A Non-Linear Theory: Summation Meaningful in Einstein Gravitation?

[empdee4]

TL;DR Summary

In the Gravitation book by Misner, Thorne and Wheeler. The total stress-energy of a swarm of particles is summed up from all categories. Is summation meaningful in non-linear theory? Thanks.

In the famous book, Gravitation, by Misner, Thorne and Wheeler, it talks about the stress-energy tensor of a swarm of particles (p.138). The total stress-energy is summed up from all categories of particles. Is summation meaningful in the non-linear theory of Einstein gravitation? Thanks.

 

[PeterDonis]

Is summation meaningful in the non-linear theory of Einstein gravitation?

It is the way MTW are doing it in that particular example. They are not summing over a finite volume of spacetime, which would require taking into account the non-linearities you mention. They are summing over different categories of particles in the same infinitesimal volume of spacetime, which is not affected by the non-linearities you mention.

 

[empdee4]

Thanks very much. In MTW book, the stress-energy tensor of each category of particles is written as a vector product. Does it mean this also can happen in an infinitesimal region only? That is, in a finite region, it cannot be written as vector product

 

[Orodruin]

There is no such thing as the stress energy tensor for a finite region. Tensors are locally defined quantities.

 

[pervect]

In a manifold, each point in the manifold is associated with its own tangent space, and the stress-energy tensor (and other tensors, for that matter ) are defined in the tangent space of a particular point.

It is a common approximation, though, to lump together a bunch of nearby points into a small regoin as if they shared a common tangent space. Strictly speaking it's not correct, but it's a reasonable approximation in many cases.

As far as the non-linearites go, the right hand side of Einstein's field equations, $T_{\mu\nu}$ is perfectly linear. The left hand side , the Einstein tensor $G_{\mu\nu$ is where the non-linarities are.

 

[PeterDonis]

the right hand side of Einstein's field equations, $T_{\mu\nu}$ is perfectly linear.

To be clear, the SET will be linear in the metric; but it does not have to be linear in the matter fields or their derivatives.

 

[empdee4]

Thanks very much.

 

[empdee4]

But if one side of the equation is linear and the other side is non-linear, how can the two sides be equal?

 

[PeterDonis]

if one side of the equation is linear and the other side is non-linear, how can the two sides be equal?

The nonlinearities in the Einstein tensor are in derivatives of the metric, not the metric itself.

 

[martinbn]

To be clear, the SET will be linear in the metric; but it does not have to be linear in the matter fields or their derivatives.

He means that the equations are linear in T.

 

[Orodruin]

But if one side of the equation is linear and the other side is non-linear, how can the two sides be equal?

The stress energy tensor is the source of the Einstein equations. It need not depend on the metric at all. In general, the EFEs are a set of non-linear differential equations. This by no means imply that they cannot be solved.

What it does mean is that the solutions do not allow superpositioning. In other words, if you have two solutions with SETs T1 and T2, the sum of the solutions will not solve the EFEs with SET T1+T2.

 

[PeroK]

But if one side of the equation is linear and the other side is non-linear, how can the two sides be equal?

$$x^2 + y^2 = 4y$$ is a perfectly valid equation.

 

[vanhees71]

To be clear, the SET will be linear in the metric; but it does not have to be linear in the matter fields or their derivatives.

Indeed, e.g., for the em. field you have
$$T^{\mu \nu} =F^{\mu \rho} {F_{\rho }}^{\nu} + \frac{1}{4} F_{\rho \sigma} F^{\rho \sigma} g^{\mu \nu}.$$

 

[PeterDonis]

He means that the equations are linear in T.

But they're also linear in the Einstein tensor $G$, so if we just look to that level of detail, the equation is linear on both sides.

When people say the EFE is nonlinear, they mean that the Einstein tensor $G$ is nonlinear in the metric and its derivatives.

 

[martinbn]

But they're also linear in the Einstein tensor $G$, so if we just look to that level of detail, the equation is linear on both sides.

When people say the EFE is nonlinear, they mean that the Einstein tensor $G$ is nonlinear in the metric and its derivatives.

I agree, i was just saying how i understood his post.

 

[empdee4]

Thanks very much!