Author Archives: growescience

Spontaneous emission from the zero point field

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 9.9.

Viewed in quantum electrodynamics, spontaneous emission of radiation is actually emission stimulated by the zero point electromagnetic field. The transition rate for stimulated emission is given by Griffiths (his equation 9.47) and is

\displaystyle  R_{a\rightarrow b}=\frac{\pi\left|\boldsymbol{\mathfrak{p}}\right|^{2}}{3\epsilon_{0}\hbar^{2}}\rho\left(\omega_{0}\right) \ \ \ \ \ (1)

where

\displaystyle  \boldsymbol{\mathfrak{p}}\equiv q\left\langle a\left|\mathbf{r}\right|b\right\rangle \ \ \ \ \ (2)

is the dipole moment averaged between the two states {a} and {b}. If the stimulating radiation is due to thermal processes, we can use Planck’s formula for {\rho\left(\omega_{0}\right)}:

\displaystyle  \rho\left(\omega_{0}\right)=\frac{1}{\left(e^{\hbar\omega_{0}/k_{B}T}-1\right)}\frac{\hbar\omega_{0}^{3}}{\pi^{2}c^{3}} \ \ \ \ \ (3)

However, for the zero point field, Planck’s formula does not apply. To get the formula properly, we’d need to use quantum electrodynamics, but following Griffiths, we can assume that the number of photons per available state is 1. That is, given that the number of states in the shell from wave number {k} to {k+dk} is (using {k=\omega/c}):

\displaystyle   d_{k} \displaystyle  = \displaystyle  \frac{k^{2}V}{\pi^{2}}dk\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \frac{\omega^{2}V}{\pi^{2}c^{2}}d\omega \ \ \ \ \ (5)

where {V} is the volume of the box in which we put the photons, then the number of photons in that range of wave numbers is just

\displaystyle  N_{\omega}=d_{k} \ \ \ \ \ (6)

rather than the formula we had used earlier: {N_{\omega}=\frac{d_{k}}{e^{\hbar\omega/k_{B}T}-1}}.

The energy density is found by using the fact that the energy of a photon is {h\nu=\hbar\omega}, so {\rho\left(\omega\right)d\omega=N_{\omega}\hbar\omega/V}, so

\displaystyle   \rho\left(\omega\right)d\omega \displaystyle  = \displaystyle  \frac{d_{k}\hbar\omega}{V}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{\hbar\omega^{3}}{\pi^{2}c^{2}}d\omega \ \ \ \ \ (8)

For the resonant frequency {\omega_{0}=\left(E_{b}-E_{a}\right)/\hbar} we thus have

\displaystyle  \rho\left(\omega_{0}\right)=\frac{\hbar\omega_{0}^{3}}{\pi^{2}c^{2}} \ \ \ \ \ (9)

so from 1 we have

\displaystyle  R_{a\rightarrow b}=\frac{\omega_{0}^{3}\left|\boldsymbol{\mathfrak{p}}\right|^{2}}{3\pi\epsilon_{0}\hbar c^{3}} \ \ \ \ \ (10)

which is the same result as we got using Einstein’s method.

Spontaneous emission: Einstein’s argument

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 9.8.

We’ve seen that external radiation can stimulate the emission and absorption of photons by the electron in an atom. However, even without any external radiation, an electron in an excited state will spontaneously decay to a lower state. Although a proper theory of spontaenous emission requires quantum electrodynamics, in 1917 Einstein devised a simple argument that gave a formula for the spontaneous decay rate without using quantum electrodynamics (which didn’t exist then).

The idea is based on a counting argument. Suppose we have a collection of atoms in thermal equilibrium, which means that the populations of the various energy states are all constant. Restricting ourselves to a 2-state system, let {N_{a}} be the number of atoms in the lower state and {N_{b}} the number in the excited state. Then {\frac{dN_{b}}{dt}=0} (because of thermal equilibrium), but the rate of change of {N_{b}} must be due to stimulated absorption, stimulated emission and spontaneous emission. The sum of these three rates must therefore be zero. The rates of stimulated absorption and emission are each proportional to {\rho\left(\omega_{0}\right)}, the density of stimulating radiation at frequency {\omega_{0}=\left(E_{b}-E_{a}\right)/\hbar}, so let {B_{ab}} and {B_{ba}} be the constants of proportionality for these two rates (from our earlier analysis, we know that {B_{ab}=B_{ba}}, but we’ll keep them separate for now), and let {A} be the rate of spontaneous emission. Then

\displaystyle  N_{a}B_{ab}\rho\left(\omega_{0}\right)-N_{b}B_{ba}\rho\left(\omega_{0}\right)-N_{b}A=0 \ \ \ \ \ (1)

One of the results from statistical mechanics is that, in thermal equilibrium, the number of particles with energy {E_{i}} is proportional to the Boltzmann factor {e^{-E_{i}/k_{B}T}}, where {k_{B}} is the Boltzmann constant and {T} is the absolute temperature. Therefore

\displaystyle   \frac{N_{a}}{N_{b}} \displaystyle  = \displaystyle  \frac{e^{-E_{a}/k_{B}T}}{e^{-E_{b}/k_{B}T}}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  e^{\left(E_{b}-E_{a}\right)/k_{B}T}\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  e^{\hbar\omega_{0}/k_{B}T} \ \ \ \ \ (4)

Dividing 1 through by {N_{b}} we get

\displaystyle  \rho\left(\omega_{0}\right)=\frac{A}{e^{\hbar\omega_{0}/k_{B}T}B_{ab}-B_{ba}} \ \ \ \ \ (5)

However, we know from our study of bosons that the density of radiation is given by Planck’s formula:

\displaystyle  \rho\left(\omega_{0}\right)=\frac{1}{\left(e^{\hbar\omega_{0}/k_{B}T}-1\right)}\frac{\hbar\omega_{0}^{3}}{\pi^{2}c^{3}} \ \ \ \ \ (6)

Comparing these two formulas, we see that the condition {B_{ab}=B_{ba}} is required if they are to be equal, and also that the rate {A} of spontaneous emission must be proportional to {B_{ab}}:

\displaystyle  A=\frac{\hbar\omega_{0}^{3}}{\pi^{2}c^{3}}B_{ab} \ \ \ \ \ (7)

For the lower frequency case, where we can approximate the electric field over the extent of the atom by a constant, we can follow a similar derivation to the one we did earlier (this derivation is done by Griffiths in his section 9.3) to find that, for radiation incident along the {z} direction we get for the stimulated absorption/emission rate:

\displaystyle  R_{a\rightarrow b}=\frac{\pi\left|\mathfrak{p}\right|^{2}}{\epsilon_{0}\hbar^{2}}\rho\left(\omega_{0}\right) \ \ \ \ \ (8)

where

\displaystyle  \mathfrak{p}\equiv q\left\langle a\left|z\right|b\right\rangle \ \ \ \ \ (9)

Griffiths also shows that if we average the radiation over all incident directions and all polarization directions we introduce a factor of {\frac{1}{3}} into the rate, so we get

\displaystyle  R_{a\rightarrow b}=\frac{\pi\left|\boldsymbol{\mathfrak{p}}\right|^{2}}{3\epsilon_{0}\hbar^{2}}\rho\left(\omega_{0}\right) \ \ \ \ \ (10)

where

\displaystyle  \boldsymbol{\mathfrak{p}}\equiv q\left\langle a\left|\mathbf{r}\right|b\right\rangle \ \ \ \ \ (11)

The coefficient {B_{ab}=B_{ba}} is therefore

\displaystyle  B_{ab}=\frac{\pi\left|\boldsymbol{\mathfrak{p}}\right|^{2}}{3\epsilon_{0}\hbar^{2}} \ \ \ \ \ (12)

for general incoherent radiation (all directions and polarizations). The spontaneous emission rate is then

\displaystyle  A=\frac{\omega_{0}^{3}\left|\boldsymbol{\mathfrak{p}}\right|^{2}}{3\epsilon_{0}\pi\hbar c^{3}} \ \ \ \ \ (13)

The ratio of spontaneous to stimulated emission is therefore

\displaystyle  \frac{A}{R_{b\rightarrow a}}=\frac{\omega_{0}^{3}\hbar}{\pi^{2}c^{3}\rho\left(\omega_{0}\right)} \ \ \ \ \ (14)

If the stimulating radiation comes from the thermal environment of the atom, then from 6 we get

\displaystyle  \frac{A}{R_{b\rightarrow a}}=e^{\hbar\omega_{0}/k_{B}T}-1 \ \ \ \ \ (15)

At room temperature {T=300\mbox{ K}} and with {k_{b}=1.38\times10^{-23}\mbox{m}^{2}\mbox{kg s}^{-2}\mbox{ K}^{-1}} then for a frequency of {5\times10^{12}\mbox{ Hz}\implies\omega_{0}=3.14\times10^{12}\mbox{ s}^{-1}} we get

\displaystyle  \frac{A}{R_{b\rightarrow a}}=0.083 \ \ \ \ \ (16)

For frequencies much lower than this the exponential gets very close to 1, so the ratio gets very small indicating that stimulated emission dominates. For larger frequencies, spontaneous emission dominates. For visible light, the frequency is in the region of {10^{15}\mbox{ s}^{-1}} so spontaneous emission dominates.

