# Corrections to Narlikar paper

1. Aug 23, 2014

### andrewkirk

Corrections to Narlikar paper

The paper by Jayant N Narlikar ’Spectral shifts
in General Relativity’
in the American Journal of Physics October 1994 is extremely useful in explaining the 'meaning' of cosmological and gravitational redshift, as well as making sense of the difference between peculiar motion and 'motion' arising from the expansion of space.

However the calculations in the paper have a few technical errors that can be confusing for beginners like me. These calculations correct the invalidity of Narlikar’s proof arising from his use of bases that do not exist, as well as an error in Equation [12] that generates errors in equations 13-17. 20 and 21.

Equation [11] for the case i=0 should be as follows:
\begin{align*}
0 &= \frac{d^2t}{du^2} - \frac{1}{2}g_{kl,0}\frac{dx^k}{du}\frac{dx^l}{du} \\
&= \frac{d^2t}{du^2} - \frac{1}{2}g_{00,0}(\frac{dt}{du})^2 - \frac{1}{2}g_{11,0}(\frac{dr}{du})^2\text{ [since }\frac{d\theta}{du}\text{ and }\frac{d\phi}{du}\text{ must be zero]}\\
&= \frac{d^2t}{du^2} - \frac{1}{2}\frac{\partial(\frac{-a(t)^2}{1-kr^2})}{\partial t}(\frac{dr}{du})^2 \text{ [since }g_{00}\text{ is constant at 1] }\\
&= \frac{d^2t}{du^2} + (\frac{dr}{du})^2 \frac{a(t)a'(t)}{1-kr^2}\ \ \ \ \textbf{[12]}\\
&= \frac{d^2t}{du^2} + (\frac{dr}{dt})^2 (\frac{dt}{du})^2 \frac{a(t)a'(t)}{1-kr^2}\\
&= \frac{d^2t}{du^2} + \frac{1-kr^2}{a(t)^2}(\frac{dt}{du})^2 \frac{a(t)a'(t)}{1-kr^2}\text{ [by 10] }\\
&= \frac{d^2t}{du^2} + (\frac{dt}{du})^2 \frac{a'(t)}{a(t)}\\
&= \frac{1}{a(t)} \frac{d(a(t)\frac{dt}{du})}{du}\\
\end{align*}

Hence, as $a(t)>0$ we have $0 = \frac{d(a(t)\frac{dt}{du})}{du}$, whence:
$a(t)\frac{dt}{du} = A$ for some constant $A$.[13]
Hence:
\begin{align*}\frac{dr}{du} &= \frac{dr}{dt}\frac{dt}{du}\\
&= \frac{A}{a(t)} \frac{\sqrt{1-kr^2}}{a(t)}\ \ \text{[by 13 and 10]}\\
&= \frac{A\sqrt{1-kr^2}}{a(t)^2} \ \ \ \textbf{[14]}\\
\end{align*}
This enables u to be determined as a function of r (as the light moves from S to O) and hence of t. It is convenient and permissible to set u=0 at S and u=1 at O.
Integrating [13] we obtain:
$A = \int_0^1A\ du = \int_{t(O)}^{t(S)}a(t)dt$ [15]
Let the tangent vector to the geodesic $\Lambda$ at $\lambda(u)\text{ be }U(u)$. In the comoving coordinates this has components $[\frac{dt}{du}, \frac{dr}{du}, \frac{d\theta}{du}, \frac{d\phi}{du}]$. The last two are zero and the first two are given by [13] and [14], hence the components are:
$U(u) = [\frac{A}{a(t)}, \frac{A\sqrt{1-kr^2}}{a(t)^2}, 0, 0]$ [16a]
and at O we have:
$U(1) = [\frac{A}{a_O}, \frac{A\sqrt{1-kr_O^2}}{a_O^2}, 0, 0]$ [16b]
where $r_O$ is the radial comoving coordinate of $O$. This is the first equation that assumes a specific centre ($S$) for the FLRW system.

