Blog #26: Distance and Velocity at a Frame of Reference

1) Before we dive into the Hubble Flow, let's do a thought experiment. Pretend that there is an infinitely long series of balls sitting in a row. Imagine that during a time interval $\Delta t$ the space between each ball increases by $\Delta x$. 

(a) Look at the shadd ball, Ball C, in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time $\Delta t$? What about Ball B?

Based on the picture above, Ball B and Ball D are moving away from Ball C.

The distance between Ball B and Ball C at time $t = 0$ is $X_{{CB}_0}$.

The distance between Ball C and Ball D at time $t = 0$ is $X_{{CD}_0}$

The distance between Ball B and Ball C at time $t = \Delta t$ is:
$$\begin{align*} X_{CB}(\Delta t) = X_{{CB}_0} + \Delta x \end{align*}$$
The distance between Ball C and Ball D at time $t = \Delta t$ is:
$$\begin{align*} X_{CD}(\Delta t) = X_{{CD}_0} + \Delta x \end{align*}$$

(b) What are the distances from Ball C to Ball A and Ball E?

The distance between Ball C and Ball A in the time span $\Delta t$ is the same as the distance between Ball C to Ball B and Ball B  to Ball A, as shown below:

$$\begin{align*} X_{CA}(\Delta t) &= X_{CB}(\Delta t) + X_(BA)(\Delta t)\\ X_{CA}(\Delta t) &= (X_{{CB}_0} + \Delta x) + (X_{{BA}_0} + \Delta x)\\ X_{CA}(\Delta t) &= X_{{CA}_0} + 2 \Delta x  \end{align*}$$

The same logic can be used for the distance between Ball A and Ball E, as follows:

$$\begin{align*} X_{CE}(\Delta t) &= X_{CD}(\Delta t) + X_(DE)(\Delta t)\\ X_{CE}(\Delta t) &= (X_{{CD}_0} + \Delta x) + (X_{{DE}_0} + \Delta x)\\ X_{CE}(\Delta t) &= X_{{CE}_0} + 2 \Delta x  \end{align*}$$

(c) Write a general expression for the distance to a ball $N$ balls away from Ball C after time $\Delta t$. Interpret your findings.

Based on the answer to part (b), we can see that the distance between Ball C and a ball N balls away is the sum of the distance between the individual balls between the two balls. Therefore, we can generalize the distance between Ball C and another ball $N$ balls away during time $\Delta t$ as follows:

$$\begin{align*} X_{CN}(\Delta t) = X_{{CN}_0} + N\Delta x \end{align*}$$

(d) Write the velocity of a ball $N$ balls away from Ball C during $\Delta t$. Interpret your finding.

Velocity is described as a change in distance over a time, $t$, as described by the equation $v(t) = \frac{\Delta x}{t}$. The change in distance between two balls over a time $\Delta t$ is given by the answer in part (b) as $N\Delta x$. Therefore, the velocity of a ball $N$ balls away from Ball C is:

$$\begin{align*} v(t) &= \frac{\Delta x}{t}\\\ v(\Delta t) &= \frac{N\Delta x}{\Delta t} \end{align*}$$

This shows that balls that are further away from Ball C are moving faster in the frame of reference for Ball C than balls that are closer to Ball C.