Stimulated emission of radiation at high frequencies

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Section 9.2; Problem 9.21a.

In our treatment of stimulated emission of radiation, we assumed that the wavelength of the perturbing radiation was much greater than the size of the atom. For high frequency radiation, such as X-rays, this assumption is no longer valid, so we need to include the spatial variation of the wave in the calculation. In that case, we need to return to the original formula for the electric field:

\displaystyle  \mathbf{E}=E_{0}\hat{\mathbf{n}}\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right) \ \ \ \ \ (1)

where {\mathbf{k}} is the direction of propagation, {\hat{\mathbf{n}}} is the direction of polarization and {\omega=kc} is the angular frequency. If we still assume that the wavelength is fairly large, so that {kr=2\pi r/\lambda} is fairly small, we can use a first order approximation:

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  E_{0}\hat{\mathbf{n}}\left[\cos\left(\mathbf{k}\cdot\mathbf{r}\right)\cos\left(\omega t\right)+\sin\left(\mathbf{k}\cdot\mathbf{r}\right)\sin\left(\omega t\right)\right]\ \ \ \ \ (2)
\displaystyle  \displaystyle  \approx \displaystyle  E_{0}\hat{\mathbf{n}}\left[\cos\left(\omega t\right)+\mathbf{k}\cdot\mathbf{r}\sin\left(\omega t\right)\right] \ \ \ \ \ (3)

where the last line uses {\sin x\approx x} for small {x}. The first term gives the transition rates we treated in the earlier post, so we’ll look at the second term here.

To get the perturbation {H'} to the hamiltonian, we can use the formula for the energy of a point charge in an electric field. The work done in moving a charge {q} from point {\mathbf{a}} to point {\mathbf{b}} is

\displaystyle  W=-q\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{E}\cdot d\mathbf{r}' \ \ \ \ \ (4)

If we take {\mathbf{a}=0} to be the centre of the atom and {\mathbf{b}} to be the field point {\mathbf{r}} then the energy due to the second term in 3 is

\displaystyle  H'=-qE_{0}\sin\left(\omega t\right)\int_{0}^{\mathbf{r}}\left(\mathbf{k}\cdot\mathbf{r}'\right)\hat{\mathbf{n}}\cdot d\mathbf{r}' \ \ \ \ \ (5)

Since the integral doesn’t depend on the path, we can use a straight line from the origin out to point {\mathbf{r}}. In that case, the direction of {\mathbf{r}'} is constant, so the angle {\theta_{k}} between {\mathbf{k}} and {\mathbf{r}}‘ is also constant, as is the angle {\theta_{n}} between {\hat{\mathbf{n}}} and {\mathbf{r}'}, so

\displaystyle   \mathbf{k}\cdot\mathbf{r}' \displaystyle  = \displaystyle  \left(k\cos\theta_{k}\right)r'\ \ \ \ \ (6)
\displaystyle  \hat{\mathbf{n}}\cdot d\mathbf{r}' \displaystyle  = \displaystyle  \left(\cos\theta_{n}\right)dr' \ \ \ \ \ (7)

We then get

\displaystyle   H' \displaystyle  = \displaystyle  -qE_{0}k\sin\left(\omega t\right)\cos\theta_{k}\cos\theta_{n}\int_{0}^{r}r'dr'\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{2}qE_{0}k\sin\left(\omega t\right)\cos\theta_{k}\cos\theta_{n}r^{2}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{2}qE_{0}\sin\left(\omega t\right)\left[kr\cos\theta_{k}\right]\left[r\cos\theta_{n}\right]\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{2}qE_{0}\sin\left(\omega t\right)\left(\mathbf{k}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{2}qE_{0}\frac{\omega}{c}\sin\left(\omega t\right)\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right) \ \ \ \ \ (12)

In the last line we used the relation {k=\omega/c}. [Note that {\hat{\mathbf{k}}} is a unit vector in the {k} direction which is not necessarily the {z} direction. The notation in Griffiths is confusing because he uses {\hat{\mathbf{k}}} earlier in the chapter to refer to the {z} direction, but in this problem it refers to the {k} direction.]

The matrix element {H'_{ba}} is

\displaystyle  H'_{ba}=-\frac{1}{2}qE_{0}\frac{\omega}{c}\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle \sin\left(\omega t\right) \ \ \ \ \ (13)

This is the same situation as the one we looked at earlier, except here we have {\sin\left(\omega t\right)} instead of {\cos\left(\omega t\right)} and

\displaystyle  V_{ba}=-\frac{1}{2}qE_{0}\frac{\omega}{c}\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle \ \ \ \ \ (14)

We can work through the calculations, replacing the cosine by the sine to get

\displaystyle   c_{b}\left(t\right) \displaystyle  = \displaystyle  -\frac{i}{\hbar}\int_{0}^{t}H'_{ba}\left(t'\right)e^{i\omega_{0}t}dt'\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  -\frac{i}{\hbar}V_{ba}\int_{0}^{t}\sin\left(\omega t'\right)e^{i\omega_{0}t'}dt'\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{2\hbar}V_{ba}\int_{0}^{t}\left[e^{i\left(\omega_{0}+\omega\right)t'}-e^{i\left(\omega_{0}-\omega\right)t'}\right]\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  -\frac{V_{ba}}{2i\hbar}\left[\frac{e^{i\left(\omega_{0}+\omega\right)t}-1}{\omega_{0}+\omega}-\frac{e^{i\left(\omega_{0}-\omega\right)t}-1}{\omega_{0}-\omega}\right] \ \ \ \ \ (18)

Keeping only the second term (because we’re considering only frequencies {\omega\approx\omega_{0}}), we get

\displaystyle   c_{b}\left(t\right) \displaystyle  = \displaystyle  \frac{V_{ba}}{2i\hbar}\frac{e^{i\left(\omega_{0}-\omega\right)t}-1}{\omega_{0}-\omega}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  \frac{V_{ba}e^{i\left(\omega_{0}-\omega\right)t/2}}{\hbar}\frac{\sin\frac{\left(\omega_{0}-\omega\right)t}{2}}{\omega_{0}-\omega} \ \ \ \ \ (20)

The transition probability for stimulation by a monochromatic plane wave is therefore

\displaystyle   P_{a\rightarrow b} \displaystyle  = \displaystyle  \left|c_{b}\right|^{2}\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{1}{2}qE_{0}\frac{\omega}{\hbar c}\right)^{2}\left|\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle \right|^{2}\frac{\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2}}{\left(\omega_{0}-\omega\right)^{2}} \ \ \ \ \ (22)

In the problem, however, we’re interested in the transition probability for an atom subject to radiation over a range of frequencies. The energy density in an electromagnetic wave can be found from the average magnitude of the Poynting vector {\mathbf{S}}, which gives the amount of energy per unit area per unit time crossing a surface. In a unit time over a unit area, a volume of {c} crosses the surface, so the energy density {u} is {\left\langle S\right\rangle }:

\displaystyle  u=\frac{1}{2}\epsilon_{0}E_{0}^{2} \ \ \ \ \ (23)

If we have a range of frequencies, then we need to replace {u} (which is for a single frequency) by {\rho\left(\omega\right)d\omega}, where {\rho\left(\omega\right)} is the density of radiation with frequencies in the range {\left[\omega,\omega+d\omega\right]}. We can then integrate over all frequencies to get the new transition probability:

