- Summary
- In an expanding metric an observer fixed at the origin at time ##\eta=0##, who uses a conformal coordinate system, calculates that the mass of comoving particles are proportional to the scale factor ##a(\eta)##. This implies that the observer calculates an effective Newton's ##G \propto 1/a^2##.

Assume that we have a flat FRW metric expressed in conformal time ##\eta## so that the line element is

$$ds^2=a^2(\eta)(d\eta^2-dx^2-dy^2-dz^2)\tag{1}$$

where ##a=1## at the present time ##\eta=0## and the speed of light ##c=1##.

This metric has the following non-zero Christoffel symbols:

\begin{eqnarray*}

\Gamma^0_{\alpha \beta} &=& \frac{\dot{a}}{a} \delta_{\alpha \beta}\tag{2} \\

\Gamma^i_{0j} &=& \Gamma^i_{j0} = \frac{\dot{a}}{a} \delta^i_j

\end{eqnarray*}

where ##\alpha,\beta=\{0,1,2,3\}## and ##i,j=\{1,2,3\}##.

Let us assume that a massive particle travels on the trajectory ##x^\mu(\tau)## with tangent vector ##V^\mu=dx^\mu/d\tau## where ##V^\mu V_\mu=1##. The four-momentum of the particle is given by

$$P_\mu=m V^\mu.\tag{3}$$

Now the four-velocity of a stationary observer at the origin ##(\eta,x,y,z)=(0,0,0,0)## is given by

$$U^\mu = (1,0,0,0).\tag{4}$$

The energy ##\epsilon## of the particle at the origin, as measured by the stationary observer who is also at the origin, is given by

$$\epsilon=U_\mu P^\mu.\tag{5}$$

Now the particle moves into the future along its trajectory ##x^\mu(\tau)##.

How does the observer's view of its energy change given that by definition he is fixed at the origin where ##\eta=0##?

We calculate the covariant derivative of ##\epsilon## as the particle moves along ##x^\mu(\tau)## using

\begin{eqnarray*}

\frac{D\epsilon}{d\tau} &=& V^\mu \nabla_\mu(U_\nu P^\nu)\tag{6} \\

&=& m\ V^\mu \nabla_\mu(U_\nu V^\nu) \\

&=& m\ U_\nu V^\mu \nabla_\mu V^\nu + m\ V^\mu V^\nu \nabla_\mu U_\nu.

\end{eqnarray*}

Now let us assume that the trajectory ##x^\mu(\tau)## is a geodesic so that the first term in the third line of equation ##(6)## vanishes according to the geodesic equation ##V^\mu \nabla_\mu V^\nu=0##. Thus we have

\begin{eqnarray*}

\frac{D\epsilon}{d\tau} &=& m\ V^\mu V^\nu \nabla_\mu U_\nu \tag{7}\\

&=& m \frac{\dot a}{a} V^\mu V^\nu g_{\mu \nu} \\

&=& m \frac{\dot a}{a} V^\mu V_\mu \\

&=& m \frac{\dot a}{a}

\end{eqnarray*}

where ##\nabla_\mu U_\nu= \partial_\mu U_\nu - \Gamma^\lambda_{\mu \nu}U_\lambda=(\dot{a}/a)g_{\mu\nu}## and ##V^\mu V_\mu=1## for a massive particle.

Now let us assume that the particle is comoving with respect to the coordinate system. Therefore it is stationary in space but moves forward in time. According to the line element ##(1)## the interval of proper time ##d\tau=ds=a\ d\eta##. Substituting into ##(7)## we find

\begin{eqnarray*}

\frac{D\epsilon}{d\tau} &=& m \frac{da/d\eta}{a} \tag{8}\\

&=& m \frac{da}{d\tau}.

\end{eqnarray*}

Integrating equation ##(8)## we find

$$\epsilon = m\ a + C.\tag{9}$$

We assume that at the present time ##\eta=0## we have ##\epsilon=m## and ##a=1## so that ##C=0##.

Therefore, finally, we find that the energy ##\epsilon(\eta)## of a comoving mass ##m## with respect to an observer who stays fixed at the origin ##(\eta,x,y,z)=(0,0,0,0)## is given by

$$\epsilon(\eta) = m\ a(\eta).\tag{10}$$

Now let us suppose that the observer wishes to use Einstein's Field Equations:

$$G_{\mu\nu}=8\pi G\ T_{\mu\nu}\tag{11}$$

in order to determine the scale factor function ##a(\eta)##.

Newton's constant ##G## can be measured by balancing the electrical repulsion between two charged masses with their gravitational attraction. Imagine, at the present time ##\eta=0##, the observer has two particles of mass ##m## and charge ##e##. The electrical and gravitational forces between these particles balance if we have

$$\frac{Gm^2}{d^2} =\frac{e^2}{4\pi \epsilon_0d^2}.\tag{12}$$

Let us use natural units so that ##\hbar=c=\epsilon_0=1## and the fine structure constant ##\alpha=e^2/4\pi##. Thus, at time ##\eta=0##, the observer measures Newton's constant to be

$$G = \frac{\alpha}{m^2}.\tag{13}$$

But according to equation ##(10)## an observer fixed at the origin of the coordinate system at ##\eta=0## calculates that the mass of comoving particles, relative to himself at the origin, increase with the scale factor ##a(\eta)## as the time ##\eta## increases. Thus according to him Newton's constant ##G## is no longer a constant but rather it is a parameter ##G(\eta)## given by

$$G(\eta) = \frac{\alpha}{m^2a^2(\eta)}.\tag{14}$$

Thus, should the observer, who is fixed at the origin ##(\eta,x,y,z)=(0,0,0,0)## in an expanding metric, use the expression ##(14)## for ##G## in Einstein's field equations rather than the constant in equation ##(13)##?

$$ds^2=a^2(\eta)(d\eta^2-dx^2-dy^2-dz^2)\tag{1}$$

where ##a=1## at the present time ##\eta=0## and the speed of light ##c=1##.

