^{\prime }$, equality implies that those ratios are independent of the relative velocity of the frames, hence equal to a ‘universal’ constant $\lambda$. Given $\alpha =\lambda \gamma v$, dividing $\left( b \right)$ by $\left( a \right)$ then yields the composition law for velocities: $v^{{\prime }{\prime }}=\frac{v+v^{\prime }}{1-\lambda vv^{\prime }}$ .

From $\left( a \right)$ with ${v}^{\prime }= -v$ and $v^{{\prime }{\prime }}=0~: \gamma \left( v \right) \gamma \left( {-v} \right) =\frac{\gamma \left( 0 \right) }{1+\lambda v^{2}}$, and letting $v=0\, (\gamma \ne 0)$ implies $\gamma \left( 0 \right) =1$. Since $\gamma$ is an even function of v, $\gamma \left( v \right) \gamma \left( {-v} \right) =\gamma \left( v \right) ^{2}$, hence $\gamma \left( v \right) =\frac{1}{\sqrt{1+\lambda v^{2}}}$ . This completes the derivation of $T_v$ : $\left\{ {{\begin{array}{l} {x^{\prime }=\frac{x-vt}{\sqrt{1+\lambda v^{2}}}} \\ {t^{\prime }=\frac{t+\lambda vx}{\sqrt{1+\lambda v^{2}}}} \\ \end{array} }} \right.$

• $\lambda =0$ yields the Galilei transformation: $\left\{ {{\begin{array}{l} {x^{\prime }=x-vt} \\ {t^{\prime }=t} \\ \end{array} }} \right.$.

• If $\lambda < 0$, then with $\lambda =-\frac{1}{\kappa ^{2}}, T_{R\rightarrow {R}^{\prime }} \equiv T_{v,\kappa } :\left\{ {{\begin{array}{l} {x^{\prime }=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{\kappa ^{2}}}}} \\ {t^{\prime }=\frac{t-\frac{v}{\kappa ^{2}}x}{\sqrt{1-\frac{v^{2}}{\kappa ^{2}}}}} \\ \end{array} }} \right.$,

which is such that $x^{{\prime }{2}}-\kappa ^{2}t^{{\prime }{2}}=x^{2}-\kappa ^{2}t^{2}$, a non-Euclidean metric invariant that suggests formulating the (S)RF in terms of a four-dimensional space-time, in contrast with the ‘3+1’ formulation of Galilei-Newtonian mechanics. $T_{v,c}$ is the Lorentz transformation given identification of $\kappa$ with the speed of light c.

As developed in Sect. 3, dynamics in the $\lambda < 0$ case is subject to an energy-momentum change-of-frame transformation, the form of which is a transposition of $T_{v,\kappa }$:

$$\begin{aligned} \left\{ {{\begin{array}{l} {p^{\prime }=\gamma \left( {p-\frac{v}{\kappa ^{2}}E} \right) } \\ {{E}^{\prime }}=\gamma (E-vp)\\ \end{array} }} \right. , \end{aligned}$$

The corresponding invariant $E^{{\prime }2}-p^{{\prime }{2}}\kappa ^{2}=E^{2}-p^{2}\kappa ^{2}$ is $E_0 ^{2}=\mu \kappa ^{2}$, where $\mu$ can be interpreted as a Proper Inertial Mass (PIM) and $E_0$ is the total energy in the rest i.e. ‘zero-momentum’ frame of the object.