[{"content":" This translation is generated by GPT-5 and may contain errors. Please refer to the Chinese version for accuracy. The Wigner rotation states that the successive application of two Lorentz boosts in arbitrary directions to a reference frame is equivalent to a single Lorentz boost together with a spatial rotation.\nNamely $$ \\hat{Q}\\left(\\boldsymbol{\\beta}_{2}\\right) \\hat{Q}\\left(\\boldsymbol{\\beta}_{1}\\right) = \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} \\tag{1} $$This article aims to treat the general case of the Wigner rotation.\nVector Form of the Relativistic Transformation #We first derive the matrix form of an arbitrary Lorentz boost in vector notation. Let \\(\\boldsymbol{\\beta}\\) characterize the Lorentz boost. For an event with coordinates \\(( \\boldsymbol{x},\\ x^{0} = ct)\\) in frame \\(S\\), the spatial vector \\(\\boldsymbol{x}\\) can be decomposed into components parallel and perpendicular to the boost direction\n$$ \\begin{align} \\boldsymbol{x} \u0026 = x_{\\parallel } \\boldsymbol{e_{\\parallel}} + x_{\\bot} \\boldsymbol{e_{\\bot}} \\notag \\\\ \u0026 = \\boldsymbol{x} \\cdot \\frac{\\boldsymbol{\\beta}}{\\beta} \\frac{\\boldsymbol{\\beta}}{\\beta} + \\left( \\boldsymbol{x} - \\boldsymbol{x} \\cdot \\frac{\\boldsymbol{\\beta}}{\\beta} \\frac{\\boldsymbol{\\beta}}{\\beta} \\right) \\tag{2} \\end{align} $$Using the Lorentz transformation formulas we obtain\n$$ \\begin{align} \\boldsymbol{x'} \u0026 = \\gamma \\left( \\boldsymbol{x} \\cdot \\frac{\\boldsymbol{\\beta}}{\\beta} - \\beta x^0 \\right) \\frac{\\boldsymbol{\\beta}}{\\beta} + \\left( \\boldsymbol{x} - \\boldsymbol{x} \\cdot \\frac{\\boldsymbol{\\beta}}{\\beta} \\frac{\\boldsymbol{\\beta}}{\\beta} \\right) \\notag \\\\ \u0026 = \\left[ \\mathbf{I} + (\\gamma - 1) \\frac{\\boldsymbol{\\beta}}{\\beta} \\frac{\\boldsymbol{\\beta}}{\\beta} \\right] \\cdot \\boldsymbol{x} - \\gamma \\boldsymbol{\\beta} x^0 \\tag{3a} \\\\ x'^0 \u0026 = \\gamma x^0 - \\boldsymbol{\\beta} \\gamma \\cdot \\boldsymbol{x} \\tag{3b} \\end{align} $$This can be written in matrix form as\n$$ \\begin{bmatrix} \\boldsymbol{x'} \\\\ x'^0 \\end{bmatrix} = \\begin{bmatrix} \\mathbf{I} + (\\gamma - 1) \\frac{\\boldsymbol{\\beta}}{\\beta} \\frac{\\boldsymbol{\\beta}}{\\beta} \u0026 -\\boldsymbol{\\beta} \\gamma \\\\ -\\boldsymbol{\\beta} \\gamma \u0026 \\gamma \\end{bmatrix} \\begin{bmatrix} \\boldsymbol{x} \\\\ x^0 \\end{bmatrix} \\tag{4} $$Let \\(\\gamma = \\ch \\theta\\), then \\(\\beta \\gamma = \\sh \\theta\\).\nLet \\(\\frac{\\boldsymbol{\\beta}}{\\beta} = \\boldsymbol{e}\\) be the unit vector along the boost direction. Then\n$$ \\hat{Q}(\\boldsymbol{\\beta}) = \\begin{bmatrix} \\mathbf{I} + (\\ch \\theta - 1) \\boldsymbol{e} \\boldsymbol{e} \u0026 -\\sh \\theta \\boldsymbol{e} \\\\ -\\sh \\theta \\boldsymbol{e} \u0026 \\ch \\theta \\end{bmatrix} \\tag{5} $$Determining the Equivalent Lorentz Boost \\(\\boldsymbol{\\beta}\\) #For two successive Lorentz boosts in arbitrary directions we write\n$$ \\begin{align} \u0026 \\hat{Q}\\left(\\boldsymbol{\\beta}_{2}\\right) \\hat{Q}\\left(\\boldsymbol{\\beta}_{1}\\right) \\notag \\\\ \u0026 = \\begin{bmatrix} \\mathbf{I} + (\\ch \\theta_2 - 1) \\boldsymbol{e}_2 \\boldsymbol{e}_2 \u0026 -\\sh \\theta_2 \\boldsymbol{e}_2 \\\\ -\\sh \\theta_2 \\boldsymbol{e}_2 \u0026 \\ch \\theta_2 \\end{bmatrix} \\begin{bmatrix} \\mathbf{I} + (\\ch \\theta_1 - 1) \\boldsymbol{e}_1 \\boldsymbol{e}_1 \u0026 -\\sh \\theta_1 \\boldsymbol{e}_1 \\\\ -\\sh \\theta_1 \\boldsymbol{e}_1 \u0026 \\ch \\theta_1 \\end{bmatrix} \\notag \\\\ \u0026 = \\begin{bmatrix} \\mathbf{A}_{11} \u0026 \\boldsymbol{a}_{12} \\\\ \\boldsymbol{a}_{21} \u0026 a_{22} \\end{bmatrix} \\tag{6} \\end{align} $$ where $$ \\begin{align} \\mathbf{A}_{11} \u0026 = \\mathbf{I} + (\\ch\\theta_1 - 1)\\boldsymbol{e}_1\\boldsymbol{e}_1 + (\\ch\\theta_2 - 1)\\boldsymbol{e}_2\\boldsymbol{e}_2 + \\left[(\\ch\\theta_1 - 1)(\\ch\\theta_2 - 1)\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 + \\sh\\theta_1\\sh\\theta_2 \\right]\\boldsymbol{e}_2 \\boldsymbol{e}_1 \\notag \\quad \\text{(7a)} \\\\ \\boldsymbol{a}_{12} \u0026 = -\\sh\\theta_1\\boldsymbol{e}_1 - \\left[\\sh\\theta_1(\\ch\\theta_2 - 1)\\boldsymbol{e}_1\\cdot \\boldsymbol{e}_2 + \\sh\\theta_2\\ch\\theta_1\\right]\\boldsymbol{e}_2 \\tag{7b} \\\\ \\boldsymbol{a}_{21} \u0026 = -\\sh\\theta_2\\boldsymbol{e}_2 - \\left[\\sh\\theta_2(\\ch\\theta_1 - 1)\\boldsymbol{e}_1\\cdot \\boldsymbol{e}_2 + \\ch\\theta_2\\sh\\theta_1\\right]\\boldsymbol{e}_1 \\tag{7c} \\\\ a_{22} \u0026 = \\sh\\theta_1\\sh\\theta_2\\boldsymbol{e}_1\\cdot \\boldsymbol{e}_2 + \\ch\\theta_1\\ch\\theta_2 \\tag{7d} \\end{align} $$ We first look ahead at the relation to be established $$ \\hat{Q}\\left(\\boldsymbol{\\beta}_{2}\\right) \\hat{Q}\\left(\\boldsymbol{\\beta}_{1}\\right) = \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} \\tag{8} $$For the left-hand side $$ \\hat{Q}\\left(\\boldsymbol{\\beta}_{2}\\right) \\hat{Q}\\left(\\boldsymbol{\\beta}_{1}\\right) = \\begin{bmatrix} \\mathbf{A}_{11} \u0026 \\boldsymbol{a}_{12} \\\\ \\boldsymbol{a}_{21} \u0026 a_{22} \\end{bmatrix} \\tag{9} $$For the right-hand side $$ \\begin{align} \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} \u0026 = \\begin{bmatrix} \\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e} \u0026 -\\sh\\theta\\boldsymbol{e} \\\\ -\\sh\\theta\\boldsymbol{e} \u0026 \\ch\\theta \\end{bmatrix} \\begin{bmatrix} \\mathbf{R}_{11} \u0026 0 \\\\ 0 \u0026 1 \\end{bmatrix} \\notag \\\\ \u0026 = \\begin{bmatrix} (\\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e})\\cdot \\mathbf{R}_{11} \u0026 -\\sh\\theta\\boldsymbol{e} \\\\ -\\sh\\theta\\boldsymbol{e}\\cdot \\mathbf{R}_{11} \u0026 \\ch\\theta \\end{bmatrix} \\tag{10} \\end{align} $$ Note that here (and below) \\(\\boldsymbol{e} \\cdot \\mathbf{R}_{11}\\) denotes \\(e_i R_{ij}\\), i.e. \\(\\boldsymbol{e}^T \\mathbf{R}_{11}\\).\nMoreover, \\(\\mathbf{R}_{11}\\) is a (pure) rotation matrix.\nThus to verify equality of the two sides we proceed:\nVerify \\(a_{22}^2 - |\\boldsymbol{a}_{12}|^2 = 1\\), which allows extraction of \\n\u0026gt; \\(\\boldsymbol{\\beta}\\). Make an ansatz for \\(\\boldsymbol{\\beta}\\). Using the ansatz compute \\(\\hat{R} = \\hat{Q}^{-1}(\\boldsymbol{\\beta})\\hat{Q}(\\boldsymbol{\\beta_2})\\hat{Q}(\\boldsymbol{\\beta_1})\\). Show \\(\\hat{R}\\) is a pure spatial rotation; the ansatz is then confirmed. We can verify\n$$ \\begin{cases} a_{22}^2 - |\\boldsymbol{a}_{12}|^2 = 1 \\\\ a_{22}^2 - |\\boldsymbol{a}_{21}|^2 = 1 \\tag{11} \\end{cases} $$Hence define\n$$ \\begin{align} \\ch\\theta \u0026= a_{22} = \\sh\\theta_1\\sh\\theta_2\\boldsymbol{e}_1\\cdot\\boldsymbol{e}_2 + \\ch\\theta_1\\ch\\theta_2 \\tag{12a} \\\\ \\sh\\theta \u0026= |\\boldsymbol{a}_{12}| = |\\boldsymbol{a}_{21}| \\tag{12b} \\end{align} $$For the direction \\(\\boldsymbol{e}\\) of \\(\\boldsymbol{\\beta}\\) we posit $$ -\\sh\\theta\\ \\ \\boldsymbol{e} = \\boldsymbol{a}_{12} \\tag{13} $$ which gives $$ \\boldsymbol{e} = \\frac{\\sh\\theta_1\\boldsymbol{e}_1 + \\left[\\sh\\theta_1(\\ch\\theta_2 - 1)\\boldsymbol{e}_1\\cdot \\boldsymbol{e}_2 + \\sh\\theta_2\\ch\\theta_1\\right]\\boldsymbol{e}_2}{\\sqrt{\\sh^2\\theta_1 \\sh^2\\theta_2(\\boldsymbol{e}_1\\cdot \\boldsymbol{e}_2)^2 + \\ch^2\\theta_1\\ch^2\\theta_2 + 2\\sh\\theta_1\\sh\\theta_2\\ch\\theta_1\\ch\\theta_2(\\boldsymbol{e}_1\\cdot \\boldsymbol{e}_2) - 1}} \\quad \\text{(14)} $$Therefore $$ \\hat{Q}(\\boldsymbol{\\beta}) = \\begin{bmatrix} \\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e} \u0026 -\\sh\\theta\\boldsymbol{e} \\\\ -\\sh\\theta\\boldsymbol{e} \u0026 \\ch\\theta \\end{bmatrix} \\tag{15} $$Verifying the Spatial Rotation \\(\\hat{R}\\) #$$ \\begin{align} \\hat{R} \u0026 = \\hat{Q}^{-1}(\\boldsymbol{\\beta})\\hat{Q}(\\boldsymbol{\\beta_2})\\hat{Q}(\\boldsymbol{\\beta_1}) \\notag \\\\ \u0026 = \\hat{Q}(-\\boldsymbol{\\beta})\\hat{Q}(\\boldsymbol{\\beta_2})\\hat{Q}(\\boldsymbol{\\beta_1}) \\notag \\\\ \u0026 = \\begin{bmatrix} 1 + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e} \u0026 \\sh\\theta\\boldsymbol{e} \\\\ \\sh\\theta\\boldsymbol{e} \u0026 \\ch\\theta \\end{bmatrix} \\begin{bmatrix} \\mathbf{A}_{11} \u0026 \\boldsymbol{a}_{12} \\\\ \\boldsymbol{a}_{21} \u0026 a_{22} \\end{bmatrix} \\notag \\\\ \u0026 = \\begin{bmatrix} \\mathbf{R}_{11} \u0026 \\boldsymbol{r}_{12} \\\\ \\boldsymbol{r}_{21} \u0026 r_{22} \\end{bmatrix} \\tag{16} \\end{align} $$with\n$$ \\begin{align} r_{22} \u0026 = \\sh\\theta\\boldsymbol{e}\\cdot\\boldsymbol{a}_{12} + \\ch\\theta \\ a_{22} \\notag \\\\ \u0026 = \\sh\\theta\\boldsymbol{e} \\cdot (-\\sh\\theta\\boldsymbol{e}) + \\ch^2\\theta \\notag \\\\ \u0026 = -\\sh^2\\theta|\\boldsymbol{e}|^2 + \\ch^2\\theta = 1 \\tag{17a} \\\\ \\boldsymbol{r}_{12} \u0026 = \\left[ \\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e} \\right] \\cdot \\boldsymbol{a}_{12} + \\sh\\theta \\boldsymbol{e}\\ a_{22} \\notag \\\\ \u0026 = \\left[ \\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e} \\right] \\cdot (-\\sh\\theta \\boldsymbol{e}) + \\sh\\theta \\ch\\theta \\boldsymbol{e} \\notag \\\\ \u0026 = \\boldsymbol{0} \\tag{17b} \\\\ \\boldsymbol{r}_{21} \u0026 = \\sh\\theta \\boldsymbol{e} \\cdot \\mathbf{A}_{11} + \\ch\\theta \\boldsymbol{a}_{21} \\notag \\\\ \u0026 = -\\boldsymbol{a}_{12} \\cdot \\mathbf{A}_{11} + a_{22} \\boldsymbol{a}_{21} \\notag \\\\ \u0026 = \\boldsymbol{0} \\tag{17c} \\\\ \\mathbf{R}_{11} \u0026 = \\left[ \\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e} \\right] \\cdot \\mathbf{A}_{11} + \\sh\\theta \\boldsymbol{e} \\boldsymbol{a}_{21} \\tag{17d} \\\\ \\end{align} $$ We observe that \\(r_{22} = 1\\) and \\(\\boldsymbol{r}_{12}=\\boldsymbol{0}\\); thus indeed \\(-\\sh\\theta\\ \\ \\boldsymbol{e} = \\boldsymbol{a}_{12}\\), confirming the ansatz (13).\nNext we must show that \\(\\mathbf{R}_{11}\\) is a pure rotation matrix.\nSince $$ \\boldsymbol{r}_{21} = \\sh\\theta \\boldsymbol{e} \\cdot \\mathbf{A}_{11} + \\ch\\theta \\boldsymbol{a}_{21} = \\boldsymbol{0} \\tag{18} $$ we have $$ \\begin{align} \\mathbf{R}_{11} \u0026 = \\left[ \\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e} \\right] \\cdot \\mathbf{A}_{11} + \\sh\\theta \\boldsymbol{e} \\frac{ - \\sh\\theta\\boldsymbol{e} \\cdot \\mathbf{A}_{11}}{\\ch\\theta} \\notag \\\\ \u0026 = \\left[ \\mathbf{I} - \\frac{\\ch\\theta - 1}{\\ch\\theta} \\boldsymbol{e}\\boldsymbol{e} \\right] \\cdot \\mathbf{A}_{11} \\notag \\\\ \u0026 = \\left[ \\mathbf{I} - \\frac{1}{\\ch\\theta (\\ch\\theta + 1)} \\boldsymbol{a}_{12} \\boldsymbol{a}_{12} \\right] \\cdot \\mathbf{A}_{11} \\notag \\\\ \u0026 = \\mathbf{I} + b_{11} \\boldsymbol{e}_1 \\boldsymbol{e}_1 + b_{12} \\boldsymbol{e}_1 \\boldsymbol{e}_2 + b_{21} \\boldsymbol{e}_2 \\boldsymbol{e}_1 + b_{22} \\boldsymbol{e}_2 \\boldsymbol{e}_2 \\tag{19} \\end{align} $$ We have written \\(\\mathbf{R}_{11}\\) in the form (19), i.e. $$ \\begin{align} \\mathbf{R}_{11} \u0026 = \\mathbf{I} + \\sum b_{ij} \\boldsymbol{e}_i \\boldsymbol{e}_j \\notag \\\\ \u0026 = \\mathbf{I} + \\begin{bmatrix} \\boldsymbol{e}_1 \u0026 \\boldsymbol{e}_2 \\end{bmatrix} \\begin{bmatrix} b_{11} \u0026 b_{12} \\\\ b_{21} \u0026 b_{22} \\end{bmatrix} \\begin{bmatrix} \\boldsymbol{e}_1 \\\\ \\boldsymbol{e}_2 \\end{bmatrix} \\tag{20} \\end{align} $$ To show it is a rotation matrix we must verify $$ \\mathbf{R}_{11} \\mathbf{R}_{11}^T = \\mathbf{I} \\tag{21} $$ $$ \\det \\mathbf{R}_{11} = 1 \\tag{22} $$Using (20), equation (21) can be simplified as\n$$ \\begin{bmatrix} b_{11} \u0026 b_{12} \\\\ b_{21} \u0026 b_{22} \\end{bmatrix} + \\begin{bmatrix} b_{11} \u0026 b_{21} \\\\ b_{12} \u0026 b_{22} \\end{bmatrix} + \\begin{bmatrix} b_{11} \u0026 b_{12} \\\\ b_{21} \u0026 b_{22} \\end{bmatrix} \\begin{bmatrix} 1 \u0026 \\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 \\\\ \\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 \u0026 1 \\end{bmatrix} \\begin{bmatrix} b_{11} \u0026 b_{21} \\\\ b_{12} \u0026 b_{22} \\end{bmatrix} = 0 \\tag{23} $$For (22), multiplying (20) on the right by \\(\\begin{bmatrix} \\boldsymbol{e}_1 \u0026 \\boldsymbol{e}_2 \\end{bmatrix}\\) gives $$ \\mathbf{R}_{11} \\begin{bmatrix} \\boldsymbol{e}_1 \u0026 \\boldsymbol{e}_2 \\end{bmatrix} = \\begin{bmatrix} \\boldsymbol{e}_1 \u0026 \\boldsymbol{e}_2 \\end{bmatrix} \\left( \\mathbf{I} + \\begin{bmatrix} b_{11} \u0026 b_{12} \\\\ b_{21} \u0026 b_{22} \\end{bmatrix} \\begin{bmatrix} 1 \u0026 \\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 \\\\ \\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 \u0026 1 \\end{bmatrix} \\tag{24} \\right) $$Since $$ \\det \\begin{bmatrix} \\boldsymbol{e}_1 \u0026 \\boldsymbol{e}_2 \\end{bmatrix} = \\sqrt{1 - (\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2)^2} \\neq 0 \\tag{25} $$ we obtain $$ \\det \\mathbf{R}_{11} = \\det \\left( \\mathbf{I} + \\begin{bmatrix} b_{11} \u0026 b_{12} \\\\ b_{21} \u0026 b_{22} \\end{bmatrix} \\begin{bmatrix} 1 \u0026 \\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 \\\\ \\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2 \u0026 1 \\end{bmatrix} \\right) = 1 \\tag{26} $$ Thus, we need to verify (23) and (26).\nThe coefficients are $$ \\begin{align} \u0026 b_{11} = b_{22} = - \\frac{(\\ch\\theta_1 - 1)(\\ch\\theta_2 - 1)}{\\ch\\theta + 1} \\tag{27a} \\\\ \u0026 b_{12} = -\\frac{\\sh\\theta_1 \\sh\\theta_2}{\\ch\\theta + 1} \\tag{27b} \\\\ \u0026 b_{21} = \\frac{2(\\ch\\theta_1 - 1)(\\ch\\theta_2 - 1)(\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2) + \\sh\\theta_1 \\sh\\theta_2}{\\ch\\theta + 1} \\tag{27c} \\\\ \\end{align} $$They satisfy $$ \\begin{cases} b_{11} = b_{22}\\\\ (b_{11}+1)(b_{22}+1) - b_{12}b_{21} = 1 \\\\ b_{12} + b_{21} + 2b_{22}(\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2) = 0 \\end{cases} \\tag{28} $$ Hence \\(\\mathbf{R}_{11}\\) is a rotation matrix.\nTherefore $$ \\hat{R}= \\begin{bmatrix} \\mathbf{R}_{11} \u0026 0 \\\\ 0 \u0026 1 \\end{bmatrix} \\tag{29} $$ corresponding to a spatial rotation.\nVector Form of the Rotation Matrix \\(\\mathbf{R}_{11}\\) # We first recall the matrix representation of an arbitrary rotation to guide simplification and analysis.\nAny rotation can be described by a unit rotation axis \\(\\boldsymbol{n}\\) and rotation angle \\(\\phi\\).\nAs shown in the figure\nFigure 1 We have $$ \\boldsymbol{r}' = \\boldsymbol{r}_\\parallel + \\boldsymbol{r}'_\\bot \\tag{30}$$where $$ \\boldsymbol{r}_\\parallel = (\\boldsymbol{r} \\cdot \\boldsymbol{n}) \\boldsymbol{n} \\tag{31} $$ $$ \\boldsymbol{r}'_\\bot = \\boldsymbol{r}_\\bot \\cos\\phi + (\\boldsymbol{n} \\times \\boldsymbol{r}_\\bot) \\sin\\phi \\tag{32} $$Thus $$ \\boldsymbol{r}' = (\\boldsymbol{r} \\cdot \\boldsymbol{n}) \\boldsymbol{n} + (\\boldsymbol{r} - (\\boldsymbol{r} \\cdot \\boldsymbol{n}) \\boldsymbol{n}) \\cos\\phi + (\\boldsymbol{n} \\times \\boldsymbol{r}) \\sin\\phi \\tag{33} $$i.e. $$ r'_i = \\left[ n_i n_j + (\\delta_{ij} - n_i n_j) \\cos\\phi - \\varepsilon_{ijk} n_k \\sin\\phi \\right] r_j \\tag{34} $$yielding the rotation matrix $$ R_{ij} = n_i n_j + (\\delta_{ij} - n_i n_j) \\cos\\phi - \\varepsilon_{ijk} n_k \\sin\\phi \\tag{35} $$ We now decompose the identity \\(\\mathbf{I}\\).\nOur goal is to decompose the identity appearing in (19). The ambient space is \\(\\mathbb{R}^3\\); hence \\( (\\operatorname{span}\\{\\boldsymbol{e}_1, \\boldsymbol{e}_2\\})^\\perp \\cong \\mathbb{R}^1\\). Let \\(\\boldsymbol{e}_{\\bot}\\) be a unit vector spanning this orthogonal complement.