The no hair theorem says that a black hole can have only three properties: mass, charge and angular momentum. But why don't we say that linear momentum is one of its properties? If we throw an object into it at a certain velocity, it's clear the linear momentum will increase. So why isn't it a fundamental property? Is it because its velocity and hence linear momentum depends on the reference frame? But then doesn't its angular momentum as well if the frame is non-inertial?

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    $\begingroup$ "The no hair theorem says that a black hole can have only three properties: mass, charge and angular momentum." - This is incorrect. See: en.wikipedia.org/wiki/No-hair_theorem - "stable black holes can be completely described (in a Cartesian coordinate system) at any moment in time by these eleven numbers: mass-energy, linear momentum (three components), angular momentum (three components, position (three components), electric charge." $\endgroup$
    – safesphere
    Nov 19, 2021 at 9:55
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    $\begingroup$ Recent related PBS Space Time video: Are black holes actually fuzzballs? $\endgroup$ Nov 20, 2021 at 12:26
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    $\begingroup$ The crux of your question lies in the last sentence: Is angular momentum frame-invariant just like linear momentum? i.e.: How can you distinguish a rotating sphere from a stationary one that has the world revolving around it in the opposite direction? I've asked this question before and I think the answer I got basically reduced to "try tossing a ball up", so to speak. It'll land on the same spot if and only if your sphere/planet isn't rotating. And the landing spot clearly doesn't depend on the reference frame. $\endgroup$
    – user541686
    Nov 21, 2021 at 9:31
  • $\begingroup$ Wait, linear momentum is not frame invariant right? Even in newtonian mechanics, would the angular momentum of the Earth from a geostationary satellite be zero? $\endgroup$ Nov 22, 2021 at 4:05

1 Answer 1


In relativity the covariant properties are tensors. The linear momentum you are referring to is a 3-vector and therefore is not covariant. In particular its magnitude is not a scalar invariant and therefore cannot be a fundamental property of a black hole.

In plainer terms the value of the momentum depends on the frame of reference of the observer. Observers moving at different velocities would observe the black hole to have different values of momentum. A fundamental property needs to have the same value for all observers.

The relativistic version of momentum is the four-momentum, and a black hole can have this property. However the associated invariant, i.e. the magnitude of the four-momentum, is just the rest mass of the black hole, and as you say the rest mass is one of the three fundamental properties a black hole can have.

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    $\begingroup$ This is more complicated than you (probably) think! The calculation of the angular momentum is done differently for the geostationary satellite and for an inertial observer because their spacetime metrics are different. The relevant covariant property is the four-vector version of angular momentum. I have to admit I'm not sure how to calculate the associated scalar invariant. $\endgroup$ Nov 19, 2021 at 7:15
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    $\begingroup$ Aha, the scalar invariant is the four-spin. In the rest frame of the Earth this is just the angular momentum (give or take a few factors). For the satellite the individual components of the four-spin would be different but the magnitude would be the same as calculated in the Earth rest frame. $\endgroup$ Nov 19, 2021 at 7:47
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    $\begingroup$ Very interesting that angular momentum becomes something frame invariant but linear momentum doesn't. I guess I have a lot of reading to do. $\endgroup$ Nov 19, 2021 at 7:50
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    $\begingroup$ @RohitPandey He said in the last paragraph of the answer that momentum is indeed associated with an invariant. You have to use relativistic version of linear momentum, that is, the four-momentum. The invariant quantity here is just the mass. $\endgroup$
    – Egg Man
    Nov 19, 2021 at 8:38
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    $\begingroup$ I see. In newtonian physics, the analogue of mass is the moment of inertia just as that of linear momentum is angular momentum and that's what's confusing me. $\endgroup$ Nov 19, 2021 at 10:30

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