\displaystyle  P_{a\rightarrow b}=\left(\frac{1}{2}q\frac{\omega}{\hbar c}\right)^{2}\frac{2}{\epsilon_{0}}\left|\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle \right|^{2}\int_{0}^{\infty}\rho\left(\omega\right)\frac{\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2}}{\left(\omega_{0}-\omega\right)^{2}}d\omega \ \ \ \ \ (24)

We can make a couple of approximations to evaluate the integral. First, since the integrand peaks around {\omega=\omega_{0}}, we can take {\rho\left(\omega\right)\approx\rho\left(\omega_{0}\right)} over the region where the integrand is significantly greater than zero. This means we can take {\rho\left(\omega_{0}\right)} as a constant and take it outside the integral. Second, again because the remaining term in the integrand peaks around {\omega=\omega_{0}}, we can extend the lower limit on the integrand to {-\infty}. Then we can transform the integral using the substitution {x\equiv\frac{\left(\omega_{0}-\omega\right)t}{2}}, so that {d\omega=-2\frac{dx}{t}} and {\left(\omega_{0}-\omega\right)^{2}=\frac{4x^{2}}{t^{2}}}. Therefore

\displaystyle   \int_{0}^{\infty}\rho\left(\omega\right)\frac{\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2}}{\left(\omega_{0}-\omega\right)^{2}}d\omega \displaystyle  \approx \displaystyle  \rho\left(\omega_{0}\right)\frac{2}{t}\frac{t^{2}}{4}\int_{-\infty}^{\infty}\frac{\sin^{2}x}{x^{2}}dx\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \rho\left(\omega_{0}\right)\frac{\pi t}{2} \ \ \ \ \ (26)

Plugging this back into 24 we get

\displaystyle  P_{a\rightarrow b}=\left(\frac{1}{2}q\frac{\omega}{\hbar c}\right)^{2}\frac{\pi}{\epsilon_{0}}\rho\left(\omega_{0}\right)\left|\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle \right|^{2}t \ \ \ \ \ (27)

The transition rate for stimulated absorption is

\displaystyle  R_{a\rightarrow b}=\frac{dP_{a\rightarrow b}}{dt}=\left(\frac{1}{2}q\frac{\omega}{\hbar c}\right)^{2}\frac{\pi}{\epsilon_{0}}\rho\left(\omega_{0}\right)\left|\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle \right|^{2} \ \ \ \ \ (28)

Because the term {\left|\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle \right|^{2}} is the same as {\left|\left\langle a\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|b\right\rangle \right|^{2}} (the operators {\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)} are just real multipliers so {\left\langle b\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|a\right\rangle =\left\langle a\left|\left(\hat{\mathbf{k}}\cdot\mathbf{r}\right)\left(\hat{\mathbf{n}}\cdot\mathbf{r}\right)\right|b\right\rangle ^*} so the moduli are equal), the transition rate is the same for both absorption and emission:

\displaystyle  R_{a\rightarrow b}=R_{b\rightarrow a} \ \ \ \ \ (29)

[We'll see how to get the transition rate for spontaneous emission in a future post. However, the result 27 leads to a different result than that given in Griffiths, in that there is an extra factor of {\frac{1}{4}} in my answer. This appears to have crept in while doing the integral in 9, and Griffiths's answer would seem to regard {\mathbf{k}\cdot\mathbf{r}} as a constant in the integration. I can't see why you would assume this, but any comments are welcome.]

Stimulated emission of radiation: lasers

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Section 9.2; Problem 9.19.

A particularly important case that we can analyze using time-dependent perturbation theory is that of the perturbation of an atom by an electromagnetic wave. A proper quantum treatment of the problem requires, of course, a proper quantum theory of electromagnetism (that is, quantum electrodynamics) but we haven’t got there yet, so we will follow Griffiths and use classical electromagnetism in combination with quantum mechanics.

We’ll look at what happens if a monochromatic plane wave encounters an electron in an atom, where the electron can be in one of two states (for example, the ground state and first excited state). To start, we’ll assume that the wave is travelling in the {+y} direction and is polarized in the {z} direction (that is, the electric field points in the {\pm z} direction, which is perpendicular to the magnetic field). We can write the electric field as

\displaystyle  \mathbf{E}=E_{0}\hat{\mathbf{z}}\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right) \ \ \ \ \ (1)

where {\mathbf{k}} is the direction of propagation and {\omega=kc} is the angular frequency. If we assume that the wavelength of the light is much larger than the atom (true for visible light and longer wavelengths), then we can drop the {\mathbf{k}\cdot\mathbf{r}} term since it’s effectively constant over the region of the atom, and we get

\displaystyle  \mathbf{E}=E_{0}\hat{\mathbf{z}}\cos\left(\omega t\right) \ \ \ \ \ (2)

To get the perturbation {H'} to the hamiltonian, we can use the formula for the energy of a point charge in an electric field. The work done in moving a charge {q} from point {\mathbf{a}} to point {\mathbf{b}} is

\displaystyle  W=-q\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{E}\cdot d\mathbf{l} \ \ \ \ \ (3)

If we take {\mathbf{a}=0} to be the centre of the atom and {\mathbf{b}} to be the field point {\mathbf{r}} then since {\mathbf{E}} points in the {z} direction and, at a given time, is constant over the region of the atom, we get

\displaystyle  W=-qE_{0}\hat{\mathbf{z}}\cdot\mathbf{r}\cos\left(\omega t\right)=-qE_{0}z\cos\left(\omega t\right) \ \ \ \ \ (4)

The perturbation is therefore

\displaystyle  H'=-qE_{0}z\cos\left(\omega t\right) \ \ \ \ \ (5)

and the matrix elements are

\displaystyle   H'_{ba} \displaystyle  = \displaystyle  -\left\langle \psi_{b}\left|z\right|\psi_{a}\right\rangle E_{0}\cos\omega t\ \ \ \ \ (6)
\displaystyle  \displaystyle  \equiv \displaystyle  -\mathfrak{p}E_{0}\cos\omega t\ \ \ \ \ (7)
\displaystyle  \mathfrak{p} \displaystyle  \equiv \displaystyle  q\left\langle \psi_{b}\left|z\right|\psi_{a}\right\rangle \ \ \ \ \ (8)

The quantity {\mathfrak{p}} is effectively a dipole moment, as the bracket factor is the average displacement of the electron from the origin. Assuming that the diagonal elements of {H'} are zero (as is true for hydrogen) we have a system with sinusoidal perturbations in time that we treated earlier, with {V_{ab}=-\mathfrak{p}E_{0}}, so we can write down the solution. (Actually, we’ll use the perturbation theory result – equation 9.28 in Griffiths – rather than Rabi’s more accurate result).

\displaystyle   P_{a\rightarrow b}\left(t\right) \displaystyle  = \displaystyle  \frac{\left|V_{ab}\right|^{2}}{\hbar^{2}\left(\omega-\omega_{0}\right)^{2}}\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \left(\frac{\left|\mathfrak{p}\right|E_{0}}{\hbar}\right)^{2}\frac{\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2}}{\left(\omega-\omega_{0}\right)^{2}} \ \ \ \ \ (10)

where {\omega_{0}=\left(E_{b}-E_{a}\right)/\hbar} is the frequency corresponding to the difference in energy levels between states {a} and {b}. That is, the probability of a transition from state {a} (the ground state) to state {b} (the excited state) oscillates in time. This is the phenomenon of stimulated absorption, in which the electric field causes the electron to jump to a higher energy level. This actually happens by absorbing a photon, although using a non-quantum theory of electrodynamics, we can’t predict that.

