Keplerian Frequency of Schwarzschild Black hole

Question

The Keplerian frequency/ Orbital frequency is the inverse of orbital period and for Schwarzschild black hole it is given by $$\frac{1}{2\pi}\sqrt{\frac{M}{r^3}}.$$ its unit is Hertz. Now To express the frequencies measured by a distant observer in physical units, one must extend the above form by multiplying by the factor $\frac{c^3}{GM}.$ So $$v_\phi=\frac{1}{2\pi}\frac{c^3}{GM}\sqrt{\frac{M}{r^3}}.$$ Am i doing it correctly? Also what will be its unit then? I am confused about the units. Also I will need $G$ in form of solar mass and not kilograms I guess. radius $r$ will be in form of $M$ (for example for ISCO it should be $6M$) or in kilometers?

Answer 1

In physical units, the period of a circular orbit of a massive particle in Schwarzschild is $T = 2 \pi \sqrt{\frac{r^3}{G M}}$ (see, e.g., Poisson and Will, Chapter 5), so: $$ f = \frac{1}{2\pi} \sqrt{\frac{GM}{r^3}} $$ Note that this quantity has SI units of $\text{s}^{-1}$.

The use of units in GR is often sloppy and made mysterious by imprecise statements. I think a good way to think about it is that one chooses a new "particular" system of units, e.g., $\left( \ell_p, t_p, m_p \right)$, and then non-dimensionalizes each variable using those units.

So we can non-dimensionalize the variables in the frequency equation as: $$ \tilde{f} := \frac{f}{1/t_p} \quad; \quad \tilde{M} := \frac{M}{m_p} \quad; \quad \tilde{r} := \frac{r}{\ell_p} $$ and, plugging those in, we find that the frequency equation in those non-dimensionalized variables becomes: $$ \frac{\tilde{f}}{t_p} = \frac{1}{2\pi} \sqrt{\frac{ G \,\tilde{M} \, m_p}{\left(\tilde{r} \ell_p \right)^3}} \quad \rightarrow \quad \tilde{f} = \frac{1}{2\pi} \sqrt{\frac{ \tilde{M}}{\tilde{r}^3}} \cdot \sqrt{ \frac{G \, m_p \, t_p^2}{\ell_p^3}} $$ We have said nothing yet about which "particular" system of units we have chosen. In GR, the standard choice is a system of units which "$G = c = 1$", or more precisely, these physical constants have the value one in those units: $$ G = 1 \frac{\ell_p^3}{t_p^2 m_p} \quad \text{and} \quad c = 1 \frac{\ell_p}{t_p} $$ If we write $G$ in that way, you see that the whole multiplicative term on the right-hand side cancels out and you arrive at your equation, which, again, should be thought of in terms of non-dimensionalized variables: $$ \tilde{f} = \frac{1}{2\pi} \sqrt{\frac{\tilde{M}}{\tilde{r}^3}} $$

To go back to physical units from a non-dimensionalized equation, you use non-dimensional variable definitions and the choices made in defining the units (relationships to $G$ and $c$): $$ \begin{align} \tilde{f} &= \frac{1}{2\pi} \sqrt{\frac{\tilde{M}}{\tilde{r}^3}}\\ \frac{f}{1/t_p} & = \frac{1}{2 \pi} \sqrt{\frac{M / m_p}{\left(r / \ell_p \right)^3}} \\ f &= \frac{1}{2 \pi} \sqrt{\frac{M}{r^3}} \sqrt{\frac{\ell_p^3}{t_p^2 \, m_p}}\\ f &= \frac{1}{2 \pi} \sqrt{\frac{M}{r^3}} \sqrt{G} \end{align} $$

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An alternative, but equivalent, prescription for getting dimensional quantities from non-dimensionalized (geometrized) quantities is given by Wald (Appendix F):

A quantity with dimension $L^n\, T^m \, M^p$ in ordinary units has dimension $L^{n+m+p}$ in geometrized units, and the conversion factor is $c^m \left( \frac{G}{c^2} \right)^p$.

So for the non-dimensionalized frequency equation we would write: $$ \begin{align} \tilde{f} &= \frac{1}{2\pi} \sqrt{\frac{\tilde{M}}{\tilde{r}^3}}\\ f \cdot c^{-1} & = \frac{1}{2 \pi} \sqrt{\frac{M \cdot \left(\frac{G}{c^2}\right)^1}{\left(r \cdot 1 \right)^3}} \\ f &= \frac{1}{2 \pi} \sqrt{\frac{G\,M}{r^3}} \end{align} $$ The reasoning behind Wald's statment is that we choose some $\ell_p = L$, and then our other unit specifications follow from "$G = c = 1$", i.e., $$ c = \frac{\ell_p}{t_p} \quad \rightarrow \quad t_p = \frac{L}{c} $$ and $$ G = \frac{l_p^3}{t_p^2 \, m_p} = c^2 \frac{\ell_p}{m_p} \quad \rightarrow \quad m_p = \frac{c^2}{G} \, L $$ So if $c$'s and $G$'s become one, then the only thing left are powers of $L$.