\nLet $$ \\mathbf{I} = C_{11} \\boldsymbol{e}_1\\boldsymbol{e}_1 + C_{12} \\boldsymbol{e}_1\\boldsymbol{e}_2 + C_{21} \\boldsymbol{e}_2\\boldsymbol{e}_1 + C_{22} \\boldsymbol{e}_2\\boldsymbol{e}_2 + \\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} \\tag{36} $$ Then $$ C_{11} = C_{22} = \\frac{1}{1 - (\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2)^2} \\tag{37} $$ $$ C_{12} = C_{21} = -\\frac{\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2}{1 - (\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2)^2} \\tag{38} $$Thus $$ \\begin{align} \\mathbf{R}_{11} \u0026= \\left( 1 + \\frac{b_{11}}{C_{11}} \\right) \\mathbf{I} - \\frac{b_{11}}{C_{11}} \\mathbf{I} + b_{11} \\boldsymbol{e}_1 \\boldsymbol{e}_1 + b_{12} \\boldsymbol{e}_1 \\boldsymbol{e}_2 + b_{21} \\boldsymbol{e}_2 \\boldsymbol{e}_1 + b_{22} \\boldsymbol{e}_2 \\boldsymbol{e}_2 \\notag \\\\ \u0026= \\left( 1 + \\frac{b_{11}}{C_{11}} \\right) \\mathbf{I} + \\left( b_{11} - b_{11}\\frac{C_{11}}{C_{11}} \\right) \\boldsymbol{e}_1\\boldsymbol{e}_1 + \\left( b_{12} - b_{11}\\frac{C_{12}}{C_{11}} \\right) \\boldsymbol{e}_1\\boldsymbol{e}_2 \\notag \\\\ \u0026\\phantom{{}= \\left( 1 + \\frac{b_{11}}{C_{11}} \\right) \\mathbf{I}} + \\left( b_{21} - b_{11}\\frac{C_{21}}{C_{11}} \\right) \\boldsymbol{e}_2\\boldsymbol{e}_1 + \\left( b_{22} - b_{11}\\frac{C_{22}}{C_{11}} \\right) \\boldsymbol{e}_2\\boldsymbol{e}_2 \\notag \\\\ \u0026\\phantom{{}= \\left( 1 + \\frac{b_{11}}{C_{11}} \\right) \\mathbf{I}} - \\frac{b_{11}}{C_{11}} \\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} \\notag \\\\ \u0026= \\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} + \\left( \\mathbf{I} -\\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} \\right) \\left( 1 + \\frac{b_{11}}{C_{11}} \\right) + \\left( b_{12} - b_{11}\\frac{C_{12}}{C_{11}} \\right) \\boldsymbol{e}_1\\boldsymbol{e}_2 \\notag \\\\ \u0026\\phantom{= \\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} + \\left( \\mathbf{I} -\\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} \\right) \\left( 1 + \\frac{b_{11}}{C_{11}} \\right)} + \\left( b_{21} - b_{11}\\frac{C_{21}}{C_{11}} \\right) \\boldsymbol{e}_2\\boldsymbol{e}_1 \\notag \\\\ \u0026= \\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} + \\left( \\mathbf{I} -\\boldsymbol{e}_{\\bot} \\boldsymbol{e}_{\\bot} \\right) \\left[ \\alpha (\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2) + \\beta \\right] + \\alpha \\left[ \\boldsymbol{e}_2\\boldsymbol{e}_1 - \\boldsymbol{e}_1\\boldsymbol{e}_2 \\right] \\tag{39} \\end{align} $$Or $$ (\\mathbf{R}_{11})_{ij} = (\\boldsymbol{e}_{\\bot})_i (\\boldsymbol{e}_{\\bot})_j + \\left[ \\delta_{ij} - (\\boldsymbol{e}_{\\bot})_i (\\boldsymbol{e}_{\\bot})_j \\right] \\left[ \\alpha (\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2) + \\beta \\right] - \\alpha \\varepsilon_{ijk} (\\boldsymbol{e}_1 \\times \\boldsymbol{e}_2)_k \\quad \\text{(40)} $$ with $$ \\begin{cases} \\alpha = \\frac{\\sh\\theta_1\\sh\\theta_2 + (\\ch\\theta_1-1)(\\ch\\theta_2-1)(\\boldsymbol{e_1} \\cdot \\boldsymbol{e_2})}{\\ch\\theta+1} \\\\ \\beta = \\frac{\\ch\\theta_1+\\ch\\theta_2}{\\ch\\theta+1} \\end{cases} \\tag{41} $$Denote the rotation angle by \\(\\phi\\) and its axis by \\(\\boldsymbol{n}\\). Comparing with (35) we identify $$ \\begin{aligned} \\boldsymbol{n} \\sin\\phi \u0026 = \\alpha (\\boldsymbol{e}_1 \\times \\boldsymbol{e}_2) \\notag \\\\ \u0026 = \\frac{\\sh\\theta_1 \\sh\\theta_2 + (\\ch\\theta_1 - 1)(\\ch\\theta_2 - 1)(\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2)}{\\ch\\theta + 1} \\boldsymbol{e}_1 \\times \\boldsymbol{e}_2 \\notag \\\\ \u0026 = \\frac{\\sh\\theta_1 \\sh\\theta_2 + (\\ch\\theta_1 - 1)(\\ch\\theta_2 - 1)(\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2)}{\\sh\\theta_1 \\sh\\theta_2 (\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2) + \\ch\\theta_1 \\ch\\theta_2 + 1} \\boldsymbol{e}_1 \\times \\boldsymbol{e}_2 \\quad \\text{(42)} \\end{aligned} $$ Supplementary Material #(i) Order of the Equivalent Boost and Rotation #The equivalent transformation considered above is $$ \\hat{Q}\\left(\\boldsymbol{\\beta}_{2}\\right) \\hat{Q}\\left(\\boldsymbol{\\beta}_{1}\\right) = \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} \\tag{43} $$ That is: rotation first, then the (net) Lorentz boost.\nIn fact there are two equivalent forms $$ \\hat{Q}\\left(\\boldsymbol{\\beta}_{2}\\right) \\hat{Q}\\left(\\boldsymbol{\\beta}_{1}\\right) = \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} = \\hat{R}' \\hat{Q}\\left(\\boldsymbol{\\beta}' \\right) \\tag{44} $$We compare the two. Interchanging the order changes the off-diagonal block elements $$ \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} = \\begin{bmatrix} (\\mathbf{I} + (\\ch\\theta - 1)\\boldsymbol{e}\\boldsymbol{e})\\cdot \\mathbf{R}_{11} \u0026 -\\sh\\theta\\boldsymbol{e} \\\\ -\\sh\\theta\\boldsymbol{e}\\cdot \\mathbf{R}_{11} \u0026 \\ch\\theta \\end{bmatrix} \\tag{45} $$ $$ \\hat{R}' \\hat{Q}\\left(\\boldsymbol{\\beta}' \\right) = \\begin{bmatrix} \\mathbf{R}'_{11} \\cdot (\\mathbf{I} + (\\ch\\theta' - 1)\\boldsymbol{e}'\\boldsymbol{e}') \u0026 -\\sh\\theta'\\ \\mathbf{R}'_{11} \\cdot \\boldsymbol{e}' \\\\ -\\sh\\theta'\\ \\boldsymbol{e}' \u0026 \\ch\\theta' \\end{bmatrix} \\tag{46} $$We see that \\(|\\boldsymbol{\\beta}'| = |\\boldsymbol{\\beta}|\\) but the direction is rotated $$ \\begin{cases} \\boldsymbol{\\beta} = \\mathbf{R}'_{11} \\cdot \\boldsymbol{\\beta}'\\\\ \\boldsymbol{\\beta}' = \\boldsymbol{\\beta} \\cdot \\mathbf{R}_{11} \\end{cases} \\tag{47} $$ or $$ \\begin{cases} \\beta_i = R'_{ij} \\beta'_j \\\\ \\beta'_i = \\beta_j R_{ji} \\end{cases} \\tag{48} $$ Thus $$ \\beta_i = R'_{ij} \\beta_k R_{kj} \\tag{49}$$ so that $$ R'_{ij} R_{kj} = \\delta_{ik} \\tag{50}$$ or $$ \\mathbf{R}'_{11} \\mathbf{R}_{11}^T = \\mathbf{I} \\tag{51}$$ which implies $$ \\mathbf{R}'_{11} = \\mathbf{R}_{11} \\tag{52}$$ Therefore $$ \\hat{Q}\\left(\\boldsymbol{\\beta}_{2}\\right) \\hat{Q}\\left(\\boldsymbol{\\beta}_{1}\\right) = \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} = \\hat{R} \\hat{Q}\\left( \\mathbf{R}_{11}^T \\cdot \\boldsymbol{\\beta} \\right) \\tag{53} $$ Alternatively, $$\\hat{Q}\\left( \\mathbf{R}_{11}^T \\cdot \\boldsymbol{\\beta} \\right) = \\hat{R}^T \\hat{Q}\\left(\\boldsymbol{\\beta} \\right) \\hat{R} \\tag{54}$$ which is immediate from spatial rotational symmetry of the Lorentz group.\n(ii) Low-Velocity Approximation #Consider the limit \\(\\beta_1, \\beta_2 \\to 0\\). Then $$ \\begin{align} \\boldsymbol{n} \\sin\\phi \u0026 = \\frac{\\gamma_1 \\gamma_2 \\beta_1 \\beta_2 + (\\gamma_1 - 1)(\\gamma_2 - 1)(\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2)}{\\gamma_1 \\gamma_2 \\beta_1 \\beta_2 (\\boldsymbol{e}_1 \\cdot \\boldsymbol{e}_2) + \\gamma_1 \\gamma_2 + 1} \\boldsymbol{e}_1 \\times \\boldsymbol{e}_2 \\notag \\\\ \u0026 \\approx \\frac{1}{2} \\beta_1 \\beta_2 \\boldsymbol{e}_1 \\times \\boldsymbol{e}_2 \\notag \\\\ \u0026 = \\frac{1}{2} \\boldsymbol{\\beta}_1 \\times \\boldsymbol{\\beta}_2 \\tag{55} \\end{align} $$Thus \\(\\sin\\phi \\approx \\phi\\) and $$ \\boldsymbol{n} \\phi = \\frac{1}{2} \\boldsymbol{\\beta}_1 \\times \\boldsymbol{\\beta}_2 \\tag{56} $$(iii) Thomas Precession #Thomas precession states that the proper frame of a particle moving along a curved trajectory (i.e. with acceleration having a component perpendicular to its velocity) undergoes a purely relativistic precession relative to the laboratory frame.\nAt times \\(t\\) and \\(t+dt\\) let the instantaneous rest frames be \\(S\\) and \\(S'\\), respectively. The transformation between them can be expressed as the composition of two boosts\n$$ S \\to S' :\\quad \\hat{\\Lambda} = \\hat{Q}\\!\\left(\\boldsymbol{\\beta}+d\\boldsymbol{\\beta}\\right)\\hat{Q}\\!\\left(\\boldsymbol{- \\beta}\\right) \\tag{57} $$Here \\(\\hat{Q}\\!\\left(\\boldsymbol{- \\beta}\\right)\\) takes \\(S\\) back to the lab frame, then \\(\\hat{Q}\\!