Why is spacetime curved by mass but not charge?

It is written everywhere that gravity is curvature of spacetime caused by the mass of the objects or something to the same effect. This raises a question with me: why isn't spacetime curved due to other forces or aspects of bodies?

Why isn't it that there are curvatures related to the charge of a body or the spin of particles or any other characteristics?

• I don't know why this question was closed. None of the answers are responsive to the question. – Jiminion Apr 1 '16 at 21:13
• @Jiminion: it isn't closed, it's protected – Rijul Gupta Apr 2 '16 at 8:58
• I don't know why this question is protected. None of the answers, and none of the links cited by the answers and related questions, seem to be responsive. Charge curving spacetime (in a general sense) does not seem to be a part of the Standard Model, whereas gravity curving spacetime is. – Jiminion Apr 4 '16 at 13:59

Charge does curve spacetime. The metric for a charged black hole is different to an uncharged black hole. Charged (non-spinning) black holes are described by the Reissner–Nordström metric. This has some fascinating features, including acting as a portal to other universes, though sadly these are unlikely to be physically relevant. There is some discussion of this in the answers to the question Do objects have energy because of their charge?, though it isn't a duplicate. Anything that appears in the stress-energy tensor will curve spacetime.

Spin also has an effect, though I have to confess I'm out of my comfort zone here. To take spin into account we have to extend GR to Einstein-Cartan theory. However on the large scale the net spin is effectively zero, and we wouldn't expect spin to have any significant effect until we get down to quantum length scales.

• ... or to speeds approaching $c$. – ratchet freak May 8 '14 at 23:47
• If charge curves spacetime then why we cannot have geometric theory of electromagnetism? I know electric charges conflict with theory of equivalence principle. But why we cannot assume that more charge curves more spacetime,and thus it will be accelerating more quickly as compared to an object with less charge? – Inder Gill Aug 19 '15 at 23:11
• @InderGill: we can. See Can all fundamental forces be fictitious forces? – John Rennie Aug 20 '15 at 5:08

Gravitation couples to anything within the stress-energy tensor, as dictated by the field equations,

$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R = 8\pi G T_{\mu\nu}$$

Charge and angular momentum both affect the curvature of spacetime, as they affect the metric. For example, consider a spinning charged black hole, desribed by the Kerr-Newman metric,

$$\mathrm{d}s^2 = -\left( {\mathrm{d}r^2 \over \Delta + \mathrm{d}\theta^2} \right)\rho^2 + \left(\mathrm{d}t-\alpha \sin^2 \theta \mathrm{d}\phi\right)^2 \frac{\Delta}{\rho^2}-\left( (r^2+\alpha^2)\mathrm{d}\phi -\alpha \mathrm{d}t\right)^2 \frac{\sin^2 \theta}{\rho^2}$$

The parameters $\alpha$ and $\rho$ depend on the angular momentum, and $\Delta$ in fact does depend on the charge of the black hole. Evidently, the curvature forms are also dependent on these.

Why isn't spacetime curved due to other forces or aspects of bodies?

Spacetime is affected by the presence of other fields. For example, electric and magnetic fields are described by the Maxwell Lagrangian,

$$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$

where $F=\mathrm{d}A$ is the field-strength, a closed $2$-form. The field theory has a non-vanishing stress-energy tensor (derived by applying Noether's theorem to spacetime translations) which sits on the right-hand side of the field equations, and will induce curvature. Another example: the Kaluza-Klein metric in Kaluza-Klein theory is given by,

$$\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu \mathrm{d}x^\nu -e^{2\sigma(x)}\left[ \mathrm{d}\psi + A_\mu \mathrm{d}x^\mu\right]^2$$ Hence, in this $5D$ model spacetime is influenced by a scalar field $\sigma(x)$, and a four-potential $A_\mu$.

For completeness, the action which the Kaluza-Klein metric gives rise to is,

$$S=-\frac{1}{16\pi G}\int \mathrm{d}^4 x \, \sqrt{g_4} \, \mathrm{d}\psi \,e^{\sigma} \left[ R^{(4)} + \frac{1}{4}e^{2\sigma}F_{\mu\nu}F^{\mu\nu} -2e^{-\sigma} \square e^{\sigma}\right]$$

which reduces to Einstein-Maxwell theory if $\psi \sim \psi + L$, for some period $L$ and the dilaton $\sigma=\mathrm{const}$.

The unique property why mass and no charge respectively spin (at least not strongly, only by "side-effects") curves space-time is the equivalence principle. The equivalence principle says that gravity mass == inertial mass. After thinking hard about this property (experimentally proven by Galilei) Einstein found out that as a consequence space-time must be curved by mass.

On the other hand there no compulsory relationship between the charge (or spin) and the inertial mass, better said, there is no relation at all. Therefore charge or spin have a priori no effect on space-time, at least not a direct one. As other said, the electromagnetic field carries energy and via its energy it contributes to the curvature of space-time. But charge is not the source of space-time curvature, this is reserved to mass respectively energy due to the equivalence principle