Momentum and spacetime

First, something we need to get out of the way: Kinetic energy as $\frac{1}{2} m v^2$ is not a precise formula; it is merely a good approximation for anything that is traveling slowly compared to the speed of light. In fact, more precisely, the energy is \begin{equation} E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + \frac{3}{8} m \frac{v^4}{c^2} + \frac{5}{16} m \frac{v^6}{c^4} + \ldots \end{equation} On the right-hand side, I have done a little Taylor expansion; assuming that $v$ is much smaller than $c$, this expression will be a very good approximation to the precise expression given just before it. You see that $mc^2$ is just the first term, and the familiar kinetic energy formula from basic physics is the second term. The first term can be ignored in basic physics, because it's constant (independent of $v$), and in basic physics all you care about is changes in energy related to changes in $v$. If $v$ is really much much smaller than $c$, the remaining terms are really tiny, which is why we only use that second term for familiar physics. But as you bring $v$ closer and closer to $c$, you would need to keep more and more of those terms. And when $v=c$, the expansion is just completely wrong; it's meaningless (though it happens to give an infinite value, just like the precise definition).

So your premise regarding this formula is off-base. Plugging $c$ into the basic KE formula and getting $(1/2)mc^2$ to be the same as one half of the mass-energy formula $m c^2$ is only a matter of coincidence. When your velocity approaches the speed of light, that basic formula breaks down, and in fact you would approach infinite energy as you approach the speed of light. (Standard physics doesn't let you actually get to the speed of light.)

Now, this "kinetic energy" (which might more accurately be defined as $E - mc^2$) is really the energy that must be given off by the particle in order for the particle that is moving with that energy to change its state of motion so that it has no spatial velocity relative to you. But you're asking how much energy needs to be given off such that the particle has no temporal velocity relative to you. And I don't think this is a meaningful question, at least in the sense that standard physics has no such definition. In nuclear reactions, mass can transform into energy or vice versa, but always in such a way that the total energy-momentum is conserved -- so nothing really stops. Even "annihilation" between matter and antimatter is really just a transformation of particles into a different form; the energy is radiated as gamma rays, for example, which continue to move through time.

Your understanding of time as another dimension through which we travel is basically right, in the sense that most physicists think vaguely along those lines. (Though I can't say precisely along those lines, because your statements are a bit squishy -- as are most such statements physicists would make.) But the question I would ask you is: what does it mean for a particle to stop traveling through time? Though I can't imagine what it might mean, I'll point out what seem to be a couple contradictions in your thinking. You said that you think things travel through time at a constant pace, but then you turn around and say that something can stop. You also seem to be happy with notions of conservation of energy-momentum, but they would seemingly suggest that something else would have to travel faster through time -- whatever that means -- which also contradicts the "constant pace" idea.

So I think you've touched on some interesting ideas, but it seems like you might benefit from making your ideas a bit more precise. I hope these facts about the standard understanding of relativity help in that effort. :)

A few quick clarifications: a particle cannot just annihilate. It disappears when it interacts with something else. The obvious example of this is an electron and positron annihilating to turn into two photons.

Also, the total energy of a particle (this applies to electrons, positrons and photons) is given by:

$$ E^2 = p^2c^2 + m^2c^4 $$

where $p$ is the relativistic momentum:

$$ p = \frac{mv}{\sqrt{1-v^2/c^2}} $$

Note that photons carry momentum as well. The photon momentum is given by:

$$ p = \frac{h}{\lambda} $$

At low speeds, where relativistic effects can be ignored, the total energy can be written as the sum of kinetic and rest energy, but this is an approximation and we normally use the full expression above.

So let's take the example of an electron and positron annihilating to produce two photons. We know energy must be conserved, so $E$ must be the same before and after. That means:

$$ p_e^2c^2 + m_e^2c^4 + p_p^2c^2 + m_p^2c^4 = p_{\gamma 1}^2c^2 + p_{\gamma 2}^2c^2 $$

where the $E$ subscript refers to the electron, $p$ refers to the positron and $\gamma_1$ and $\gamma_2$ refer to the two photons.

We also know momentum is conserved, so:

$$ p_e + p_p = p_{\gamma 1} + p_{\gamma 2} $$

(Actually the expression I've written is only true under non-relativistic conditions, but let's gloss over this for now.)

So if we know the initial momenta we can solve these two equations to calculate the momenta of the two photons $p_{\gamma 1}$ and $p_{\gamma 2}$. So the calculation you're talking about can be done, and indeed it's a standard exercise for students.

Re your last question: have a look at my answer to this question. Stationary objects are indeed travelling in the time dimension at a speed of $c$. Moving objects aren't travelling up the time dimension at $c$, but the magnitude of their velocity in spacetime remains $c$.