Curiously, if we analyze the situation where the electron starts out in state {b}, we get the same result. To see this, look back at the derivation of {P_{a\rightarrow b}} that we did earlier. If we start with initial conditions {c_{a}\left(0\right)=0} and {c_{b}\left(0\right)=1} so that the electron starts in state {b}, we can just interchange {a} and {b} all the way through the derivation. This results in {\omega_{0}} being replaced by {-\omega_{0}=\left(E_{a}-E_{b}\right)/\hbar} so we get

\displaystyle  c_{a}\left(t\right)=-\frac{V_{ab}}{2\hbar}\left[\frac{e^{i\left(-\omega_{0}+\omega\right)t}-1}{-\omega_{0}+\omega}-\frac{e^{i\left(-\omega_{0}-\omega\right)t}-1}{\omega_{0}+\omega}\right] \ \ \ \ \ (11)

In the approximation, we now keep only the first term (since {\omega\approx\omega_{0}}) so we now get

\displaystyle   c_{a}\left(t\right) \displaystyle  = \displaystyle  -\frac{iV_{ab}}{\hbar\left(\omega-\omega_{0}\right)}e^{i\left(\omega-\omega_{0}\right)t/2}\sin\frac{\left(\omega-\omega_{0}\right)t}{2}\ \ \ \ \ (12)
\displaystyle  P_{b\rightarrow a}\left(t\right)=\left|c_{a}\left(t\right)\right|^{2} \displaystyle  = \displaystyle  \frac{\left|V_{ab}\right|^{2}}{\hbar^{2}\left(\omega-\omega_{0}\right)^{2}}\sin^{2}\frac{\left(\omega-\omega_{0}\right)t}{2} \ \ \ \ \ (13)

In our case this gives

\displaystyle  P_{b\rightarrow a}\left(t\right)=\left|\left\langle \psi_{a}\left|z\right|\psi_{b}\right\rangle \right|^{2}\left(\frac{E_{0}}{\hbar}\right)^{2}\frac{\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2}}{\left(\omega-\omega_{0}\right)^{2}} \ \ \ \ \ (14)

The first term is

\displaystyle   \left|\left\langle \psi_{a}\left|z\right|\psi_{b}\right\rangle \right|^{2} \displaystyle  = \displaystyle  \left|\left\langle \psi_{b}\left|z\right|\psi_{a}\right\rangle ^*\right|^{2}\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \left|\left\langle \psi_{b}\left|z\right|\psi_{a}\right\rangle \right|^{2}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \left|\mathfrak{p}\right|^{2} \ \ \ \ \ (17)

so

\displaystyle  P_{b\rightarrow a}\left(t\right)=P_{a\rightarrow b}\left(t\right) \ \ \ \ \ (18)

That is, it is just as likely for an electromagnetic wave to stimulate the emission of a photon as it is to stimulate the absorption of one. This has a profound application in modern technology: the laser (which is an acronym for Light Amplification by Stimulated Emission of Radiation). If we can arrange to have a collection of atoms initially in the excited state {b}, then a small input of electromagnetic radiation will stimulate a few atoms to emit photons. These photons, in turn, will stimulate other atoms to emit photons giving a cascading effect. In other words, the initial light pulse is amplified by stimulated emission.

As we’ll see, in addition to stimulated absorption and emission, there is also the phenomenon of spontaneous emission, in which an atom in an excited state decays to a lower state and emits a photon without the intervention of an external field. Actually, it turns out that spontaneous emission is due to the interaction of an excited atom with a ‘zero-point’ electrodynamic field, which is something that arises only in quantum electrodynamics.

By the way, there is no such thing as ‘spontaneous absorption’ (absorption without an external field), since this would essentially require the acquisition of energy (a photon) from nothing.

Sinusoidal perturbations in time

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 9.7.

Using time-dependent perturbation theory we can get estimates of the coefficients in the solution of the Schrödinger equation with a time-dependent potential:

\displaystyle  \Psi\left(x,t\right)=c_{a}\left(t\right)\psi_{a}\left(x\right)e^{-iE_{a}t/\hbar}+c_{b}\left(t\right)\psi_{b}\left(x\right)e^{-iE_{b}t/\hbar} \ \ \ \ \ (1)


where the complete hamiltonian is {H=H^{0}+H'}, {\psi_{a}} and {\psi_{b}} are the two eigenstates of {H^{0}} and the coefficients are solutions of the coupled ODEs

\displaystyle   \dot{c}_{a} \displaystyle  = \displaystyle  -\frac{i}{\hbar}\left[c_{a}H'_{aa}+c_{b}H'_{ab}e^{-i\left(E_{b}-E_{a}\right)t/\hbar}\right]\ \ \ \ \ (2)
\displaystyle  \dot{c}_{b} \displaystyle  = \displaystyle  -\frac{i}{\hbar}\left[c_{b}H'_{bb}+c_{a}H'_{ba}e^{i\left(E_{b}-E_{a}\right)t/\hbar}\right] \ \ \ \ \ (3)

where

\displaystyle  H'_{ij}\equiv\left\langle \psi_{i}\right|H'\left|\psi_{j}\right\rangle \ \ \ \ \ (4)

If the diagonal matrix elements are zero, then we get the simpler ODEs:

\displaystyle   \dot{c}_{a} \displaystyle  = \displaystyle  -\frac{i}{\hbar}H'_{ab}e^{-i\omega_{0}t}c_{b}\ \ \ \ \ (5)
\displaystyle  \dot{c}_{b} \displaystyle  = \displaystyle  -\frac{i}{\hbar}H'_{ba}e^{i\omega_{0}t}c_{a} \ \ \ \ \ (6)

where

\displaystyle  \omega_{0}\equiv\frac{E_{b}-E_{a}}{\hbar} \ \ \ \ \ (7)

If the perturbation has a sinusoidal dependence on time such as

\displaystyle  H'\left(\mathbf{r},t\right)=V\left(\mathbf{r}\right)\cos\omega t \ \ \ \ \ (8)

then

\displaystyle   H'_{ab} \displaystyle  = \displaystyle  V_{ab}\cos\omega t\ \ \ \ \ (9)
\displaystyle  V_{ab} \displaystyle  \equiv \displaystyle  \left\langle \psi_{a}\right|V\left(\mathbf{r}\right)\left|\psi_{b}\right\rangle \ \ \ \ \ (10)

If we stop after the first-order step in the iterative solution we have

\displaystyle   c_{a}^{\left(1\right)}\left(t\right) \displaystyle  = \displaystyle  1-\frac{i}{\hbar}\int_{0}^{t}H'_{aa}\left(t'\right)dt'\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  1\ \ \ \ \ (12)
\displaystyle  c_{b}^{\left(1\right)}\left(t\right) \displaystyle  = \displaystyle  -\frac{i}{\hbar}\int_{0}^{t}H'_{ba}\left(t'\right)e^{i\omega_{0}t}dt'\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  -\frac{i}{\hbar}V_{ba}\int_{0}^{t}\cos\left(\omega t'\right)e^{i\omega_{0}t'}dt'\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  -\frac{i}{2\hbar}V_{ba}\int_{0}^{t}\left[e^{i\left(\omega_{0}+\omega\right)t'}+e^{i\left(\omega_{0}-\omega\right)t'}\right]\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  -\frac{V_{ba}}{2\hbar}\left[\frac{e^{i\left(\omega_{0}+\omega\right)t}-1}{\omega_{0}+\omega}+\frac{e^{i\left(\omega_{0}-\omega\right)t}-1}{\omega_{0}-\omega}\right] \ \ \ \ \ (16)

At this point, we can study only frequencies {\omega\approx\omega_{0}} and Griffiths does this in his section 9.1.3. The idea is that the first term can be dropped as the second term is much larger, due to the {\omega_{0}-\omega} in the denominator.

A different approach was taken by the Polish-American physicist Isidor Isaac Rabi (1898 – 1988). He noticed that this approximation is mathematically the same as taking the original perturbation to be

\displaystyle  H'=\frac{V}{2}e^{-i\omega t} \ \ \ \ \ (17)

so that

\displaystyle   H'_{ba} \displaystyle  = \displaystyle  \frac{V_{ba}}{2}e^{-i\omega t}\ \ \ \ \ (18)
\displaystyle  H'_{ab} \displaystyle  = \displaystyle  \frac{V_{ab}}{2}e^{i\omega t} \ \ \ \ \ (19)

In this case, we can solve the original ODEs 5 and 6 exactly without the need for perturbation theory. The equations become

\displaystyle   \dot{c}_{a} \displaystyle  = \displaystyle  -\frac{i}{2\hbar}V_{ab}e^{i\left(\omega-\omega_{0}\right)t}c_{b}\ \ \ \ \ (20)
\displaystyle  \dot{c}_{b} \displaystyle  = \displaystyle  -\frac{i}{2\hbar}V_{ba}e^{-i\left(\omega-\omega_{0}\right)t}c_{a} \ \ \ \ \ (21)

With the initial conditions {c_{a}\left(0\right)=1} and {c_{b}\left(0\right)=0}, these equations are formally the same as those we solved earlier with {H'_{ab}} replaced by {V_{ab}/2} and {\omega_{0}} replaced by {-\left(\omega-\omega_{0}\right)}. We can therefore write down the solution as

