Markozeta

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Mechanical Engineer graduate from Cal Poly Pomona.

Top/Bottom points of an ellipse rotated by an angle theta:

$x=(a^{2}-b^{2})sin(\theta )cos(\theta )\left({\frac {a^{2}cos(\theta )^{2}+b^{2}sin(\theta )^{2}}{(4a^{2}b^{2}+(a^{2}-b^{2})^{2}cos(\theta )^{2}sin(\theta )^{2})}}\right)^{1/2}$

Relativistic Spring-Mass System

The typical spring mass system can be found -somewhere on wikipedia-. However, this does not cover the case when the spring mass system has velocities approaching the speed of light. Indeed, if a mass has a displacement of a 1 light-seconds and a natural period of 1 second, the system described would clearly violate the laws of special relativity.

The relativistic corrected acceleration is, with one-dimensional movement:

${\frac {m{\ddot {\mathbf {x} }}}{\left(1-{\frac {{\dot {\mathbf {x} }}^{2}}{c^{2}}}\right)^{3/2}}}=-k\mathbf {x}$

The triple Lorentz factor correction stems from the four-acceleration. Recognizing this as an autonomous differential equation, set the velocity to y, then

${\ddot {\mathbf {x} }}=y'(x)y(x)$

Where y is the velocity. The equation is easily separable:

${\frac {\mathbf {y} d\mathbf {y} }{\left(1-{\frac {\mathbf {y} ^{2}}{c^{2}}}\right)^{3/2}}}=-{\frac {k}{m}}\mathbf {x} d\mathbf {x}$

Integrate both sides for the (not-so) surprising result:

${\frac {c^{2}}{\left(1-{\frac {\mathbf {y} ^{2}}{c^{2}}}\right)^{1/2}}}=-{\frac {k}{2m}}\mathbf {x^{2}} +\mathbf {c_{1}}$

Recognizing k/m as the square of the natural frequency from classical physics, and the lorentzian from relativistic physics, we see a clash of the titans emerge:

$\gamma =\mathbf {c_{1}} -{\frac {\omega _{n}^{2}}{2c^{2}}}\mathbf {x^{2}}$

The lorentzian is in fact a function of distance squared. This constant (C1) is indeed a constant - it turns out to be the Hamiltonian (or total energy) of the system. Taking the moment to plug in boundary conditions, $x_{m}$ is the maximum displacement of the system. At this distance, the velocity of the system is momentarily zero, and therefore the lorentzian is 1.

$1=\mathbf {c_{1}} -{\frac {\omega _{n}^{2}}{2c^{2}}}x_{m}^{2}$

Therefore:

$\gamma =1+{\frac {\omega _{n}^{2}}{2c^{2}}}{\left(x_{m}^{2}-\mathbf {x^{2}} \right)}$

This equation demonstrates what is happening in this system. As the displacement varies, from negative $x_{m}$ to positive $x_{m}$, the lorentzian changes.

$1=\left(1+{\frac {\omega _{n}^{2}}{2c^{2}}}{\left(x_{m}^{2}-\mathbf {x} ^{2}\right)}\right)\left(1-{\frac {\mathbf {\dot {x}} ^{2}}{c^{2}}}\right)^{1/2}$

Square both sides to obtain:

$\left({\frac {\frac {dx}{dt}}{c}}\right)^{2}={\frac {\omega _{n}^{4}}{4c^{4}}}\left((x_{m}^{2}-\mathbf {x} ^{2})({\frac {4c^{2}}{\omega _{n}^{2}}}+x_{m}^{2}-\mathbf {x} ^{2})\right)\left(1+{\frac {\omega _{n}^{2}}{2c^{2}}}{\left(x_{m}^{2}-\mathbf {x} ^{2}\right)}\right)^{-2}$

This gives rise to the introduction of a new constant,

${\frac {4c^{2}}{\omega _{n}^{2}}}+x_{m}^{2}=\mu ^{-2}x_{m}^{2}$

So $\mu ^{2}$ is:

$\mu ^{2}={\frac {\omega _{n}^{2}x_{m}^{2}}{4c^{2}+\omega _{n}^{2}x_{m}^{2}}}$

Squaring both sides of the original equation, we can now separate the equation:

${\frac {\left(1+{\frac {\omega _{n}^{2}}{2c^{2}}}{\left(x_{m}^{2}-\mathbf {x} ^{2}\right)}\right)}{((x_{m}^{2}-\mathbf {x} ^{2})(x_{m}^{2}\mu ^{-2}-\mathbf {x} ^{2}))^{1/2}}}d\mathbf {x} ={\frac {\omega _{n}^{2}}{2c}}d\mathbf {t}$

Calling: $\mathbf {\delta } =\mathbf {x} /x_{m}$

${\frac {\left(1+{\frac {\omega _{n}^{2}x_{m}^{2}}{2c^{2}}}{\left(1-\mathbf {\delta } ^{2}\right)}\right)}{((1-\mathbf {\delta } ^{2})(1-\mu ^{2}\mathbf {\delta } ^{2}))^{1/2}}}d\mathbf {\delta } ={\frac {\omega _{n}^{2}x_{m}}{2c\mu }}d\mathbf {t}$

Since ${\frac {\omega _{n}^{2}x_{m}^{2}}{2c^{2}}}={\frac {2\mu ^{2}}{1-\mu ^{2}}}$

${\frac {\left(\mu ^{2}-1\right)+2{\left(1-\mu ^{2}\mathbf {\delta } ^{2}\right)}}{((1-\mathbf {\delta } ^{2})(1-\mu ^{2}\mathbf {\delta } ^{2}))^{1/2}}}d\mathbf {\delta } =\omega _{n}(1-\mu ^{2})^{1/2}d\mathbf {t}$

Focusing on $\mu$, this constant can be interpreted as the sine of an angle, $\phi$. A triangle with one side as $\omega _{n}x_{m}$ and the other side of length 2c will have this angle.

Such an equation is not easy to integrate, but with elliptical functions it can be accomplished. Fortunately, the work with $\mu$ has put the integral in cannon forms for the elliptical integrals of the first and second kind.

$(\mu ^{2}-1)F(sin^{-1}({\frac {x}{x_{m}}}),\mu )+2E(sin^{-1}({\frac {x}{x_{m}}}),\mu )=\omega _{n}(1-\mu ^{2})^{1/2}\mathbf {t} +C_{2}$

Where F and E are the elliptical integrals.