# LOG#105. Einstein’s equations.

In 1905,  one of Einstein’s achievements was to establish the theory of Special Relativity from 2 single postulates and correctly deduce their physical consequences (some of them time later).  The essence of Special Relativity, as we have seen, is that  all the inertial observers must agree on the speed of light “in vacuum”, and that the physical laws (those from Mechanics and Electromagnetism) are the same for all of them.  Different observers will measure (and then they see) different wavelengths and frequencies, but the product of wavelength with the frequency is the same.  The wavelength and frequency are thus Lorentz covariant, meaning that they change for different observers according some fixed mathematical prescription depending on its tensorial character (scalar, vector, tensor,…) respect to Lorentz transformations.  The speed of light is Lorentz invariant.

By the other hand, Newton’s law of gravity describes the motion of planets and terrrestrial bodies.  It is all that we need in contemporary rocket ships unless those devices also carry atomic clocks or other tools of exceptional accuracy.  Here is Newton’s law in potential form:

$4\pi G\rho = \nabla ^2 \phi$

In the special relativity framework, this equation has a terrible problem: if there is a change in the mass density $\rho$, then it must propagate everywhere instantaneously.  If you believe in the Special Relativity rules and in the speed of light invariance, it is impossible. Therefore, “Houston, we have a problem”.

Einstein was aware of it and he tried to solve this inconsistency.  The final solution took him ten years .

The apparent silly and easy problem is to develop and describe all physics in the the same way irrespectively one is accelerating or not. However, it is not easy or silly at all. It requires deep physical insight and a high-end mathematical language.  Indeed,  what is the most difficult part are  the details of Riemann geometry and tensor calculus on manifolds.  Einstein got  private aid from a friend called  Marcel Grossmann. In fact, Einstein knew that SR was not compatible with Newton’s law of gravity. He (re)discovered the equivalence principle, stated by Galileo himself much before than him, but he interpreted deeper and seeked the proper language to incorporante that principle in such a way it were compatible (at least locally) with special relativity! His  “journey” from 1907 to 1915 was a hard job and a continuous struggle with tensorial methods…

Today, we are going to derive the Einstein field equations for gravity, a set of equations for the “metric field” $g_{\mu \nu}(x)$. Hilbert in fact arrived at Einstein’s field equations with the use of the variational method we are going to use here, but Einstein’s methods were more physical and based on physical intuitions. They are in fact “complementary” approaches. I urge you to read “The meaning of Relativity” by A.Einstein in order to read a summary of his discoveries.

We now proceed to derive Einstein’s Field Equations (EFE) for General Relativity (more properly, a relativistic theory of gravity):

Step 1. Let us begin with the so-called Einstein-Hilbert action (an ansatz).

$S = \int d^4x \sqrt{-g} \left( \dfrac{c^4}{16 \pi G} R + \mathcal{L}_{\mathcal{M}} \right)$

Be aware of  the square root of the determinant of the metric as part of the volume element.  It is important since the volume element has to be invariant in curved spacetime (i.e.,in the presence of a metric).  It also plays a critical role in the derivation.

Step 2. We perform the variational variation with respect to the metric field $g^{\mu \nu}$:

$\delta S = \int d^4 x \left( \dfrac{c^4}{16 \pi G} \dfrac{\delta \sqrt{-g}}{\delta g^{\mu \nu}} + \dfrac{\delta (\sqrt{-g}\mathcal{L}_{\mathcal{M}})}{\delta g^{\mu \nu}} \right) \delta g^{\mu \nu}$

Step 3. Extract out  the square root of the metric as a common factor and use the product rule on the term with the Ricci scalar R:

$\delta S = \int d^4 x \sqrt{-g} \left( \dfrac{c^4}{16 \pi G} \left ( \dfrac{\delta R}{\delta g^{\mu \nu}} +\dfrac{R}{\sqrt{-g}}\dfrac{\delta \sqrt{-g}}{\delta g^{\mu \nu}} \right) +\dfrac{1}{\sqrt{-g}}\dfrac{\delta ( \sqrt{-g}\mathcal{L}_{\mathcal{M}})}{\delta g^{\mu\nu}}\right) \delta g^{\mu \nu}$

Step 4.  Use the definition of a Ricci scalar as a contraction of the Ricci tensor to calculate the first term:

