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# Tracelessness of the EM tensor in a closed radiation dominated FRW universe?

2018-02-23 08:04:57

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 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}

2018-02-23 09:17:27