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- Question about Tong's cosmology lecture notes eqn. 1.19 (integral identity used?).
Dear all,
I have a rather basic question about an equation in David Tong's lecture notes on cosmology; see
http://www.damtp.cam.ac.uk/user/tong/cosmo.html
My question is about eqn. 1.19 (page 14), in which the cosmological redshift is derived. It's not about the physics, but about some basic integral identity I apparently don't see: how does he arrive at the conclusion after the arrow in that eqn. 1.19? To reproduce that eqn.,
$$\int_{t_1 + \delta t_1}^{t_0 + \delta t_0} \frac{dt}{a(t)} - \int_{t_1}^{t_0} \frac{dt}{a(t)} = 0 \rightarrow \frac{\delta t_1}{a(t_1)} = \frac{\delta t_0}{a(t_0)} \ \ \ \ (1.19) $$
I do understand that we can write the LHS as
$$\int_{t_0}^{t_0 + \delta t_0} \frac{dt}{a(t)} - \int_{t_1}^{t_1 + \delta t_1} \frac{dt}{a(t)} = 0$$
and I've drawn the integral equation geometrically, but I don't see how Tong arrives at the RHS of eqn. 1.19. Is there some mean value theorem involved? Differentiating the equation also leads me nowhere. I feel a bit silly. Thanks!
I have a rather basic question about an equation in David Tong's lecture notes on cosmology; see
http://www.damtp.cam.ac.uk/user/tong/cosmo.html
My question is about eqn. 1.19 (page 14), in which the cosmological redshift is derived. It's not about the physics, but about some basic integral identity I apparently don't see: how does he arrive at the conclusion after the arrow in that eqn. 1.19? To reproduce that eqn.,
$$\int_{t_1 + \delta t_1}^{t_0 + \delta t_0} \frac{dt}{a(t)} - \int_{t_1}^{t_0} \frac{dt}{a(t)} = 0 \rightarrow \frac{\delta t_1}{a(t_1)} = \frac{\delta t_0}{a(t_0)} \ \ \ \ (1.19) $$
I do understand that we can write the LHS as
$$\int_{t_0}^{t_0 + \delta t_0} \frac{dt}{a(t)} - \int_{t_1}^{t_1 + \delta t_1} \frac{dt}{a(t)} = 0$$
and I've drawn the integral equation geometrically, but I don't see how Tong arrives at the RHS of eqn. 1.19. Is there some mean value theorem involved? Differentiating the equation also leads me nowhere. I feel a bit silly. Thanks!