$\dfrac{\delta R}{\delta g^{\mu \nu}} = \dfrac{\delta (g^{\mu \nu}R_{\mu \nu})}{\delta g^{\mu \nu} }= R_{\mu\nu} + g^{\mu \nu}\dfrac{\delta R_{\mu \nu}}{\delta g^{\mu \nu}} = R_{\mu \nu} + \mbox{total derivative}$

A total derivative does not make a contribution to the variation of the action principle, so can be neglected to find the extremal point.  Indeed, this is the Stokes theorem in action. To show that the variation in the Ricci tensor is a total derivative, in case you don’t believe this fact, we can proceed as follows:

Check 1. Write  the Riemann curvature tensor:

$R^{\rho}_{\, \sigma \mu \nu} = \partial _{\mu} \Gamma ^{\rho}_{\, \sigma \nu} - \partial_{\nu} \Gamma^{\rho}_{\, \sigma \mu}+ \Gamma^{\rho}_{\, \lambda \mu} \Gamma^{\lambda}_{\, \sigma \nu} - \Gamma^{\rho}_{\, \lambda \nu} \Gamma^{\lambda}_{\, \sigma \mu}$

Note the striking resemblance with the non-abelian YM field strength curvature two-form

$F=dA+A \wedge A = \partial _{\mu} A_{\nu} - \partial _{\nu} A_{\mu} + k \left[ A_\mu , A_{\nu} \right]$.

There are many terms with indices in the Riemann tensor calculation, but we can simplify stuff.

Check 2. We have to calculate the variation of the Riemann curvature tensor with respect to the metric tensor:

$\delta R^{\rho}_{\, \sigma \mu \nu} = \partial _{\mu} \delta \Gamma^{\rho}_{\, \sigma \nu} - \partial_\nu \delta \Gamma^{\rho}_{\, \sigma \mu} + \delta \Gamma ^{\rho}_{\, \lambda \mu} \Gamma^{\lambda}_{\, \sigma \nu} - \delta \Gamma^{\rho}_{\lambda \nu}\Gamma^{\lambda}_{\, \sigma \mu} + \Gamma^{\rho}_{\, \lambda \mu}\delta \Gamma^{\lambda}_{\sigma \nu} - \Gamma^{\rho}_{\lambda \nu} \delta \Gamma^{\lambda}_{\, \sigma \mu}$

One cannot calculate the covariant derivative of a connection since it does not transform like a tensor.  However, the difference of two connections does transform like a tensor.

Check 3. Calculate the covariant derivative of the variation of the connection:

$\nabla_{\mu} ( \delta \Gamma^{\rho}_{\sigma \nu}) = \partial _{\mu} (\delta \Gamma^{\rho}_{\, \sigma \nu}) + \Gamma^{\rho}_{\, \lambda \mu} \delta \Gamma^{\lambda}_{\, \sigma \nu} - \delta \Gamma^{\rho}_{\, \lambda \sigma}\Gamma^{\lambda}_{\mu \nu} - \delta \Gamma^{\rho}_{\, \lambda \nu}\Gamma^{\lambda}_{\, \sigma \mu}$

$\nabla_{\nu} ( \delta \Gamma^{\rho}_{\sigma \mu}) = \partial _\nu (\delta \Gamma^{\rho}_{\, \sigma \mu}) + \Gamma^{\rho}_{\, \lambda \nu} \delta \Gamma^{\lambda}_{\, \sigma \mu} - \delta \Gamma^{\rho}_{\, \lambda \sigma}\Gamma^{\lambda}_{\mu \nu} - \delta \Gamma^{\rho}_{\, \lambda \mu}\Gamma^{\lambda}_{\, \sigma \nu}$

Check 4. Rewrite the variation of the Riemann curvature tensor as the difference of two covariant derivatives of the variation of the connection written in Check 3, that is, substract the previous two terms in check 3.

$\delta R^{\rho}_{\, \sigma \mu \nu} = \nabla_{\mu} \left( \delta \Gamma^{\rho}_{\, \sigma \nu}\right) - \nabla _{\nu} \left(\delta \Gamma^{\rho}_{\, \sigma \mu}\right)$

Check 5. Contract the result of Check 4.

