Can Newton's G and the "Einstein Scalar" Be Related?

[kent davidge]

Is it possible to take the "Eistein Scalar" from the Einstein Tensor, like one can take the Ricci Scalar from the Ricci Tensor? If so, is the G of Newton's law the same as this "Einstein Scalar" or is it just the same symbol used in very different things.

(Sorry for my bad English. I think Meaning of G would be a better title for the thread.)

 

[pervect]

Is it possible to take the "Eistein Scalar" from the Einstein Tensor, like one can take the Ricci Scalar from the Ricci Tensor? If so, is the G of Newton's law the same as this "Einstein Scalar" or is it just the same symbol used in very different things.

(Sorry for my bad English. I think Meaning of G would be a better title for the thread.)

The term isn't used much, but given the Ricci scalar $R = R^a{}_a$ , we can write the equivalent for the Einstein tensor, $G^a{}_a = -R^a{}_a$. See for instance http://math.ucr.edu/home/baez/gr/outline2.html.

In particular, note that $R = R^a{}_a = -T^a{}_a$, while $G^a{}_a = T^a{}_a$, thus $G^a{}_a = -R^a{}_a$

Because one is just the negative of the other, people usually talk about the Ricci scalar $R^a{}_a$ and there's no real need to talk about $G^a{}_a$.

Neither one is related to the G in Newton's law, which can be regarded as a unit conversion factor. Physically, the significance of R would be that if we choose a local orthonormal frame, $R = \rho - P_{xx} - P_{yy} - P_{zz}$ or in the case where the pressure is isotropic, $R = \rho - 3P$. Here $\rho$ is the density (energy per unit volume), and P is the pressure (or the x, y, and z components of the pressure if they're not all the same).

 

[kent davidge]

The term isn't used much, but given the Ricci scalar $R = R^a{}_a$ , we can write the equivalent for the Einstein tensor, $G^a{}_a = -R^a{}_a$. See for instance http://math.ucr.edu/home/baez/gr/outline2.html.

In particular, note that $R = R^a{}_a = -T^a{}_a$, while $G^a{}_a = T^a{}_a$, thus $G^a{}_a = -R^a{}_a$

Because one is just the negative of the other, people usually talk about the Ricci scalar $R^a{}_a$ and there's no real need to talk about $G^a{}_a$.

Neither one is related to the G in Newton's law, which can be regarded as a unit conversion factor. Physically, the significance of R would be that if we choose a local orthonormal frame, $R = \rho - P_{xx} - P_{yy} - P_{zz}$ or in the case where the pressure is isotropic, $R = \rho - 3P$. Here $\rho$ is the density (energy per unit volume), and P is the pressure (or the x, y, and z components of the pressure if they're not all the same).

Many thanks.

 

[pervect]

You're welcome - I should probably add that I followed Baez's lead in glossing over some constant conversions factors when I wrote $G_{aa} = T_{aa}$ rather than $G_{aa} = 8 \pi T_{aa}$ or even $G_{aa} = (8 \pi G / c^4) T_{aa}$.