# reperiendi

## Imaginary time

Posted in General physics by Mike Stay on 2009 January 9
 Statics (geometric = no time): $\displaystyle x$ [x] x coordinate $\displaystyle y(x)$ [y] y coordinate $\displaystyle k$ [k] proportionality constant $\displaystyle y'(x) = \frac{dy(x)}{dx}$ [y/x] slope $\displaystyle s(x) = ky'(x)$ [k y/x] proportional to slope $\displaystyle T(x) = \frac{1}{2} ky'(x)^2$ [k y^2/x^2] distortion $\displaystyle V(y(x))$ [k y^2/x^2] original shape $\displaystyle S(y) = \int (T + V\circ y)(x) dx$ $\displaystyle = \int \left[ \frac{k}{2} \left(\frac{dy(x)}{dx}\right)^2 + V(y(x)) \right] dx$ [k y^2/x] least S at equilibrium Statics (with energy): $\displaystyle x$ [x] parameterization of curve $\displaystyle y(x)$ [y] y coordinate $\displaystyle k$ [kg x/s^2] spring constant at x $\displaystyle v(x) = \frac{dy(x)}{dx}$ [y/x] slope $\displaystyle F(x) = kv(x)$ [kg y/s^2] force due to stretching $\displaystyle T(x) = \frac{1}{2} kv(x)^2$ [kg y^2/s^2 x = J/x] stretching energy density $\displaystyle V(y(x))$ [kg y^2/s^2 x= J/x] gravitational energy density $\displaystyle E(y) = \int (T + V\circ y)(x) dx$ $\displaystyle = \int \left[ \frac{k}{2} \left(\frac{dy(x)}{dx}\right)^2 + V(y(x)) \right] dx$ [kg y^2/s^2 = J] energy (least energy at equilibrium) Statics (unitless distance): $\displaystyle x$ [1] parameterization of curve $\displaystyle y(x)$ [m] y coordinate $\displaystyle k$ [kg/s^2] spring constant $\displaystyle v(x) = \frac{dy(x)}{dx}$ [m] relative displacement $\displaystyle F(x) = kv(x)$ [kg m/s^2 = N] force at x due to stretching $\displaystyle T(x) = \frac{1}{2} kv(x)^2$ [kg m^2 / s^2 = J] stretching energy at x $\displaystyle V(y(x))$ [kg m^2 / s^2 = J] gravitational energy at x $\displaystyle E(y) = \int (T + V\circ y)(x) dx$ $\displaystyle = \int \left[ \frac{k}{2} \left(\frac{dy(x)}{dx}\right)^2 + V(y(x)) \right] dx$ [kg m^2 / s^2 = J] energy (least energy at equilibrium) Dynamics ($\displaystyle \underline{\lambda x.y(x) \mapsto \lambda t.y(it)}$): $\displaystyle t$ [s] time $\displaystyle y(it)$ [m] y coordinate $\displaystyle m$ [kg] mass $\displaystyle v(t) = \frac{dy(it)}{dt} = i\frac{dy(it)}{dit}$ [m/s] velocity $\displaystyle p(t) = m v(t)$ [kg m/s] momentum $\displaystyle T(t) = \frac{1}{2}m\;v(t)^2 = -\frac{m}{2}\left(\frac{dy(it)}{dit}\right)^2$ [-kg m^2/s^2 = -J] -kinetic energy $\displaystyle V(y(it))$ [kg m^2 / s^2 = J] potential energy $\displaystyle iS(y) = \int (T + V\circ y \circ i)(t) dt$ $\displaystyle = \int \left[ -\frac{m}{2}\left(\frac{dy(it)}{dit}\right)^2 + V(y(it)) \right] dt$ $\displaystyle = i \int \left[ \frac{m}{2}\left(\frac{dy(it)}{dit}\right)^2 - V(y(it)) \right] d it$ [kg m^2/s] i * action