Is there a frequency limit for the pendulum near a black hole, and is this related to photons and the UV-cutoff?

Question

I imagine a swinging pendulum being held outside a black hole, supported by the normal force of a rocket. The rocket is hovering the swinging pendulum right above the event horizon in the strong gravitational potential outside of the black hole.

My first question is: what would the frequency be of this pendulum and would this be the natural limit to swinging frequencies? How could the frequency possibly increase (besides by shortening length)?

And, following the consequences of this: I believe photons and pendulums behave identically under the influence of a gravitational potential; does this then imply photons also have a natural cut-off limit for their frequency?

Answer 1

A hovering observer will feel a local gravitational acceleration $$g_{\rm eff} = \frac{GM}{r^2}\left(1 - \frac{r_s}{r}\right)^{-1/2}\ ,$$ where $r_s$ is the Schwarzschild radius. This expression diverges to infinity as $r \rightarrow r_s$ (as does the force required to maintain your observer at fixed $r$) and so the period of the pendulum, which goes as $\sqrt{l/g_{\rm eff}}$ will become arbitrarily small according to an observer with the pendulum. There is no upper limit or cut-off to the pendulum frequency. NB: The expression above assumes that $l$ is small compared with $r-r_s$, so that $g_{\rm eff}$ can be assumed constant.

However, a distant observer viewing the pendulum would disagree. There would be an additional time dilation factor of $(1- r_s/r)^{-1/2}$, which means that a distant observer would see the pendulum period become arbitrarily longer as $r \rightarrow r_s$.

Objects with mass (a pendulum) do not behave in the same way as massless particles (photons) near a black hole. In particular, a laser on board the rocket would not undergo any change in frequency according to an observer on the rocket; whereas according to a distant observer, the laser light will appear redshifted by an amount increasing towards infinity as $r\rightarrow r_s$.

Answer 2

The pendulum will librate with the proper angular frequency $\omega_{\rm prop} =\sqrt{a_{\rm prop}/l}$, where $ a_{\rm prop}$ is the proper acceleration of the rocket hovering above the horizon, and $l$ is the (proper) length of the pendulum. Proper refers to quantities measured by rulers and stopwatches in the frame of the rocket. For a rocket hovering above a Schwarzschild black hole of mass $M$ at Schwarzschild radius $r$, this acceleration is $$a_{\rm prop} = \left(1 - \frac{2GM}{c^2r} \right)^{-1/2} \frac{GM}{r^2}$$ This gives us $$\omega_{\rm prop} = \left(1 - \frac{2GM}{c^2r} \right)^{-1/4} \sqrt{\frac{GM}{l r^2}}$$ In other words, the proper frequency of the pendulum diverges as one approaches the horizon $r=2M$ for any finite length of the pendulum. Note that one cannot have a rocket at the horizon, this is only a statement about approaching the horizon from above.

There is only one limit for this I can possibly think of, and that is the string tension of the pendulum. Quite generally, pressure waves in the string cannot travel faster than the speed of light. Then, when the theoretical period of the pendulum $2 \pi/\omega_{\rm prop}$ reaches the crossing time of the length of the pendulum $l/c$, the pendulum must break. That is, $$2 \pi \left(1 - \frac{2GM}{c^2r} \right)^{1/4} \sqrt{\frac{l r^2}{GM}} \gtrsim \frac{l}{c} $$ from which you can get $$l \lesssim \frac{4\pi^2 c^2}{a_{\rm prop}}\;,\;\; \omega_{\rm prop} \gtrsim \frac{a_{\rm prop}}{2\pi c}$$ This means that there is a lower limit on the oscillations frequencies of pendula (an infrared cut-off)! Note that this is also a statement about the general integrity of structures in the hovering state above the horizon. They have to necessarily be smaller than $4\pi^2 c^2/a_{\rm prop}$ to even have the chance of being structurally stable against small oscillations.

Now let's talk about photons. Photons are not limited in frequency anywhere near the black hole. There is no particular cut-off that needs to be applied here, apart from cut-offs such as the Planck energy etc. However, one can find some sort of relationship between photons and the pendulum. Imagine you want to keep a photon localized by bouncing it around in a cavity. In order for this to work, you need the cavity to have the size of at least a single wavelength of the photon $\lambda$. But the cavity must also be smaller than $4\pi^2 c^2/a_{\rm prop}$ to be stable without violating causality. The wavelength is related to the angular frequency as $\omega_{\rm phot} = 2\pi c/\lambda$. You then see that one is unable to localize photons in the rocket frame unless their frequency is $$\omega_{\rm phot} \gtrsim \frac{a_{\rm prop}}{2\pi c}$$ which is the same lower limit (infrared cut-off) as you have for the pendulum.