\\left(\\boldsymbol{\\beta}+d\\boldsymbol{\\beta}\\right)\\) carries the lab to the new instantaneous rest frame \\(S'\\). With $$\\boldsymbol{\\beta} = \\frac{\\boldsymbol{v}}{c}, \\quad d\\boldsymbol{\\beta} = \\frac{\\boldsymbol{a} dt}{c} \\tag{58}$$ Using (42) and \\(d\\boldsymbol{\\beta} \\ll \\boldsymbol{\\beta}\\), keeping first order $$ \\boldsymbol{n} \\sin d\\phi \\approx \\frac{\\gamma^2 \\beta^2 - (\\gamma - 1)^2}{ - \\gamma^2 \\beta^2 + \\gamma^2 + 1} \\frac{ - \\boldsymbol{\\beta} \\times d\\boldsymbol{\\beta}}{\\beta^2} = \\frac{ 1- \\gamma }{\\beta^2} \\boldsymbol{\\beta} \\times d\\boldsymbol{\\beta} = \\frac{\\gamma^2}{\\gamma + 1} d\\boldsymbol{\\beta} \\times \\boldsymbol{\\beta} \\tag{59} $$ Thus $$\\boldsymbol{n} d{\\phi} = \\frac{\\gamma^2}{\\gamma + 1} d\\boldsymbol{\\beta} \\times \\boldsymbol{\\beta} = \\frac{\\gamma^2}{\\gamma + 1} \\frac{dt}{c^2} \\boldsymbol{a} \\times \\boldsymbol{v} \\tag{60}$$ The Thomas precession angular velocity is $$ \\boldsymbol{\\omega}_T = \\frac{\\gamma^2}{\\gamma + 1} \\frac{ \\boldsymbol{a} \\times \\boldsymbol{v} }{c^2} \\tag{61} $$ For \\(v \\ll c\\) $$ \\boldsymbol{\\omega}_T \\approx \\frac{1}{2} \\frac{ \\boldsymbol{a} \\times \\boldsymbol{v} }{c^2} \\tag{62} $$ In the Bohr model of hydrogen this yields $$ \\omega_T = \\frac{\\mu_0}{8\\pi} \\frac{m e^8}{(4\\pi \\epsilon_0)^3 (n \\hbar)^5} \\tag{63} $$(iv) Algebraic Structure #The Wigner rotation originates from the non-vanishing commutator of the Lorentz boost generators \\(\\hat{\\boldsymbol{K}}\\).\nLet the rapidities of the two boosts be \\(\\boldsymbol{\\eta}^1, \\boldsymbol{\\eta}^2\\)\n$$ \\boldsymbol{\\eta}^1 = (\\sh^{-1} \\beta_{1}\\gamma_1) \\frac{\\boldsymbol{\\beta}_1}{\\beta_1} \\tag{64} $$ $$ \\boldsymbol{\\eta}^2 = (\\sh^{-1} \\beta_{2}\\gamma_2) \\frac{\\boldsymbol{\\beta}_2}{\\beta_2} \\tag{65} $$Hence $$ \\hat{Q}(\\boldsymbol{\\beta}_1) = \\exp\\left(i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}} \\right) \\tag{66} $$ $$ \\hat{Q}(\\boldsymbol{\\beta}_2) = \\exp\\left(i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}\\right) \\tag{67} $$Since $$ \\hat{Q}(\\boldsymbol{\\beta}_2) \\hat{Q}(\\boldsymbol{\\beta}_1) = \\hat{Q}(\\boldsymbol{\\beta}) \\hat{R} \\tag{68} $$ we have $$ \\exp(i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}) \\exp(i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}}) = \\exp(i \\boldsymbol{\\eta} \\cdot \\hat{\\boldsymbol{K}}) \\hat{R} \\tag{69} $$ By symmetry, $$ \\exp(i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}}) \\exp(i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}) = \\exp(i \\boldsymbol{\\eta} \\cdot \\hat{\\boldsymbol{K}}) \\hat{R}^{-1} \\tag{70} $$ Thus $$ \\exp(i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}) \\exp(i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}}) = \\exp(i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}}) \\exp(i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}) \\hat{R}^2 \\tag{71} $$Using the Baker-Campbell-Hausdorff formula\n$$ \\hat{R}^2 = \\exp\\left( \\left[i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}, i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}}\\right] + \\frac{1}{2}\\left[ \\left[i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}, i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}} \\right], i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}} + i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}}\\right] + \\cdots \\right) \\quad \\text{(72)} $$In the low-velocity limit \\(\\beta_1, \\beta_2 \\to 0\\)\n$$ \\begin{align} \\hat{R}^2 \u0026 \\approx \\exp\\left(\\left[i \\boldsymbol{\\eta}^2 \\cdot \\hat{\\boldsymbol{K}}, i \\boldsymbol{\\eta}^1 \\cdot \\hat{\\boldsymbol{K}}\\right]\\right) \\notag \\\\ \u0026 =\\exp\\left(\\left[i {\\eta}^2_i \\hat{K}_i, i {\\eta}^1_j \\hat{K}_j \\right]\\right) \\notag \\\\ \u0026 =\\exp\\left(- {\\eta}^2_i {\\eta}^1_j \\left[\\hat{K}_i, \\hat{K}_j \\right]\\right) \\notag \\\\ \u0026 = \\exp\\left( i {\\eta}^2_i {\\eta}^1_j \\varepsilon_{ijk} \\hat{J}_k \\right) \\notag \\\\ \u0026 = \\exp\\left( - i (\\boldsymbol{\\eta}^1 \\times \\boldsymbol{\\eta}^2 ) \\cdot \\hat{\\boldsymbol{J}} \\right) \\tag{73} \\end{align} $$ Hence $$ \\hat{R} \\approx \\exp\\left( - \\frac{i}{2} (\\boldsymbol{\\eta}^1 \\times \\boldsymbol{\\eta}^2 ) \\cdot \\hat{\\boldsymbol{J}} \\right) \\tag{74} $$Rotation angle $$ \\boldsymbol{n} \\phi \\approx \\frac{1}{2} \\boldsymbol{\\eta}^1 \\times \\boldsymbol{\\eta}^2 \\approx \\frac{1}{2} \\boldsymbol{\\beta}_1 \\times \\boldsymbol{\\beta}_2 \\tag{75} $$","date":"19 August 2025","permalink":"https://trojanpt.github.io/en/jottings/wigner-rotation/","section":"Jottings","summary":"The Wigner rotation states that the successive action of two Lorentz boosts in arbitrary directions is equivalent to a single Lorentz boost followed by a spatial rotation. This article discusses the general case.","title":"General Case of the Wigner Rotation"},{"content":"Welcome to Trojan TeaHouse! Come on in, enjoy a cup of tea before you go. ο(=•ω＜=)ρ ~☆\n","date":null,"permalink":"https://trojanpt.github.io/en/","section":"Home","summary":"","title":"Home"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/jottings/","section":"Jottings","summary":"","title":"Jottings"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/tags/physics/","section":"Tags","summary":"","title":"Physics"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/tags/relativity/","section":"Tags","summary":"","title":"Relativity"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/tags/","section":"Tags","summary":"","title":"Tags"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/tags/mechanics/","section":"Tags","summary":"","title":"Mechanics"},{"content":" This translation is generated by DeepSeek and may contain errors. Please refer to the Chinese version for accuracy. Statement of the Problem #The treatment of first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system (e.g., with \\(n\\) degrees of freedom) involves solving the system of differential equations $$ \\bm{\\ddot{x}} = -\\mathbf{A} \\bm{x} \\tag{1} $$ The solution can be formally written as $$ \\bm{x}=\\bm{x_1}\\exp\\left(i\\mathbf{A^\\frac{1}{2}}\\ t \\right) + \\bm{x_2}\\exp\\left(-i \\mathbf{A^\\frac{1}{2}}\\ t \\right) \\tag{2} $$ Assuming \\(\\mathbf{A}\\) is diagonalizable, we can set $$ \\mathbf{A} = \\mathbf{P}\\mathbf{\\Lambda}\\mathbf{P^{-1}} \\tag{3} $$ Using the Taylor expansion, equation (2) transforms into $$ \\bm{x}=\\bm{x_1}\\mathbf{P}\\exp\\left(i\\mathbf{\\Lambda^\\frac{1}{2}}\\ t \\right)\\mathbf{P^{-1}} + \\bm{x_2}\\mathbf{P}\\exp\\left(- i \\mathbf{\\Lambda^\\frac{1}{2}}\\ t \\right)\\mathbf{P^{-1}} \\tag{4} $$ where $$ \\exp(i\\mathbf{\\Lambda^{\\frac{1}{2}}}\\ t) = \\begin{bmatrix} \\exp(i\\lambda_1^{\\frac{1}{2}}\\ t) \u0026 0 \u0026 0 \u0026 \\\\ 0 \u0026 \\exp(i\\lambda_2^{\\frac{1}{2}}\\ t) \u0026 0 \u0026 \\cdots \\\\ 0 \u0026 0 \u0026 \\exp(i\\lambda_3^{\\frac{1}{2}}\\ t) \u0026 \\\\ \u0026 \\cdots \u0026 \u0026 \\end{bmatrix} \\tag{5} $$ Thus, equation (4) represents the solution to the system of differential equations.\nAnother more common approach found in many textbooks is to introduce a trial solution for (1) $$ \\bm{x} = \\bm{x_0} \\cos (\\omega t + \\varphi) \\tag{6} $$ Substituting yields $$ (\\mathbf{A}-\\omega^2\\mathbf{I})\\bm{x_0}=0 \\tag{7} $$ For non-trivial solutions \\(\\bm{x_0}\\), we require $$ \\det(\\mathbf{A} - \\mathbf{\\omega^2} \\mathbf{I}) = 0 \\tag{8} $$ This solves for the normal mode frequencies \\(\\omega_i\\) and the corresponding eigenvectors \\(\\bm{x_i}\\) \\((i=1,2,\\cdots, n)\\). The solution to (1) is then $$ \\bm{x}=\\sum_i \\ \\bm{x_i} \\cos (\\omega_i t + \\varphi_i) \\tag{9} $$ In many mechanics textbooks, the following is stated without proof\nThe eigenvalues \\(\\lambda_i\\) of the coefficient matrix \\(\\mathbf{A}\\) are positive (\\(\\lambda_i \u003e0\\)). \\(\\mathbf{A}\\) is diagonalizable, meaning its geometric multiplicity equals \\(n\\). This implies that for first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system, there exist normal coordinates equal in number to its degrees of freedom.