\displaystyle   c_{a}\left(t\right) \displaystyle  = \displaystyle  e^{-i\left(\omega-\omega_{0}\right)t/2}\left[\cos\left(\omega_{r}t\right)-\frac{i\left(\omega-\omega_{0}\right)}{2\omega_{r}}\sin\left(\omega_{r}t\right)\right]\ \ \ \ \ (22)
\displaystyle  c_{b}\left(t\right) \displaystyle  = \displaystyle  -\frac{i\left|V_{ab}\right|}{2\hbar\omega_{r}}e^{-i\left(\omega-\omega_{0}\right)t/2}\sin\left(\omega_{r}t\right) \ \ \ \ \ (23)

where {Q} in the earlier problem now becomes the Rabi flopping frequency {\omega_{r}}:

\displaystyle  \omega_{r}=\frac{1}{2}\sqrt{\left(\omega-\omega_{0}\right)^{2}+\frac{\left|V_{ab}\right|^{2}}{\hbar^{2}}} \ \ \ \ \ (24)

Note that

\displaystyle   \left|c_{a}\right|^{2}+\left|c_{b}\right|^{2} \displaystyle  = \displaystyle  \cos^{2}\left(\omega_{r}t\right)+\frac{1}{4\omega_{r}^{2}}\sin^{2}\left(\omega_{r}t\right)\left[\left(\omega-\omega_{0}\right)^{2}+\frac{\left|V_{ab}\right|^{2}}{\hbar^{2}}\right]\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  1 \ \ \ \ \ (26)

The transition probability (for a flip from state {a} to state {b}) is {\left|c_{b}\right|^{2}}:

\displaystyle  P_{a\rightarrow b}=\left|c_{b}\right|^{2}=\frac{\left|V_{ab}\right|^{2}}{4\hbar^{2}\omega_{r}^{2}}\sin^{2}\left(\omega_{r}t\right) \ \ \ \ \ (27)

The maximum probability occurs when {t=\pi/2\omega_{r}} which gives

\displaystyle   P_{a\rightarrow b} \displaystyle  \le \displaystyle  \frac{\left|V_{ab}\right|^{2}}{4\hbar^{2}\omega_{r}^{2}}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  \left[\frac{\hbar^{2}\left(\omega-\omega_{0}\right)^{2}}{\left|V_{ab}\right|^{2}}+1\right]^{-1}<1 \ \ \ \ \ (29)

For small perturbations, {\left|V_{ab}\right|\ll\hbar\left|\left(\omega-\omega_{0}\right)\right|} and if we expand 27 in a Taylor series about {V_{ab}=0} we get

\displaystyle   \frac{\left|V_{ab}\right|^{2}}{4\hbar^{2}\omega_{r}^{2}}\sin^{2}\left(\omega_{r}t\right) \displaystyle  = \displaystyle  \frac{\left|V_{ab}\right|^{2}}{4\hbar^{2}}\left[\frac{4}{\left(\omega-\omega_{0}\right)^{2}}\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2}+...\right]\ \ \ \ \ (30)
\displaystyle  \displaystyle  \approx \displaystyle  \frac{\left|V_{ab}\right|^{2}}{\hbar^{2}\left(\omega-\omega_{0}\right)^{2}}\sin^{2}\frac{\left(\omega_{0}-\omega\right)t}{2} \ \ \ \ \ (31)

which is the same as equation 9.28 in Griffiths.

The probability of the system being in its original state {a} is 1 at {t=0} and next at {t=\pi/\omega_{r}}.

Vacuum stress-energy and the cosmological constant

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.9.

The Einstein equation is

\displaystyle  R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)

Up to now, we’ve usually taken {\Lambda=0}, since we know from the Newtonian limit that {\Lambda} must be very small. If {\Lambda\ne0}, the Newtonian limit becomes

\displaystyle  \nabla^{2}\Phi=4\pi G\rho-\Lambda \ \ \ \ \ (2)

so {\Lambda} acts as a negative mass density, that is, it adds a repulsive term into the gravitational force. Einstein originally introduced it to counter the attractive force of gravity on a cosmological scale, since at the time it was believed that the universe was static (neither expanding nor contracting) and if gravity were purely attractive, the universe would be contracting.

At the moment, the universe is believed to be expanding so {\Lambda} is believed to be non-zero and positive, although still small enough that its effects are not noticeable on the scale of the solar system (or indeed on a galactic scale). Because of this, {\Lambda} is called the cosmological constant.

We can include {\Lambda} within the stress-energy tensor by defining a vacuum stress-energy as

\displaystyle  T_{vac}^{ij}=-\frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (3)

We can define a vacuum stress-energy scalar:

\displaystyle   T_{vac} \displaystyle  \equiv \displaystyle  g_{ij}T_{vac}^{ij}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\Lambda}{8\pi G}g_{ij}g^{ij}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  -\frac{4\Lambda}{8\pi G} \ \ \ \ \ (6)

Therefore

\displaystyle   T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac} \displaystyle  = \displaystyle  -\frac{\Lambda}{8\pi G}g^{ij}+\frac{2\Lambda}{8\pi G}g^{ij}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{\Lambda}{8\pi G}g^{ij} \ \ \ \ \ (8)

and we can write 1 as

\displaystyle   R^{ij} \displaystyle  = \displaystyle  8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T+T_{vac}^{ij}-\frac{1}{2}g^{ij}T_{vac}\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  \equiv \displaystyle  8\pi G\left(T_{all}^{ij}-\frac{1}{2}g^{ij}T_{all}\right) \ \ \ \ \ (10)

where {T_{all}^{ij}} includes the stress-energy from the mass-energy density and the vacuum.

The dominant energy condition is a constraint placed on the stress-energy tensor so that observers in any local orthogonal frame will measure the fluid’s speed to be less than the speed of light. The condition is that if {a^{i}} is any four-vector that is causal, that is, it satisfies the conditions

\displaystyle   \mathbf{a}\cdot\mathbf{a} \displaystyle  \le \displaystyle  0\ \ \ \ \ (11)
\displaystyle  a^{t} \displaystyle  > \displaystyle  0 \ \ \ \ \ (12)

then we require the stress-energy tensor {T^{ij}} to satisfy the condition that if

\displaystyle  b^{i}=-T^{ij}g_{jk}a^{k} \ \ \ \ \ (13)


then {\mathbf{b}} is also a causal four-vector. For the vacuum stress-energy this condition says

\displaystyle   b^{i} \displaystyle  = \displaystyle  -T_{vac}^{ij}g_{jk}a^{k}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \frac{\Lambda}{8\pi G}g^{ij}g_{jk}a^{k}\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \frac{\Lambda}{8\pi G}\delta_{\; k}^{i}a^{k}\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \frac{\Lambda}{8\pi G}a^{i} \ \ \ \ \ (17)

That is, {b^{i}} is just a positive (if {\Lambda>0}) constant multiplied by {a^{i}}, so if {a^{i}} is causal, then {b^{i}} must also be causal. Thus {T_{vac}^{ij}} satisfies the dominant energy condition.

Einstein equation for an exponential metric

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.8.

Consider the metric:

\displaystyle  ds^{2}=-e^{2gx}dt^{2}+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (1)

We’ll have a look at what the Einstein equation has to say about gravity in a spacetime using this metric. First, we’ll find the Christoffel symbols using the method of comparing the two forms of the geodesic equation. The geodesic equation is

\displaystyle  \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial x^{a}}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (2)

The following equation is formally equivalent to this (where a dot above a symbol means the derivative with respect to {\tau}):

\displaystyle  \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (3)

Since the metric is diagonal and none of its components depends on {y} or {z}, the LHS of 2 is identically zero for {a=y} or {a=z}, so all Christoffel symbols with any index being {y} or {z} is zero.