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If we chose, as is done, e.g., in Misner, Thorne, and Wheeler (MTW), to let $\ell_p = L = 1\,\text{cm}$, then we would find: $$ \ell_p = 1\, \text{cm} \quad ; \quad t_p = \frac{\ell_p}{c} \approx 3.33 \times 10^{-11}\, \text{cm} \quad ; \quad m_p = \frac{c^2}{G} \ell_p \approx 1.35 \times 10^{25} \, \text{kg} $$ And we could express the solar mass as: $$ 1 \, \text{solar mass} = 1.99 \times 10^{30} \, \text{kg} \cdot \frac{1\, m_p}{1.35 \times 10^{25} \, \text{kg}} \approx 147,500 \, m_p $$ But, maddeningly, as both MTW and Wald say (this is Wald):

all quantities which in ordinary units have dimension expressible in terms of length, time, and mass, are given a dimension of a power of length in geometrized units

So the solar mass is actually written as: $$ 1 \, \text{solar mass} \approx 147,500 \, m_p = 147,500 \, \frac{c^2}{G} \ell_p = 147,500 \, \text{cm} $$ where the last equality follows by "$c = G = 1$" and the unwritten rule that we will suppress all $\left( \ell_p, t_p, m_p \right)$ except for the one that we chose to be a particular physical unit, $\ell_p = 1 \, \text{cm}$ (and we will suppress that as well, if it appears within $c$ or $G$). That last equality should really say: $$ 1 \, \text{solar mass} \approx 147,500 \, m_p = 147,500 \, \frac{c^2}{G} \ell_p = 147,500 \, \left(1 \frac{m_p}{\ell_p} \right) \text{cm} $$ because while "$c=G=1$", $$ \frac{c^2}{G} = \frac{\left(1 \frac{\ell_p}{t_p} \right)^2}{\left(\frac{\ell_p^3}{t_p^2 \,m_p}\right)} = 1 \frac{m_p}{\ell_p} $$

Answer 2

The $\sqrt{\frac{M}{r^3}}$ part of your expression is in geometric units, where $G=c=1$. In this system you pick one dimension mass, length, or time and express all quantities in that unit. Since you are trying calculate a period, it makes most sense to express $M$ and $r$ in units of time (e.g. seconds). Then $\sqrt{\frac{M}{r^3}}$ has units of per-time (e.g. Hertz). If your input quantities are in the correct units, then you don't need to convert your final answer. It already has the correct dimensions.

The part you added in $\frac{c^3}{GM}$ includes the $c$'s and $G$ explicitly. This expression also has dimensions of per-time, which you can verify by plugging in $c$ and $G$ in S.I. units and any $M$ in kg. This is not a unit conversion factor. A unit conversion factor should have units of [something]-per-[something else], so it can cancel the units of a quantity and replace them with new ones.

Had you input $M$ and $r$ in geometric time units, the full expression would then have dimensions of per-time-squared (e.g. Hz$^2$), which isn't what you want.

Had you input $M$ and $r$ in S.I. units, the full expression would also have the wrong units, $\mathrm{kg}^{1/2}\mathrm{m}^{-3/2}\mathrm{s}^{-1}$.

convert before

I think it is easier to do the unit conversions before you start by expressing both $M$ and $r$ in units of time. The most common conversion factors in GR are $G/c^2$ (mass-per-length) and $G/c^3$ (mass-per-time).

It's also common to define $T_\odot$ and $R_\odot$, mass of the sun expressed in units of time and length, respectively. You can then use those to more easily convert astronomical masses expressed in $M_\odot$.

As an example consider an orbit at $r=6M$ around a $M=10M_\odot$ black hole. $$M = 10 M_\odot \left[ \frac{2\times 10^{30}\,\mathrm{kg}}{M_\odot} \cdot \frac{6.7\times 10^{-11}\,\mathrm{m}^3}{\mathrm{kg\,s^2}} \cdot \left(\frac{\mathrm{s}}{3\times 10^8\,\mathrm{m}}\right)^3 \right]$$ The term in brackets has units of $\frac{\mathrm{s}}{M_\odot}$. This is the same unit conversion to calculate $T_\odot$. Once we know $M$ in units of time we can easily calculate $r=6M$.

convert after

We can plug the $M=10M_\odot$ and $r=6M$ to your expression: $$\sqrt{\frac{M}{r^3}} = \sqrt{\frac{10 M_\odot}{(6\cdot 10M_\odot)^3}}$$

This will result in something with units of $\frac{1}{M_\odot}$. The unit conversion becomes: $$f = \frac{[\mathrm{number}]}{M_\odot} \left[ \frac{M_\odot}{2\times 10^{30}\,\mathrm{kg}} \cdot \frac{\mathrm{kg\,s^2}}{6.7\times 10^{-11}\,\mathrm{m}^3} \cdot \left(\frac{3\times 10^8\,\mathrm{m}}{\mathrm{s}}\right)^3 \right]$$

The correct unit conversion factor is just $c^3/G$. The mass of the black hole, $M$, doesn't enter into the unit conversion.