# B Does The Presence of Charge Add To Relativistic Mass-Energy?

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1. Jan 11, 2019

### 1977ub

Does The Presence of Charge Add To Relativistic Mass-Energy, or does it take two particles attracting or repelling one another to generate charge-related mass-energy?

2. Jan 11, 2019

### Staff: Mentor

Yes. The presence of charge means there is an electromagnetic field, and the electromagnetic field has stress-energy.

No.

3. Jan 16, 2019 at 4:21 AM

### stevebd1

While PeterDonis has answered the question, you can see this at work in the Reissner–Nordström metric which is the vacuum solution for a charged black hole. Gravity for a charged black hole is-

$$a_g=\frac{M}{r^2\sqrt{1-\frac{2M}{r}-\frac{Q^2}{r^2}}}$$

where $M=Gm/c^2$ (mass in geometric units) and $Q=C\sqrt(Gk_e)/c^2$ (charge in geometric units) where $G$ is the gravitational constant, $c$ is the speed of light, $C$ is the charge in Coulombs and $k_e$ is Coulomb's constant (as a rule, $M\geq Q$). Multiply the answer by $c^2$ for S.I. units of gravity (m/s2).

If charge is removed and the black hole is reduced to it's irreducible mass (the same black hole becoming a Schwarzschild black hole) where-

$$M_{\text{ir}}=\frac{r_+}{2}\ \ \rightarrow\ \ M=\frac{Q^2}{4M_{\text{ir}}}+M_{\text{ir}}$$

where $M_{\text{ir}}$ is the irreducible mass and $r_+=M+\sqrt{M^2-Q^2}$ is the outer event horizon, the equation for gravity becomes-

$$a_g=\frac{M{\text{ir}}}{r^2\sqrt{1-\frac{2M{\text{ir}}}{r}}}$$

you'll notice a distinct drop in the gravity field at a specific r some distance from the black hole. It might also be worth looking at Kerr-Newman metric which also includes for spin. Up to 50% of RN black hole's 'mass' can be attributed to charge though charged black holes are not considered realistic.