Now look at {a=t}. We get from 2, since {g_{ij}} doesn’t depend on {t}:

\displaystyle   \frac{d}{d\tau}\left(g_{tj}\frac{dx^{j}}{d\tau}\right) \displaystyle  = \displaystyle  0\ \ \ \ \ (4)
\displaystyle  -2ge^{2gx}\dot{x}\dot{t}-e^{2gx}\ddot{t} \displaystyle  = \displaystyle  0\ \ \ \ \ (5)
\displaystyle  \ddot{t}+2g\dot{x}\dot{t} \displaystyle  = \displaystyle  0 \ \ \ \ \ (6)

Comparing with 3 we see that

\displaystyle  \Gamma_{\; tx}^{t}=\Gamma_{\; xt}^{t}=g \ \ \ \ \ (7)

Now for {a=x}:

\displaystyle   \ddot{x}-\frac{1}{2}\left(\partial_{x}g_{tt}\right)\dot{t}^{2} \displaystyle  = \displaystyle  0\ \ \ \ \ (8)
\displaystyle  \ddot{x}+ge^{2gx}\dot{t}^{2} \displaystyle  = \displaystyle  0\ \ \ \ \ (9)
\displaystyle  \Gamma_{\; tt}^{x} \displaystyle  = \displaystyle  ge^{2gx} \ \ \ \ \ (10)

These are the only non-zero Christoffel symbols. We can get an expression for the acceleration of a particle in its rest frame by noting that

\displaystyle  \ddot{x}=-ge^{2gx}\dot{t}^{2} \ \ \ \ \ (11)

The four velocity of a particle at rest (so {dx=dy=dz=0}) must satisfy

\displaystyle   u^{i}u_{i} \displaystyle  = \displaystyle  -1\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  g_{ij}u^{i}u^{j}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  g_{tt}\dot{t}^{2}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  -e^{2gx}\dot{t}^{2}\ \ \ \ \ (15)
\displaystyle  \dot{t}^{2} \displaystyle  = \displaystyle  e^{-2gx} \ \ \ \ \ (16)

Plugging into 11 we get

\displaystyle  \ddot{x}=-g \ \ \ \ \ (17)

That is, the acceleration in the {x} direction is a constant, so this would seem to indicate a uniform gravitational field. However, let’s apply the Einstein equation and see what we get. We’ll need the Riemann tensor in order to get the Ricci tensor, so we’ll use the definition of the former:

\displaystyle  R_{\; j\ell m}^{i}\equiv-\left[\partial_{m}\Gamma_{\;\ell j}^{i}-\partial_{\ell}\Gamma_{\; mj}^{i}+\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i}-\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}\right] \ \ \ \ \ (18)

Since all Christoffel symbols with a {y} or {z} index are zero, the only possibly non-zero components of {R_{ij\ell m}} are those containing only {x} or {t}, and due to the symmetry relations, the only independent component is

\displaystyle   R_{txtx} \displaystyle  = \displaystyle  g_{ti}R_{\; xtx}^{i}\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  e^{2gx}\left[\partial_{x}\Gamma_{\; tx}^{t}-\partial_{t}\Gamma_{\; xx}^{t}+\Gamma_{\; tx}^{k}\Gamma_{\; kx}^{t}-\Gamma_{\; xx}^{k}\Gamma_{\; tk}^{t}\right]\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  e^{2gx}\left[0-0+\Gamma_{\; tx}^{t}\Gamma_{\; tx}^{t}-0\right]\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  g^{2}e^{2gx} \ \ \ \ \ (22)

Now for the Ricci tensor. We have

\displaystyle   R_{ab} \displaystyle  = \displaystyle  R_{\; aib}^{i}\ \ \ \ \ (23)
\displaystyle  \displaystyle  = \displaystyle  g^{ij}R_{jaib} \ \ \ \ \ (24)

Since the metric is diagonal, the upstairs components are just the reciprocals of the downstairs ones, so

\displaystyle   g^{tt} \displaystyle  = \displaystyle  -e^{-2gx}\ \ \ \ \ (25)
\displaystyle  g^{xx} \displaystyle  = \displaystyle  g^{yy}=g^{zz}=1t \ \ \ \ \ (26)

and we get

\displaystyle   R_{tt} \displaystyle  = \displaystyle  g^{xx}R_{xtxt}=-g^{2}e^{2gx}\ \ \ \ \ (27)
\displaystyle  R_{xx} \displaystyle  = \displaystyle  g^{tt}R_{txtx}=g^{tt}R_{xtxt}=g^{2}\ \ \ \ \ (28)
\displaystyle  R_{xt}=R_{tx} \displaystyle  = \displaystyle  g^{ab}R_{bxat}=0 \ \ \ \ \ (29)

where in the last line we see that {R_{bxat}} can be non-zero only if {b=t} and {a=x} but since {g^{ab}} is diagonal, {g^{xt}=0}. All components of {R_{ab}} involving {y} or {z} indices are zero because all components of {R_{abcd}} involving {y} or {z} indices are zero. To use the Ricci tensor in the Einstein equation, we need the upstairs version, which is

\displaystyle   R^{tt} \displaystyle  = \displaystyle  g^{ta}g^{tb}R_{ab}=e^{-4gx}\left(-g^{2}e^{2gx}\right)=-g^{2}e^{-2gx}\ \ \ \ \ (30)
\displaystyle  R^{xx} \displaystyle  = \displaystyle  g^{xa}g^{xb}R_{ab}=R_{xx}=g^{2} \ \ \ \ \ (31)

In a vacuum (assuming {\Lambda=0}) the stress-energy tensor {T^{ij}=0} and the Einstein equation says that

\displaystyle  R^{ab}=0 \ \ \ \ \ (32)

The only way this can be true is if {g=0}, meaning that there is no gravitational field.

More generally, suppose that there is some fluid in the region so that {T^{ij}\ne0}. In that case

\displaystyle   R^{tt} \displaystyle  = \displaystyle  -g^{2}e^{-2gx}=8\pi G\left(T^{tt}+\frac{1}{2}e^{-2gx}T\right)\ \ \ \ \ (33)
\displaystyle  R^{xx} \displaystyle  = \displaystyle  g^{2}=8\pi G\left(T^{xx}-\frac{1}{2}T\right)\ \ \ \ \ (34)
\displaystyle  R^{yy} \displaystyle  = \displaystyle  0=8\pi G\left(T^{yy}-\frac{1}{2}T\right)\ \ \ \ \ (35)
\displaystyle  R^{zz} \displaystyle  = \displaystyle  0=8\pi G\left(T^{zz}-\frac{1}{2}T\right) \ \ \ \ \ (36)

where the stress-energy scalar is

\displaystyle   T \displaystyle  = \displaystyle  g_{ij}T^{ij}\ \ \ \ \ (37)
\displaystyle  \displaystyle  = \displaystyle  -e^{2gx}T^{tt}+T^{xx}+T^{yy}+T^{zz} \ \ \ \ \ (38)

From 35 and 36 we have

\displaystyle  T^{yy}=T^{zz}=\frac{T}{2} \ \ \ \ \ (39)


so from 38 we have

\displaystyle  -e^{2gx}T^{tt}+T^{xx}=0 \ \ \ \ \ (40)

However, if we multiply 33 by {e^{2gx}} and add it to 34 we get

\displaystyle  0=8\pi G\left(T^{tt}e^{2gx}+T^{xx}\right) \ \ \ \ \ (41)

Combining the last two equations we get

\displaystyle  T^{tt}=T^{xx}=0 \ \ \ \ \ (42)

Plugging 39 into 34 we get

\displaystyle  T^{yy}=T^{zz}=-\frac{g^{2}}{8\pi G} \ \ \ \ \ (43)

In a perfect fluid at rest, these components represent the pressure of the fluid in the {y} and {z} directions, so this result implies that the pressure would have to be negative, which doesn’t make sense.

Einstein equation on the surface of a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21, Problem 21.7.

According to the Einstein equation, the Riemann tensor in 2D must be zero in empty space, implying that gravitational fields cannot exist in 2D. Another consequence of the Einstein equation is that the stress-energy must be zero on the surface of a sphere. That is, even though a 2D surface is manifestly curved, the curvature is not the result of any mass or energy. This is another example of how general relativity breaks down in two dimensions.

The Einstein equation is

\displaystyle  R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)

The Ricci tensor for a spherical surface is

\displaystyle  R^{ij}=\left[\begin{array}{cc} \frac{1}{r^{4}} & 0\\ 0 & \frac{1}{r^{4}\sin^{2}\theta} \end{array}\right] \ \ \ \ \ (2)

The metric for a sphere is (in both forms):

\displaystyle   g^{ij} \displaystyle  = \displaystyle  \left[\begin{array}{cc} \frac{1}{r^{2}} & 0\\ 0 & \frac{1}{r^{2}\sin^{2}\theta} \end{array}\right]\ \ \ \ \ (3)
\displaystyle  g_{ij} \displaystyle  = \displaystyle  \left[\begin{array}{cc} r^{2} & 0\\ 0 & r^{2}\sin^{2}\theta \end{array}\right] \ \ \ \ \ (4)

Since the off-diagonal elements of {g^{ij}} and {R^{ij}} are all zero, 1 tells us that

\displaystyle  T^{\theta\phi}=T^{\phi\theta}=0 \ \ \ \ \ (5)

To deal with the diagonal elements, we first need the stress-energy scalar.

