Momentum and spacetime

Question

My apologies up-front for the naive question and my rudimentary understanding, but I should be delighted if someone would enlighten me :)

The question has to do with the application of the conservation of momentum, when considering time as a fourth spatial dimension. I'll explain my understanding of time first, then pose the question in that context.

Time as a Fourth Spatial Dimension

I understand time to be a fourth spatial dimension through which we travel, in one direction only, at a constant pace^†, such that in one second's time we will be 3x10⁸ metres further along the time dimension than we are now.

Momentum vs Annihilation

It strikes me that although the pace of travel along such a time dimension is seemingly fixed at '$c$', it could also be zero. In other words, an object that is annihilated will stop travelling through the time dimension.

In classical physics, the kinetic energy released from a loss of momentum is given by:

$$e = \frac{1}{2} mv^2$$

As our hypothetical object had previously been travelling at $c$ through the time dimension, and is now travelling at zero, the kinetic energy released would be:

$$e = \frac{1}{2} mc^2$$

However, in relativistic physics the energy associated with annihilation is given by:

$$e = mc^2$$

Questions

A couple of questions:

• Why isn't the loss of kinetic energy ($1/2 mc^2$) equal to the energy released on annihilation ($mc^2$) ?
• I am clearly missing something in my rudimentary knowledge, but the two equations are intriguingly close, does that lend weight to my opening premise, that we are travelling at a fixed pace ($c$) along a fourth spatial dimension which we perceive as time?

Thank you for listening. I looked elsewhere on SE Physics and couldn't find anything similar, but I apologise if this has been asked before.

Stuart

_{† - I have deliberately avoided the terms 'speed' and 'velocity', as they are with respect to time, and hence I struggle with applying them in this context.}

Answer 1

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. :)

Answer 2

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 $c$.