Let the vector $\vec{V}_S$ parallel transported from $S$ to $\lambda(u)$ be $\vec{V}^*(u)$, and let us denote $\vec{V}^*(1)$ by $\vec{\bar{V}}_S$. Then, as parallel transport preserves magnitude and direction, both $\lvert\lvert\vec{V}^*(u)\rvert\rvert = g(\vec{V}^*(u),\vec{V}^*(u))\text{ and }g(\vec{V}^*(u),\vec{U}(u))$ must be constant over u. [17]

Hence $\lvert\lvert\vec{\bar{V}}_S\rvert\rvert = \lvert\lvert\vec{V}^*(1)\rvert\rvert = \lvert\lvert\vec{V}^*(0)\rvert\rvert = \lvert\lvert\vec{V}_S\rvert\rvert = 1$ [by 9a]. [18a]

And $g(\vec{\bar{V}}_S,\vec{U}(1)) = g(\vec{V}^*(1),\vec{U}(1)) = g(V^*(0),U(0)) = g(\vec{V}_S,\vec{U}(0))$ [18b]

Now we must have $V^{*2}(u) = V^{*3}(u) = 0$ because otherwise the parallel transportation establishes a preferred direction in space, circumferential to $S$, which contradicts the isotropy assumption. [This is a bit hand-wavy. Seek to make it more rigorous]

From [18a] we get:
$1 = \lVert\vec{\bar{V}}_S\rVert = g(O)(\vec{\bar{V}}_S, \vec{\bar{V}}_S) = g_{ik}(O)\bar{V}^i_S\bar{V}^k_S= g_{00}(O)(\bar{V}^0_S)^2 + g_{11}(O)(\bar{V}^0_S)^2\\$
[as off-diagonal elements of $g$ are zero everywhere in spacetime [from 7] and
$\bar{V}^2_S = \bar{V}^3_S = V^{*2}(1) = V^{*3}(1) = 0] = (\bar{V}^0_S)^2 - \frac{a_O^2}{1-kr_O^2} (\bar{V}^1_S)^2$

Hence $(\bar{V}^0_S)^2 - \frac{a_O^2}{1-kr_O^2} (\bar{V}^1_S)^2 = 1$ [19]
From [18b] we get $g_O(\vec{\bar{V}}_S, \vec{U}(1)) = g_S(\vec{V}_S,\vec{U}(0))$
hence $g_{ik}(O) \bar{V}^i_S U^k(1) = g_{ik}(S) V^i_S U^k(0)$

The left-hand side is:
$= g_{00}(O)\bar{V}^0_S\ U^0(1) + g_{11}(O)\bar{V}^1_S\ U^1(1)$
[since $\bar{V}^2_S = \bar{V}^3_S = 0$]
$= \bar{V}^0_S\ U^0(1) - \frac{a_O^2}{1-kr_O^2} \bar{V}^1_S\ U^1(1) = \bar{V}^0_S \frac{A}{a_O} - \frac{a_O^2}{1-kr_O^2}\frac{A\sqrt{1-kr_O^2}}{a_O^2} \bar{V}^1_S$ [by 16b]
$= A(\frac{\bar{V}^0_S}{a_O} - \frac{\bar{V}^1_S}{\sqrt{1-kr_O^2}})$

The right-hand side is (note that, since we are operating in $T_SM$ here, we have to switch to $FO$, the FLRW basis centred at $O$, in order for the components to be well-defined):

$= {}_{FO}g_{00}(S) {}_{FO}V^0_S {}_{FO}U(0)^0 + {}_{FO}g_{11}(S) {}_{FO}V^1_S {}_{FO}U(0)^1$

[since $U(0)^2=U(0)^3=0$ because the light ray is radial. \hl{This is also a bit hand-wavy}]
$\ = {}_{FO}g_{00}(S) {}_{FO}U(0)^0$