Answer 3

The Newtonian small angle approximation for the frequency of a pendulum is $$f = \frac{1}{2 \pi} \sqrt{\frac{g}{l}} = \frac{1}{2 \pi} \sqrt{\frac{GM}{l R^2} } $$

Now if we assume that the Newtonian approximation is valid infinitely far from a black hole in Schwarzschild coordinates, then the coordinate frequency of the pendulum lowered very slowly into the gravitational field would be:

$$f_{coord} \approx \frac{1}{2 \pi } \sqrt{\frac{GM}{l R^2}} \left(1-\frac{2GM}{Rc^2}\right)^{1/4},$$

according to the Schwarzschild observer at infinity. Since frequency is the inverse of the period (f = 1/t) the locally measured frequency taking gravitational time dilation into account is:

$$f_{local} \approx \frac{1}{2 \pi } \sqrt{\frac{GM}{l R^2}} \left(1-\frac{2GM}{Rc^2}\right)^{-1/4},$$

Now, as has been noted, this local frequency diverges to infinity as R approaches $2GM/c^2$. However, I think it is worth noting that the oscillation speed of a pendulum is largely due to the acceleration of gravity at the top of the swing and at the lowest point of the swing the downward acceleration has no effect on the horizontal velocity. Therefore a more realistic approximation of the pendulum equation is:

$$f_{local} \approx \frac{1}{2 \pi } \sqrt{\frac{GM}{l (R+h)^2}} \left(1-\frac{2GM}{(R+h)c^2}\right)^{-1/4},$$

where h is the difference in height between the top of the swing and the bottom of the swing. This modified equation does not go to infinity unless h=0 or l=0 or $\theta_{max} = 0$ in which case the pendulum cannot be described as oscillating, or the bottom of the swing is below the event horizon while the pivot is held stationary, which is impossible.

As to whether the horizontal velocity of the pendulum exceeds the speed of light, we can analyse the situation taking potential energy into account. A pendulum converts potential energy to horizontal kinetic energy and vice versa. Starting with the equation for gravitation force in the metric:

$$F=\frac{GMm}{R^2 \sqrt{1-2GM/(Rc^2)}} ,$$

then the potential energy at a given radius, is the indefinite integral of the above equation:

$$PE = m c^2 \sqrt{1-2GM/(R c^2)} .$$

The kinetic energy at the lowest point of the pendulum swing is equal to the difference in potential energy between the highest point and the lowest point so:

$$KE = m c^2 \left(\sqrt{1-\frac{2GM}{(R+h)c^2}} -\sqrt{1-\frac{2GM}{R}} \right),$$

where (R+h) is the maximum height of the swing and R is the lowest point. Using the fairly well known formula $KE = m c^2(\gamma -1)$ which derives from the relativistic momentum energy relationship, where $\gamma = 1/\sqrt{1-v^2/c^2} $ the local maximum velocity of the pendulum at its lowest point can be found:

$$v_{max} = c \ \sqrt{\frac{(KE^2 + 2 {KE} \times mc^2)}{(KE^2 + 2 KE \times mc^2) + mc^2}}.$$

Note that the parts in brackets are identical, so the extra term in the denominator ensures that v is never greater than c.

Now I have established that the local frequency does not become infinite and the local velocity does not exceed the speed of light, lets consider what happens to the frequency of a photon near the event horizon. Imagine an observer (Anne) is in an elevator that is slowly lowered into the gravitational field and she is carrying a laser. As the elevator descends, the observer at infinity sees the frequency of the laser get slower, but Anne sees no change at all in the frequency of the light from her own laser, no matter how close she gets to the event horizon. No one is seeing the frequency increasing. However, if a second observer (Bob) higher up has a downward velocity of v and shines a laser down towards Anne, she will see the frequency of the descending light as increased proportional to the relativistic Doppler factor:

$$ f_{_{Anne}} = f_{_{Bob}} \sqrt{\frac{1+v/c}{1-v/c}} \sqrt{\frac{1-2GM/(R_{Bob} \ c^2)}{1-2GM/(R_{Anne} \ c^2)}}$$

The frequency observed by Anne never becomes infinite, unless Anne is stationary at the event horizon (impossible) or Bob has a downward velocity equal to the speed of light (also impossible), so an ultra-violet cutoff is not implied or necessary.