The motivation for studying relativistic dynamics comes from thinking about conservation of the standard forms of energy and momentum with our new relativistic dynamics. It is easy to demonstrate that $mv$ cannot be conserved in all inertial frames of reference in special relativity. Consider two balls of equal mass colliding inelastically with equal speed $v$ in opposite directions, $+v$ and $-v$. They smash into each other and remain stationary.
Now boost into one of the balls' frames, say $v$. Now the velocity of the other ball is $2v/(1+v^2)$, so the total initial momentum is $-2mv/(1+v^2)$. But after the collision, we see the thing moving at a velocity of $-v$ (we know this because it was 0 in the original frame), which means the final total momentum is $-2mv$, so momentum is not conserved.
But we don't like this! If this expression isn't conserved, we can't use it so nicely in calculations and stuff. We want to define momentum in a way that it is conserved. Similar arguments can be used to show that $mv^2/2$ is not conserved, either.
You may try to derive a conserved expression via similar arguments as the symmetry-based arguments we use in non-relativistic mechanics, swapping Galilean symmetry with Lorentz symmetry where appropriate. The resulting functional equations would be ludicrously complicated, though, and we'd much rather use a different symmetric argument.
We've made several arguments so far based on known properties of light, and it would make sense to assume other, quantum mechanical properties of light as well. Two such properties are:
$$\begin{array}{l}p = hf/c\\E = hf\end{array}$$
This means that we know the behavior of $p$ and $E$ at low velocities, as well as at velocities close to the speed of light. Surely, we're smart enough to fill in the stuff in between?
Consider the following set-up: a stationary mass m lets out two equal flashes of light in opposite directions, each with energy = momentum (since $c=1$) E/2. We then analyse the same set-up from a boosted reference frame with velocity $v$. This involves a doppler shift in the frequency of each light beam.
We'll consider this set-up in the following three examples:
(a) v is small, momentum conservation
We first consider the case where v is small enough to allow the usage of non-relativistic mechanics. Formally, this means taking the limit as $v\to0$.
Then the doppler shift factor $\sqrt{\frac{1+v}{1-v}}$ approaches $1+v$ and $\sqrt{\frac{1-v}{1+v}}$ approaches $1-v$. Both energy and momentum are scaled by the same factor since they're proportional to frequency. Now you know why we choose momentum conservation instead of energy conservation -- the total energy is clearly conserved anyway.
The reason we consider low velocities is that we know the formula for momentum must reduce to the Newtonian $p=mv$, i.e. the initial momentum of the system was $-mv$. The total momentum of the two flashes of light is $((1+v)E/2-(1-v)E/2)=vE$. Since momentum must be conserved, this means the momentum of the mass itself is no longer $-mv$. But its velocity is constant, and still low, so this means some of the mass must have been converted into the energy of the photons. Specifically,
$$-m_fv-(-m_iv)=vE$$
Giving us the celebrated equation
$$E=m$$
Where $m$ is the amount of mass that was converted into energy. You could, of course, write this in inelegant ways such as $E=c^2m$ or even $E=mc^2$.
this change in mass is not linked to the whole "relativistic mass" thing we'll be doing later. This decrease in mass is absolute, mass is not conserved, it is also seen in the rest frame, and is required to produce that bit of energy. It's only the derivation that requires boosting into another reference frame, to ensure conservation in all reference frames.
On a related note, note that conserved and invariant are by no means the same thing, or even related. A quantity is conserved if it doesn't change with time when taken of the whole system. It is invariant if it is the same from all reference frames. The difference isn't even subtle -- proper mass is an invariant in special relativity, but Energy and momentum are conserved.
Something to think about: why doesn't our argument work in a non-relativistic frame? I mean, we even assumed that v is small. Try to perform the same arguments without relativity -- you will see that since there is no relativistic doppler shift, the result will have a unit of mass being worth an infinite amount of energy -- something you get in the limit $c\to\infty$ -- useless anyway.
(b) v is not small, energy conservation
We said the decrease in mass exists in all reference frames. If we found what exactly the decrease in mass $\Delta m$ is in each reference frame, then we'd be able to see how mass transforms under a Lorentz transformation.
In the rest frame, energy $E$ is released, therefore by energy conservation the energy (or equivalently, the mass) of the object decreases by $E$.
In the moving frame, one of the beams transforms as $\sqrt {\frac{{1 + v}}{{1 - v}}} \frac{E}{2}$ while the other transforms as $\sqrt {\frac{{1 - v}}{{1 + v}}} \frac{E}{2}$. So the total energy released (i.e. the energy loss of the object) is:
$$\left( {\sqrt {\frac{{1 - v}}{{1 + v}}} + \sqrt {\frac{{1 + v}}{{1 - v}}} } \right)\frac{E}{2} = \gamma E$$
So the mass has transformed as $\gamma m$ under a Lorentz boost of significant velocity.
We call this mass the "relativistic mass" $M$, and distinguish it from the rest mass $m$.
Then the following are immediately true:
- $E = m$ is only true when an object is at rest. In general, $E = \gamma m$. We may call $E_0=m$ the rest energy.
- $E=M$
- $M=\gamma m$
- The increase in mass is essentially the kinetic energy. One may Taylor (or Newton's Binomial) expand out $m/\sqrt{1-v^2}$ to see that the terms start as $m+\frac12mv^2+3/8mv^4+...$, and the higher-order terms vanish at low speeds. Therefore the relativistic kinetic energy is generally $M-m=(\gamma-1)c^2m$.
In general, we will denote the relativistic mass as $E$ and the rest mass as $m$ unless otherwise stated.
It is a fad among modern relativity textbooks to claim the phrase "relativistic mass" is a misnomer or even a mnemonic to help kids understand relativity and simply call it the energy, reserving the word "mass" to mean the rest mass. However, this obscures some of the best analogies between spacetime and momentum-energy, as we will soon see -- for instance, the relativistic mass is actually analogous to the co-ordinate time and the rest mass to the proper time/spacetime interval.
Therefore, we will use the word "mass" to refer to the relativistic mass $E$ and "proper mass" and "momentum-energy interval" to refer to the rest mass $m$. This is a convention in our course only.
(c) v is not small, momentum conservation
We may do a similar analysis as above with momentum to arrive at the expression for relativistic momentum.
The total/net momentum of the light beams in the boosted frame is
$$\left( {\sqrt {\frac{{1 + v}}{{1 - v}}} - \sqrt {\frac{{1 - v}}{{1 + v}}} } \right)\frac{E}{2} = \gamma vE$$
(Note that $E$ represents the total rest energy of the light beams here, as was defined in the question.)
Therefore $p=\gamma mv$, or $p=vE$.
You may use this to calculate the relativistic calculation for $F=dp/dt$, but it's simply computation from this point, so I'll just direct you to wikipedia. Come up with an expression for a general directional inertia (simple).
some people are surprised by the relation $p=vE$, or even remember it wrongly as $E=vp$ because of the seeming resemblance with $E=pc$ at the speed of light (this confusion is because of people not getting the hang of $c=1$ natural units). But it's really nothing new. $E$ is simply the mass. We know momentum equals mass times velocity. This is not new.
Continued in Minkowski everything -- spacetime vectors, rapidity.
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