\n\\(\\lambda_i \u003e0\\) means the motion described is oscillatory (normal modes). The geometric multiplicity of \\(\\mathbf{A}\\) equaling \\(n\\) implies the number of normal coordinates matches the system\u0026rsquo;s degrees of freedom. Hence the statement: there exist normal coordinates equal in number to its degrees of freedom.\nThis article aims to discuss the existence of these conditions.\nSimilarly, under the alternative approach, the statement \u0026lsquo;solves for the normal mode frequencies \\(\\omega_i\\) and the corresponding eigenvectors \\(\\bm{x_i}\\) \\((i=1,2,\\cdots, n)\\)\u0026rsquo; is not self-evident. Potential issues are\n\\(\\omega_i\\) being imaginary ( \\(\\omega_i^2 \u003c 0\\) ). The geometric multiplicity of \\(\\mathbf{A}\\) being less than \\(n\\). Even over \\(\\mathbb{C}\\), \\(n\\) linearly independent eigenvectors \\(\\bm{x_i}\\) may not exist if there are degenerate eigenvalues. Consequently, \\(\\bm{x}\\) could have particular solutions not expressible as trigonometric functions (or finite sums thereof). These two points are equivalent to the two points above. We need to discuss whether such situations can occur.\nExistence of Normal Coordinates #For first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system, the equations of motion can be written as $$ \\mathbf{M}\\bm{\\ddot{x}}+\\mathbf{K}\\bm{x}=0 \\tag{10} $$ where the matrices \\(\\mathbf{M}\\) and \\(\\mathbf{K}\\) are both real, symmetric, and positive definite. The coefficient matrix \\(\\mathbf{A}\\) in (1) can then be expressed as $$ \\mathbf{A} = \\mathbf{M^{-1}K} \\tag{11} $$ Given that \\(\\mathbf{M}\\) is positive definite, this can be rewritten as $$ \\mathbf{A} = \\mathbf{M^{-\\frac{1}{2}}\\ M^{-\\frac{1}{2}}\\ K\\ M^{-\\frac{1}{2}}\\ M^{\\frac{1}{2}}} \\tag{12} $$ Here, \\(\\mathbf{M^{-\\frac{1}{2}}\\ K\\ M^{-\\frac{1}{2}}}\\) is a real symmetric matrix. Therefore, it can be diagonalized $$ \\mathbf{M^{-\\frac{1}{2}}\\ K\\ M^{-\\frac{1}{2}}} = \\mathbf{P\\Lambda P^{-1}} \\tag{13} $$ Substituting back gives $$ \\mathbf{A} = \\mathbf{M^{-\\frac{1}{2}} P\\Lambda P^{-1} M^{\\frac{1}{2}}} \\tag{14} $$ This demonstrates that $$ \\boxed{\\text{\\textbf{The coefficient matrix \\(\\mathbf{A}\\) is diagonalizable}}} \\notag $$ Consider the eigenvalue equation for \\(\\mathbf{A}\\) $$ \\mathbf{A}\\bm{x} = \\mathbf{M^{-1}K}\\bm{x} = \\lambda \\bm{x} \\tag{15} $$ which is equivalent to the generalized eigenvalue problem $$ \\mathbf{K}\\bm{x} = \\lambda\\ \\mathbf{M}\\bm{x} \\tag{16} $$ Left-multiplying both sides by \\(\\bm{x^T}\\) yields $$ \\bm{x^T} \\mathbf{K} \\bm{x} = \\lambda\\ \\bm{x^T} \\mathbf{M}\\bm{x} \\tag{17} $$ Since \\(\\mathbf{M}\\) and \\(\\mathbf{K}\\) are positive definite matrices, we have $$ \\bm{x^T} \\mathbf{K} \\bm{x} \u003e 0 \\quad \\text{and} \\quad \\bm{x^T} \\mathbf{M}\\bm{x} \u003e 0 \\quad \\text{(for } \\bm{x} \\neq \\bm{0}\\text{)} \\tag{18} $$ Therefore $$ \\lambda = \\frac{\\bm{x^T} \\mathbf{K} \\bm{x}}{\\bm{x^T} \\mathbf{M}\\bm{x}} \u003e 0 \\tag{19} $$ This proves that $$ \\boxed{\\text{\\textbf{The eigenvalues \\(\\lambda\\) of the coefficient matrix \\(\\mathbf{A}\\) are positive}}} \\notag $$ That means, for first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system, there exist normal coordinates equal in number to its degrees of freedom.\n","date":"23 July 2025","permalink":"https://trojanpt.github.io/en/jottings/normal-mode/","section":"Jottings","summary":"For the first-order small vibrations of a mechanical system near its stable equilibrium position, there exist normal modes equal in number to its degrees of freedom.","title":"Verification of the Existence of Normal Coordinates for Small Vibrations"},{"content":" This translation is generated by DeepSeek and may contain errors. Please refer to the Chinese version for accuracy. ​\tFeynman mentions in his lectures on physics Vol. I, §39-4 that if the pressures on both sides of a smooth adiabatic piston are equal but the temperatures are unequal, energy will transfer from the higher-temperature gas to the lower-temperature gas through microscopic vibrations of the piston, even when all heat conduction is neglected.\n​\tThis article attempts to examine microscopically whether the piston can achieve equilibrium and the process of reaching the final equilibrium state.\nWe begin our discussion with a simple model: As shown in Fig. 1, a smooth adiabatic piston divides an adiabatic container into two compartments. For the piston, the cross-sectional area is denoted by \\(A\\) and its mass by \\(M\\). For the gas, the mass of a single molecule is denoted by \\(m ({\\ll} M)\\), the constant-pressure molar heat capacity by \\(C^{mol}_p\\), and the gas can be treated as ideal.\nIn the initial state, the amount of substance of the left gas is \\(\\nu_1\\), pressure \\(p_0\\), temperature \\(T_{10}\\); the right gas has amount of substance \\(\\nu_2\\), pressure \\(p_0\\), and temperature \\(T_{20} (\u003e T_{10})\\).\nAt some time \\(t\\) during the process, let the pressure of the left gas be \\(p\\), temperature \\(T_{1}\\); the right gas has pressure \\(p\\), temperature \\(T_{2}\\). The velocity of the piston is \\(u\\) (defined as positive to the right).\nFig. 1 Collision Between Gas Molecules and the Piston #At a moment when the piston\u0026rsquo;s velocity is \\(u\\), consider a gas molecule on the left with velocity component perpendicular to the piston \\(v_x\\).\nAfter collision with the piston, its velocity becomes $$ v_x' =\\frac{\\left ( m-M \\right )v_x+2Mu }{M+m} \\tag{1} $$Energy Change #The energy change of this gas molecule before and after collision is $$ \\begin{align} \\triangle \\varepsilon\\left ( v_x \\right )=\u0026\\frac{1}2{} m\\left ( v_x'^2 -v_x^2 \\right ) \\notag \\\\ =\u0026 2Mm\\frac{Mu^2-\\left ( M-m \\right )uv_x-mv_x^2 }{\\left ( M+m \\right )^2 } \\tag{2} \\end{align} $$Considering the randomness of \\(u\\). Taking the average, we obtain $$ \\left \\langle \\triangle \\varepsilon \\right \\rangle _{(v_x)}=\\frac{2Mm}{\\left ( M+m \\right )^2}\\left ( M\\left \\langle u^2 \\right \\rangle - (M - m) v_x \\left \\langle u \\right \\rangle - mv_x^2 \\right ) \\tag{3} $$Given that \\(\\sqrt{\\frac{kT_1}{m} }\\gg u\\), we integrate over \\(v_x\\in[0,+\\infty]\\) to find the energy change per unit time for the left gas $$ \\begin{align} P_1 \u0026=\\int_{0}^{\\infty }\\triangle \\varepsilon _{\\left ( v_x \\right )}v_xAn_1\\sqrt{\\frac{m}{2\\pi kT_1} }e^{-\\frac{m}{2kT_1}v_x^2}dv_x \\notag \\\\ \u0026=\\frac{2Mm}{\\left ( M+m \\right )^2 }An_1\\sqrt{\\frac{m}{2\\pi kT_1} }\\frac{kT_1}{m}\\left ( M\\left \\langle u^2 \\right \\rangle - \\sqrt{\\frac{\\pi k T_1}{2 m}} (M-m) \\left \\langle u \\right \\rangle - 2kT_1 \\right ) \\notag \\quad \\text{(4)} \\end{align} $$Using \\(p=n_1kT_1\\), we get $$ P_1=\\frac{2Mm}{\\left ( M+m \\right )^2 }Ap \\sqrt{\\frac{m}{2\\pi kT_1} } \\frac{1}{m} \\left ( M\\left \\langle u^2 \\right \\rangle - \\sqrt{\\frac{\\pi k T_1}{2 m}} (M-m) \\left \\langle u \\right \\rangle - 2kT_1 \\right ) \\quad \\text{(5)} $$Effective Temperature of the Piston #Define $$ \\left \\langle u \\right \\rangle=0\\quad,\\quad\\frac{1}2M\\left \\langle u^2 \\right \\rangle = \\frac{1}2kT_0 \\tag{6} $$ where \\(T_0\\) is the effective temperature of the piston. Note that \\(\\left \\langle u \\right \\rangle \\neq 0\\) in reality, but it will be shown later that \\(\\left \\langle u^2 \\right \\rangle \\gg \\sqrt{\\frac{\\pi k T_1}{2 m}} \\left \\langle u \\right \\rangle\\), allowing \\(\\left \\langle u \\right \\rangle\\) to be neglected. Thus $$ P_1=\\frac{2Mm}{\\left ( M+m \\right )^2 }Ap \\sqrt{\\frac{m}{2\\pi kT_1} } \\frac{1}{m} \\left ( kT_0-2kT_1 \\right ) \\tag{7} $$ Similarly, the energy change per unit time for the right gas is $$ P_2=\\frac{2Mm}{\\left ( M+m \\right )^2 }Ap \\sqrt{\\frac{m}{2\\pi kT_2} } \\frac{1}{m} \\left ( kT_0-2kT_2 \\right ) \\tag{8} $$Given \\(\\frac{1}{2}M\\left \\langle u^2 \\right \\rangle\\ll \\nu_{1,2}C^{mol}_V T_{1,2}\\), energy conservation for the system can be written as $$ P_1+P_2=0 \\tag{9} $$ Therefore $$ \\frac{T_0}{\\sqrt{T_1}}-2\\sqrt{T_1}+\\frac{T_0}{\\sqrt{T_2} }-2\\sqrt{T_2}=0 \\tag{10} $$ $$ T_0=2\\sqrt{T_1T_2} \\tag{11} $$Energy Flux #Substituting (11) into (7), the energy flux from the right gas to the left gas is $$ P=\\frac{4Mm}{\\left ( M+m \\right )^2 }A\\sqrt{\\frac{k}{2\\pi m} }p\\left ( \\sqrt{T_2}-\\sqrt{T_1} \\right ) \\tag{12} $$Thermodynamic Quantities #Initial state $$ p_0V_{10}={\\nu} _1RT_{10}, \\quad p_0V_{20}=\\nu_2RT_{20} \\tag{13} $$State at time \\(t\\) $$ pV_1=\\nu_1 RT_1, \\quad pV_2=\\nu_2 RT_2 \\tag{14} $$Again, given \\(\\frac{1}{2}M\\left \\langle u^2 \\right \\rangle\\ll \\nu_{1,2}C^{mol}_V T_{1,2}\\), energy conservation for the system is $$ \\nu _1C^{mol}_V T_{10}+\\nu _2C^{mol}_V T_{20}=\\nu _1C^{mol}_V T_1+\\nu _2C^{mol}_V T_2 \\tag{15} $$From $$ V_{10}+V_{20}=V_1+V_2 \\tag{16} $$ it follows that $$ p=p_0 \\tag{17} $$Differential Equations for Thermodynamic Quantities #For a time element \\(t\\) ~ \\(t+dt\\), suppose the piston moves right by \\(dx\\). The energy conservation equation for the left gas is $$ \\nu _1C_V^{mol}dT_1=Pdt-p_0Adx \\tag{18} $$Since pressure is constant by (17), we have $$ \\frac{dT_1}{T_1}-\\frac{Adx}{V_1}=0 \\tag{19} $$ i.e., $$ p_0Adx=\\nu _1RdT_1 \\tag{20} $$Substituting into (18) gives $$ \\nu _1C_p^{mol}dT_1=Pdt \\tag{21} $$Substituting (12) and eliminating \\(P\\), we obtain $$ \\nu _1C_p^{mol}dT_1=\\frac{4Mm}{\\left ( M+m \\right )^2 }A\\sqrt{\\frac{k}{2\\pi m} }p_0\\left ( \\sqrt{T_2}-\\sqrt{T_1} \\right )dt \\tag{22} $$Combining with (15) yields the system of differential equations for \\(T_1(t)\\), \\(T_2(t)\\) $$ \\begin{cases} \\ \\nu _1T_{10}+\\nu _2T_{20}=\\nu _1T_1+\\nu _2T_2 \\\\\\\\ \\ \\nu _1C_P^{mol}\\frac{dT_1}{dt}=\\frac{4Mm}{\\left ( M+m \\right )^2 }A\\sqrt{\\frac{k}{2\\pi m} }P_0\\left ( \\sqrt{T_2}-\\sqrt{T_1} \\right ) \\end{cases} \\tag{23} $$Approximate Solution #(23) cannot be solved analytically. Consider the case where \\(T_{10}-T_{20}=\\delta T_0 \\ll T_1, T_2\\). The final equilibrium temperature is $$ T_f=\\frac{\\nu _1T_{10}+\\nu _2T_{20}}{\\nu _1+\\nu _2} \\tag{24} $$During the process, let \\(\\delta T=T_1-T_2\\), then $$ {\\begin{cases}T_1=T_f+\\frac{\\nu _2}{\\nu _1+\\nu _2}\\delta T \\\\T_2=T_f-\\frac{\\nu _1}{\\nu _1+\\nu _2}\\delta T\\end{cases}} \\tag{25} $$Substituting into (22) and retaining first-order small quantities gives $$ \\frac{d\\left ( \\delta T \\right ) }{\\delta T}=-\\frac{2Mm}{\\left ( M+m \\right )^2 }A\\sqrt{\\frac{kT_f}{2\\pi m} }p_0\\frac{1}{T_f}\\frac{1}{C_p^{mol}}\\frac{\\nu_1+\\nu _2}{\\nu _1\\nu _2}dt \\tag{26} $$ i.e., $$ \\delta T=\\delta T_0 e^{-\\frac{t}{\\tau }} \\tag{27} $$ $$ T_1-T_2=(T_{10}-T_{20}) e^{-\\frac{t}{\\tau }} \\tag{28} $$ where the characteristic time constant \\(\\tau\\) is $$ \\begin{align} \\tau \u0026=\\frac{\\left ( M+m \\right )^2 }{2Mm}\\frac{T_f}{Ap_0}\\sqrt{\\frac{2\\pi m}{kT_f} }C_p^{mol}\\frac{\\nu _1\\nu _2}{\\nu _1+\\nu _2} \\notag \\\\ \u0026=\\frac{\\left ( M+m \\right )^2 }{2Mm}\\frac{1}{Ap_0}\\sqrt{\\frac{2\\pi m}{k} \\cdot \\frac{\\nu _1T_{10}+\\nu _2T_{20}}{\\nu _1+\\nu _2}}C_p^{mol}\\frac{\\nu _1\\nu _2}{\\nu _1+\\nu _2} \\tag{29} \\end{align} $$As a reference, using values \\(M=10^{-3} \\text{kg}, m=5\\times 10^{-26} \\text{kg}, A=10^{-2} \\text{m}^2, p_0=1 \\text{atm}, T_{10}=310 \\text{K}, T_{20}=300 \\text{K}, \\nu_1=\\nu_2=10 \\text{mol}, C^{mol}_p=\\frac{7}{2}R\\), we find $$ \\tau \\approx 1.2\\times10^{14} \\text{a (years)} \\tag{30} $$Thus, this effect is negligible in macroscopic systems. (Additionally, \\(\\left \\langle u \\right \\rangle \\sim \\frac{l}{\\tau}\\), where \\(l \\sim \\frac{\\nu R T}{p A}\\) is the characteristic length of the container. This validates the earlier assumption \\(\\left \\langle u^2 \\right \\rangle \\gg \\sqrt{\\frac{\\pi k T_1}{2 m}} \\left \\langle u \\right \\rangle\\) )\n","date":"26 June 2025","permalink":"https://trojanpt.github.io/en/jottings/smooth-insulated-piston/","section":"Jottings","summary":"If the pressures on both sides of a smooth adiabatic piston are equal but the temperatures are unequal, energy will transfer from the higher-temperature gas to the lower-temperature gas through microscopic vibrations of the piston, even when all heat conduction is neglected.","title":"Equilibrium Conditions for a Smooth Adiabatic Piston"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/tags/thermodynamics/","section":"Tags","summary":"","title":"Thermodynamics"},{"content":" This translation is generated by DeepSeek and may contain errors. Please refer to the Chinese version for accuracy. Crunching numbers on paper is way easier than bungee jumping (￣﹃￣')\nConsider an elastic rope with elastic modulus \\(Y\\), unstretched length \\(l\\) (gravity-free state), cross-sectional area \\(S_0\\), linear mass density \\(\\lambda\\), and Poisson\u0026rsquo;s ratio \\(\\frac{1}{2}\\) (i.e., the medium is incompressible). One end is fixed at point O, while the other end is attached to a mass \\(M\\). The entire system hangs vertically under gravitational acceleration \\(g\\). Only longitudinal vibrations of the elastic rope are considered.\nFig. 1 Define the unstretched coordinate \\(x\\) as the distance from O to a point on the rope in its free state. The elongation coordinate \\(\\xi{(x)}\\) is the displacement (positive downward) at the point where the unstretched coordinate is \\(x\\).\nTension Distribution \\(F(x)\\) on the Rope #The elastic rope is incompressible, hence $$ S\\cdot(\\mathrm{d}x+\\mathrm{d}\\xi)=S_0\\cdot\\mathrm{d}x \\tag{1} $$ From Hooke\u0026rsquo;s law $$ \\frac{F}{S}=Y\\frac{\\partial \\xi}{\\partial x} \\tag{2} $$ Therefore $$ F(x)=S_0 Y\\frac{\\frac{\\partial \\xi}{\\partial x}}{1+\\frac{\\partial \\xi}{\\partial x}} \\tag{3} $$Wave Equation #For an infinitesimal element \\(x\\,{\\backsim}\\,x+dx\\), we have $$ \\lambda dx\\cdot\\frac{\\partial^2\\xi}{\\partial t^2}=\\lambda dx\\cdot g+F(x+dx)-F(x) \\tag{4} $$Substituting (3) yields $$ \\lambda\\frac{\\partial^2\\xi}{\\partial t^2}=\\lambda g+\\frac{S_0Y}{(1+\\frac{\\partial\\xi}{\\partial x})^2}\\frac{\\partial^2\\xi}{\\partial x^2} \\tag{5} $$Boundary conditions $$ \\begin{cases} \\xi|_{x=0}=0\\\\ M\\cdot\\frac{\\partial^2\\xi}{\\partial t^2}|_{x=l}=Mg-S_0Y\\frac{(\\frac{d\\xi}{dx})|_{x=l}}{1+(\\frac{d\\xi}{dx})|_{x=l}} \\end{cases} \\tag{6} $$Steady-State Solution #The steady-state solution without vibration is \\(\\xi_0(x)\\), where \\(\\frac{\\partial^2\\xi _0}{\\partial t^2}=0\\). Thus, (5) simplifies to $$ \\lambda g+\\frac{S_0Y}{(1+\\frac{d\\xi _0}{dx})^2}\\frac{d^2\\xi _0}{d x^2} = 0 \\tag{7} $$Integrating gives $$ \\frac{d\\xi_0 }{dx}=\\frac{Mg+\\lambda (l-x)g }{S_0Y-Mg-\\lambda( l-x)g } \\tag{8} $$Hence $$ \\xi_0 \\left ( x \\right ) =\\frac{S_0Y}{\\lambda g} \\ln\\left ( \\frac{S_0Y-Mg-\\lambda lg+\\lambda xg}{S_0Y-Mg-\\lambda lg} \\right ) -x \\tag{9} $$Small-Amplitude Vibration Approximation #For small-amplitude vibrations, define $$ \\xi(x,t)=\\xi_0(x)+\\delta(x,t) \\quad , |\\delta|\\ll|\\xi_0| \\tag{10} $$Substituting into (5) and retaining first-order terms gives $$ \\lambda\\frac{\\partial^2\\xi}{\\partial t^2}=\\lambda g+\\frac{S_0Y}{(1+\\frac{d\\xi_0}{dx})^2}\\left[\\frac{d^2\\xi_0}{dx^2}+\\frac{\\partial^2\\delta}{\\partial