 localization and tensor product of algebras
 Uniqueness of weak solutions of the HamiltonJacobi Equation (Evans proof)
 Positive scalar curvature in dimension 4
 Transition probability depending on $\Omega$ and $n$
 Is Cartesian product of Graph and Composition of Graph the same thing?
 Consider a function on a domain
 Given $0<\lambda<1.$ Prove that if all $x,y>0,$ $f(x+y) = \lambda f(x) + (1\lambda)f(y),$ then $f$ is a constant function.
 Continuum Approach to Modeling Cell Proliferation and Differentiation (PDE)
 Forms of Alternating Skipping Sequences
 Existence of a (1)curve on a smooth nonminimal surface
 written reference that a discretetime signal cannot be differentiable
 Proof that there is a biholomorphic map from a Fatou set to the exterior of the unit circle?
 Presentation for $H^*(\mathbb{C}\mathbb{P}^1 \times \mathbb{C}\mathbb{P}^1)$
 Determine the maximal likelihood estimator for the coin probability
 Figuring out a function given a parameter
 Euclidean Domains
 Lecture videos on number theory?
 Can a physical parameter originally defined as a scalar be shown to be a tensor through its relationship to another parameter that is a tens
 Cardinality of a set of functions into $\Bbb Q$
 Irreversibility of dispersion phenomena
Tracelessness of the EM tensor in a closed radiation dominated FRW universe?
This might be a silly question. If we're considering The closed Radiation dominated FriedmannLemaitreRobertsonWalker universe It's common to consider the radiation as a perfect fluid and solve the Friedmann equations from there.
My issue is that the trace of the Stress energy Tensor of the free electromagnetic field is zero (or equivalently conformally invariant) implying zero Ricci curvature. (My understanding is this is true in a general spacetime) How can we then use this in such a model which supposes nonzero scalar Ricci curvature?
I'm sure I'm missing something simple. Thanks for reading!
You're quite correct. In a radiation dominated FLRW universe the scalar curvature is zero. This happens because the scalar curvature is given by:
$$ R = 6\left( \frac{\ddot a}{a} + \frac{{\dot a}^2}{a^2} + \frac{k}{a^2} \right) $$
and for a radiation (or matter) dominated universe $\ddot a \lt 0$. That means the terms can cancel to give zero.
For a spatially flat r

You're quite correct. In a radiation dominated FLRW universe the scalar curvature is zero. This happens because the scalar curvature is given by:
$$ R = 6\left( \frac{\ddot a}{a} + \frac{{\dot a}^2}{a^2} + \frac{k}{a^2} \right) $$
and for a radiation (or matter) dominated universe $\ddot a \lt 0$. That means the terms can cancel to give zero.
For a spatially flat radiation dominate universe we can show this easily (it gets messier when $k \ne 0$) because in this case we have:
$$ a(t) = A t^{1/2} $$
where $A$ is a constant (equal to $\sqrt{2H_0\Omega_0}$). So:
$$ \dot a(t) = A \tfrac{1}{2} t^{1/2} $$
$$ \ddot a(t) = A \tfrac{1}{4} t^{3/2} $$
So:
$$\begin{align}
R &= 6\left( \frac{\tfrac{1}{4} t^{3/2}}{t^{1/2}} + \left(\frac{{\tfrac{1}{2} t^{1/2}}}{t^{1/2}}\right)^2 \right) \\
&= 6\left( \tfrac{1}{4} t^{2} + \left({\tfrac{1}{2} t^{1}}\right)^2 \right) \\
&= 0
\end{align}$$
20180223 09:17:27