I would like to show that in an FRW metric the momentum of a freely falling object decays as the inverse of the scale factor. I know there are many proofs and arguments for this but I am trying to get this starting from geodesics and having some trouble.

General logic

My general approach is the following.

The object is moving along some path. I want to find the momentum of the object that would be measured by a coincident, locally inertial observer, i.e. a detector that is freely falling and locally cannot tell that spacetime is curved.

In the locally inertial frame where the observer is momentarily at rest the 4-velocity of the observer has components $u^\mu\equiv d\xi^\mu/d\sigma= (1,0,0,0)$, where $\xi^\mu$ are the coordinates of this local inertial (cartesian) frame and $\sigma$ is the observer's proper time (in this frame $d\sigma=d\xi^0$). In this same frame the object has 4-momentum $p^\mu \equiv md\xi^\mu/d\tau$, where $\tau$ is the object's proper time, and $m$ the rest mass.

Since special relativity holds in this frame the components of $p^\mu$ have the values $p^\mu=(E_\mathrm{obs},\mathbf{p}_\mathrm{obs})$, the energy and momentum that would be measured by the observer at rest in this frame. They obey $E_\mathrm{obs}^2-\mathbf{p}_\mathrm{obs}^2=m^2$. We can retrieve the energy $E$ by contracting $u^\mu$ with $p^\mu$:

$$ E_\mathrm{obs}=g_{\mu\nu}u^\mu p^\nu, $$ where $g_{\mu\nu}=\text{diag}(1,-1,-1,-1)$ are the components of the metric in this locally inertial coordinate system.

The magnitude of the 3-momentum in this system is just

$$ |\mathbf{p}_\mathrm{obs}| = \sqrt{(g_{\mu\nu}u^\mu p^\nu)^2 - m^2}. $$

The right hand side is invariant under general coordinate transformations. Therefore, I think we can calculate the RHS in any coordinate system we want and the value will be the magnitude of the 3-momentum as measured by the observer corresponding to the path $u^\mu$ in a locally inertial frame.

Is this ok so far?

FRW metric

$x^\mu = (t, \mathbf{x})$ are the coordinates of the FRW metric, defined by:

$$ d\tau^2 = dt^2 - a(t)^2 \left(d\mathbf{x}^2 + K \frac{(\mathbf{x}\cdot d\mathbf{x})^2}{1-K\mathbf{x}^2} \right), $$ where $K =0$, $+1$, or $−1$.

The geodesic paths can be found using the Euler-Lagrange equations (see this post). The resulting equations are

\begin{align} 0 &= \frac{d^2 t}{d\tau^2} + a\dot{a} \left(\mathbf{x}'^2 +\frac{K(\mathbf{x}\cdot \mathbf{x}')^2}{1-K \mathbf{x}^2}\right)\\ \mathbf{0} &= \frac{d}{d\tau} \left[ a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right) \right] - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} a^2 \left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right), \end{align} where a prime means $d/d\tau$ and $\dot{a}=da/dt$ is a function of $t$.

The simple solutions to these equations are $t=\tau, \mathbf{x}=\text{const}$. These solutions correspond to the "comoving" observers that move along with the cosmic expansion. The world lines of these observers all have 4-velocity $u^\mu=(1,0,0,0)$ in this coordinate system. When the object is at some location $\mathbf{x}$ at time $t$ I want to get the momentum as measured in a locally inertial frame in which the comoving observer that sits at position $\mathbf{x}$ is momentarily at rest.

To get the equations of motion for an arbitrary freely falling object you have to integrate the above equations. I was able to do this only for the second equation but I think that's enough.

Define $\mathbf{f}$ by $$ \mathbf{f} \equiv a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right), $$ to write the second equation as $$ \mathbf{0} = \frac{d \mathbf{f}}{d\tau} - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} \mathbf{f}. $$ Multiply through by the integrating factor $\sqrt{1-K\mathbf{x}^2}$ to get

\begin{align} \mathbf{0} &= \sqrt{1-K\mathbf{x}^2}\frac{d \mathbf{f}}{d\tau} - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{\sqrt{1-K\mathbf{x}^2}} \mathbf{f} \\ &= \sqrt{1-K\mathbf{x}^2}\frac{d \mathbf{f}}{d\tau} + \mathbf{f} \frac{d}{d\tau}\sqrt{1-K\mathbf{x}^2} \\ &= \frac{d}{d\tau}\left(\mathbf{f} \sqrt{1-K\mathbf{x}^2}\right). \end{align}

The solution is \begin{align} \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} = a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right), \end{align} for some constant 3-vector $\mathbf{c}$.

The term $(\mathbf{x}\cdot\mathbf{x}')$ can be isolated by dotting this equation with $\mathbf{x}$:

\begin{align} \frac{\mathbf{c}\cdot\mathbf{x}}{\sqrt{1-K\mathbf{x}^2}} &= a^2\left((\mathbf{x} \cdot \mathbf{x}') + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}^2}{1-K\mathbf{x}^2}\right)\\ &= a^2 (\mathbf{x} \cdot \mathbf{x}') \left(1+ \frac{K\mathbf{x}^2}{1-K\mathbf{x}^2}\right)\\ &= a^2 \frac{\mathbf{x} \cdot \mathbf{x}'}{1-K\mathbf{x}^2}. \end{align}

Now $\mathbf{x}'$ can be written in terms of $\mathbf{x}$: \begin{align} a^2 \mathbf{x}' &= \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} - a^2 \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\\ &= \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} - \frac{K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{\sqrt{1-K\mathbf{x}^2}}\\ &= \frac{\mathbf{c} - K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{\sqrt{1-K\mathbf{x}^2}}. \end{align}

Finally the spatial part of the object's 4-momentum is \begin{align} p^i = m\frac{d\mathbf{x}}{d\tau} = m \frac{\mathbf{c} - K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{a^2\sqrt{1-K\mathbf{x}^2}}. \end{align}

Now to try to use the $|\mathbf{p}_\mathrm{obs}| = \sqrt{(g_{\mu\nu}u^\mu p^\nu)^2 - m^2}$ formula I had earlier.