\displaystyle   T \displaystyle  = \displaystyle  g_{ij}T^{ij}\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  r^{2}T^{\theta\theta}+r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (7)

We have

\displaystyle   R^{\theta\theta} \displaystyle  = \displaystyle  \frac{1}{r^{4}}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \kappa\left(T^{\theta\theta}-\frac{1}{2r^{2}}T\right)+\frac{\Lambda}{r^{2}}\ \ \ \ \ (9)
\displaystyle  R^{\phi\phi} \displaystyle  = \displaystyle  \frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \kappa\left(T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T\right)+\frac{\Lambda}{r^{2}\sin^{2}\theta} \ \ \ \ \ (11)

Combining these we get

\displaystyle   \frac{R^{\theta\theta}}{\sin^{2}\theta}-R^{\phi\phi} \displaystyle  = \displaystyle  \kappa\left(\frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi}\right)=0\ \ \ \ \ (12)
\displaystyle  \frac{T^{\theta\theta}}{\sin^{2}\theta}-T^{\phi\phi} \displaystyle  = \displaystyle  0\ \ \ \ \ (13)
\displaystyle  T^{\theta\theta} \displaystyle  = \displaystyle  T^{\phi\phi}\sin^{2}\theta\ \ \ \ \ (14)
\displaystyle  T \displaystyle  = \displaystyle  2r^{2}\sin^{2}\theta T^{\phi\phi} \ \ \ \ \ (15)

Therefore

\displaystyle   T^{\theta\theta}-\frac{1}{2r^{2}}T \displaystyle  = \displaystyle  T^{\phi\phi}-\frac{1}{2r^{2}\sin^{2}\theta}T=0 \ \ \ \ \ (16)

holds identically. Thus the stress-energy contribution to the Einstein equation is always zero on a sphere (although the stress-energy tensor may have two non-zero components, these two components always combine to give zero contribution to the Einstein equation).

Ricci tensor and curvature scalar for a sphere

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 19; Box 19.6.

As an example of calculating the Ricci tensor and curvature scalar we’ll find them for the 2-d surface of a sphere. The Ricci tensor is calculated from the Riemann tensor, and that in turn depends on the Christoffel symbols, so we’ll need them first. It’s easiest to find them from the geodesic equation

\displaystyle  g_{aj}\ddot{x}^{j}+\left(\partial_{i}g_{aj}-\frac{1}{2}\partial_{a}g_{ij}\right)\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (1)

which is formally equivalent to

\displaystyle  \ddot{x}^{m}+\Gamma_{\; ij}^{m}\dot{x}^{j}\dot{x}^{i}=0 \ \ \ \ \ (2)

The metric for a sphere is

\displaystyle  ds^{2}=r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (3)

where {r} is the constant radius of the sphere, so

\displaystyle   g_{\theta\theta} \displaystyle  = \displaystyle  r^{2}\ \ \ \ \ (4)
\displaystyle  g_{\phi\phi} \displaystyle  = \displaystyle  r^{2}\sin^{2}\theta \ \ \ \ \ (5)

We have two equations arising from 1. For {a=\theta}

\displaystyle  r^{2}\ddot{\theta}-r^{2}\sin\theta\cos\theta\dot{\phi}^{2}=0 \ \ \ \ \ (6)

Comparing with 2 we get, after dividing out the {r^{2}}:

\displaystyle  \Gamma_{\;\phi\phi}^{\theta}=-\sin\theta\cos\theta=-\frac{1}{2}\sin2\theta \ \ \ \ \ (7)

For {a=\phi}:

\displaystyle  r^{2}\sin^{2}\theta\ddot{\phi}^{2}+2r^{2}\sin\theta\cos\theta\dot{\theta}\dot{\phi}=0 \ \ \ \ \ (8)

Dividing through by {r^{2}\sin^{2}\theta} and comparing with 2 we get (remember that the second term is {\left(\Gamma_{\;\theta\phi}^{\phi}+\Gamma_{\;\phi\theta}^{\phi}\right)\dot{\theta}\dot{\phi}} and that {\Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}}):

\displaystyle  \Gamma_{\;\theta\phi}^{\phi}=\Gamma_{\;\phi\theta}^{\phi}=\cot\theta \ \ \ \ \ (9)

All other Christoffel symbols are zero.

In 2D, the Riemann tensor has only one independent component, which we can take to be {R_{\theta\phi\theta\phi}}, which can be calculated from

\displaystyle  R_{\; j\ell m}^{i}\equiv\partial_{\ell}\Gamma_{\; mj}^{i}-\partial_{m}\Gamma_{\;\ell j}^{i}+\Gamma_{\; mj}^{k}\Gamma_{\;\ell k}^{i}-\Gamma_{\;\ell j}^{k}\Gamma_{\; km}^{i} \ \ \ \ \ (10)

Lowering the first index, we have

\displaystyle   R_{\theta\phi\theta\phi} \displaystyle  = \displaystyle  g_{\theta i}R_{\;\phi\theta\phi}^{i}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  g_{\theta\theta}R_{\;\phi\theta\phi}^{\theta}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  r^{2}\left(\partial_{\theta}\Gamma_{\;\phi\phi}^{\theta}-\partial_{\phi}\Gamma_{\;\theta\phi}^{\theta}+\Gamma_{\;\phi\phi}^{k}\Gamma_{\;\theta k}^{\theta}-\Gamma_{\;\theta\phi}^{k}\Gamma_{\; k\phi}^{\theta}\right)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  r^{2}\left(-\cos2\theta-0+0-\Gamma_{\;\theta\phi}^{\phi}\Gamma_{\;\phi\phi}^{\theta}\right)\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  r^{2}\left(\sin^{2}\theta-\cos^{2}\theta+\cos^{2}\theta\right)\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  r^{2}\sin^{2}\theta \ \ \ \ \ (16)

We can now find the Ricci tensor.

\displaystyle  R_{ij}=g^{ab}R_{aibj} \ \ \ \ \ (17)

Since the metric is diagonal and {g^{ab}} is the inverse of {g_{ab}}, we have

\displaystyle   g^{\theta\theta} \displaystyle  = \displaystyle  \frac{1}{g^{\theta\theta}}=\frac{1}{r^{2}}\ \ \ \ \ (18)
\displaystyle  g^{\phi\phi} \displaystyle  = \displaystyle  \frac{1}{g^{\phi\phi}}=\frac{1}{r^{2}\sin^{2}\theta} \ \ \ \ \ (19)

so, using the symmetries of the Riemann tensor,

\displaystyle   R_{\theta\theta} \displaystyle  = \displaystyle  g^{ab}R_{a\theta b\theta}\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  g^{\phi\phi}R_{\phi\theta\phi\theta}\ \ \ \ \ (21)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{r^{2}\sin^{2}\theta}r^{2}\sin^{2}\theta\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  1\ \ \ \ \ (23)
\displaystyle  R_{\phi\phi} \displaystyle  = \displaystyle  g^{ab}R_{a\phi b\phi}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  g^{\theta\theta}R_{\theta\phi\theta\phi}\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \sin^{2}\theta\ \ \ \ \ (26)
\displaystyle  R_{\theta\phi} \displaystyle  = \displaystyle  R_{\phi\theta}=0 \ \ \ \ \ (27)

We can get the upstairs version of the Ricci tensor as well:

\displaystyle   R^{\theta\theta} \displaystyle  = \displaystyle  g^{\theta i}g^{\theta j}R_{ij}\ \ \ \ \ (28)
\displaystyle  \displaystyle  = \displaystyle  g^{\theta\theta}g^{\theta\theta}R_{\theta\theta}\ \ \ \ \ (29)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{r^{4}}\ \ \ \ \ (30)
\displaystyle  R^{\phi\phi} \displaystyle  = \displaystyle  g^{\phi\phi}g^{\phi\phi}R_{\phi\phi}\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{r^{4}\sin^{2}\theta} \ \ \ \ \ (32)

The curvature scalar is

\displaystyle   R \displaystyle  = \displaystyle  g_{ij}R^{ij}\ \ \ \ \ (33)
\displaystyle  \displaystyle  = \displaystyle  r^{2}\frac{1}{r^{4}}+r^{2}\sin^{2}\theta\frac{1}{r^{4}\sin^{2}\theta}\ \ \ \ \ (34)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{r^{2}} \ \ \ \ \ (35)

As we would expect, the curvature of a sphere decreases as its radius gets larger.

