# Energy equation

### Projectile motion

Recall, we derived Hubble's constant by considering the distance, speed etc, of an aribitrary pair of galaxies. We do the same here and, for simplicity, we choose our galaxy and any galaxy receding from us in our model of the universe.

We shall make the assumption that the motion of a galaxy away from us is equivalent to Newtonian projectile motion, and in particular, the motion of an object fired from the surface of the earth.

If the object has mass $m$ and the earth has mass $M$ then the energy of the projectile will satisfy the equation $$E = \frac{1}{2}mv^2 - \frac{m M G}{r}$$ where $v = v(t)$ is the speed of the object away - or towards - the earth, at any given time and $r = r(t)$ is the distance from the centre of the earth, not from the surface - even though the object was fired from the surface.

This is a feature of gravity in Newtonian mechanics, namely that we can consider the gravitational force of the earth acting on an object on or above the surface as equivalent to the force due a particle with the same mass as the earth and located at the centre of mass of the earth.

Figure 9.2 - Mass affecting receding galaxy

We can use this feature in our model of a galaxy receding from us, where now, $m$ is the mass of the galaxy and $M$ is the total mass enclosed by a spherical surface, $S$, centred on us, and of radius $$R = a(t)\Delta x$$ which is the physical distance between us and the receding galaxy.

We only need to consider the mass inside $S$ because Newton's law also implies that the mass outside of $S$ cancels out and doesn't contribute at all the gravitational force on the receding galaxy.

Nb. Another way to argue this, and completely equivalent to using Newton's law, is to use Gauss' law for gravitation, which states that the gravitational flux over a closed surface is proportional to the amount of mass enclosed by that surface.

### Energy of recession

Given that $R = a(t)\Delta x$, we can also write, as before, the speed of recession is $$V = \dot{R} = \dot{a}\Delta x = \frac{\dot{a}}{a}R = HR$$

Now suppose the density of matter, $\rho$, in the universe, which at any given instant of time and on large enough scales, we can consider to be a constant, and given that the volume enclosed by a sphere of radius $R$ is $\frac{4}{3}\pi R^3$, then the mass inside $S$ is $$M = \frac{4}{3}\rho \pi R^3$$

Thus, the energy equation for the motion of a galaxy receding from us is \begin{align*}E &= \frac{1}{2}mV^2 - \frac{m M G}{R}\\&= \frac{1}{2}m\left(HR\right)^2 - \frac{m G}{R} \frac{4}{3}\rho \pi R^3\\&= \frac{1}{2}m R^2 \left (H^2 - \frac{8 \pi G}{3}\rho \right )\end{align*}

### Zero energy

In projectile motion, we define the escape velocity to be that velocity which implies that the total energy - kinetic and potential - is zero. If the receding galaxy is at the escape velocity, then $$0 = \frac{1}{2}m R^2 \left (H^2 - \frac{8 \pi G}{3}\rho \right )$$ or $$H^2 = \frac{8 \pi G}{3}\rho$$

Now, both Hubble's constant, $H = \frac{\dot{a}}{a}$ and the density, $\rho = \rho(t)$ are time-dependent, $$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \rho(t)$$

So, is $V = HR$ really an escape velocity - or equivalently - is the recession energy zero?

The evidence is that, if we measure $H$ and $\rho$ independently, where $\rho$ takes into consideration all of the matter and energy in the universe - galaxies, intergalactic dust, dark matter, vacuum energy etc - then the equation holds to within about a $1$% accuracy.

### Total energy and curvature in general relativity

We have seen before that, under the assumptions of homogeneity, isotropy and expansion, space, and so space-time, can have

• Positive curvature
• Zero curvatuve (that is, it's flat)
• Negative curvature

Now, we choose the energy to be exactly zero, but we could also chose it to be either positive or negative. Since, the scale factor and the Hubble parameter are both constant across space at a given instant of time, there are also three possibilities for the total, summed up, energy in the universe. That is,

• Positive total energy
• Zero total energy
• Negative total energy

Using Einstein's equations of general relativity, we can show (but won't here) that there is a simple relationship between the total energy in the universe and the curvature of space-time, which is $$\begin{matrix}+\text{ve energy} &\Leftrightarrow & -\text{ve curvature} & \text{ ever-increasing expansion (time and space)}\\\text{zero energy} &\Leftrightarrow & \text{zero curvature} & \text{ever-decreasing expansion, but no collapse}\\-\text{ve energy} &\Leftrightarrow & +\text{ve curvature} & \text{ initial expansion but eventual collapse}\end{matrix}$$