Space-Time curvature? the units?

[MoonAlex]

What would the units be on the curvature of spacetime? G(curvature)=8πGT/c^4

 

[CompuChip]

LMGTFY...

 

[phyzguy]

CompuChip's answer is not right. T in the Einstein equation refers to the stress-energy tensor, not temperature.

In fact, the units of curvature are 1/length^2. The metric tensor is dimensionless, and the curvature tensor, being the second derivative of the metric tensor, has units 1/L^2. T has units of energy density (M/(L*T^2)), and 8*pi*G/c^4 has units (T^2/(M*L)). Note that the cosmological constant also has units of 1/L^2.

 

[bcrowell]

There is no uniquely well defined answer to this question. You can do GR in units where c=1 and G=1, or units where c=1 but $G\ne 1$, or in SI units whether neither of these equals 1. Also, there is no requirement in GR that coordinates have any particular units. For instance, you could have spherical coordinates $(t,r,\theta,\phi)$, where t and r might have units of meters (with c=1), but the angles would be unitless. So a curvature component like $R_{tr}$ wouldn't have the same units as one like $R_{\theta\phi}$.

 

[MoonAlex]

Would you then take then distance² and multiply it by the value of the tensor and get the curverature in radians? Or how would you get the curverature using the tensor?

 

[bcrowell]

Would you then take then distance² and multiply it by the value of the tensor and get the curverature in radians? Or how would you get the curverature using the tensor?

Not sure what you mean by this. There are various measures of curvature, which are all tensors of some kind. In GR, the mother of all curvature tensors is the Riemann tensor, which is rank 4. From it, you can derive various rank-4, -2 and -0 (scalar) curvature tensors. All of these could have any units (or components with any mixture of units) you like, depending on how you choose your coordinates.