Einstein equation in the Newtonian limit

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 21; Problem 21.1.

The Einstein equation is

\displaystyle  R^{ij}=\kappa\left(T^{ij}-\frac{1}{2}g^{ij}T\right)+\Lambda g^{ij} \ \ \ \ \ (1)


where we have yet to determine the constant {\kappa}. To do this, we need to show that the Einstein equation reduces to Newton’s law of gravity for weak gravitational fields. Actually, there are three conditions that should hold in the Newtonian limit. First, as we’ve said, the gravitational field is weak, meaning that spacetime is nearly flat. Second, objects should travel with a speed much less than the speed of light (the spatial four-velocity components {u^{i}\ll1} for {i=x,y,z}). The second condition implies that the only non-negligible component of the stress-energy tensor {T^{ij}} is {T^{tt}}. For example, for a perfect fluid, we can assume that it’s effectively at rest, so the off-diagonal elements are all zero. For the diagonal spatial compoments, we have (for {T^{zz}}; the other 2 components have the analogous formulas):

\displaystyle  T^{zz}=\frac{1}{L^{3}}\int dp^{x}\int dp^{y}\int\left(p^{z}\right)^{2}\frac{N\left(p\right)}{p^{t}}dp^{z} \ \ \ \ \ (2)

This equation is for a cubic volume of side length {L} containing {N\left(p\right)} particles of momentum {p}. Since {p^{z}=mu^{z}} and {u^{z}\ll1}, {T^{zz}\approx0} (the requirement of a weak gravitational field means that {N\left(p\right)} can’t be very large, since we can’t have that much mass). As the spatial diagonal elements {T^{ii}=P}, the pressure and {T^{tt}=\rho}, the energy density, this condition translates to {\rho\gg P}.

With these assumptions, we can try to show that the relativistic equation of geodesic deviation reduces to the Newtonian version. That is, we want to show that

\displaystyle  \ddot{\mathbf{n}}^{i}=-R_{\; j\ell m}^{i}u^{m}u^{j}n^{\ell} \ \ \ \ \ (3)


reduces to

\displaystyle  \ddot{n}^{i}=-\eta^{ij}\left(\partial_{k}\partial_{j}\Phi\right)n^{k}\mbox{ (for }i,j,k=x,y,z\mbox{)} \ \ \ \ \ (4)


where {n^{i}} is the separation of two infinitesimally close geodesics (this is the tidal force.) and {\Phi} is the Newtonian gravitational potential.

Starting with 3, we can eliminate terms where {m\ne t} or {j\ne t} (because {u^{i}\approx0} for {i\ne t}) to get

\displaystyle  \ddot{n}^{i}=-R_{\; t\ell t}^{i}n^{\ell} \ \ \ \ \ (5)

where we’re now considering only the spatial components: {i,\ell=x,y,z}. Note that summing {\ell} over {x,y,z} is the same as summing it over {t,x,y,z} since due to the anti-symmetry of the Riemann tensor under interchange of its last two indices, {R_{\; ttt}^{i}=-R_{\; ttt}^{i}=0}. Also, because we’re in the non-relativistic limit, the proper time and coordinate time are essentially the same thing: {\tau\approx t}, so the time derivative is with respect to {t}.

Comparing this to 4, we have (renaming {\ell} to {k} in the last equation):

\displaystyle  R_{\; tkt}^{i}\approx\eta^{ij}\left(\partial_{k}\partial_{j}\Phi\right) \ \ \ \ \ (6)

Newton’s law of gravity in differential form is

\displaystyle   \nabla^{2}\Phi \displaystyle  = \displaystyle  4\pi G\rho\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \eta^{ij}\partial_{i}\partial_{j}\Phi\ \ \ \ \ (8)
\displaystyle  \displaystyle  \approx \displaystyle  R_{\; tit}^{i}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  R_{tt} \ \ \ \ \ (10)

(Again, the contraction over index {i} in the second to last line can be taken over 3 or 4 coordinates, since {R_{\; ttt}^{i}=0}.) We can raise both indices on the Ricci tensor in the usual way:

\displaystyle   R^{tt} \displaystyle  = \displaystyle  g^{ti}g^{tj}R_{ij}\ \ \ \ \ (11)
\displaystyle  \displaystyle  \approx \displaystyle  \eta^{ti}\eta^{tj}R_{ij}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \left(-1\right)^{2}R_{tt}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  R_{tt} \ \ \ \ \ (14)

so we can write 10 in upper index form as

\displaystyle  \nabla^{2}\Phi\approx R^{tt} \ \ \ \ \ (15)

Now looking back at 1 and using the condition that {T^{tt}=\rho} is the only significant entry in the stress-energy tensor, we have

\displaystyle  T=g_{ij}T^{ij}=\eta_{ij}T^{ij}=-\rho \ \ \ \ \ (16)

so

\displaystyle   R^{tt} \displaystyle  = \displaystyle  \kappa\left(T^{tt}-\frac{1}{2}\eta^{tt}T\right)+\Lambda\eta^{tt}\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  \frac{\kappa}{2}\rho-\Lambda\ \ \ \ \ (18)
\displaystyle  \displaystyle  \approx \displaystyle  \nabla^{2}\Phi\ \ \ \ \ (19)
\displaystyle  \displaystyle  = \displaystyle  4\pi G\rho \ \ \ \ \ (20)

Comparing the second and fourth lines, we see that {\Lambda\approx0} and

\displaystyle  \kappa=8\pi G \ \ \ \ \ (21)

and the Einstein equation becomes

\displaystyle  R^{ij}=8\pi G\left(T^{ij}-\frac{1}{2}g^{ij}T\right) \ \ \ \ \ (22)


or in its original form

\displaystyle  G^{ij}=8\pi GT^{ij} \ \ \ \ \ (23)

Actually, we can’t take {\Lambda=0}; all we can say is that for gravitational systems on the scale of the solar system (where Newtonian theory works well, except in the case of the orbit of Mercury) {\Lambda\ll4\pi G\rho}. To get a feel for how small {\Lambda} needs to be, suppose we have a spherical gravitational potential in empty space ({\rho=0}). Then in spherical coordinates

\displaystyle   \nabla^{2}\Phi \displaystyle  = \displaystyle  \frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{d\Phi}{dr}\right)\ \ \ \ \ (24)
\displaystyle  \displaystyle  \approx \displaystyle  -\Lambda\ \ \ \ \ (25)
\displaystyle  \frac{d}{dr}\left(r^{2}\frac{d\Phi}{dr}\right) \displaystyle  \approx \displaystyle  -\Lambda r^{2}\ \ \ \ \ (26)
\displaystyle  \frac{d\Phi}{dr} \displaystyle  \approx \displaystyle  -\frac{\Lambda}{3}r \ \ \ \ \ (27)

This is the radial component (the only non-zero component) of the gradient of the potential, and the negative gradient of the gravitational potential is the gravitational field, which is the acceleration of gravity. In the solar system, Newton’s theory says that the acceleration due to the sun is

\displaystyle  g=\frac{GM_{s}}{r^{2}} \ \ \ \ \ (28)

so if {\Lambda\ne0} but its effect is not felt in Newton’s theory, we must have

\displaystyle  \frac{\Lambda}{3}r\ll\frac{GM_{s}}{r^{2}} \ \ \ \ \ (29)

in order for Newton’s theory to be valid in the solar system. Distances in the solar system are around {r\approx10^{12}\mbox{ m}}, {GM_{s}\approx1500\mbox{ m}} and {G} in general relativistic units is {7.426\times10^{-28}\mbox{ m kg}^{-1}} so this means

\displaystyle  \frac{\Lambda}{8\pi G}\ll\frac{1500}{8\pi\left(7.426\times10^{-28}\right)\left(10^{12}\right)^{3}}=2.4\times10^{-7}\mbox{ kg m}^{-3} \ \ \ \ \ (30)

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