Lagrangian

The Lagrangian is a function that summarises the dynamics of a system. In pilitron physics, the Lagrangian can be expressed as an amount of energy, from which all information about the system can be obtained.

Basic Definition
For an object that moves in a perfectly straight line, in flat (non-curved) space, with no forces acting on it, the Lagrangian is:


 * $$\mathcal{L} = pv$$

Where $$p$$ is the momentum, and $$v$$ is the velocity of the object.

Deriving properties
From the Lagrangian, you can find all non-constant properties of an object.

Velocity
Velocity is given by:


 * $$v = \frac{\mathcal{L}}{p}$$

A rough approximation of the momentum is given by $$mv$$ for objects with mass, and we can substitute that in:


 * $$v = \frac{\mathcal{L}}{mv}$$

And this then continues:


 * $$v = \frac{\mathcal{L}}{m\frac{\mathcal{L}}{m\frac{\mathcal{L}}{m...}}}$$

However, this is not a good method to find the velocity. A better way is if we substitute this value of $$p$$ into the Lagrangian itself:


 * $$\mathcal{L} = mv^2$$

Which leads to:


 * $$v = \sqrt{\frac{\mathcal{L}}{m}}$$

Time-relative Lagrangian
A time-relative Lagrangian is a Lagrangian that is calculated based on time. An example time-based Lagrangian is:


 * $$\mathcal{L}(t) = \frac{pd}{t}$$

Where $$d$$ is the distance from the starting point.

Power
The derivative of a time-based Lagrangian is the power of the object at the given point in time.


 * $$\mathcal{L}^\prime(t) = \lim_{h \rightarrow 0} \frac{\frac{pd}{t+h} - \frac{pd}{t}}{h}$$

If we make the two fractions on top have a common denominator, we get:


 * $$\mathcal{L}^\prime(t) = \lim_{h \rightarrow 0} \frac{\frac{pdt}{t^2+th} - \frac{pd(t+h)}{t^2+th}}{h}$$

Then we get:


 * $$\mathcal{L}^\prime(t) = \lim_{h \rightarrow 0} \frac{\frac{pdt}{t^2+th} - \frac{pdt+pdh}{t^2+th}}{h}$$

Which simplifies to:


 * $$\mathcal{L}^\prime(t) = \lim_{h \rightarrow 0} \frac{\frac{pdh}{t^2+th}}{h}$$

Which further simplifies to:


 * $$\mathcal{L}^\prime(t) = \lim_{h \rightarrow 0} \frac{pdh}{t^2h + th^2}$$

Now factor out the $$h$$ on the bottom:


 * $$\mathcal{L}^\prime(t) = \lim_{h \rightarrow 0} \frac{pdh}{h\left(t^2 + th\right)}$$

And we can simplify that to:


 * $$\mathcal{L}^\prime(t) = \lim_{h \rightarrow 0} \frac{pd}{t^2 + th}$$

Now, if we let $$h = 0$$, we get:


 * $$\mathcal{L}^\prime(t) = \frac{pd}{t^2}$$

If $$d$$ is a function, then its derivative must be found as well, and multiplied by a small difference in its argument.