x^2}-2\\frac{d^2\\xi_0}{dx^2}\\frac{\\frac{\\partial\\delta}{\\partial x}}{1+\\frac{d\\xi_0}{dx}}\\right] \\tag{11} $$Substituting (9) and simplifying yields the first-order approximated equation for small vibrations $$ \\lambda\\frac{\\partial^2\\xi}{\\partial t^2}=\\frac{[S_0Y-Mg-\\lambda g(l-x)]^2}{S_0Y}\\frac{\\partial^2\\delta}{\\partial x^2}+2\\lambda g\\frac{S_0Y-Mg-\\lambda g(l-x)}{S_0Y}\\frac{\\partial\\delta}{\\partial x} \\quad \\text{(12)} $$Separate variables by letting \\(\\delta(x,t) =A(x)B(t)\\) $$ \\frac{1}{B} \\lambda \\frac{d^2B}{dt^2} =\\left \\{ \\frac{\\left [ S_0Y-Mg-\\lambda \\left ( l-x \\right )g \\right ]^2}{S_0Y} \\frac{d^2A}{dx^2} +2\\lambda g\\frac{S_0Y-Mg-\\lambda \\left ( l-x \\right )g }{S_0Y} \\frac{dA}{dx} \\right \\} \\frac{1}{A} \\quad \\text{(13)} $$The left side depends only on \\(t\\), the right side only on \\(x\\). Thus, both sides equal a constant, denoted \\(-\\lambda \\omega^2\\). The left side gives $$ \\frac{1}{B} \\lambda \\frac{d^2B}{dt^2}=-\\lambda \\omega^2 \\tag{14} $$ $$ B\\left ( t\\right )=B_o\\cos \\left ( \\omega t+\\varphi \\right ) \\tag{15} $$The right side gives $$ \\frac{\\left [ S_0Y-Mg-\\lambda \\left ( l-x \\right )g \\right ]^2 }{S_0Y} \\frac{d^2A}{dx^2} +2\\lambda g\\frac{S_0Y-Mg-\\lambda \\left ( l-x \\right )g}{S_0Y}\\frac{dA}{dx}+\\lambda \\omega^2A=0 \\quad \\text{(16)} $$This is an Euler equation. Let $$ t=\\ln\\left ( S_0Y-Mg-\\lambda lg+\\lambda gx \\right ) \\tag{17} $$Substituting back into (16) gives $$ \\frac{d^2A}{dt^2}+\\frac{dA}{dt}+\\frac{\\omega^2S_0Y }{\\lambda g^2}A=0 \\tag{18} $$The characteristic equation is $$ \\lambda ^2+\\lambda +\\frac{\\omega^2S_0Y }{\\lambda g^2}=0 \\tag{19} $$ $$ \\lambda_{1,2}=\\frac{-1\\pm \\sqrt{1-4\\frac{\\omega^2S_0Y}{\\lambda g^2} } }{2} \\tag{20} $$Thus $$ A=A_1e^{\\lambda_1t}+A_2e^{\\lambda_2t} \\tag{21} $$ $$ \\delta =\\left [ A_1\\left ( S_0Y-Mg-\\lambda lg+\\lambda xg \\right )^{\\lambda_1}+A_2\\left ( S_0Y-Mg-\\lambda lg+\\lambda xg \\right ) ^{\\lambda_2} \\right ]\\cos \\left ( \\omega t+\\varphi \\right ) \\quad \\text{(22)} $$ (The constant \\(B_0\\) is absorbed into \\(A_1,A_2\\).)\nApplying the first-order approximation to boundary conditions (6) $$ \\begin{cases} \\delta|_{x=0}=0\\\\ M\\frac{\\partial ^2\\delta }{\\partial t^2}|_{x=l}=Mg-S_0Y\\frac{\\frac{d\\xi_0}{dx}}{1+\\frac{d\\xi_0}{dx}}|_{x=l}-S_0Y\\frac{\\frac{\\partial \\delta}{\\partial x}}{1+\\frac{d\\xi_0}{dx}}|_{x=l} \\end{cases} \\tag{23} $$Substituting (9) to eliminate \\(\\xi_0\\) $$ \\begin{cases} \\delta|_{x=0}=0\\\\ M\\frac{\\partial ^2\\delta }{\\partial t^2}|_{x=l}=-\\frac{(S_0Y-Mg)^2}{S_0Y}\\frac{\\partial\\delta}{\\partial x}|_{x=l} \\end{cases} \\tag{24} $$Substituting (22) to eliminate \\(\\delta\\) $$ {\\begin{cases} A_1\\left ( S_0Y-Mg-\\lambda lg \\right ) ^{\\lambda _1-\\lambda _2}+A_2=0\\\\ w^2M\\left [ A_1\\left ( S_0Y-Mg \\right )^{\\lambda _1-\\lambda _2} +A_2 \\right ]=\\frac{S_0Y-Mg}{S_0Y}\\lambda g [ A_1\\lambda _1\\left ( S_0Y-Mg )^{\\lambda _1-\\lambda _2}+A_2\\lambda _2 \\right ] \\end{cases}} \\quad \\text{(25)} $$This is a homogeneous linear system in \\(A_1,\\ A_2\\). For non-trivial solutions, the determinant must be zero, yielding the equation for \\(\\omega\\) $$ \\omega^2M\\left [ \\left ( \\frac{S_0Y-Mg}{S_0Y-Mg-\\lambda lg} \\right )^{\\lambda _1-\\lambda _2} -1 \\right ] =\\frac{S_0Y-Mg}{S_0Y}\\lambda g \\left[ \\lambda _1\\left(\\frac{S_0Y-Mg}{S_0Y-Mg-\\lambda lg}\\right)^{\\lambda _1-\\lambda _2} -\\lambda _2 \\right ] \\quad \\text{(26)} $$If $$ 4\\frac{\\omega^2S_0Y}{\\lambda g^2} \u003e 1 \\tag{27} $$ then \\(\\lambda_1,\\lambda_2\\) are complex. Let $$ \\lambda _{1,2}=\\alpha \\pm \\beta i \\tag{28} $$ where $$ \\alpha =-\\frac{1}{2} \\tag{29} $$ $$ \\beta =\\sqrt{\\frac{\\omega^2S_0Y}{\\lambda g^2}-\\frac{1}{4}} \\tag{30} $$Substituting into (26) gives $$ \\begin{align} \\omega^2M =\u0026 \\frac{S_0Y-Mg}{S_0Y}\\lambda g\\left \\{ \\alpha +\\beta \\cot \\left [ \\beta \\ln\\left ( \\frac{S_0Y-Mg}{S_0Y-Mg-\\lambda lg} \\right ) \\right ] \\right \\} \\notag \\\\ =\u0026 \\frac{S_0Y-Mg}{S_0Y}\\lambda g\\left \\{ -\\frac{1}{2} +\\sqrt{\\frac{\\omega^2S_0Y}{\\lambda g^2}-\\frac{1}{4}}\\ \\cot \\left [ \\sqrt{\\frac{\\omega^2S_0Y}{\\lambda g^2}-\\frac{1}{4}} \\ \\ln\\left ( \\frac{S_0Y-Mg}{S_0Y-Mg-\\lambda lg} \\right ) \\right ] \\right \\} \\notag \\quad \\text{(31)} \\end{align} $$ This is the equation satisfied by \\(\\omega\\).\n","date":"26 June 2025","permalink":"https://trojanpt.github.io/en/jottings/vibration-of-elasticbody/","section":"Jottings","summary":"Small-amplitude vibrations of an incompressible elastic rope in a gravitational field.","title":"Vibrations of an Elastic Body in a Gravitational Field"},{"content":"Under Construction\u0026hellip;\n","date":"22 June 2025","permalink":"https://trojanpt.github.io/en/about/","section":"Home","summary":"","title":"About"},{"content":" This site contains some images sourced from the internet. If you believe that any content on this site infringes your copyright, please contact me, and I will address it promptly. Unless otherwise noted or specified, all content on this site is licensed under the Attribution-NonCommercial 4.0 International (CC BY-NC 4.0) license.\nCC BY-NC 4.0 ","date":"22 June 2025","permalink":"https://trojanpt.github.io/en/copyright/","section":"Home","summary":"","title":"Copyright"},{"content":" This translation is generated by DeepSeek and may contain errors. Please refer to the Chinese version for accuracy. Ampere’s Law provides a method for calculating the force between two current sources. However, the modern form of Ampere’s Law is not the one originally proposed by Ampere. While they differ mathematically, there is no measurable physical distinction under steady-state conditions.\nThis article will introduce the derivation process of the original form of Ampere’s Law and discuss its relationship with the modern form.\nI. Derivation of the Original Ampere’s Law #Experimental Basis # Force reverses when current reverses $$ d\\vec{F}_{12} \\propto i_1 i_2 \\tag{1} $$ Current elements possess vector superposition property $$ d\\vec{F}_{12} = d\\vec{F}_{12} (i_1, i_2, d\\vec{l}_1, d\\vec{l}_2, \\vec{r}_{12}) \\tag{2} $$ Force on a current element is perpendicular to the element itself $$ \\left( \\oint_{L_1}{d\\vec{F}_{12}} \\right) \\cdot d\\vec{l}_2 = 0 \\tag{3} $$ Force magnitude is independent of geometric scale $$ |d\\vec{F}_{12}| \\propto \\frac{|d\\vec{l}_1| |d\\vec{l}_2|}{|\\vec{r}_{12}|^2} \\tag{4} $$ Assumption #The force between two current elements \\( d\\vec{l}_1 \\) and \\( d\\vec{l}_2 \\) acts along the direction of \\( \\vec{r}_{12} \\).\nDerivation Process #Based on the assumption and considering experimental basis (1) and (2), we can write $$ d\\vec{F}_{12} = k i_1 i_2 \\varphi(d\\vec{l}_1, d\\vec{l}_2, \\vec{r}_{12}) \\hat{r}_{12} \\tag{5} $$ where \\( \\varphi \\) is a scalar-valued function.\nConsidering experiment (2), the path element \\( d\\vec{l} \\) characterizing the current element exhibits vector properties.\nA simple consideration is that \\( \\varphi \\) depends linearly on \\( d\\vec{l}_1 \\) and \\( d\\vec{l}_2 \\).\nThus, we assume $$ \\varphi = \\varphi_1(\\vec{r}_{12})(d\\vec{l}_1 \\cdot d\\vec{l}_2) + \\varphi_2(\\vec{r}_{12})(d\\vec{l}_1 \\cdot \\vec{r}_{12})(d\\vec{l}_2 \\cdot \\vec{r}_{12}) + \\varphi_3(\\vec{r}_{12})(d\\vec{l}_1 \\times d\\vec{l}_2) \\cdot \\vec{r}_{12} \\quad \\text{(6)} $$ The rationale for this assumption is that terms within \\( \\varphi \\) must take forms such as\n$$ (d{l}_1)_\\alpha (d{l}_2)_\\beta ({r}_{12})_\\gamma ({r}_{12})_\\mu \\cdots \\tag{7} $$① \\( \\alpha = \\beta \\): This corresponds to the \\( (d\\vec{l}_1 \\cdot d\\vec{l}_2) \\) term.