# Do massless particles have gravity? [duplicate]

What I know so far is that according to the general theory of relativity, gravity is not the force of attraction between two objects; instead, it is a consequence of the curvature of space-time. And the curvature of space-time is due to massive objects. For example, a massive star will curve the spacetime fabric surrounding it, giving rise to gravity. So if this is true, then if a single massless particle is present in spacetime, as this particle is massless, it should not bend the curvature of the spacetime and hence should not possess gravity. Also, on this basis, a vacuum should also not have gravity. But I have heard that gravity is present everywhere in spacetime. So how is this possible?

• It helps to provide sources because sometimes it's just wrong or there is some nuanced interpretation being omitted. Commented Aug 8, 2022 at 17:26
• as this particle is massless, it should not bend the curvature of the spacetime. You are forgetting famous $E=mc^2$ equation. In short, mass is just a very well compressed energy (by factor of $c^2$). So "energy concentrated balls",- stars,planets gives rise to strong gravity forces or strong curvature of spacetime, in terms of general relativity. Everything that has energy can do it. Commented Aug 8, 2022 at 17:42
• Possible duplicates: physics.stackexchange.com/q/481557/2451 , physics.stackexchange.com/q/130552/2451 and links therein. Commented Aug 8, 2022 at 18:18

And the curvature of space-time is due to massive objects.

This is not exactly correct. The source of space-time curvature is the stress-energy tensor. The stress-energy tensor includes energy density, momentum density, and stress. While a massless particle has no mass it does have energy and momentum and therefore can indeed act as a source of gravity.

Also, on this basis, a vacuum should also not have gravity.

The relationship between the stress-energy tensor and spacetime curvature is not so simple. It is given by the Einstein field equations $$R_{\mu \nu} + \frac{1}{2}R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$ where $$R$$ is the curvature tensor and scalar, $$\Lambda$$ is the cosmological constant, $$g$$ is the metric tensor, and $$T$$ is the stress energy tensor. Because the relationship between the curvature $$R$$ and the stress-energy $$T$$ is non-trivial, it is possible to have vacuum solutions. This means that you can have curvature even where there are no sources, similar to how you can have electromagnetic fields in vacuum also.

Any energy or momentum-containing particle generates curvature in space-time. This is due to the fact that Einstein's field equation depends on the energy-momentum tensor $$T_{\mu \nu}$$ as

$$G_{\mu \nu} = T_{\mu \nu} - \Lambda g_{\mu \nu},$$

and the energy-momentum tensor is dependent, of course, of the total energy and momentum of the system.

Edit: the left side of this equation is interpreted as the "geometry" (gravity) of space-time. Even if the energy-momentum tensor is identically zero, there can be a contribution by this lambda-term - i.e., empty space is not empty.

• Does Einstein tensor G$_{\mu \nu}$ directly represent curvature? Nonzero G$_{\mu \nu}$ means nonzero curvature?
– apk
Commented Aug 8, 2022 at 18:27
• If you have Λ the energy tensor will not be 0 since the Λ is also in it, if the energy tensor could be 0 despite a positive Λ you'd need to neutralize Λ with some negative energy density, see the energy tensor for the FLRW which includes Λ, while the Einstein tensor doesn't: f.yukterez.net/einstein.equations/files/23.html#9 Commented Aug 8, 2022 at 19:13

It is not strictly true that the curvature of space-time is due to massive objects. The curvature of spacetime depends on the stress-energy tensor, which includes both mass-energy and momentum. The equation central to general relativity is $$G^{\mu \nu}=\frac{8\pi G}{c^4} T^{\mu \nu}$$

$$G_{\mu\nu}$$ represents the curvature of spacetime, $$G$$ is Newton's gravitational constant, and $$T_{\mu\nu}$$ is known as the stress-energy tensor. For a perfect fluid sitting still, we have: $$T^{\mu \nu}=\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0\\ 0 & 0 & p & 0\\ 0 & 0 & 0 & p\\ \end{bmatrix}$$

where $$p$$ is the pressure of the fluid and $$\rho$$ is the mass-energy density. But I had to write 16 numbers for that matrix (really a tensor). It is symmetric so we really can only play with $$10$$ numbers, but you can imagine plugging in any number of things for $$T$$ and then solving the lefthand side for $$G$$ (a monumental task!)

Some common choices include an ultrarelativistic gas, with all the particles moving near the speed of light: $$$$T^{\mu\nu}= \begin{bmatrix} \rho & 0 &0 &0\\ 0 & \frac{\rho}{3} &0 &0\\ 0 & 0 &\frac{\rho}{3} &0\\ 0 & 0 &0 &\frac{\rho}{3}\\ \end{bmatrix}$$$$

Or, for dark energy $$T^{\mu \nu}=-\Lambda g^{\mu\nu}$$: $$$$T^{\mu\nu}= \begin{bmatrix} -\Lambda & 0 &0 &0\\ 0 & \Lambda &0 &0\\ 0 & 0 &\Lambda &0\\ 0 & 0 &0 &\Lambda\\ \end{bmatrix}$$$$

(plugging this in for $$T$$ or writing it as a separate term like in Lucas's answer are both OK ways of going about it). That negative sign in the top left is the "negative energy" of dark energy.

Even the study of empty spacetime $$T^{\mu\nu}=0$$ has very interesting solutions, black hole solutions among them!