From the Lorentz transformation equations, show that if time intervals between two events,$\mathbf{∆}\mathit{t}$ and$\mathbf{∆}\mathit{t}\mathbf{\text{'}}$ , in two frames are of opposite sign, then the events are too far apart in either frame for light to travel from one to the other. Argue that therefore they cannot be casually related.

The given data can be listed below as:

• The time interval in the first event is $∆t$.
• The time interval in the second event is$∆t\text{'}$ .

The Lorentz transformation is referred to as the relationship amongst two different frames of coordinates which moves at uniform velocity. The coordinates are also relative against each other.

The equation of the time interval in the first event is expressed as:

$$\begin{gathered} ∆t\text{'}={\gamma }_v\left(∆t+\frac{v∆x}{c^2}\right) \\ \frac{∆t\text{'}}{∆t}={\gamma }_v\left(1+\frac{v∆x}{∆t{c}^2}\right) \end{gathered}$$

Here,$\Delta t^{\text{'}}$ is the time interval in the second event, $∆t$is the time interval in the first event, v is the velocity,$∆x$ is the change in the displacement in the first event, c is the speed of the light,${\gamma }_v$ is the gamma factor.

Here, it can be identified that$\frac{\Delta t^{\text{'}}}{∆t}<0$ and ${\gamma }_v\ge 1.$.

Then one solution of the above equation can be expressed as:

$$\begin{gathered} 1-\frac{V∆x}{c^2∆t}<0 \\ \frac{V∆x}{c^2∆t}<0 \end{gathered}$$

The above equation shows that $\frac{\Delta x}{\Delta t}>c.$.

The equation of the time interval in the second event is expressed as:

role="math" localid="1659178144922" $$\begin{gathered} ∆t={\gamma }_v\left(∆t\text{'}+\frac{v∆x\text{'}}{c^2}\right) \\ \frac{∆t}{∆t\text{'}}={\gamma }_v\left(1+\frac{v∆x\text{'}}{∆t\text{'}{c}^2}\right) \end{gathered}$$

Here, $\Delta x^{\text{'}}$is the change in the displacement in the second event.

Here, it can be identified that $\frac{\Delta t}{\Delta t^{\text{'}}}<0$and ${\gamma }_v\ge 1$.

Then one solution of the above equation can be expressed as:

$$\begin{gathered} 1+\frac{V\Delta x\text{'}}{C^2\Delta t\text{'}}<0 \\ \frac{V\Delta x\text{'}}{C^2\Delta t\text{'}}>1 \end{gathered}$$

Thus, in the Lorentz transformation, it is given that $\frac{\Delta t\text{'}}{\Delta t}>0$, then the propagation velocity is$\frac{\Delta x}{\Delta t}>c$that shows that there are casually disconnected events.