Parametric Equations

This one will be a short post, it's mostly intuition as opposed to a definite proof.

So I was musing over finding the area under a parametric curve. And I considered how we'd usually find the area under a curve, integration. Which in 2D with an x and y axis can be expressed generally as:
\[ \int_a^b y \ \mathrm{d}x\]
Parametric equations define a curve in terms of a parameter, usually t. So let's consider an arbitrary set of parametric equations:
\[\begin{align*} x=x(t) \\ y = y(t) \end{align*} \]
So to change the dx to a dt to allow integration one must differentiate x with respect to t and solve for dx.
\[ \begin{align*} x &= x(t) \\ \frac{\mathrm{d}x}{\mathrm{d}t} &=x'(t)\\ \mathrm{d}x &= x'(t) \mathrm{d}t \end{align*}\]
The value for dx must then be substituted back into the original expression.
\[\int_a^b y(t)x'(t) \ \mathrm{d}t\]
And that's it, how to find the area under a curve defined by a parametric equation.

Kinetic Energy

So I started writing this a while ago but never got around to finishing it, this post is fairly calculus heavy so be warned.

Another equation we so often use in physics is $E = \frac{1}{2}mv^2$, this is the equation for the kinetic energy of an object. Kinetic energy is also used pretty vaguely at GCSE and A level. A more formal definition is "The energy possessed by an object because of its motion" or "The energy required to accelerate a mass to a velocity". In an ideal system, i.e. one with no external forces acting the accelerated body will possess this energy and associated velocity until collision.

Clearly kinetic energy is related to $E = \int F \mathrm{d}s$_[1] as this is work equation and is the very definition of mechanical work in physics. The unit of energy we use in physics is the joule. When considering a mechanical situation such as this one, 1 joule is defined as the amount of energy used in applying a force of 1 newton over 1 metre.

However this knowledge alone is not enough to derive the expression for kinetic energy, we also need to know what force is. Force is defined mathematically in Newton's second law as $F=\frac{\mathrm{d}p}{\mathrm{d}t}$ or more simply in the case of constant mass as $F=m\frac{\mathrm{d}v}{dt}=ma$

From Newton's second law and the work equation the equation for kinetic energy can easily be found using some calculus.

First we substitute $F=m\frac{\mathrm{d}v}{dt}$ into the work equation: \[F\mathrm{d}s = m \frac{\mathrm{d}v}{\mathrm{d}t}\mathrm{d}s \] Second we need to rewrite the work equation into a different form to remove the ds. s, is just distance and can be expressed as velocity multiplied by time, $\mathrm{d}s = v \ \mathrm{d}t$. So using this knowledge the work equation can be rewritten as: \[F\mathrm{d}s = m \frac{\mathrm{d}v}{\mathrm{d}t}v \ \mathrm{d}t \] Cancelling the dt's and simplifying leaves: \[F\mathrm{d}s = mv \ \mathrm{d}v \] This can now be substituted back into the work equation: \[E = \int \! mv \ \mathrm{d}v \] The equation for kinetic energy assumes that mass is constant, so this can be factored out of the integral as a constant. \[E = m \int v \ \mathrm{d}v \] Integrating this yields: \[E = m \left[ \frac{v^2}{2} + C\right] \] Putting limits on this integral will eliminate the +C and provide the energy associated with a particular change in velocity. Assuming the particle was always accelerated from 0 to a velocity v this simplifies to: \[E = m \left[ \frac{v^2}{2} + C\right]_{0}^{v} = \frac{1}{2}mv^2\] Which is the equation for kinetic energy!

[1] strictly speaking the equation is $E = \int F\cdot \mathrm{d}s$, this includes use of the dot product. In this post have assumed that the force is always acting in the same direction as the motion, however this is often not the case and hence the dot product is required.

So why does E = mc2?

I'm sure you've seen this equation before, $E = mc^2$ , it's probably the most famous equation in all of Physics. While you may have heard of it you probably don't know what it means or how it came about. Einstein didn't just decide one day that E must equal mc^2 and it magically fitted into place with all of physics so far; it was actually a by-product of special relativity that just happens to be exceptionally useful!

So what does it mean? Well $E = mc^2$ means that mass and energy are equivalent. Basically that means that you can convert them into each other, which you can express mathematically as $E \propto m $. It just so happens that $c^2$ is the conversion factor; for those of you unfamiliar with physical constants c is the speed of light in a vacuum. So you will appreciate that $c^2$ is a phenomenally massive number, in fact it is $c^2 = 9 \times 10^{16}$, or 9 followed by 16 zero's!

The best way to see mass energy equivalence is through an example:
If I had an average cat, it probably has a mass of about 5kg, so if I put that back into our expression we'll have: \[E = 5 \times 9\times 10^{16} = 45 \times 10^{16} Joules\] so you can appreciate that written down it's:
450,000,000,000,000,000 Joules of energy, that's quite a lot!

In fact "Fat Man", the bomb dropped on Nagasaki had a yield of only 88,000,000,000 Joules. That means all the energy contained in your pet cat is about 5 millions times greater!

*(note: the total energy contained in the bomb was much larger than the cat as it is of greater mass; however its yield energy was not as high as its intrinsic energy)

So that's what $E = mc^2$ means, but how did Einstein arrive at it?

I'm going to have to set some axioms for this, else this blog post could span on a lot and get rather boring, so you'll have to take my word on a few things.

Firstly:

$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2} }}$

Secondly:

$m = \gamma m_o$

All that means is that as you move faster your mass increases, even at slow speeds. This is due to the fact that the speed of light is a constant, but that is another story entirely.

So to derive $E = mc^2$ we just have to write the second expression in an alternate form. \[m = m_0(1 - \frac{v^2}{c^2})^{-\frac{1}{2}} \]And those of you familiar with A level maths should recognise this as a binomial expansion. It expands to: \[m = m_0(1 + \frac{v^2}{2c^2} + \frac{3v^4}{8c^4} + \cdots) \]This sequence converges very rapidly at low velocities as c is so high so: \[m \simeq m_0 + \frac{1}{2}m_0 v^2(\frac{1}{c^2}) \]If we multiply throughout by $c^2$ the sequence simplifies to: \[mc^2 \simeq m_0 c^2 + \frac{1}{2}m_0v^2 + \cdots \]If we assume an object is not moving then its velocity is 0 so: \[mc^2 = m_0c^2\]As this is an energy term so we can then say: \[E = mc^2 = m_0c^2\] as long as the object in question is not moving.

In actual fact we can use this binomially expanded equation while v is much less than c, even if the object is in motion. However in its usual form $E = mc^2$ is referring to the intrinsic energy of the object, i.e. its energy when it is not moving.

And there you have it, with nothing more than an understanding of binomial expansion it is possible to show that $E = mc^2$.