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Four-momentum conservation gives $P={\sum }_i{Q}_i$, while the mass-shell condition gives $P\cdot P=M^2{c}^2$. ThereforeIn the parent rest frame, $M{c}^2={\sum }_i{E}_i$, and each relativistic energy obeys $E_i\ge m_i{c}^2$. Thus

The four-momentum isThusMultiplication by $c$ identifies the first two terms as the rest energy $m{c}^2$ and the Newtonian kinetic energy $m|\mathbf{u}{|}^2/2$.

For the fixed-target reaction, the initial invariant isAt threshold the four final particles are at rest in their centre-of-momentum frame, giving invariant $(4mc{)}^2$. Hence $\gamma =7$ and the least laboratory speed is

Dualize the antisymmetric tensor to a vector,An isotropic symmetric rank-two tensor depending quadratically on one vector can only be a linear combination of ${\delta }_{ij}{d}_k{d}_k$ and $d_i{d}_j$: after rotating $d$ to the third axis, invariance under all rotations about that axis forces the matrix to be $\mathrm{diag}(a,a,b)$.

Since $D_{ik}{D}_{jk}={\delta }_{ij}{d}^2-d_i{d}_j$ and $D_{kl}{D}_{kl}=2d^2$, the most general answer may be writtenThe two displayed tensors are independent, for example by taking $d$ along a coordinate axis and comparing the components parallel and perpendicular to it. Hence there are exactly two independent terms.

The relation isTaking its trace, symmetric traceless part, and antisymmetric part separately givesEquivalently,

If $\beta =-\gamma \ne 0$, $\alpha \ne 0$, and $A_{ij}=0$, its symmetric part forces $B_{kk}=0$ and its antisymmetric part forces $B_{[ij]}=0$. Thus the solutions are exactly the symmetric traceless tensors $B_{ij}$.

In three dimensions, the most general isotropic tensors of ranks zero through three areThe scalar and zero vector are unchanged by rotations. For every $R\in SO(3)$, $R_{ia}{R}_{jb}{\delta }_{ab}={\delta }_{ij}$ and $R_{ia}{R}_{jb}{R}_{kc}{\epsilon }_{abc}=(\det R){\epsilon }_{ijk}={\epsilon }_{ijk}$, proving isotropy.

Taking the symmetric part givesContracting the antisymmetric part with the Levi-Civita symbol givesIndeed, $\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{k\ell m}{T}_{\ell m}=\frac{1}{2}(T_{ij}-T_{ji})$, so $S_{ij}+{\epsilon }_{ijk}{V}_k=T_{ij}$.

Applying the functional equation at $x^{{\alpha }^{-1}}$ repeatedly givesSince $x^{{\alpha }^{-n}}\to 1$, continuity at $1$ gives $f(x)=f(1)$. Thus $f$ is constant.

Without continuity the claim is false. The function $f(1)=1$ and $f(x)=0$ for $x\ne 1$ satisfies $f(x)=f(x^{\alpha })$ because $x^{\alpha }=1$ for positive $x$ exactly when $x=1$, but it is not constant.

Let the nonconstant polynomial have degree $d$ and leading coefficient $c$. If $d$ is odd, its limits at $+\infty$ and $-\infty$ have opposite signs, so the intermediate value theorem shows that its image is $\mathbb{R}$. If $d$ is even and $c>0$, it tends to $+\infty$ at both ends. Outside a sufficiently large compact interval it exceeds $f(0)$, while on that interval the extreme value theorem gives a global minimum $a$. Continuity and the intermediate value theorem then give image $[a,\infty )$. For $c<0$, applying this argument to $-f$ gives $(-\infty ,a]$. These are exactly the three possible images.

The function is continuous at $x$ when, for every $\epsilon >0$, there is $\delta >0$ such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\epsilon$.

If this holds and $x_n\to x$, eventually $|x_n-x|<\delta$, so $f(x_n)\to f(x)$. Conversely, if continuity fails, some ${\epsilon }_0>0$ allows, for every $n$, a point $x_n$ with $|x_n-x|<1/n$ but $|f(x_n)-f(x)|\ge {\epsilon }_0$. Then $x_n\to x$ while $f(x_n)\nrightarrow f(x)$, a contradiction. This is the sequential characterization of continuity in metric spaces.

The generalized characteristic polynomial isThe generalized Rayleigh quotient is maximized by the largest generalized eigenvalue, $4$. Solving $(A-4B)\mathbf{v}=0$ gives

The quadratic form has symmetric matrixwhose characteristic polynomial is $(\lambda -2)(\lambda -3)(\lambda -8)$. All its eigenvalues are positive, so $A$ is positive definite and ${\mathbf{x}}^{T}A\mathbf{x}=1$ is an ellipsoid. The spectral theorem for real symmetric matrices gives semi-axis lengthsThe nearest points lie in the eigendirection of the largest eigenvalue. Since $(5,2,1{)}^T$ is an $8$-eigenvector, they are

The characteristic polynomial of the displayed rotation matrix is $(1-\lambda )({\lambda }^2-2\lambda \cos \theta +1)$, so its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, all of modulus one. The real eigenvector $(0,0,1{)}^T$ spans the rotation axis, while the arguments $\pm \theta$ of the conjugate eigenvalues record the rotation angle on its perpendicular plane.

Applying first the $z$-rotation and then the $x$-rotation givesIts $1$-eigenspace is spanned by $(1,-1,1{)}^T$, so $\mathbf{n}=(1,-1,1{)}^T/\sqrt{3}$. The matrix trace identity $\mathrm{tr}R=1+2\cos \varphi$ gives $0=1+2\cos \varphi$. Thus $\varphi =2\pi /3$.

In Einstein notation, $(\mathbf{n}\cdot \mathbf{x})n_i=n_i{n}_j{x}_j$, so the matrix and, for $\alpha \ne 0$, its matrix inverse are

The only noninvertible case is $\alpha =0$. Then $\Phi$ is the orthogonal projection onto ${\mathbf{n}}^{\perp }$, soThese dimensions agree with the rank-nullity theorem.