\n② \\( \\alpha \\neq \\beta \\): Since \\( \\varphi \\) is a scalar function, there are two possible forms\n\\( (d{l}_1)_\\alpha (d{l}_2)_\\beta ({r}_{12})_\\alpha ({r}_{12})_\\beta \\cdots \\) i.e., the \\( (d\\vec{l}_1 \\cdot \\vec{r}_{12})(d\\vec{l}_2 \\cdot \\vec{r}_{12}) \\) term.\n\\( \\varepsilon_{\\gamma\\alpha\\beta} (d{l}_1)_\\alpha (d{l}_2)_\\beta ({r}_{12})_\\gamma \\cdots \\) i.e., the \\( (d\\vec{l}_1 \\times d\\vec{l}_2) \\cdot \\vec{r}_{12} \\) term.\nConsidering (4), we have $$ d\\vec{F}_{12} = i_1 i_2 \\vec{r}_{12} \\left[ (d\\vec{l}_1 \\cdot d\\vec{l}_2) \\frac{A}{r_{12}^3} + (d\\vec{l}_1 \\cdot \\vec{r}_{12})(d\\vec{l}_2 \\cdot \\vec{r}_{12}) \\frac{B}{r_{12}^5} + (d\\vec{l}_1 \\times d\\vec{l}_2) \\cdot \\vec{r}_{12} \\frac{C}{r_{12}^4} \\right] \\quad \\text{(8)} $$ Considering (3), as shown in the figure,\nFig. 1 where \\(d\\vec{l}_1 = - d\\vec{r}_{12}\\).\nNote equation (3) $$ \\left( \\oint_{L_1} d\\vec{F}_{12} \\right) \\cdot d\\vec{l}_{2} = \\oint_{L_1} (d\\vec{F}_{12} \\cdot d\\vec{l}_{2}) = 0 \\tag{9} $$ This equality is independent of the path \\(L_1\\), implying that \\(d\\vec{F}_{12} \\cdot d\\vec{r}_{12}\\) is the differential of some potential function \\(\\phi\\),\ni.e., \\( d\\phi = d\\vec{F}_{12} \\cdot d\\vec{l}_{2} \\).\nFor \\(\\phi\\), considering the above, we have $$ d\\phi \\propto \\frac{|d\\vec{r}_{12}| |d\\vec{l}_2|^2}{|\\vec{r}_{12}|^2} \\tag{10} $$ $$ \\phi \\propto \\frac{|d\\vec{l}_2|^2}{|\\vec{r}_{12}|} \\tag{11} $$ Thus, we formally express \\(\\phi\\) as $$ \\phi = (d\\vec{l}_2 \\cdot d\\vec{l}_2) \\frac{a}{r_{12}} + (d\\vec{l}_2 \\cdot \\vec{r}_{12})^2 \\frac{b}{r_{12}^3} + (d\\vec{l}_2 \\times \\vec{r}_{12})^2 \\frac{c}{r_{12}^3} \\tag{12} $$ Taking the differential, we obtain $$ \\begin{align} d\\phi = \u0026 - (d\\vec{l}_2 \\cdot d\\vec{l}_2) \\frac{a \\vec{r}_{12} \\cdot d\\vec{r}_{12}}{r_{12}^3} + \\notag \\\\ \u0026\\ 2(d\\vec{l}_2 \\cdot \\vec{r}_{12})(d\\vec{l}_2 \\cdot d\\vec{r}_{12}) \\frac{b}{r_{12}^3} - 3(d\\vec{l}_2 \\cdot \\vec{r}_{12})^2 \\frac{b \\vec{r}_{12} \\cdot d\\vec{r}_{12}}{r_{12}^5} + \\notag \\\\ \u0026\\ 2(d\\vec{l}_2 \\times \\vec{r}_{12}) \\cdot (d\\vec{l}_2 \\times d\\vec{r}_{12}) \\frac{c}{r_{12}^3} - 3(d\\vec{l}_2 \\times \\vec{r}_{12})^2 \\frac{c \\vec{r}_{12} \\cdot d\\vec{r}_{12}}{r_{12}^5} \\tag{13} \\end{align} $$ From (5), we also have $$ d\\vec{F}_{12} \\cdot d\\vec{l}_2 = -i_1 i_2 (\\vec{r}_{12} \\cdot d\\vec{l}_2) \\left[ (d\\vec{r}_{12} \\cdot d\\vec{l}_2) \\frac{A}{r_{12}^3} + (d\\vec{r}_{12} \\cdot \\vec{r}_{12}) (d\\vec{l}_2 \\cdot \\vec{r}_{12}) \\frac{B}{r_{12}^5} + (d\\vec{r}_{12} \\times d\\vec{l}_2) \\cdot \\vec{r}_{12} \\frac{C}{r_{12}^4} \\right] \\quad \\text{(14)} $$ And since \\( d\\phi = d\\vec{F}_{12} \\cdot d\\vec{l}_2 \\), comparing coefficients yields, we have $$ \\begin{cases} a = 0 \\\\ -A = 2b \\\\ -B = -3b \\\\ c = 0 \\\\ C = 0 \\end{cases} \\tag{15} $$ Thus, we can set \\( A = -2k, B = 3k, C = 0\\). Substituting into (9) gives $$ d\\vec{F}_{12} = -k i_1 i_2 \\vec{r}_{12} \\left[ (d\\vec{l}_{1} \\cdot d\\vec{l}_2) \\frac{2}{r_{12}^3} - (d\\vec{l}_{1} \\cdot \\vec{r}_{12}) (d\\vec{l}_2 \\cdot \\vec{r}_{12}) \\frac{3}{r_{12}^5} \\right] \\tag{16} $$ This is the original form of Ampere’s Law.\nII. Discussion on the Applicability of the Original Ampere’s Law # Under steady-state conditions, the force between a pair of current elements cannot be measured in practice.\nThe physically measurable quantity is at least the force exerted by a closed loop on a current element.\n$$ \\begin{align} d\\vec{F}_{L_1,2} =\u0026 \\oint_{L_1} d\\vec{F}_{12} \\notag \\\\ =\u0026 \\oint_{L_1} k i_1 i_2 \\vec{r}_{12} \\left[ \\frac{2 (d\\vec{r}_{12} \\cdot d\\vec{l}_2)}{r_{12}^3} - \\frac{3 (d\\vec{r}_{12} \\cdot \\vec{r}_{12}) (d\\vec{l}_2 \\cdot \\vec{r}_{12})}{r_{12}^5} \\right] \\notag \\\\ =\u0026 \\oint_{L_1} k i_1 i_2 \\left[ d\\left(\\vec{r}_{12} \\frac{(d\\vec{l}_2 \\cdot \\vec{r}_{12})}{r_{12}^3} \\right) + \\frac{1}{r_{12}^3} \\left[ \\vec{r}_{12} (d\\vec{r}_{12} \\cdot d\\vec{l}_2) - d\\vec{r}_{12} (d\\vec{r}_{12} \\cdot d\\vec{l}_2) \\right] \\right] \\notag \\quad \\text{(17)} \\end{align} $$Here, $$ \\oint_{L_1} k i_1 i_2 d\\left( \\frac{\\vec{r}_{12} (d\\vec{l}_2 \\cdot \\vec{r}_{12})}{r_{12}^3} \\right) = 0 \\tag{18} $$$$ \\vec{r}_{12} (d\\vec{r}_{12} \\cdot d\\vec{l}_2) - d\\vec{r}_{12} (d\\vec{r}_{12} \\cdot d\\vec{l}_2) = d\\vec{l}_2 \\times (\\vec{r}_{12} \\times d\\vec{r}_{12}) \\tag{19} $$ This shows that the modern Ampere’s Law differs from the original Ampere’s Law by an exact differential term.\nTherefore, for closed loops under steady-state conditions, there is no practical distinction between them.\nThus, $$ d\\vec{F}_{L_1,2} = \\oint_{L_1} k i_1 i_2 \\frac{1}{r_{12}^3} d\\vec{l}_2 \\times (\\vec{r}_{12} \\times d\\vec{l}_1) \\tag{20} $$ Considering that \\( d\\vec{r}_{12} = -d\\vec{l}_1 \\), we have $$ d\\vec{F}_{L_1,2} = \\oint_{L_1} k i_1 i_2 \\frac{1}{r_{12}^3} d\\vec{l}_2 \\times (d\\vec{l}_1 \\times \\vec{r}_{12}) \\tag{21} $$ This is consistent with the result obtained from the modern Ampere’s Law.\nIII. Why the Modern Form of Ampere’s Law is Used # While stable, isolated current elements do not exist in practice,\nthey can exist under non-steady conditions.\nFor example, a moving point charge can be treated as an independent current element $$ q \\vec{v} = i d\\vec{l} \\tag{22} $$ Consider the scenario depicted.\nFig. 2 Using the relativistic transformation of electromagnetic fields, the electromagnetic field produced by \\( q_1 \\) at position \\( \\vec{r}_{12} \\) at that instant satisfies $$ \\begin{cases} \\vec{E}(\\vec{r}_{12}) = \\left[ \\gamma \\vec{\\vec{I}} - (\\gamma - 1) \\frac{\\vec{v}_1 \\vec{v}_1}{|\\vec{v}_1|^2} \\right] \\cdot \\frac{1}{4\\pi\\varepsilon_0} \\frac{q_1 \\vec{r}_{12}'}{|\\vec{r}_{12}'|^3} \\\\\\\\ \\vec{B}(\\vec{r}_{12}) = \\gamma \\frac{\\vec{v}_1}{c^2} \\times \\frac{1}{4\\pi\\varepsilon_0} \\frac{q_1 \\vec{r}_{12}'}{|\\vec{r}_{12}'|^3} \\end{cases} \\tag{23} $$ where $$ \\vec{r}_{12}' = \\left[ \\vec{\\vec{I}} + (\\gamma - 1) \\frac{\\vec{v}_1 \\vec{v}_1}{|\\vec{v}_1|^2} \\right] \\vec{r}_{12} \\tag{24} $$Note that for a real current element, besides the moving charge carriers, there should be stationary charges maintaining overall electrical neutrality. Therefore, the actual electric field, after accounting for the stationary charges, should be negligible and is not considered here.\nThus, under the approximation \\( v \\ll c \\), the net force on \\( q_2 \\) is $$ \\begin{align} \\vec{F}_{12} =\u0026\\ q_2 \\vec{v}_2 \\times \\vec{B} \\\\ =\u0026\\ q_2 \\vec{v}_2 \\times \\left( \\frac{\\vec{v}_1}{c^2} \\times \\frac{1}{4\\pi\\varepsilon_0} \\cdot \\frac{q_1 \\vec{r}_{12}}{|\\vec{r}_{12}|^3} \\right) \\\\ =\u0026\\ \\frac{\\mu_0}{4\\pi} \\frac{q_2 \\vec{v}_2 \\times \\left( q_1 \\vec{v}_1 \\times \\vec{r}_{12} \\right)}{|\\vec{r}_{12}|^3} \\end{align} \\tag{25} $$ If we treat the moving charge \\( q\\vec{v} \\) as a current element \\( i d\\vec{l} \\), then $$ \\vec{F}_{12} = \\frac{\\mu_0}{4\\pi} \\frac{(i_2 d\\vec{l}_2) \\times \\left( (i_1 d\\vec{l}_1) \\times \\vec{r}_{12} \\right)}{|\\vec{r}_{12}|^3} \\tag{26} $$ This is consistent with the form of the modern Ampere’s Law.\n","date":"22 June 2025","permalink":"https://trojanpt.github.io/en/jottings/amperes-law/","section":"Jottings","summary":"This article introduces the derivation process of the original form of Ampere’s Law and discusses its relationship with the modern form.","title":"Discussions on the Original Form of Ampere’s Law"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/tags/electromagnetism/","section":"Tags","summary":"","title":"Electromagnetism"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/categories/","section":"Categories","summary":"","title":"Categories"},{"content":"","date":null,"permalink":"https://trojanpt.github.io/en/notes/","section":"Notes","summary":"","title":"Notes"}]