Can Einstein Tensor be the Product of Two 4-Vectors?

[empdee4]

TL;DR Summary

For particles of uniform mass and uniform momentum, stress-energy tensor can be written as product of two 4-vectors, can Einstein tensor be written in the same way?

In Gravitation by Misner, Thorne and Wheeler (p.139), stress-energy tensor for a single type of particles with uniform mass m and uniform momentum p (and E = p2 +m2) ½ ) can be written as a product of two 4-vectors,T(E,p) = (E,p)×(E,p)/[V(E2 – p2 )½ ]
Since Einstein equation is G = 8πGT, I am wondering if the left hand side, Einstein tensor, can also be written in the same way,
G(T,X) = (T,X)×(T,X)/[V (T2 – X2 )½ ]
If not, in what special case, or in what approximation, it can be expressed this way.
Thanks very much,

 

[PeterDonis]

In Gravitation by Misner, Thorne and Wheeler (p.139), stress-energy tensor for a single type of particles with uniform mass m and uniform momentum p (and E = p2 +m2) ½ ) can be written as a product of two 4-vectors

Basically this is the SET of a perfect fluid with zero pressure, usually called "dust" in the literature. But this is a highly idealized situation. As soon as you add nonzero pressure, the SET can no longer be expressed purely as a product of 4-vectors; see the very next page of MTW (equation 5.21).

Since Einstein equation is G = 8πGT, I am wondering if the left hand side, Einstein tensor, can also be written in the same way

Of course, just divide multiply $T$ by $8 \pi$. That's what the equation says. But, as noted above, this is a highly idealized situation.

 

[empdee4]

Basically this is the SET of a perfect fluid with zero pressure, usually called "dust" in the literature. But this is a highly idealized situation. As soon as you add nonzero pressure, the SET can no longer be expressed purely as a product of 4-vectors; see the very next page of MTW (equation 5.21).Of course, just divide $T$ by $8 \pi$. That's what the equation says. But, as noted above, this is a highly idealized situation.

Thanks very much for explanation.

Just not clear what T / 8π means.

 

[Ibix]

Thanks very much for explanation.

Just not clear what T / 8π means.

It just means that each element of the Einstein tensor is equal to the equivalent element of the stress-energy tensor multiplied by $8\pi G/c^4$ (not divided - @PeterDonis made a typo). You still have a bunch of nasty simultaneous differential equations to solve to extract the metric tensor.

 

[PeterDonis]

It just means that each element of the Einstein tensor is equal to the equivalent element of the stress-energy tensor multiplied by $8\pi G/c^4$ (not divided - @PeterDonis made a typo).

Oops, yes. I've fixed the post now.

 

[empdee4]

Thanks for clarification. Does it mean Einstein equation in this very special case can be reduced to a vector equation, as follows:

G = 8πGT
T(E,p) = (E,p)×(E,p)/[V (E2 – p2 )½ ]
G(T,X) = (T,X)×(T,X)/[V (T2 – X2 )½ ]

Thus, Einstein equation becomes

(T,X)x(T,X)/[V(T–X2)1/2]
=8πG(E,p)×8πG(E,p)/[V8πG(E2–p2)½]

which can be reduced to a 4-vector equation,

(T,X) =8πG(E,p)

thanks very much