$\delta R^{\rho}_{\, \mu \rho \nu} = \delta R_{\mu \nu} = \nabla_{\rho} \left( \delta \Gamma^{\rho}_{\, \mu \nu}\right) - \nabla _{\nu} \left(\delta \Gamma^{\rho}_{\, \rho \mu}\right)$

Check 6. Contract the result of Check 5:

$g^{\mu \nu}\delta R_{\mu \nu} = \nabla_\rho (g^{\mu \nu} \delta \Gamma^{\rho}_{\mu\nu})-\nabla_\nu (g^{\mu \nu}\delta \Gamma^{\rho}_{\rho \mu}) = \nabla _\sigma (g^{\mu \nu}\delta \Gamma^{\sigma}_{\mu \nu}) - \nabla_\sigma (g^{\mu \sigma}\delta \Gamma ^{\rho}_{\rho \mu})$

Therefore, we have

$g^{\mu \nu}\delta R_{\mu \nu} = \nabla_\sigma (g^{\mu \nu}\delta \Gamma^{\sigma}_{\mu\nu}- g^{\mu \sigma}\delta \Gamma^{\rho}_{\rho\mu})=\nabla_\sigma K^\sigma$

Q.E.D.

Step 5. The variation of the second term in the action is the next step.  Transform the coordinate system to one where the metric is diagonal and use the product rule:

$\dfrac{R}{\sqrt{-g}} \dfrac{\delta \sqrt{-g}}{\delta g^{\mu \nu}}=\dfrac{R}{\sqrt{-g}} \dfrac{-1}{2 \sqrt{-g}}(-1) g g_{\mu \nu}\dfrac{\delta g^{\mu \nu}}{\delta g^{\mu \nu}} =- \dfrac{1}{2}g_{\mu \nu} R$

The reason of the last equalities is that $g^{\alpha\mu}g_{\mu \beta}=\delta^{\alpha}_{\; \beta}$, and then its variation is

$\delta (g^{\alpha\mu}g_{\mu \nu}) = (\delta g^{\alpha\mu}) g_{\mu \nu} + g^{\alpha\mu}(\delta g_{\mu \nu}) = 0$

Thus, multiplication by the inverse metric $g^{\beta \nu}$ produces

$\delta g^{\alpha \beta} = - g^{\alpha \mu}g^{\beta \nu}\delta g_{\mu \nu}$

that is,

$\dfrac{\delta g^{\alpha \beta}}{\delta g_{\mu \nu}}= -g^{\alpha \mu} g^{\beta \nu}$

By the other hand, using the theorem for the derivation of a determinant we get that:

$\delta g = \delta g_{\mu \nu} g g^{\mu \nu}$

since

$\dfrac{\delta g}{\delta g^{\alpha \beta}}= g g^{\alpha \beta}$

because of the classical identity

$g^{\alpha \beta}=(g_{\alpha \beta})^{-1}=\left( \det g \right)^{-1} Cof (g)$

Indeed

$Cof (g) = \dfrac{\delta g}{\delta g^{\alpha \beta}}$

and moreover

$\delta \sqrt{-g}=-\dfrac{\delta g}{2 \sqrt{-g}}= -g\dfrac{ \delta g_{\mu \nu} g^{\mu \nu}}{2 \sqrt{-g}}$

so

$\delta \sqrt{-g}=\dfrac{1}{2}\sqrt{-g}g^{\mu \nu}\delta g_{\mu \nu}=\dfrac{1}{2}\sqrt{-g}g_{\mu \nu}\delta g^{\mu \nu}$

Q.E.D.

Step 6. Define the stress energy-momentum tensor as the third term in the action (that coming from the matter lagrangian):

$T_{\mu \nu} = - \dfrac{2}{\sqrt{-g}}\dfrac{(\sqrt{-g} \mathcal{L}_{\mathcal{M}})}{\delta g^{\mu \nu}}$

or equivalently

$-\dfrac{1}{2}T_{\mu \nu} = \dfrac{1}{\sqrt{-g}}\dfrac{(\sqrt{-g} \mathcal{L}_{\mathcal{M}})}{\delta g^{\mu \nu}}$

Step 7. The extremal principle. The variation of the Hilbert action will be  an extremum when the integrand is equal to zero:

$\dfrac{c^4}{16\pi G}\left( R_{\mu \nu} - \dfrac{1}{2} g_{\mu \nu}R\right) - \dfrac{1}{2} T_{\mu \nu} = 0$

i.e.,

$\boxed{R_{\mu \nu} - \dfrac{1}{2}g_{\mu \nu} R = \dfrac{8\pi G}{c^4}T_{\mu\nu}}$