This metric has the following non-zero Christoffel symbols:

\begin{eqnarray*}

\Gamma^0_{\alpha \beta} &=& \frac{\dot{a}}{a} \delta_{\alpha \beta}\tag{2} \\

\Gamma^i_{0j} &=& \Gamma^i_{j0} = \frac{\dot{a}}{a} \delta^i_j

\end{eqnarray*}

where ##\alpha,\beta=\{0,1,2,3\}## and ##i,j=\{1,2,3\}##.

Let us assume that a massive particle travels on the trajectory ##x^\mu(\tau)## with tangent vector ##V^\mu=dx^\mu/d\tau## where ##V^\mu V_\mu=1##. The four-momentum of the particle is given by

$$P_\mu=m V^\mu.\tag{3}$$

Now the four-velocity of a stationary observer at the origin ##(\eta,x,y,z)=(0,0,0,0)## is given by

$$U^\mu = (1,0,0,0).\tag{4}$$

The energy ##\epsilon## of the particle at the origin, as measured by the stationary observer who is also at the origin, is given by

$$\epsilon=U_\mu P^\mu.\tag{5}$$

Now the particle moves into the future along its trajectory ##x^\mu(\tau)##.

How does the observer's view of its energy change given that by definition he is fixed at the origin where ##\eta=0##?

We calculate the covariant derivative of ##\epsilon## as the particle moves along ##x^\mu(\tau)## using

\begin{eqnarray*}

\frac{D\epsilon}{d\tau} &=& V^\mu \nabla_\mu(U_\nu P^\nu)\tag{6} \\

&=& m\ V^\mu \nabla_\mu(U_\nu V^\nu) \\

&=& m\ U_\nu V^\mu \nabla_\mu V^\nu + m\ V^\mu V^\nu \nabla_\mu U_\nu.

\end{eqnarray*}

Now let us assume that the trajectory ##x^\mu(\tau)## is a geodesic so that the first term in the third line of equation ##(6)## vanishes according to the geodesic equation ##V^\mu \nabla_\mu V^\nu=0##. Thus we have

\begin{eqnarray*}

\frac{D\epsilon}{d\tau} &=& m\ V^\mu V^\nu \nabla_\mu U_\nu \tag{7}\\

&=& m \frac{\dot a}{a} V^\mu V^\nu g_{\mu \nu} \\

&=& m \frac{\dot a}{a} V^\mu V_\mu \\

&=& m \frac{\dot a}{a}

\end{eqnarray*}

where ##\nabla_\mu U_\nu= \partial_\mu U_\nu - \Gamma^\lambda_{\mu \nu}U_\lambda=(\dot{a}/a)g_{\mu\nu}## and ##V^\mu V_\mu=1## for a massive particle.

Now let us assume that the particle is comoving with respect to the coordinate system. Therefore it is stationary in space but moves forward in time. According to the line element ##(1)## the interval of proper time ##d\tau=ds=a\ d\eta##. Substituting into ##(7)## we find

\begin{eqnarray*}

\frac{D\epsilon}{d\tau} &=& m \frac{da/d\eta}{a} \tag{8}\\

&=& m \frac{da}{d\tau}.

\end{eqnarray*}

Integrating equation ##(8)## we find

$$\epsilon = m\ a + C.\tag{9}$$

We assume that at the present time ##\eta=0## we have ##\epsilon=m## and ##a=1## so that ##C=0##.

Therefore, finally, we find that the energy ##\epsilon(\eta)## of a comoving mass ##m## with respect to an observer who stays fixed at the origin ##(\eta,x,y,z)=(0,0,0,0)## is given by

$$\epsilon(\eta) = m\ a(\eta).\tag{10}$$

Now let us suppose that the observer wishes to use Einstein's Field Equations:

$$G_{\mu\nu}=8\pi G\ T_{\mu\nu}\tag{11}$$

in order to determine the scale factor function ##a(\eta)##.

Newton's constant ##G## can be measured by balancing the electrical repulsion between two charged masses with their gravitational attraction. Imagine, at the present time ##\eta=0##, the observer has two particles of mass ##m## and charge ##e##. The electrical and gravitational forces between these particles balance if we have

$$\frac{Gm^2}{d^2} =\frac{e^2}{4\pi \epsilon_0d^2}.\tag{12}$$

Let us use natural units so that ##\hbar=c=\epsilon_0=1## and the fine structure constant ##\alpha=e^2/4\pi##. Thus, at time ##\eta=0##, the observer measures Newton's constant to be

$$G = \frac{\alpha}{m^2}.\tag{13}$$

But according to equation ##(10)## an observer fixed at the origin of the coordinate system at ##\eta=0## calculates that the mass of comoving particles, relative to himself at the origin, increase with the scale factor ##a(\eta)## as the time ##\eta## increases. Thus according to him Newton's constant ##G## is no longer a constant but rather it is a parameter ##G(\eta)## given by

$$G(\eta) = \frac{\alpha}{m^2a^2(\eta)}.\tag{14}$$

Thus, should the observer, who is fixed at the origin ##(\eta,x,y,z)=(0,0,0,0)## in an expanding metric, use the expression ##(14)## for ##G## in Einstein's field equations rather than the constant in equation ##(13)##?