In General relativity, what is the meaning of flow of $x$ momentum in $x$ direction or pressure in $x$ direction? [duplicate]

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• Layman's explanation and understanding of Einstein's field equations 3 answers
• Intuitive understanding of the elements in the stress-energy tensor 2 answers

I found this interesting paper on Arxiv devoted to explaining Einstein's field equations in simple English. The author, JC Baez, does this by considering a group of small spherical balls in gravitational field, at rest with respect to each other. He then proposes a law by which the balls would shrink in volume in presence of gravity. The idea is that when the balls shrink while still maintaining contact with each other, they must move. Also, one has to consider all the infinitely many possible ways the mutually stationary balls could initially be moving to completely understand the motion of particles in a complicated gravitational field.

Here is the law that the author describes on the rate of change of volume of such balls in gravitational field: Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball, plus the pressure in the x direction at that point, plus the pressure in the y direction, plus the pressure in the z direction.

Alternatively,

My question: what is pressure in x (or t) direction? What is the physical meaning of this pressure? What is flow of x ( or t) momentum in x ( or t) direction? I understand x and t momentum as the component of the momentum of a body in x and t direction respectively. But what is the flow of these things in these directions? What exactly we mean by flow? What is going on?

if you do not understand my question, kindly ask to clarify or read the Arxiv paper in question.

Answer 1

The elements of the stress-energy tensor are, where $(t=x^0,x=x^1,y=x^2,z=x^3)$

$$ T^{\mu\nu}=\frac{\frac{1}{2}(dP^{\mu}{dx^{\nu}}+dP^{\nu}dx^{\mu})}{dx^0dx^1dx^2dx^3} $$ Now interpret the diagonal components, which are the ones you are interested in

$$ T^{11}=\frac{dP^1dx^1}{dx^0dx^1dx^2dx^3}=\frac{dP^1}{dx^1dx^2dx^3}\frac{dx^1}{dx^0}=\frac{dP^1}{dx^0dx^2dx^3} $$

In fluid mechanics the flow of a fluid in the direction of $\vec{v}$ is equal to $\rho \vec{v}=[\frac{kg}{m^3}][\frac{m}{sec}]$. Generalize this by putting something else like [$P^1$] in for [kg]. So the second expression for $T^{11}$ says it is the flow of x-momentum in the x-direction. Rearranging again, the third expression says this is equal to the force in the x-direction ($\frac{dP^1}{dx^0}$) per the area perpendicular to $dx^1$, which is the pressure in the x-direction. The pressure could be from a gas of particles, or photons, or even a magnetic field.

$$T^{00}=\frac{dP^0dx^0}{dx^0dx^1dx^2dx^3}=\frac{dP^0}{dx^1dx^2dx^3}\frac{dx^0}{dx^0}=\frac{dE}{dx^1dx^2dx^3} $$

Now put [$P^0$] in for [kg]. So the second expression for $T^{00}$ says it is the flow of t-momentum in the t-direction. Rearranging again, the third expression says this is equal to the energy density.

You can continue in this manner to interpret the other elements of the stress-energy tensor. For fun, you could also think about the antisym components of $T^{\mu\nu}$ (which are not part of Einstein's field eqns). These correspond to [angular momentum per 4-volume] or [torque per 3-volume]. These must be good for something!