1
$\begingroup$

  1. The Schwarzschild radius $r_s = \frac{2GM}{c^2}$ increases linearly with mass.

  2. The gravitational attraction at any multiple of $r_s$ grows arbitrarily weak, namely by the inverse of the mass multiple.

Perhaps this is a convoluted way of proving point 2, but subbing $r$ with $r_s$ in Newton's formula for gravitational acceleration $$\frac{g_M}{g_m}=\frac{\frac{GM}{(\frac{2GM}{c^2})^2}}{\frac{Gm}{(\frac{2Gm}{c^2})^2}} = \frac{GM}{(\frac{4G^2M^2}{c^4})}*\frac{(\frac{4G^2m^2}{c^4})}{Gm} = \frac{M*m^2}{M^2*m} = \frac{m}{M}$$

In other words, if $M$ equals $2m$, the ratio of the gravitational acceleration at the same multiple of $r_s$ is $\frac{1}{2}$.

My question is: since the acceleration is halved, but $r_s$ is doubled, does that mean it requires the same amount of energy to move an object between the same two multiples of the Schwarzschild radius, regardless of the mass of the black hole?

Improve this question
$\endgroup$
4
  • $\begingroup$ Please explain what $g_M$ and $g_m$ represent. $\endgroup$
    – ProfRob
    Jul 25 at 21:45
  • 1
    $\begingroup$ The equivalent of potential energy in GR is the mass defect that depends only on the time dilation, which in turn depends only on $r/r_s$. Thus your conclusion is correct, as long as you stay away from the horizon itself where the gravitational acceleration is infinite. $\endgroup$
    – safesphere
    Jul 25 at 23:32
  • $\begingroup$ @ProfRob Sorry, it means "g of big mass" and "g of little mass" respectively $\endgroup$
    – Kalle Anka
    Jul 26 at 8:44
  • 1
    $\begingroup$ Note that mass is variable and there is no potential energy in GR. The conserved total energy of a free falling object is internal ($mc^2$) plus kinetic. As the kinetic energy increases, the mass of the object proportionally decreases (mass defect) when measured in the same coordinates (e.g. Schwarzschild) before and after. As the object approaches the horizon, its mass approaches zero and its speed approaches the speed of light. $\endgroup$
    – safesphere
    Jul 26 at 13:43

2 Answers 2

Reset to default
1
$\begingroup$

This can be seen from simple dimensional analysis. If we use geometric units such that $c=G=1$, all quantities have dimensions of [mass] to some power. In particular, the specific energy of an object, i.e. its energy per unit mass, has dimension $[\rm mass]^0$, i.e. its dimensionless. Consequently, it needs to be function of other dimensionless quantities. The only dimensionless variable available are the ratio of the radial position to the Schwarzschild radius and the ratio of the two masses, the specific energy has to be a function of just these two quantities. In the test-particle limit, we can set the mass-ratio to zero and the energy is a function of just of the radial position in Schwarzschild radii. When we cannot ignore the gravitational field generated by the second object, the energy will also depend on the mass-ratio.

A particular example of this, is that the energy of a particle orbiting at the innermost stable circular orbit (at $6M$) is always $\frac{2\sqrt{2}}{3}\approx0.94$ times the rest mass of the object, regardless of the mass of the Schwarzschild black hole.

Note that if the black hole is rotating, this introduces a new scale into the problem, the spin angular momentum of the BH, $S$, which has dimension $[\rm mass]^2$, and the energy will be a function of $\chi=S/M^2$ as well.

Improve this answer
$\endgroup$
3
  • $\begingroup$ “*When we cannot ignore the gravitational field generated by the second object, the energy will also depend on the mass-ratio.*$ - With no exact solution, where does this conclusion come from? Is it based on the results of numerical gravity, Newtonian limit, or mere intuition that “there is no other way”? $\endgroup$
    – safesphere
    Jul 26 at 14:17
  • $\begingroup$ @safesphere That is a standard result from gravitational self-force theory. $\endgroup$
    – TimRias
    Jul 26 at 15:03
  • $\begingroup$ I am sure you are right, but the self-force approximation is far too complex and unclear to make such conclusions without a specific reference. For example, is this a result of the first or second order calculations? Does this depend on how compact the objects are or on how close they are to each other? Also, the self-force theory typically ignores the gravitational radiation. Would this radiation break the mass ratio idea? Will this ratio work for both stellar and supermassive black holes or would the fact that the universe is not scale invariant break the ratio? +1 $\endgroup$
    – safesphere
    Jul 26 at 22:12
0
$\begingroup$

The total energy (kinetic, potential and rest) of a particle at $\rm r$ is

${\rm E}=g_{\rm tt} \ \rm \dot{t} = m c^2 \sqrt{1-r_s/r} \ / \sqrt{1-v^2/c^2}$

and the force acting on a stationary particle with $\rm v=0$

$\rm F=\partial E / \partial r=G M/r^2/\sqrt{1-r_s/r}$

At $\rm r=r_s$ we have $\rm E=0, \ F=\infty$

so you need an infinite force to stay stationary at $\rm r=r_s$,

let alone move to a higher $\rm r$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ (1) There is no potential energy in GR. Instead, mass is variable. Your $E=0$ at the horizon refers to the zero mass in the Schwarzschild coordinates while $m$ refers to the rest mass in the rest frame. (2) The OP is asking about moving between multiples of $r_s$, for example from $2r_s$ to $3r_s$ where nothing infinite is needed. Removing the reference to non-existing “potential energy” and considering multiples of $r_s$ would make your answer correct, but it is hardly helpful the way it is. $\endgroup$
    – safesphere
    Jul 26 at 14:11
  • $\begingroup$ @safesphere - the potential energy tells you how much kinetic energy you can gain in free fall, which is the total energy minus the current kinetic energy minus the rest energy. Whatever you want to call that quantity, the force required to stay stationary at the horizon would be infinite and the F=∂E/∂r is also standard, the 1/r² is only in the weak field limit. $\endgroup$
    – Yukterez
    Jul 26 at 15:57
  • $\begingroup$ I don’t think you got my point - there is no potential energy concept in GR at all. The full energy of a falling object is a sum of the kinetic energy and internal energy ($mc^2$ where $m$ is variable). The increase in the kinetic energy during the fall comes from the mass defect (similar to quantum mechanics), but not from “potential energy”. Per your own formula: $m= m_o \sqrt{1-r_s/r}$ where $m_o$ is the rest mass while the root represents the time dilation. The difference between the rest energy and internal energy is the mass defect, but not “potential energy”, which does not exist in GR. $\endgroup$
    – safesphere
    Jul 26 at 17:33
  • $\begingroup$ @safesphere - I already told you the definition: the total energy minus the current kinetic energy minus the rest energy, which tells you how much kinetic energy you can gain in free fall. If you don't understand that I can't help you, but for those interested in that numbers it's a useful tool. $\endgroup$
    – Yukterez
    Jul 26 at 19:02
  • 1
    $\begingroup$ The physical meaning here is super simple. The energy of a photon is its frequency and thus is in a direct proportion to the rate of time. The baryonic mass very roughly can be viewed as the energy of the (virtual) massless gluons. This energy, similarly to massless photons, is the (color)“frequency” of gluons, so the baryonic mass too is directly proportional to the rate of time. The nature of the interaction doesn’t matter. The point is that mass is energy and thus a process whose value is defined by the rate of time. So the mysterious Newtonian “potential” simply is the rate of time. $\endgroup$
    – safesphere
    Jul 27 at 0:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.