Graduate Units of Kerr Understanding Mass & Angular Momentum

[swampwiz]

I was reading this paper, and I got confused:

https://projecteuclid.org/journals/...ws-of-black-hole-mechanics/cmp/1103858973.pdf

It discusses the Kerr solution for the case of { M⁴ > J² } where M is mass & J is angular momentum. However it seems that angular momentum should have the units { M L² T^-1 }, which would means that M is equivalent to { L² T^-1 }. I could see how M is equivalent to { L² T^-2 }.

What am I missing here?

 

[PeroK]

I was reading this paper, and I got confused:

https://projecteuclid.org/journals/...ws-of-black-hole-mechanics/cmp/1103858973.pdf

It discusses the Kerr solution for the case of { M⁴ > J² } where M is mass & J is angular momentum. However it seems that angular momentum should have the units { M L² T^-1 }, which would means that M is equivalent to { L² T^-1 }. I could see how M is equivalent to { L² T^-2 }.

What am I missing here?

I assume the paper is using natural units, where mass length and time all have the same dimension of length.

 

[swampwiz]

I assume the paper is using natural units, where mass length and time all have the same dimension of length.

OK, so it seems that you are saying that the constants c & G are to be used in order to get the units to match up?

 

[PeterDonis]

it seems that angular momentum should have the units { M L² T^-1 }, which would means that M is equivalent to { L² T^-1 }.

In the "geometric units" commonly used in GR, where $G = c = 1$, this is true, because mass $M$ has units of length (the conversion factor is $G / c^2$ in conventional units) and so does time $T$ (the conversion factor is just $c$ in conventional units). So angular momentum $J$ has units of $M L^2 T^{-1} = L L^2 L^{-1} = L^2$, i.e., the square of the unit of mass.