Is Pressure A Source Of Gravity?

In a previous series of articles, I posed the question “Does Gravity Gravitate?” and explained how, depending on how you interpreted the terms “gravity” and “gravitate”, one could answer the question, either way, yes or no. This article will treat its title question in a similar fashion. :-)

To be sure, this case is a bit simpler than the previous one, because there is only one term that can be given more than one meaning. The term “pressure” is clear: it means the spatial diagonal components of the stress-energy tensor. (More precisely, it means those components in the rest frame of the fluid, but we won’t delve into such technical details very much in this article.) The ambiguity is in the term “source of gravity”. What does that term mean?

The standard answer to that question in GR is that the source of gravity is the stress-energy tensor, the thing that appears on the RHS of the Einstein Field Equation. So anything that appears in the stress-energy tensor is a source of gravity. That seems to lead to a clear, simple “yes” answer to our title question: pressure appears in the stress-energy tensor, therefore pressure is a source of gravity. QED. (In the previous article, a similar argument leads to a “no” answer to its title question, since “gravity” itself–spacetime curvature–does not appear in the stress-energy tensor; it appears on the LHS of the Einstein Field Equation, not the RHS.)

Even though this answer is clear and simple, it is worth unpacking it a bit. This is done, for example, in the article “The Meaning of Einstein’s Equation”, by John Baez and Emory Bunn [1]. That article rearranges the equation somewhat in order to obtain an equation for the Ricci tensor, which has a simpler, more straightforward physical interpretation than the Einstein tensor (the tensor that appears on the LHS of the Einstein Field Equation) does. This rearrangement is similar to the one that was done in the second article of the previous series on gravity gravitating, and for the simple case of a perfect fluid, it looks like this:

$$
R_{0 0} = 4 \pi \left( \rho + 3 p \right)
$$

where $\rho$ is the energy density and $p$ is the pressure. Now $R_{0 0}$, the 0-0 (or “time-time”) component of the Ricci tensor, describes, heuristically, how attractive gravity is; the more positive $R_{0 0}$ gets, the more gravity “pulls matter inward”, heuristically speaking. So positive pressure, just like positive energy density, makes gravity more attractive.

It is worth noting that this equation also explains why “dark energy” makes the expansion of the universe accelerate. Dark energy has the equation of state $p = – \rho$, which, when plugged into the equation, gives $R_{0 0} = – 2 \rho$. This is negative, meaning, heuristically, that dark energy causes “repulsive gravity” instead of “attractive gravity”; a uniform dark energy fluid does not pull matter inward, it pushes matter outward. Applied to the universe as a whole, this becomes accelerated expansion.

So we have our “yes” answer to the title question. But those who are familiar with the Einstein Field Equation for the special case of a static, spherically symmetric spacetime, as discussed for example in this Insights article, might have spotted what looks like an inconsistency. Let’s explore that.

For the case of a static, spherically symmetric spacetime, we can define a function $m(r)$ that, physically, describes the “mass inside radius $r$” of a static, spherically symmetric body like a planet or a star. As was shown in the Insights article on this topic, this function turns out to be as follows:

$$
m(r) = \int_0^r 4 \pi \rho(r’) r’^2 dr’
$$

As was noted in that other article, this is a function of $\rho$ only. Pressure does not appear at all! What’s going on?

We can gain some understanding of this by looking at how we would construct this integral using our equation for the Ricci tensor above. This was done in the second article of the “Does Gravity Gravitate?” series, and as noted there, the result is called the Komar mass of the spacetime. For the case of a static, spherically symmetric body composed of a perfect fluid, the Komar mass inside radius $r$ looks like this:

$$
m_K(r) = \int_0^r 4 \pi \sqrt{J(r’)} \left( \rho(r’) + 3 p(r’) \right) r’^2 dr’
$$

where $J(r)$ is the “redshift factor” as a function of $r$, i.e., the absolute value of the metric coefficient $g_{tt}$ in standard Schwarzschild coordinates.

Now we can see what is going on. Since we must have $m_K(r) = m(r)$, the extra contribution from the $p$ term in $m_K$ must be exactly compensated for by the reduction due to $J(r)$, which will be less than $1$ throughout the range of integration. Physically, this means that the extra attractive gravity due to pressure is exactly compensated for by the redshift factor, which, heuristically, captures the (negative) contribution of “gravitational potential energy” to the total mass. For the highly unrealistic but analytically solvable case of $\rho$ constant, independent of $r$, in a spherically symmetric body like a star or planet, the cancellation just referred to can be mathematically derived explicitly, i.e., using the known solutions for $J(r)$ and $p(r)$ for this case, it can be shown that $\sqrt{J} \left( \rho + 3 p \right) = \rho$. Several other static cases for which an exact cancellation can also be shown explicitly are considered in a paper by Ehlers, Osvath, Schucking, and Schang. [2]

[1] https://arxiv.org/pdf/gr-qc/0103044.pdf

[2] https://arxiv.org/abs/gr-qc/0510041

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