Since $u^\mu=(1,0,0,0)$, $g_{00}=1$, and $g_{0i}=0$ at $\mathbf{x}$, $|\mathbf{p}_\mathrm{obs}| = \sqrt{(p^0)^2 - m^2}$.

Writing out $g_{\mu\nu}p^\mu p^\nu = m^2$ gives \begin{align} m^2 &= (p^0)^2 - m^2 a^2\left(\mathbf{x}'^2 + K \frac{(\mathbf{x}\cdot \mathbf{x}')^2}{1-K\mathbf{x}^2} \right) \\ &= (p^0)^2 - m^2 (\mathbf{f} \cdot \mathbf{x}') \\ &= (p^0)^2 - m^2 \frac{\mathbf{c}}{\sqrt{1-K\mathbf{x}^2}} \cdot \frac{\mathbf{c} - K(\mathbf{c}\cdot\mathbf{x})\mathbf{x}}{a^2\sqrt{1-K\mathbf{x}^2}} \\ & = (p^0)^2 - m^2 \frac{\mathbf{c}^2 - K(\mathbf{c}\cdot\mathbf{x})^2}{a^2 (1-K\mathbf{x}^2)}, \end{align}

and I get

$$ |\mathbf{p}_\mathrm{obs}| = \sqrt{(p^0)^2 - m^2} = \frac{m}{a}\sqrt{\frac{\mathbf{c}^2 - K(\mathbf{c}\cdot\mathbf{x})^2}{1-K\mathbf{x}^2}}. $$

But I don't think this is proportional to $1/a$ in general.

If the object passes through the origin then everything works as I expect. In this case $\mathbf{c}$ is parallel to $\mathbf{x}'$ when $\mathbf{x}=0$ and the equation of motion keeps it moving along a straight line parallel to this initial $\mathbf{x}'$ so $\mathbf{x}$ remains parallel to $\mathbf{c}$ and $(\mathbf{c} \cdot \mathbf{x})^2 = \mathbf{c}^2\mathbf{x}^2$ and the term in the square root is just a constant $|\mathbf{c}|$.

But for an object flying somewhere else, not passing through the origin I think the angle between $\mathbf{c}$ and $\mathbf{x}$ will change as it goes (and in particular it will not always be $0$) and so the momentum observed by a locally inertial observer will not simply scale as $1/a$.

I guess maybe the angle between $\mathbf{c}$ and $\mathbf{x}$ changes as $\mathbf{x}^2$ changes in just such a way as to keep the term in the square root constant but I don't see how to show that.

Or do I have the entire logic of the exercise wrong?


It turns out that last factor in the square root is actually a constant -- so everything works nicely and momentum decays as $1/a$ for any object moving on a geodesic as it's supposed to.

I took the derivative of the term in the square root: $$ \frac{\mathbf{c}^2-K(\mathbf{c}\cdot \mathbf{x})^2}{1-K\mathbf{x}^2} \equiv A. $$

This introduces the terms $\mathbf{c}\cdot \mathbf{x}'$ and $\mathbf{x}\cdot \mathbf{x}'$, which we can get rid of using some of the formulas in the original post:

\begin{align} \mathbf{c}\cdot \mathbf{x}' &= \frac{\mathbf{c}^2-K(\mathbf{c}\cdot \mathbf{x})^2}{a^2\sqrt{1-K\mathbf{x}^2}} = \frac{\sqrt{1-K\mathbf{x}^2}}{a^2}A \\ \mathbf{x}\cdot \mathbf{x}' &= \frac{1}{a^2}\sqrt{1-K\mathbf{x}^2}(\mathbf{c}\cdot \mathbf{x}) \end{align}

\begin{align} \frac{dA}{d\tau} &=\frac{-2K (\mathbf{c}\cdot \mathbf{x})(\mathbf{c}\cdot \mathbf{x}')}{1-K\mathbf{x}^2} + \frac{\left(\mathbf{c}^2-K(\mathbf{c}\cdot \mathbf{x})^2\right)2K(\mathbf{x}\cdot \mathbf{x}')}{(1-K\mathbf{x}^2)^2}\\ &= \frac{-2K (\mathbf{c}\cdot \mathbf{x})}{1-K\mathbf{x}^2}\frac{\sqrt{1-K\mathbf{x}^2}}{a^2}A + \frac{A}{1-K\mathbf{x}^2} \frac{2K}{a^2}\sqrt{1-K\mathbf{x}^2}(\mathbf{c}\cdot \mathbf{x})\\ &= 0. \end{align}

So $$ |\mathbf{p}_\mathrm{obs}| = \frac{m}{a}\sqrt{A}= \frac{\text{const}}{a}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.