# Units of Kerr Understanding Mass & Angular Momentum

• A
• swampwiz

#### 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?

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.

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?

• malawi_glenn
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.

• malawi_glenn