# Why expression of energy occurs in triplet?

I have basically two questions in mind,which are

1. Why expression of energy occurs in triplet?
2. Why the expressions are somewhat symmetrical?

Coming to elaborated form of 1st part, we have

$$U=\frac{C V^2}{2}$$

$$U=\frac{Q^2}{2C}$$

$$U=\frac{QV}{2}$$

The above expressions are for the energy stored in capacitors.

$$U=\frac{V^2}{Rt}$$

$$U=I^2Rt$$

$$U=VIT$$

These are for heat dissipation in resistor.

See in both cases the energy occurs in triplet. There are several other examples, too, in thermodynamics. I want to ask here: Why nature favours $$3$$ in physics?

Now the second part of the question says why the expressions are quite symmetrical?

$$U=\frac{C V^2}{2}$$

$$U=\frac{mv^2}{2}$$

$$U=\frac{I\Omega^2}{2}$$

Are both parts of the question a coincidence, or something else?

It's because you have chosen to focus your attention on three items $$C,V,Q$$ and since there is one equation relating them: $$Q = C V$$, it follows that anything to do with these items can be written down in terms of any two of the three. Since there are three ways of choosing two items from a set of three, you end up with three equations.

With the resistor you picked, quite naturally, $$V,I,R$$ and you have $$V=IR$$ so the same logic applies.

The second part of your question is a very different sort of a question. It is fundamentally to do with the fact that energy is related to an integral of force over distance, so if the force is varying linearly with distance, which is the case near any local minimum of energy, then you will get a quadratic formula for energy. So these quadratic formulae generally apply for small changes, small amounts of energy. With a real physical object serving as a capacitor, for example, the linear approximation will no longer apply at high enough voltage or charge, owing to details of the materials involved, and then the simple equation no longer holds.

For the second part, we can say that generally, in classical physics, many quantities are linked by first-order differential equations. Thus, we can often express the power P like this: $$P=aK\dfrac{dK}{dt}$$. To have the energy, we need to integrate P over time: $$\int P dt=\int aK\dot{K}dt=\frac{1}{2}aK^2$$.

For a capacitor, $$i_c=C\frac{du_c}{dt}\Rightarrow P=u_ci_c=Cu_c\frac{du_c}{dt}$$

For an inductance, $$u_L=L\frac{di_L}{dt}\Rightarrow P=u_Li_L=Li_L\frac{di_L}{dt}$$

For a mechanical system, $$P=Fv=mav=mv\frac{dv}{dt}$$

One could also argue that the form $$\frac{1}{2}aK^2$$ looks like the term of a Taylor expansion around $$0$$ (and therefore an approximation for low values): $$f(x)=f(0)+xf'(0)+\frac{1}{2}f''(0)x^2+o(x^3)$$. This is the case for the kinetic energy for example.

Einstein found that the energy of a moving body is: $$E=\dfrac{m_oc^2}{\sqrt{1-\frac{v^2}{c^2}}}$$. For $$\frac{v^2}{c^2}<<1$$, $$E\approx m_oc^2\left[1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\left(\frac{v^2}{c^2}\right)^2\right]$$. And now, you can see the $$\frac{1}{2}aK^2$$ factor appears (and others that we neglect).

The hooke's law gives an elastic potential energy: $$U_s=\frac{1}{2}Kx^2$$ because we assume that the force is linear which is a result of a Taylor expansion around the stable distance $$r_s$$ between two molecules given by the Lennard-Jones potential.

For the triplet, I don't think there's any meaning to this, but I might be wrong.

Please excuse me for my poor english, I'm learning...