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

empdee4
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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,
 
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empdee4 said:
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).

empdee4 said:
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.
 
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PeterDonis said:
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.
 
empdee4 said:
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.
 
Ibix said:
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.
 
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
 
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