Life with Physics

We proved the hyperbolic motion viewed in the lab frame of a relativistic particle with constant proper acceleration as in this post : $$\begin{equation}t(\tau) = \frac{c}{a}\sinh\left(\frac{a\tau}{c}\right)\,,\quad x(\tau) = \frac{c^2}{a}\cosh\left(\frac{a\tau}{c}\right)\,,\tag{1} \end{equation}$$ where $\tau$ is the proper time of the particle. The next question is: What is the time and space coordinates transformation between the lab frame $(t, x)$ and the particle frame? The time in the particle frame is just the proper time $\tau$ and we denote the space coordinate in the particle frame by $\xi$. The transformation between $(t, x)$ and $(\tau, \xi)$ is obviously not Lorentz transformation since the particle frame is not an inertial frame because of the acceleration. Instead, Eq. (1) is a special case of the transformation at $\xi=0$. The key insight is that although the particle frame is not an inertial frame, there exists a series of instantaneous co-moving inertial frames at each ti...

Considering special relativity, what is the world line of a moving particle with constant proper acceleration $a$ observed in the still world frame? The proper acceleration is defined as the acceleration measured by the inertial observer to which the particle looks rest during the measurement. Lorentz transformation of acceleration The problem is about the Lorentz transformation of acceleration. For this purpose, we start by the Lorentz transformations in special relativity between a still frame $(t, x)$ and a moving frame $(t', x')$ with constant speed $v$ relative to the still frame: $$\begin{eqnarray} x'&=&\frac{x-vt}{\sqrt{1-{v^2}/{c^2}}}\,,\tag{1}\\ t'&=& \frac{t-vx/c^2}{\sqrt{1-{v^2}/{c^2}}}\,.\tag{2}\end{eqnarray}$$ A motion with speeds $u_x\equiv \frac{dx}{dt}$ in the still frame and $u'_x\equiv \frac{dx'}{dt'}$ in the moving frame has the transformation: $$\begin{eqnarray} u'_x = \frac{u_x - v}{1-{vu_x}/{c^2}}\,,\tag{3} \end{eqnarray}$$ ...

 The formula of the monthly mortgage payments for a fixed-rate loan can be found online as $$\begin{equation} M = P\frac{r(1+r)^n}{(1+r)^n-1}\,,\tag{1}\end{equation}$$ where  $M$ is the mortgage payment.   $P$ is the principal, i.e., the initial amount borrowed.  $r$ is the monthly interest rate. For annual interest rate $2.5\%$, $r=2.5\% / 12$. $n$ is the number of payments. For 30-years fixed rate mortgage, $n=30 * 12 = 360$. Here we provide a derivation of the above formula. Let $b_i$ be the owned balance in the $i$-th month. Initially, $b_0=P$ and we choose the constant monthly payment $M$ such that $b_n=0$. We then have the recursion relation between two consecutive month as $$\begin{equation}b_{i+1}=b_i(1+r) - M\,.\tag{2}\end{equation}$$ A trick to solve (2) is to write Eq. (2) into the following form $$\begin{equation}b_{i+1}+C = \left(b_i+C\right)(1+r)\,,\tag{3}\end{equation}$$ so that $b_i+C$ is a geometric sequence: \begin{equation}b_i + C = (b_0 + C) (1+r)^i\,.\end{eq...