# Why can we say that $dt_{obs}=(1+z)dt$?

I'm studying how the luminosity of a distant source changes as viewed from an observer far away from it because of the effects of the expansion of the universe. Now, apparently, if a point source emits light with luminosity $$L$$, with units of energy over time, then an observer far from it will recieve $$L_{obs} \propto \frac{1}{(1+z)^2}$$, with $$z$$ being the redshift parameter.

This happens this way since the energy of the emitted photon is redshifted by a factor $$(1+z)$$ because of the universe expansion AND because $$dt_{obs}=(1+z)dt$$, so the rate at which the photons arrive also decreases and thus we have that $$L_{obs} \propto \frac{1}{(1+z)^2}$$ relation. However, I can't find any intuitive explanation for why $$dt_{obs}=(1+z)dt$$ holds, since, as far as I know, as we're working in the Friedmann-Robertson-Walker metric, it's the spatial part of space (forgive the repetition) that's expanding.

In an intuitive way, the luminosity is basicly photons. Imagine a source (like a star) which emits photons isotropicaly with a Luminosity L. In the flat space (Minkovsky) whithout expansion the flux observed is $$F = \frac{L}{4\pi d^2}$$ because photons are emited spherically and the surface of a sphere is $$S=4\pi r^2$$.

But you know that our Universe is expanding, the metric of the Universe is given by: $$ds^2 = dt^2 - a^2(t) \left\{ dr^2 + d\theta^2+\sin^2(\theta)d\phi^2\right\}$$ Remember that for photons $$ds^2=0$$ Imagine a photon is emitted at $$t_0$$ the time at which an observer get it is $$t_r$$ and we have: $$c\int_{t_0}^{t_r} \frac{dt}{a(t)} = -\int_{r}^{0}dr$$ where the minus sign appear because the photon is send to the observer. A second photon is emmited afeter a time $$\delta t_o$$ and rerceive with a time delay $$\delta t_r$$ : $$c\int_{t_0+\delta t_0}^{t_r+\delta t_r} \frac{dt}{a(t)} = -\int_{r}^{0}dr=c\int_{t_0}^{t_r} \frac{dt}{a(t)}\\ \int_{t_0+\delta t_0}^{t_r+\delta t_r} \frac{dt}{a(t)} -\int_{t_0}^{t_r} \frac{dt}{a(t)} =0$$ We can decompose: $$\int_{t_0+\delta t_0}^{t_r+\delta t_r} \frac{dt}{a(t)} = \int_{t_0+\delta t_0}^{t_r} \frac{dt}{a(t)} +\int_{t_r}^{t_r+\delta t_r} \frac{dt}{a(t)}$$ and remark that :

$$\int_{t_0+\delta t_0}^{t_r} \frac{dt}{a(t)} - \int_{t_0}^{t_r} \frac{dt}{a(t)} = \int_{t_0+\delta t_0}^{t_0} \frac{dt}{a(t)}= \frac{-\delta t_0}{a(t_0)}$$

We have: $$\int_{t_r}^{t_r+\delta t_r} \frac{dt}{a(t)} = \frac{\delta t_r}{a(t_r)}$$

If we put all the stuff together we obtain that: $$\delta t_r = \frac{a(t_r)}{a(t_0)}\delta t_0 = (1+z)\delta t_0$$

Indeed the expansion of Universe affect the time, I think the fact that the expansion factor in FLRW metric isn't put in front of time is just conventional because in General Relativity coordinates are arbitrary and in cosmology sometimes we use conformal time that is $$d\eta = dt/a$$ and allow to rewrite FLRW metric as : $$ds^2 = a^2\left\{d\eta^2 - \left( dr^2 + d\theta^2+\sin^2(\theta)d\phi^2\right)\right\}$$

I hope it will help you and I don't make mistake in my explanation!

• Yes! This was very clear and useful, thanks!! Apr 3 '21 at 10:47