Deriving the Reissner-Nordström Metric

Hello! Welcome back to our journey through the solutions of Einstein's equations.

In the last few lessons, we've thoroughly dissected the Schwarzschild metric, the unique solution describing a static, spherically symmetric, uncharged body in a vacuum. We even mapped its complete, complex causal structure using the powerful tool of the Penrose diagram.

Today, we'll take a step closer to physical reality by adding another fundamental property to our central object: electric charge. Our goal is to derive the Reissner-Nordström metric for a charged, non-rotating black hole. This requires us to move beyond the vacuum Einstein equations ( $G_{μν} = 0$ ) and solve the coupled Einstein-Maxwell equations, where the electromagnetic field acts as a source for spacetime curvature, and spacetime curvature, in turn, dictates how the electromagnetic field propagates.

This derivation will showcase how the principles we've learned can be extended to include matter and energy, and it will set the stage for discovering an even more intricate internal structure than the one we found in the Schwarzschild case.

1. The Einstein-Maxwell System

To find the metric for a charged mass, we need to solve two sets of fundamental equations simultaneously:

Einstein's Field Equations (EFE): The geometry of spacetime is determined by the distribution of energy and momentum. $G_{μν} \equiv R_{μν} - \frac{1}{2} R g_{μν} = \frac{8 π G}{c ^{4}} T_{μν}$
Maxwell's Equations in Curved Spacetime: The electromagnetic field propagates according to Maxwell's laws, now written in a form that is valid in any coordinate system on a curved manifold. $\nabla_{β} F^{α β} = μ_{0} j^{α} and \nabla_{[α} F_{β γ]} = 0$

The crucial link between these is the stress-energy tensor for the electromagnetic field, $T_{μν}$ , which describes the energy, momentum, and stress contributed by the field. Since we are looking for a solution in the space surrounding a central charge, the current density $j^{α}$ is zero.

The Reissner-Nordström metric

To begin, let's formally define the equations we need to solve. The following sections from the paper 'The Reissner-Nordström metric' introduce the electromagnetic stress-energy tensor and state the full system of Einstein-Maxwell equations in the context of our problem.

Please read section 4.3, 'Electrodynamics in General Relativity' (page 21), to see the generalization of Maxwell's equations and the formal definition of the EM stress-energy tensor. Then, glance at the beginning of Chapter 5 (pages 25-26) to see how these equations are set up for the specific problem of a spherically symmetric, charged body. Pay close attention to equation (5.1) for the stress-energy tensor and the simplified form of the EFE when the tensor is traceless.

A key property you just read about is that the electromagnetic stress-energy tensor is traceless, i.e., $T = T_{μ}^{μ} = 0$ . This provides a significant simplification. By taking the trace of the EFE, $G = κ T$ , we find that $R = - κ T$ . If $T = 0$ , then the Ricci scalar $R = 0$ . Substituting this back into the EFE gives the 'trace-reversed' form you saw in the paper:

R_{μν} = \frac{8 π G}{c ^{4}} T_{μν}

These are the specific equations we need to solve.

2. The Derivation: A Step-by-Step Guide

The derivation of the Reissner-Nordström metric follows a path very similar to our derivation of the Schwarzschild metric, with the key difference being the non-zero right-hand side of the field equations.

The following video provides an excellent, clear walkthrough of the entire calculation. We will use it as our primary guide. The logic is elegant and builds directly on our previous work.

Reissner-Nordstrom Metric Derivation | Solving The Maxwell-Einstein Equations

This video by Dietterich Labs meticulously derives the Reissner-Nordström metric from the Einstein-Maxwell equations. It's a comprehensive guide that we will follow closely.

Watch the entire video (approx. 15 minutes). I will break down the key steps and add commentary below, so you can pause and reflect as you go. Pay attention to how prior results from the Schwarzschild derivation are leveraged to simplify the process.

Let's break down the derivation you just watched into its logical components.