[since the FLRW spatial coordinates of $S$ are constant, so $\vec{V}_S = [1,0,0,0]$ in the $FO$ frame]
$\ = {}_{FO}U(0)^0 \text{ [by 7a]} = \frac{A}{a_S}$ [by 16a]

Hence, equating the right and left sides we get: $\frac{\bar{V}^0_S}{a_O} - \frac{\bar{V}^1_S}{\sqrt{1-kr_O^2}} = \frac{1}{a_S}$ [20]

Next we note that the four-velocity of the observer has components $[1,0,0,0]$ in any FLRW frame (because the observer has zero spatial coordinate velocity in that frame) and also in the observer's inertial frame at O. Hence the time basis vectors of the $MO$ and $FS$ frames must be identical: $\vec{e}_0(O) = {}_{MO}\vec{e}_0$.
Now $\vec{\bar{V}}_S = {}_{MO}\bar{V}^i_S \ {}_{MO}\vec{e}_i = \bar{V}^i_S\ \vec{e}_i(O) = \bar{V}^0_S \ \vec{e}_0(O) + \frac{a_O^2}{\sqrt{1-kr_O^2}}^1_S\ \vec{e}_1(O)\\ = \bar{V}^0_S\ {}_{MO}\vec{e}_0 + \bar{V}^1_S\ \vec{e}_1(O)$

Hence, since both bases are orthogonal, we have ${}_{MO}\bar{V}^0_S = \bar{V}^0_S$ and, by rotating the Lorentz frame appropriately around $O$, we can without loss of generality choose our basis vector ${}_{MO}\vec{e}_1$ so that it aligns with $\vec{e}_1(O)$. Then, since $\vec{\bar{V}}_S = [\bar{V}_S^0, \bar{V}_S^1, 0, 0]$ in the $FS$ basis, we can write $\vec{\bar{V}}_S = [\gamma, \gamma\ V, 0, 0]$ in the $MO$ basis, where ${}_{MO}\bar{V}^0_S = \gamma = \frac{1}{\sqrt{1-V^2}} = \bar{V}^0_S$ and $V$ has the same sign as $\bar{V}^1_S$ (because $\gamma$ is positive and the corresponding basis vectors, ${}_{MO}\vec{e}_1$ and $\vec{e}_1(O)$, point in the same direction).

Hence $\lvert\lvert\vec{\bar{V}}_S\rvert\rvert = g(\vec{\bar{V}}_S,\vec{\bar{V}}_S) = \gamma^2 - (\gamma V)^2 = \gamma^2 - \frac{a_O^2}{1-kr_O^2} (\bar{V}^1_S)^2$
where we calculate the magnitude in the $MO$ and $FS$ bases and equate the results.\\
Hence $(\gamma V)^2 = \frac{a_O^2}{1-kr_O^2} (\bar{V}^1_S)^2$ and so
$\gamma V = \frac{a_O}{\sqrt{1-kr_O^2}} \bar{V}^1_S$ [21]
where the sign is positive because $\gamma$ is positive and $V$ and $\bar{V}^1_S$ have the same sign.

Now the red shift is given by:
$1+z = \frac{a_O}{a_S}$ [8]
$= a_O(\frac{\bar{V}^0_S}{a_O} - \frac{\bar{V}^1_S}{\sqrt{1-kr_O^2}})$ [by 20]
$= \gamma - \frac{a_O}{\sqrt{1-kr_O^2}}\ \bar{V}^1_S = \gamma-\gamma V$ [by 21]
$= \gamma(1-V) = \frac{1-V}{\sqrt{1-V^2}} = \sqrt{\frac{(1-V)^2}{(1-V)(1+V)} }\\ = \sqrt{\frac{1 - V}{1+V}}\ \ \ \ \$ [22]
This is the reciprocal of the formula for a Doppler shift within a Lorentz frame.

V.425.