Usually this is recasted and simplified using the Einstein’s tensor

$G_{\mu \nu}= R_{\mu \nu} - \dfrac{1}{2}g_{\mu \nu} R$

as

$\boxed{G_{\mu\nu}=\dfrac{8\pi G}{c^4}T_{\mu\nu}}$

This deduction has been mathematical. But there is a deep physical picture behind it. Moreover,  there are a huge number of physics issues one could go into. For instance, these equations bind to particles with integral spin which is good for bosons, but there are matter fermions that also participate in gravity coupling to it. Gravity is universal.  To include those fermion fields, one can consider the metric and the connection to be independent of each other.  That is the so-called Palatini approach.

Final remark: you can add to the EFE above a “constant” times the metric tensor, since its “covariant derivative” vanishes. This constant is the cosmological constant (a.k.a. dark energy in conteporary physics). The, the most general form of EFE is:

$\boxed{G_{\mu\nu}+\Lambda g_{\mu\nu}=\dfrac{8\pi G}{c^4}T_{\mu\nu}}$

Einstein’s additional term was added in order to make the Universe “static”. After Hubble’s discovery of the expansion of the Universe, Einstein blamed himself about the introduction of such a term, since it avoided to predict the expanding Universe. However, perhaps irocanilly, in 1998 we discovered that the Universe was accelerating instead of being decelerating due to gravity, and the most simple way to understand that phenomenon is with a positive cosmological constant domining the current era in the Universe. Fascinating, and more and more due to the WMAP/Planck data. The cosmological constant/dark energy and the dark matter we seem to “observe” can not be explained with the fields of the Standard Model, and therefore…They hint to new physics. The character of this  new physics is challenging, and much work is being done in order to find some particle of model in which dark matter and dark energy fit. However, it is not easy at all!

May the Einstein’s Field Equations be with you!

# LOG#036. Action and relativity.

The hamiltonian formalism and the hamiltonian H in special relativity has some issues with the definition. In the case of the free particle one possible definition, not completely covariant, is the relativistic energy

$\boxed{E=\sqrt{m^2c^4+c^2p^2}=H}$

There are two others interesting scalars in classical relativistic theories. They are the lagrangian L and the action functional S. The lagrangian is obtained through a Legendre transformation from the hamiltonian:

$\boxed{L=pv-H}$

From the hamiltonian, we get the velocity using the so-called hamiltonian equation:

$\dot{\mathbf{q}}=\mathbf{v}=\dfrac{\partial H}{\partial \mathbf{p}}=c^2\dfrac{\mathbf{p}}{E}$

Then,

$L=\dfrac{E}{c^2}\mathbf{v}^2-E=E\left(\dfrac{v^2}{c^2}-1\right)=-\dfrac{E}{\gamma^2}=-\dfrac{m\gamma c^2}{\gamma^2}=-\dfrac{mc^2}{\gamma}$

and finally

$\boxed{L=-mc^2\sqrt{1-\dfrac{v^2}{c^2}}=-\dfrac{mc^2}{\gamma}=-mc\sqrt{-\dot{X}^2}}$

The action functional is the time integral of the lagrangian:

$\boxed{S=\int Ldt}$

However, let me point out that the above hamiltonian in SR has some difficulties in gauge field theories. Indeed, it is quite easy to derive that a more careful and reasonable election for the hamiltonian in SR should be zero!

In the case of the free relativistic particle, we obtain

$S=-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt$

Using the relation between time and proper time (the time dilation formula):

$dt=\gamma d\tau\rightarrow \dfrac{dt}{\gamma}=d\tau$

direct substitution provides

$-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt=-mc^2\int d\tau$

And defining the infinitesimal proper length in spacetime as $ds=cd\tau$, we get the simple and wonderful result:

$\boxed{S=-mc\int ds}$

Sometimes, the covariant lagrangian for the free particle is also obtained from the following argument. The proper length is defined as

$ds^2=d\mathbf{x}^2-c^2dt^2$

The invariant in spacetime is related with the proper time in this way:

