Question

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.

Short answer

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.

Step-by-step solution

Identification of given data

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$∆t\text{'}$ .

Significance of the Lorentz transformation

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.

Determination of the casual relation

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:

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

The above equation shows that $\frac{\Delta x\text{'}}{\Delta t\text{'}}>c$.

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.