Derivations of Einstein field equations

In summary: They both give you a valid conversion factor of ##G / c^4##.In summary, the constant term in Einstein's field equations has a $$c^4$$ in the denominator to correct the units between the energy density of ##T_{\mu \nu}## and the curvature of ##G_{\mu \nu}##. This conversion factor of ##G / c^4## can also be written as ##8\pi G / c^4##, which is obtained from matching the prediction for static, weak-field gravity with Poisson's equation. Choosing to use different units, such as mass density instead of energy density, will result in a different conversion factor, such as ##G / c^2##.
  • #1
cr7einstein
87
2
Hello Everyone,
I have read many derivations of Einstein field equations (done one myself), but none of them explain why the constant term should have a $$c^4$$ in the denominator. the 8πG term can be obtained from Poisson's equation, but how does c^4 pop up? Most of the books just derive it with $$8\pi G$$, and say that in units where c is not equal to 1, you get $$8πG/c^4$$, even though there is no mention of an explicit assumption that c=1. They kind of just bring it up suddenly, and there is no prior need to assume c=1 anyway. I don't want to do it with Einstein-Hilbert action, but the standard $$G_{\mu\nu}=kT_{\mu\nu}$$ approach, where I need to show that $$k=\frac{8\pi G}{c^4}$$.
Thanks in advance!
 
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  • #2
cr7einstein said:
the 8πG term can be obtained from Poisson's equation, but how does c^4 pop up?

It's just correcting the units. Normally ##T_{\mu \nu}## is in units of energy density, and ##G_{\mu \nu}## is in units of curvature, i.e., inverse length squared. The conversion factor between these units is ##G / c^4##. (The ##8 \pi## part comes from matching the prediction for static, weak-field gravity with Poisson's equation, as you say.) If you choose to use different units, that will change the conversion factor; for example, if you use units of mass density instead of energy density for ##T_{\mu \nu}##, then the factor will be ##G / c^2##.
 
  • #3
So are both (energy density and mass density) equally valid?
 
  • #4
cr7einstein said:
So are both (energy density and mass density) equally valid?

Sure, it's just a matter of which you prefer, or which is more suitable for whatever problem you're working on.
 
  • #5


Thank you for bringing up this question. The constant term with $$c^4$$ in the denominator is a fundamental part of the Einstein field equations and is derived from the principles of general relativity. To understand why it is necessary, we need to look at the role of the speed of light in the theory of general relativity.

In general relativity, the speed of light is considered to be a fundamental constant, just like the gravitational constant (G). This means that the speed of light has the same value in all inertial frames of reference, regardless of their relative motion. This is a crucial aspect of the theory, as it allows us to understand how gravity affects the propagation of light and other electromagnetic waves.

Now, when we consider the Einstein field equations, we are essentially describing the relationship between the curvature of spacetime and the distribution of matter and energy. In other words, we are looking at how gravity is affected by the presence of matter and energy. And as we know from special relativity, matter and energy are related by the famous equation $$E=mc^2$$, where c is the speed of light.

Therefore, when we are trying to describe the effects of gravity on matter and energy, we need to consider the speed of light as a fundamental constant. This is why we see the constant term with $$c^4$$ in the denominator in the Einstein field equations. It is a direct consequence of the relationship between matter, energy, and the speed of light.

As for the assumption of c=1, it is simply a matter of convenience. In some cases, it is easier to work with units where c=1, as it simplifies the equations. However, the fundamental principles of general relativity still hold, regardless of the value of c used in the equations.

I hope this helps to clarify why the constant term with $$c^4$$ is necessary in the Einstein field equations. It is a crucial aspect of the theory and is derived from the fundamental principles of general relativity.
 

1. What are the Einstein field equations?

The Einstein field equations are a set of ten nonlinear partial differential equations that describe the gravitational field in general relativity. They were developed by Albert Einstein in 1915 as part of his theory of general relativity.

2. How are the Einstein field equations derived?

The Einstein field equations are derived from the Einstein-Hilbert action, which is a mathematical expression that describes the curvature of space-time. This action is then varied with respect to the metric tensor, resulting in the ten nonlinear partial differential equations that make up the field equations.

3. What is the significance of the Einstein field equations?

The Einstein field equations are significant because they provide a mathematical framework for understanding the relationship between matter and the curvature of space-time. They also allow for the prediction of various phenomena, such as gravitational waves and black holes.

4. How have the Einstein field equations been tested and confirmed?

The predictions of the Einstein field equations have been tested and confirmed through various experiments and observations, such as the bending of starlight by the sun and the detection of gravitational waves. These confirmations have solidified the validity of general relativity and the field equations.

5. Are there any limitations to the Einstein field equations?

While the Einstein field equations are incredibly accurate in describing the behavior of gravity, they do have limitations. They do not take into account quantum effects, and they break down in extreme conditions, such as at the singularity of a black hole. It is believed that a theory of quantum gravity would be needed to fully understand these situations.

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