$ds^2=-c^2d\tau^2=d\mathbf{x}^2-c^2dt^2$

Thus, dividing by $dt^2$

$-c^2\dfrac{d\tau^2}{dt^2}=\mathbf{v}^2-c^2$

and

$d\tau^2=\gamma^{-2}dt^2=\dfrac{1}{\gamma^2}dt^2=\left(1-\dfrac{\mathbf{v}^2}{c^2}\right)dt^2$

so

$d\tau=\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}dt$

$cd\tau=ds=\sqrt{c^2-\mathbf{v}^2}dt=\sqrt{-\dot{X}^2}dt$

that is

$\boxed{ds=cd\tau=\sqrt{-\dot{X}^2}dt=\sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=\sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}$

and the free coordinate action for the free particle would be:

$\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}dt=-mc^2\int \sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}$

Note, that since the election of time “t” is “free”, we can choose $t=\tau$ to obtain the generally covariant free action:

$\boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}d\tau=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}d\tau}$

Remark: the (rest) mass is the “coupling” constant for the free particle proper lenght to guess the free lagrangian

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

Now, we can see from this covariant action that the relativistic hamiltonian should be a feynmanity! From the equations of motion,

$P_\mu=\dfrac{\partial L}{\partial \dot{X}^\mu}=mc\dfrac{\dot{X}_\mu}{\sqrt{-\dot{X}^2}}$

The covariant hamiltonian $\mathcal{H}$, different from H, can be build in the following way:

$\mathcal{H}=P_\mu \dot{X}^\mu-L=mc\dfrac{\dot{X}_\mu \dot{X}^\mu}{\sqrt{-\dot{X}^2}}-mc\sqrt{-\dot{X}^2}=0$

The meaning of this result is hidden in the the next identity ( Noether identity or “hamiltonian constraint” in some contexts):

$\boxed{\mathcal{H}=P_\mu P^\mu+m^2c^2=0}$

since

$P_\mu P^\mu=m U_\mu mU^\mu=-m^2c^2$

This strange fact that $\mathcal{H}=0$ in SR, a feynmanity as the hamiltonian, is related to the Noether identity $E^\mu \dot{X}_\mu$ for the free relativistic lagrangian, indeed, a consequence of the hamiltonian constraint and the so-called reparametrization invariance $\tau'=f (\tau)$. Note, in addition, that the free relativistic particle would also be invariant under diffeomorphisms $x^{\mu'}= f^\mu (x)=f^\mu (x^\nu)$ if we were to make the metric space-time dependent, i.e., if we make the substitution $\eta_{\mu\nu}\rightarrow g_{\mu\nu} (x)$. This last result is useful and important in general relativity, but we will not discuss it further in this moment. In summary, from the two possible hamiltonian in special relativity

$H=E=\sqrt{\mathbf{p}^2c^2+(mc^2)^2}$

$\mathcal{H}=P_\mu P^\mu+m^2c^2=0$

the natural and more elegant (due to covariance/invariance) is the second one. Moreover, the free particle lagrangian and action are:

$\boxed{L=-mc^2\sqrt{-\dot{X}^2}}$

$\boxed{S=-mc^2\int d\tau=-mc\int ds=\int L dt}$

Remark: The true covariant lagrangian dynamics in SR is a “constrained” dynamics, i.e., dynamics where we are undetermined. There are more variables that equations as a result of a large set of symmetries ( reparametrization invariance and, in the case of local metrics, we also find diffeomorphism invarince).

The dynamical equations of motion, for a first order lagrangian (e.g., the free particle we have studied here), read for the lagrangian formalism:

$\boxed{\delta S=\delta \int L (q,\dot{q};t)dt =0\leftrightarrow\begin{cases}E(L)=0,\mbox{with E(L)=Euler operator}\\ E(L)=\dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{q}}\right)=0\end{cases}}$

By the other hand, for the hamiltonian formalism, dynamical equations are:

$\boxed{\delta S=\delta \int H (q, p; t)dt =0\leftrightarrow\begin{cases}p=\dfrac{\partial L}{\partial \dot{q}},\;\mbox{with}\; \det\left(\dfrac{\partial ^2L}{\partial \dot{q}^i\partial \dot{q}^j}\right)\neq 0\\ \;\\ \dfrac{dq}{dt}=\dot{q}=\dfrac{\partial H}{\partial p}\\ \;\\\dfrac{dp}{dt}=\dot{p}=-\dfrac{\partial H}{\partial q} \\ \;\\ \dfrac{dH}{dt}=\dot{H}=\dfrac{\partial H}{\partial t}=-\dfrac{\partial L}{\partial t}\end{cases}}$