Step 1: Assumptions and the Metric Ansatz (Video 0:00 - 3:00)

Just as with Schwarzschild, we assume a static and spherically symmetric spacetime. This allows us to use the exact same metric ansatz:

d s^{2} = A (r) c^{2} d t^{2} - B (r) d r^{2} - r^{2} d Ω^{2}

where $d Ω^{2} = d θ^{2} + sin^{2} θ d ϕ^{2}$ .

Crucially, because the form of the metric is identical to the Schwarzschild case, the formulas for the Christoffel symbols and the Ricci tensor components ( $R_{μν}$ ) in terms of the unknown functions $A (r)$ and $B (r)$ are also identical. This is a powerful shortcut that saves us from re-deriving them from scratch.

Step 2: The Electromagnetic Field (Video 3:00 - 4:43 & 7:57 - 9:16)

The assumption of spherical symmetry also constrains the electromagnetic field. The electric field must be purely radial ( $E = E (r) r^$ ), and there can be no magnetic field. As the video argues, the form of this field is not altered by the spherically symmetric gravity, so it must be the familiar Coulomb field from a point charge $Q$ .

This corresponds to a four-potential $A_{μ} = (ϕ (r), 0, 0, 0)$ with $ϕ (r) = - Q / (4 π ϵ_{0} r)$ (in SI units), and a Faraday tensor $F_{μν} = \partial_{μ} A_{ν} - \partial_{ν} A_{μ}$ with the only non-zero components being $F_{t r} = - F_{r t} = \partial_{r} ϕ (r)$ . The video shows how to compute all the components of $F_{μν}$ from this potential.

Step 3: The Stress-Energy Tensor (Video 9:16 - 10:41)

With the Faraday tensor $F_{μν}$ and the metric $g_{μν}$ in hand, we can now compute the components of the stress-energy tensor:

T_{μν} = \frac{1}{μ _{0}} (F_{μα} F_{ν}^{α} - \frac{1}{4} g_{μν} F_{α β} F^{α β})

The video walks through this calculation, finding the non-zero components $T_{tt}$ , $T_{r r}$ , $T_{θ θ}$ , and $T_{ϕϕ}$ .

Test your understanding!

Using the diagonal metric ansatz and the components of the stress-energy tensor shown in the video at 10:20 (or by calculating them yourself), explicitly verify that the trace $T = g^{μν} T_{μν}$ is zero.

Hint: Remember that $g^{tt} = 1/ g_{tt}$ , $g^{r r} = 1/ g_{r r}$ , etc.

Show answer

The trace is $T = g^{μν} T_{μν} = g^{tt} T_{tt} + g^{r r} T_{r r} + g^{θ θ} T_{θ θ} + g^{ϕϕ} T_{ϕϕ}$ .
The diagonal components of $T_{μν}$ are proportional to the metric components themselves. Let's use the expressions from the Diva Portal paper (Eq. 5.9), which are equivalent to the video's result, setting $ϵ = - \frac{1}{2 μ _{0}} F_{01} F^{01}$ :
$T_{00} = ϵ g_{00}$ , $T_{11} = ϵ g_{11}$ , $T_{22} = - ϵ g_{22}$ , $T_{33} = - ϵ g_{33}$ .
(Note: The video gets a sign flip because its $T_{00}$ has a $+ 1/2$ and others a $- 1/2$ in the final form, but the principle is the same).
Let's use the video's matrix form at 10:20. Let $C = \frac{Q ^{2}}{32 π ^{2} ϵ _{0}^{2} r ^{4}}$ .
Then $T_{tt} = C \cdot A$ , $T_{r r} = C \cdot B$ , $T_{θ θ} = - C \cdot r^{2}$ , $T_{ϕϕ} = - C \cdot r^{2} sin^{2} θ$ .
The trace is:

T = \frac{1}{A} (C \cdot A) + \frac{1}{- B} (C \cdot B) + \frac{1}{- r ^{2}} (- C \cdot r^{2}) + \frac{1}{- r ^{2} sin ^{2} θ} (- C \cdot r^{2} sin^{2} θ)

