In the first semester when we studied mechanics, I mentioned that the expression we used for momentum is not valid for speeds greater than around 0.1c. I made similar mention of the expression for kinetic energy. We will now look at the correct versions of these expressions that work for all speeds.[br][br]Let me first draw attention to the failings of Newtonian physics. When we wrote [math]\vec{F}=m\vec{a}[/math] we obviously implied that so long as a force is applied on a mass, the mass would keep accelerating. Supposing we wait long enough there is nothing limiting the eventual velocity of the object. It would accelerate right up to and beyond the speed of light. Newton's second law formulated as [math]\vec{F}=m\vec{a}[/math] is not true.[br][br]It is interesting to note that Newton himself also wrote his second law in the form of momentum, albeit classical momentum which we'll fix shortly. Written in terms of momentum, the second law is[br][br][center][math][br]\vec{F}=m\vec{a} \\[br]\vec{F}=m\frac{d\vec{v}}{dt} \\[br]\vec{F}=\frac{d}{dt}(m\vec{v}) \\[br]\vec{F}=\frac{d\vec{p}}{dt}.[br][/math][/center][br][br]It turns out that when we think of force as a means of changing the momentum, that version of Newton’s second law is true in all cases so long as we "fix" the momentum expression we used in classical mechanics.

Fixing momentum is done by insertion of [math]\gamma[/math] into the expression. If you recall, [math]\gamma=1[/math] until speeds get very great. Obviously Newton would have had no means of knowing or measuring such things. The correct version of momentum is [br][br][center][math]\vec{p}=\gamma m\vec{v} = \frac{m\vec{v}}{\sqrt{1-v^2/c^2}}. [br][/math][/center][br][br]You can see that the reason we didn't use this form in first semester is mainly because the velocity is in two places in the equation, and therefore doing simple calculations becomes unnecessarily tedious for slow speeds.[br][br]It can be seen in the plot below that at slow speeds the two expressions give nearly identical results. However, as v approaches the speed of light (dotted line), the classical expression passes right through the speed of light while the relativistic expression diverges.

An object of mass m only has two forms of energy - rest energy and kinetic energy. We made the same statement in our work/energy chapter in first semester. The total energy will here be denoted [math]E[/math], the rest energy [math]E_0[/math] and the kinetic energy [math]K.[/math] As we should expect, the total is the sum of the rest and kinetic energies, or[br][br][center][math][br]\text{One expression for the total energy is: } \\[br]E=E_0+K \\[br]\text{where rest energy is } \\[br]E_0=mc^2 \\[br]\text{Another way total energy may be expressed is: } \\[br]E=\gamma E_0 = \gamma mc^2. [br][/math][/center][br][br]If you are thinking there are other forms of energy like those associated with temperature, or excited electron states in a material, etc, they are not really different in a fundamental sense. Rather, the mass of a system rises on account of them. [br][br]In other words, a hotter cup of coffee has more mass than a cooler cup even if we take the time to ensure that no evaporation took place and thus no molecules of coffee were lost. You will notice that cooling still implies a transfer of energy either via conduction or via radiation. Even in vacuum there will be radiation from a hot cup of coffee. We can see that the energy is lost as photons, and while photons have no mass, they do contain energy. Energy is equivalent to mass. In fact, perhaps a good way to define mass is as localized, condensed energy. What is the equation that defines the mass to energy conversion? [math]E=mc^2[/math] of course.[br][br]Likewise an atom with an excited electron is more massive than one in the ground state. How much more massive? If you know the excitation energy [math]\Delta E[/math] then [math]\Delta E = \Delta m c^2[/math] where [math]\Delta m[/math] is the associated increase in mass.[br][br]If you every wondered in what form potential energy is stored, it is in the form of mass as well. In other words, every time we see systems with energy standing still or remaining localized, it manifests as mass! Conversion from mass to pure energy (light) has been observed countless times in laboratories. Recently scientists have made particles from nothing but light for the first time as well.

Given the expression for total energy, we can get the kinetic energy by algebra. Solving for K, we get [br][br][center][math][br]E=E_0+K\\ [br]K=E-E_0 \\[br]K=\gamma mc^2-mc^2 \\[br]K=(\gamma -1)mc^2.[br][/math][/center][br][br]Admittedly this looks nothing like [math]K=\frac{1}{2}mv^2[/math] from first semester, but we should expect that somehow this is equivalent. If we expand the relativistic expression for K in a Taylor series expanded about v=0 and keep the first term, we will find the classical kinetic energy expression.[br][br]To make things simple we can expand [math]\gamma[/math] while substituting [math]x=v^2/c^2.[/math] The expansion of [math]1/\sqrt{1-x} = 1 + \frac{1}{2} \; x + \frac{3}{8} \; x^{2} +...[/math] Substituting back in for x gives [math]\gamma = 1/\sqrt{1-v^2/c^2} = 1 + \frac{1}{2} \; v^2/c^2 + \frac{3}{8} \; v^4/c^4 +...[/math] To get K, we need to subtract one and multiply by [math]mc^2[/math]. That gives us[br] [br][center][math]K = \frac{1}{2} \; mv^2 + \frac{3}{8} \; mv^4/c^2 + ...[/math][/center] [br][br]As you can see, the first non-zero term in the Taylor expansion is the classical expression for the kinetic energy. The next correction term shown is small whenever v is quite a lot smaller than c.