(h) Assume that must be smaller than a finite value. · OlympiadBench #970

Physics · modern-physics

Problem

(h) Assume that vv must be smaller than a finite value. Show that 1/κ1 / \sqrt{\kappa} is the maximum allowable speed, and that this speed is frame invariant, i.e., dx′dt′=dxdt\frac{d x^{\prime}}{d t^{\prime}}=\frac{d x}{d t} for something moving at speed 1/κ1 / \sqrt{\kappa}. Experiment has shown that this speed is cc, the speed of light. Setting κ=1/c2\kappa=1 / c^{2}, show that you recover the Lorentz boost. Given constants: 4. Lorentz Boost In Newtonian kinematics, inertial frames moving relatively to each other are related by the following transformations called Galilean boosts: x′=x−vtt′=t\begin{aligned} x^{\prime} & =x-v t \\ t^{\prime} & =t \end{aligned} In relativistic kinematics, inertial frames are similarly related by the Lorentz boosts: x′=11−v2/c2(x−vt)t′=11−v2/c2(t−vc2x)\begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-v^{2} / c^{2}}}\left(t-\frac{v}{c^{2}} x\right) \end{aligned} In this problem you will derive the Lorentz transformations from a minimal set of postulates: the homogeneity of space and time, the isotropy of space, and the principle of relativity. You will show that these assumptions about the structure of space-time imply either (a) there is a universal "speed limit" which is frame invariant, which results in the Lorentz boost, or (b) there is no universal "speed limit," which results in the Galilean boost. For simplicity, consider a one-dimensional problem only. Let two frames FF and F′F^{\prime} be such that the frame F′F^{\prime} moves at relative velocity vv in the positive- xx direction compared to frame FF. Denote the coordinates of FF as (x,t)(x, t) and the coordinates of F′F^{\prime} as (x′,t′)\left(x^{\prime}, t^{\prime}\right). The most general coordinate transformations between FF and F′F^{\prime} are given by functions X,TX, T, x′=X(x,t,v)t′=T(x,t,v)\begin{aligned} x^{\prime} & =X(x, t, v) \\ t^{\prime} & =T(x, t, v) \end{aligned} which we will refer to as the generalized boost. Context question: (a) The homogeneity of space and time imply that the laws of physics are the same no matter where in space and time you are. In other words, they do not depend on a choice of origin for coordinates xx and tt. Use this fact to show that ∂X∂x\frac{\partial X}{\partial x} is independent of the position xx and ∂T∂t\frac{\partial T}{\partial t} is independent of the time tt. (Hint: Recall the definition of the partial derivative.) Context answer: 证明题 Extra Supplementary Reading Materials: Analogously, we can conclude additionally that ∂X∂x\frac{\partial X}{\partial x} is independent of both xx and tt and ∂T∂t\frac{\partial T}{\partial t} is independent of xx and tt. It can be shown that X,TX, T may be given in the form X(x,t,v)=A(v)x+B(v)tT(x,t,v)=C(v)x+D(v)t\begin{aligned} X(x, t, v) & =A(v) x+B(v) t \\ T(x, t, v) & =C(v) x+D(v) t \end{aligned} where A,B,C,DA, B, C, D are functions of vv. In other words, the generalized boost is a linear transformation of coordinates. Context question: (b) The isotropy of space implies that there is no preferred direction in the universe, i.e., that the laws of physics are the same in all directions. Use this to study the general coordinate transformations X,TX, T after setting x→−xx \rightarrow-x and x′→−x′x^{\prime} \rightarrow-x^{\prime} and conclude that A(v),D(v)A(v), D(v) are even functions of vv and B(v),C(v)B(v), C(v) are odd functions of vv. (Hint: the relative velocity vv is a number which is measured by the FF frame using v=dxdtv=\frac{d x}{d t}.) Context answer: 证明题 Context question: (c) The principle of relativity implies that the laws of physics are agreed upon by observers in inertial frames. This implies that the general coordinate transformations X,TX, T are invertible and their inverses have the same functional form as X,TX, T after setting v→−vv \rightarrow-v. Use this fact to show the following system of equations hold: A(v)2−B(v)C(v)=1D(v)2−B(v)C(v)=1C(v)(A(v)−D(v))=0B(v)(A(v)−D(v))=0.\begin{aligned} A(v)^{2}-B(v) C(v) & =1 \\ D(v)^{2}-B(v) C(v) & =1 \\ C(v)(A(v)-D(v)) & =0 \\ B(v)(A(v)-D(v)) & =0 . \end{aligned} (Hint: It's convenient to write X,TX, T as matrices and recall the definition of matrix inverses.) Physically, we must have that B(v)B(v) and C(v)C(v) are not both identically zero for nonzero vv. So, we can conclude from the above that D(v)=A(v)D(v)=A(v) and C(v)=A(v)2−1B(v)C(v)=\frac{A(v)^{2}-1}{B(v)}. Context answer: 证明题 Context question: (d) Use the previous results and the fact that the location of the F′F^{\prime} frame may be given by x=vtx=v t in the FF frame to conclude that the coordinate transformations have the following form: x′=A(v)x−vA(v)tt′=−(A(v)2−1vA(v))x+A(v)t\begin{aligned} x^{\prime} & =A(v) x-v A(v) t \\ t^{\prime} & =-\left(\frac{A(v)^{2}-1}{v A(v)}\right) x+A(v) t \end{aligned} Context answer: 证明题 Context question: (e) Assume that a composition of boosts results in a boost of the same functional form. Use this fact and all the previous results you have derived about these generalized boosts to conclude that A(v)2−1v2A(v)=κ.\frac{A(v)^{2}-1}{v^{2} A(v)}=\kappa . for an arbitrary constant κ\kappa. Context answer: 证明题 Context question: (f) (1 point) Show that κ\kappa has dimensions of (velocity )−2)^{-2}, and show that the generalized boost now has the form x′=11−κv2(x−vt)t′=11−κv2(t−κvx)\begin{aligned} x^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(x-v t) \\ t^{\prime} & =\frac{1}{\sqrt{1-\kappa v^{2}}}(t-\kappa v x) \end{aligned} Context answer: 证明题 Context question: (g) Assume that vv may be infinite. Argue that κ=0\kappa=0 and show that you recover the Galilean boost. Under this assumption, explain using a Galilean boost why this implies that a particle may travel arbitrarily fast. Context answer: 证明题

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