This might be a silly question. If we're considering The closed Radiation dominated Friedmann-Lemaitre-Robertson-Walker 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 non-zero 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 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:

$$ \dot a(t) = A \tfrac{1}{2} t^{-1/2} $$

$$ \ddot a(t) = A -\tfrac{1}{4} t^{-3/2} $$


$$\begin{align} R &= -6\left( \frac{-\tfrac{1}{4} t^{-3/2}}{t^{1/2}} + \left(\frac{{\tfrac{1}{2} t^{-1/2}}}{t^{1/2}}\right)^2 \right) \\ &= -6\left( -\tfrac{1}{4} t^{-2} + \left({\tfrac{1}{2} t^{-1}}\right)^2 \right) \\ &= 0 \end{align}$$

  • $\begingroup$ Great Answer! I was wondering if it was that thank you! Does this mean that in comoving coordinate system for instance the negative 00 component of the Ricci tensor balances the three spatial curvature parts (say ii), so it traces to zero (actually I guess it would have to in order to mirror the Em tensor) So it's not that space is flat but that the scalar "average of the curvatures" is zero? Spatially theres still a three-sphere $\endgroup$ – R. Rankin Feb 23 '18 at 6:57
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    $\begingroup$ @R.Rankin there is a (much underappreciated) site called The Universe in Problems and there are loads of interesting related problems on its FLRW page. You'll find calculations of the Ricci tensor and lots more on that page. $\endgroup$ – John Rennie Feb 23 '18 at 7:03

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