T = C - C + C + C = 2 C \neq = 0

Wait, something is wrong in my reasoning or the video's presentation. Let's recheck the formula for T.
$T_{β}^{α} = \frac{1}{μ _{0}} (F^{α γ} F_{β γ} - \frac{1}{4} δ_{β}^{α} F_{γ δ} F^{γ δ})$ .
The trace is $T_{α}^{α} = \frac{1}{μ _{0}} (F^{α γ} F_{α γ} - \frac{1}{4} δ_{α}^{α} F_{γ δ} F^{γ δ})$ .
Since $δ_{α}^{α} = 4$ , this is $T_{α}^{α} = \frac{1}{μ _{0}} (F^{α γ} F_{α γ} - F_{γ δ} F^{γ δ}) = 0$ .
The property is fundamental to 4D Maxwell theory. The video correctly states it at 6:45. The matrix components he writes at 10:20 must be for the covariant tensor $T_{μν}$ . The calculation of the trace requires $g^{μν} T_{μν}$ . Let's re-do it carefully.
From the video at 9:30, $F_{α β} F^{α β} = - \frac{Q ^{2}}{8 π ^{2} ϵ _{0}^{2} c ^{2} r ^{4}}$ . Let's call this scalar $S$ .
Then $T_{μν} = \frac{1}{μ _{0}} (F_{μα} F_{ν}^{α} - \frac{1}{4} g_{μν} S)$ .
Let's check $T_{tt}$ : $T_{00} = \frac{1}{μ _{0}} (F_{0 α} F_{0}^{α} - \frac{1}{4} g_{00} S)$ . The only non-zero $F_{0 α}$ is $F_{01}$ . So $F_{0 α} F_{0}^{α} = F_{01} F_{0}^{1} = F_{01} g^{1 β} F_{0 β} = F_{01} g^{11} F_{01} = (F_{01})^{2} / g_{11}$ .
The components derived in the Diva Portal paper are correct. Let's use those.
They show $T_{00} = - \frac{1}{2 μ _{0}} g_{00} F_{01} F^{01}$ and $T_{11} = - \frac{1}{2 μ _{0}} g_{11} F_{01} F^{01}$ , while $T_{22} = \frac{1}{2 μ _{0}} g_{22} F_{01} F^{01}$ .
Let $K = - \frac{1}{2 μ _{0}} F_{01} F^{01}$ . This is a positive scalar.
Then $T_{00} = K g_{00}$ , $T_{11} = K g_{11}$ , $T_{22} = - K g_{22}$ , $T_{33} = - K g_{33}$ .
The trace is $T = g^{00} T_{00} + g^{11} T_{11} + g^{22} T_{22} + g^{33} T_{33}$ .
$T = g^{00} (K g_{00}) + g^{11} (K g_{11}) + g^{22} (- K g_{22}) + g^{33} (- K g_{33})$ .
$T = K (1) + K (1) - K (1) - K (1) = 0$ .
The property holds. It seems there was a minor sign error in the video's presentation of the T matrix, or my interpretation of it. The key physical principle stands.

Step 4: Solving the Field Equations (Video 10:41 - 14:47)

This is the final assault. We now have explicit expressions for $R_{μν}$ (from the Schwarzschild problem) and $T_{μν}$ (from Step 3). We set them equal: $R_{μν} = κ T_{μν}$ .

