# What does gamma tensor $\gamma_{\mu\nu}$ mean in General Relativity

I was recently going through Einstein's 'The Meaning of Relativity', where in chapter 4, he describes:

To the second approximation we must then put $$g_{\mu\nu} = -\delta_{\mu\nu} + \gamma_{\mu\nu}$$ where $$\gamma_{\mu\nu}$$ are to be regarded as small of the first order.

At first I thought that $$\gamma_{\mu\nu}$$ is just any other addition to the metric, but as I read along, I found that $$\gamma_{\mu\nu}$$ is used plenty of times, even while discussing the cosmological problem. Einstein, at one point also says that $$\frac{\gamma_{44}}{2}$$ can be identified as the gravitational potential.

My question is this: Is there a physical and/or mathematical meaning for $$\gamma_{\mu\nu}$$. And what is it's connection with the metric, and how can it be derived. Basically I want to know what $$\gamma_{\mu\nu}$$ means in the real world and how it is obtained mathematically. And I think mathematical rigour would be preferable for me to grasp the concept more fundamentally.

• In this context the gamma is a small disturbance to an otherwise flat space-time. He is applying a weak field approximation to simplify, linearize, the equations of GR. It is the same in all contexts.
– user196418
Apr 18, 2020 at 13:52
• Can you mathematically elaborate on that? Thanks
– PNS
Apr 18, 2020 at 13:53
• I don't know how much more can be added to Einstein's statement above. He has a metric tensor, that provides a local measure of distance and time intervals, he is then assuming that this tensor is described by the standard flat metric + a small deviation. The rest is plug and chug. Are you asking for elaboration on the meaning of a metric, the process of perturbation theory, or something else. It is not a fair question to ask, without more context.
– user196418
Apr 18, 2020 at 13:57
• Gamma is not "derived" it is part of the metric.
– user196418
Apr 18, 2020 at 13:57
• But what does gamma mean physically i.e. how does it account in the curvature of space-time?
– PNS
Apr 18, 2020 at 13:59

## 1 Answer

Since the universe (at any length scale) is not very symmetric, in almost all physical situations, we do not know the spacetime geometry exactly. What this means is that one can never analytically compute $$g_{\mu \nu}$$ for a given region of spacetime. Thankfully, we don't need to, since many phenomena in the universe can be described using a geometry that is approximately $$g_{\mu \nu}$$. This means that we can consider a 'small' deviation $$\gamma_{\mu \nu}$$ from a known and highly symmetric geometry $$g_{\mu \nu}$$, to describe physics to a good approximation.

One example is gravitational waves traveling large distances to reach the earth from some binary black hole collision. These waves have essentially been traveling in empty space (or flat space, approximately) for a long time. We can therefore fix $$g_{\mu \nu}$$ as flat space (= $$\eta_{\mu \nu}$$), and $$\gamma_{\mu \nu}$$ as gravitational waves traveling on this flat space.

As another example, consider a non-rotating black hole. It can be described exactly by a Schwarzschild metric. But this is a perfect scenario. In real life, black holes interact with stars, matter, and other black holes around them. So there is to be some deviation from pure Schwarzschild geometry in such dynamical processes. We then fix $$g_{\mu \nu}$$ as the Schwarzschild metric, and $$\gamma_{\mu \nu}$$ as some deviation around it. One then substitutes the approximation of the full metric $$g_{\mu \nu} + \gamma_{\mu \nu}$$ in Einstein's equations and obtains equations for $$\gamma_{\mu \nu}$$. These describe how exactly the black hole changes its shape from pure Schwarzschild geometry. This is just an example - in real life, astrophysical black holes are always rotating (because of angular momentum conservation) so you would take the Kerr metric instead of Schwarschild.

• Thanks for the explanation!
– PNS
Apr 25, 2020 at 15:05