Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 14

Take the total differential of the logarithm of the relativistic energy, with the speed of light fixed:Using the chain rule,Every term is a dimensionless fractional change. The differential relation is exact; applying it to finite small changes is a first-order Taylor expansion.

Equal energies mean $dE=0$ in the linear approximation. Evaluate the coefficients at particle $A$, where $v_A=0.90c$, and use $\Delta v=0.01c$. The fractional rest mass change isConsequentlyAs a check beyond the requested first-order approximation, exact equality of relativistic energy gives $m_B/m_A=\sqrt{(1-0.91^2)/(1-0.90^2)}=\sqrt{1719/1900}\approx 0.95118$, consistent with the sign and approximate size of the difference.

Work on an open domain with $xy>0$ and $x\ne 0$, where the displayed square root is real and differentiable. Split $u=v+h$ with $v=\varphi (xy)$ and $h=\sqrt{xy}\psi (y/x)$. For the first term, the chain rule givesso $x^2{v}_{xx}-y^2{v}_{yy}=0$.

For the second term, set $t=y/x$. Its first partial derivatives areThey show $x{h}_x+y{h}_y=h$, the degree-one case of Euler theorem for homogeneous functions. Differentiating that identity with respect to each variable yieldsThe mixed partial derivatives agree here by the chain rule. Multiplying these equations by $x$ and $y$, respectively, proves the second derivative identity for a degree-one homogeneous function:Adding the results for $v$ and $h$ verifies the partial differential equationEquivalently, $h(\lambda x,\lambda y)=\lambda h(x,y)$ for positive $\lambda$, so the same proof follows directly from its homogeneous function property. The statement is local to a domain where the given expression is defined; it does not require assigning values on $x=0$.