Show $A (r) B (r) = 1$ : The video demonstrates a beautiful trick by combining the $R_{tt}$ and $R_{r r}$ equations. It turns out that the combination $\frac{T _{tt}}{A} + \frac{T _{r r}}{B} = 0$ . This implies $\frac{R _{tt}}{A} + \frac{R _{r r}}{B} = 0$ , which leads to the differential equation $(ln (A B))^{'} = 0$ . Thus, $A (r) B (r)$ is a constant, which must be 1 to ensure the metric becomes Minkowskian at $r \to \infty$ .
Solve for $A (r)$ : With $B = 1/ A$ , the system reduces to a single ordinary differential equation for $A (r)$ . The video chooses the third field equation (the one derived from $R_{22} = κ T_{22}$ ), which simplifies to a first-order linear ODE for the function $A (r)$ .
Fix Integration Constants: The final solution has constants of integration. These are fixed by applying physical limits:
- As $Q \to 0$ , the solution must reduce to the Schwarzschild metric, $g_{tt} = 1 - \frac{2 GM}{c ^{2} r}$ . This fixes the mass term.
- The term proportional to $Q^{2}$ is determined by the differential equation itself.

3. The Reissner-Nordström Metric

The grand result of this derivation is the Reissner-Nordström line element:

d s^{2} = (1 - \frac{2 GM}{c ^{2} r} + \frac{G Q ^{2}}{4 π ϵ _{0} c ^{4} r ^{2}}) c^{2} d t^{2} - (1 - \frac{2 GM}{c ^{2} r} + \frac{G Q ^{2}}{4 π ϵ _{0} c ^{4} r ^{2}})^{- 1} d r^{2} - r^{2} d Ω^{2}

It is common to define two characteristic radii:

The Schwarzschild radius, $r_{s} = \frac{2 GM}{c ^{2}}$ .
The charge radius, $r_{Q}^{2} = \frac{G Q ^{2}}{4 π ϵ _{0} c ^{4}}$ .

Using these, the metric component $g_{tt}$ takes the compact form:

g_{tt} = 1 - \frac{r _{s}}{r} + \frac{r _{Q}^{2}}{r ^{2}}

This looks deceptively simple, but the addition of the $+ r_{Q}^{2} / r^{2}$ term has profound consequences for the spacetime structure.

For a deeper, more formal dive into the mathematics, you can read through the full derivation in the Diva Portal paper, which calculates every Christoffel symbol and Ricci tensor component explicitly.

The Reissner-Nordström metric

For a complete, unabridged derivation that aligns with your preference for mathematical rigor, I encourage you to review the following section from the paper we looked at earlier. It fills in every calculational detail that the video summarized.

Skim through the entirety of Chapter 5, 'The Reissner-Nordström Metric' (starts on page 25). You don't need to re-check every line of algebra, but appreciate the full scope of the calculation: the listing of all Christoffel symbols and Ricci components, the systematic computation of the stress-energy tensor, the solution of Maxwell's equations, and the final integration of the EFE. This complements the video's narrative with the underlying formal machinery.

Conclusion

Today, we have successfully derived the Reissner-Nordström metric, our second exact solution to the Einstein Field Equations.

Key Takeaways:

The Reissner-Nordström metric describes the spacetime of a static, spherically symmetric object with mass $M$ and charge $Q$ .
It is found by solving the coupled Einstein-Maxwell equations, with the electromagnetic stress-energy tensor acting as the source for gravity.
The derivation leverages the same symmetry arguments and metric ansatz as the Schwarzschild solution, allowing us to reuse the form of the Ricci tensor.
Key simplifications arise from the tracelessness of the EM stress-energy tensor ( $T = 0$ ) and the fact that the property $g_{tt} = - 1/ g_{r r}$ still holds.
The final metric introduces a new length scale, the charge radius $r_{Q}$ , leading to a $1/ r^{2}$ term in the metric components.

Preview of Next Lesson:

In the Schwarzschild spacetime, the single event horizon was located where $g_{tt} = 1 - r_{s} / r = 0$ . For Reissner-Nordström, the horizons are located where $g_{tt} = 1 - r_{s} / r + r_{Q}^{2} / r^{2} = 0$ . This is a quadratic equation for $r$ , which can have two, one, or zero real solutions, depending on the relative size of the mass and charge. This hints at a much richer and more complex causal structure. In our next lesson, we will analyze this structure, identify the new horizons, and construct the fascinating Penrose diagram for the Reissner-Nordström spacetime.

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