Holographic Cosmology lecture of Leonard Susskind

[MathematicalPhysicist]

In the following lectures:

Prof. Susskind writes the terms $\exp(3Ht_n)$ and $\exp(3ht_n)$.
If I understand correctly he computes $R(t)=R_0 \exp[\int^{t_n}H(t')dt']$ where $R(t)$ is the scale factor of the Universe and $H(t')$ is Hubble's parameter. What I still don't understand is why $H(t')=3H$?

 

[AndyW]

Thank you for your question. I am glad to see that you are engaged and seeking to understand the material presented in the lectures.

In order to fully understand why $H(t')=3H$ in the expression for the scale factor of the Universe, we first need to define what Hubble's parameter is. Hubble's parameter, denoted by $H$, is a measure of the rate at which the Universe is expanding at a given time. It is defined as the ratio of the change in the scale factor of the Universe over time, $\dot{R}$, to the scale factor itself, $R$:

$$H=\frac{\dot{R}}{R}$$

Now, let's look at the expression for the scale factor, $R(t)$, in more detail:

$$R(t)=R_0 \exp\left[\int^{t_n}H(t')dt'\right]$$

In this expression, $R_0$ represents the initial scale factor at time $t=0$, and $t_n$ represents the current time. The integral term represents the cumulative effect of the expansion of the Universe over time. Now, if we take the derivative of this expression with respect to time, we get:

$$\dot{R}(t)=R_0\frac{d}{dt}\left[\exp\left(\int^{t_n}H(t')dt'\right)\right]=R_0H(t)\exp\left(\int^{t_n}H(t')dt'\right)$$

Notice that the derivative of the integral term is simply $H(t)$, which is the value of Hubble's parameter at time $t$. Now, if we substitute this value into our original expression for the scale factor, we get:

$$R(t)=R_0 \exp\left[\int^{t_n}H(t')dt'\right]=R_0\exp\left[\int^{t_n}\frac{\dot{R}}{R}dt'\right]=R_0\exp\left[\int^{t_n}3Hdt'\right]$$

This is where the $3H$ comes from. Since $H(t)=3H$, we can substitute this value into the integral and simplify the expression. Therefore, we can say that $H(t')=3H$ because this is the value of H