# I Does G effectively change in an expanding metric?

#### jcap

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 $(13)$?

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#### PAllen

Can you cite any reference for your proposed approach? At a glance, it makes no sense to me. G is constant in GR, and m is invariant for all coordinates/observers for a given particle, for all time, unless it decays or emits radiation. Any contrary notion is simply not general relativity, but some other hypothetical theory.

#### Orodruin

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Now the four-velocity of a stationary observer at the origin (η,x,y,z)=(0,0,0,0)(η,x,y,z)=(0,0,0,0)(\eta,x,y,z)=(0,0,0,0) is given by
Uμ=(1,0,0,0).​
No it is not. That is not properly normalised.

#### jcap

No it is not. That is not properly normalised.
It is normalised as the observer stays fixed at the origin of the coordinate system $(\eta,x,y,z)=(0,0,0,0)$

I try to calculate the energy of the comoving particle at arbitrary time $\eta$ relative to the observer who stays fixed at the origin where $\eta=0$.

#### Orodruin

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It is normalised as the observer stays fixed at the origin of the coordinate system (η,x,y,z)=(0,0,0,0)(η,x,y,z)=(0,0,0,0)(\eta,x,y,z)=(0,0,0,0)
No it is not. It is normalised such that
$$U^2 = g_{\mu\nu} U^\mu U^\nu = g_{00} = a^2.$$

We calculate the covariant derivative of ϵϵ\epsilon as the particle moves along xμ(τ)xμ(τ)x^\mu(\tau) using
This is also not correctly interpreted. The observer is not moving with the particle and so has no way of actually measuring $U\cdot P$ locally. What you have is a field of 4-velocities (if you normalise them correctly) of a family of co-moving observers.

#### jcap

Can you cite any reference for your proposed approach? At a glance, it makes no sense to me. G is constant in GR, and m is invariant for all coordinates/observers for a given particle, for all time, unless it decays or emits radiation. Any contrary notion is simply not general relativity, but some other hypothetical theory.
I can't give any reference. I'm just trying to understand this stuff myself.

My question to a GR expert is this: Can an observer fixed at the origin of a curved spacetime calculate the energy of a particle at a distant spacetime position?

#### Orodruin

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My question to a GR expert is this: Can an observer fixed at the origin of a curved spacetime calculate the energy of a particle at a distant spacetime position?
No. Not unambiguously or without reference to some convention. The energy of a particle relative to an observer is, as you have stated, $P \cdot U$. If $P$ and $U$ are not in the same tangent space, then this product is meaningless. You need to have a prescription to relate the two.

#### jcap

No. Not unambiguously or without reference to some convention. The energy of a particle relative to an observer is, as you have stated, $P \cdot U$. If $P$ and $U$ are not in the same tangent space, then this product is meaningless. You need to have a prescription to relate the two.
I calculate $P \cdot U$ at the origin and then calculate its covariant derivative as the particle moves forward in time. By integrating I find a value for $P \cdot U$ at the later time. Does it make sense to do this? Have I unambiguously found an expression for the energy of the particle relative to the observer fixed at the origin?

#### Orodruin

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I calculate $P \cdot U$ at the origin and then calculate its covariant derivative as the particle moves forward in time. By integrating I find a value for $P \cdot U$ at the later time. Does it make sense to do this?
What you would obtain from $\nabla_V(U\cdot P)$, if you did it correctly, would be the derivative with respect to the proper time of the particle of the energy that the family of comoving observers would observe, i.e., the particular observer the particle passes at any given point along its world-line. This is not the same thing as the energy a particular single fixed observer finds and it most certainly is not the derivative with respect to that observer's proper time.

Have I unambiguously found an expression for the energy of the particle relative to the observer fixed at the origin?
No.

#### Martin Scholtz

Gold Member
Summary: In an expanding metric an observer fixed at the origin at time $\eta=0$,...
At what state of motion is the observer fixed at $\eta=0$? Ţhe world-line must be time-like curve, it cannot be neither a fixed point nor space-like curve lying in the plane of constant $\eta$. That's for the starters.

#### Dale

Mentor
Can an observer fixed at the origin of a curved spacetime calculate the energy of a particle at a distant spacetime position?
Not in any unique frame independent manner.

My personal view on the general topic of this thread is that G is purely a matter of the units used. You can choose units where G is 1 or even where G is $1/8\pi$. If you wish you can choose units such that G is a function of time or space.

#### PeterDonis

Mentor
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 $(13)$?
No.

First, the observer is not fixed at the spacetime origin; that's impossible. You can't stay at one point in spacetime. The observer is fixed at the spatial origin, $(x, y, z) = (0, 0, 0)$. But the observer's worldline includes all values of $\eta$, not just $\eta = 0$.

Second, the equation $(13)$ is an equation written in a local inertial frame, just as equation $(12)$, where it came from, is. Those equations are not valid in the global coordinate chart you are using, in which the metric $(1)$ is written. So you can't equate the $m$ in equations $(12)$ and $(13)$ with the $m$ in the other equations, which is what you are doing when you write down equation $(14)$.

In fact, the $m$ in your other equations does not represent the rest mass of the particle at all; it represents a coordinate-dependent quantity that has no direct physical meaning. At one particular event, the spacetime origin $(\eta,x,y,z)=(0,0,0,0)$, it happens to numerically equal the particle's rest mass.

In short, you are using intuitions that would be valid in flat spacetime but trying to carry them over to curved spacetime, where they do not work.

#### jcap

As I understand it energy is conserved in a metric with a time-like Killing vector.

It is well-known that FRW metrics do not have a time-like Killing vector but my above calculations confirm that they do have a conformal time-like Killing vector $K^\mu=(1,0,0,0)$ such that
$$L_K g_{\mu\nu} = \nabla_\mu K_\nu + \nabla_\nu K_\mu = 2\frac{\dot a}{a} g_{\mu\nu}$$
I believe my above calculations imply that the energy of massless particles like photons ($V^\mu V_\mu=0$) is conserved whereas the energy of stationary massive particles ($V^\mu V_\mu=1$) is proportional to the scale factor $a(\eta)$.

Thus, in reality, the cosmological redshift is not due to photon energy $\propto 1/a$ but rather it is due to atomic energy $\propto a$.

Last edited:

#### Orodruin

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Again, you are wrong.

#### PeterDonis

Mentor
As I understand it energy is conserved in a metric with a time-like Killing vector.
More precisely, there is a conserved quantity corresponding to the timelike Killing vector, in spacetimes that have one, which can be interpreted as an "energy" (the usual term is "energy at infinity").

However, FRW spacetime, which is what we are discussing, does not have a timelike Killing vector, so there is no such conserved quantity in FRW spacetime.

they do have a conformal time-like Killing vector
So what? There is no conserved quantity corresponding to a conformal timelike Killing vector. Not even for photons.

in reality, the cosmological redshift is not due to photon energy $\propto 1/a$ but rather it is due to atomic energy $\propto a$.
No, this is not correct.

#### PeterDonis

Mentor
The OP question has been answered. Thread closed.

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