91影视

Consider a distance $鈭�x$between the two stations to be 5000m.

Consider the speed of the train $\left(v\right)$to be .

Consider a train slow down at the rate of velocity ( v ) to be 0.1c .

Write the formula of length of the train from the passenger鈥檚 perspective.

$\mathit{L}\mathbf{\text{'}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{0}}}{\mathbf{纬}}$ 鈥︹�� (1)

Here, ${\mathbf{L}}_{\mathbf{0}}$ is distance between the two stations and$\mathit{纬}$ is relativistic Lorentz transformation.

Write the formula of length of the train from the observer point.

$\mathit{L}\mathit{=}\frac{{\mathit{L}}_{\mathit{0}}}{\mathit{纬}}$ 鈥︹�� (2)

Here, ${\mathit{L}}_{\mathbf{0}}$ is distance between the two stations and is relativistic Lorentz transformation.

Write the formula of the time lag between the two events occurring.

$\mathbf{鈭�}\mathit{t}\mathbf{\text{'}}\mathbf{=}\mathit{纬}\mathbf{(}\mathbf{鈭�}\mathbf{t}\mathbf{-}\frac{\mathbf{v}\mathbf{鈭�}\mathbf{x}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{)}$ 鈥︹�� (3)

Here, $\mathit{纬}$ is relativistic Lorentz transformation, $\mathbf{鈭�}\mathit{t}$ is time needed to arrive at the second station, $\mathbf{鈭�}\mathit{t}$is distance, v is speed of train and c is velocity of light.

Determine the length of the train from the passenger鈥檚 perspective.

Substitute 5000m for $L_0$and $\sqrt{1-\frac{{\left(0.8c\right)}^2}{c^2}}$for $纬$into equation (1).

$$\begin{gathered} L\text{'}=\sqrt{1-\frac{{\left(0.8c\right)}^2}{c^2}}脳5000\mathrm{m} \\ =\sqrt{1-0.8}脳5000\mathrm{m} \\ =3000\mathrm{m} \end{gathered}$$

Thus, the length of the train should be considered from the perspective of the passengers, or in other words, should be the right length. Additionally, I simply want to remind out that if we had followed the Lorentz transformation equation, we may have reached the findings.

$鈭�x=纬\left(鈭�x\text{'}+v鈭�t\text{'}\right)$

Substitute 0 for $鈭�t\text{'}$into above equation.

$鈭�x\text{'}=\frac{鈭�t\text{'}}{纬}$

Since this is how we arrived at the Lorentz contraction formula, it should come as no surprise that this is exactly the length contraction formula. I want to underline that both approaches are fundamentally equivalent, but that we often choose the one that will make our analysis simpler.

In part, we calculated the length of the train as seen by its passengers. Please don't mistake the 5 km that ground observers view as the distance between the two stations with the length of the train, since this is not the same length that they will see. If we reword the findings in section ), we will say that is the distance between the two stops as seen from the train's point of view, and we were fortunate enough to be provided that is equal to the distance between the front and back ends of the train. \bigbreak moving quickly on to our current issue, the length contraction occurs once again using the correct train length (which is known from part).

Determine the length of the train from the observer point.

Substitute 3000 m for $L_0$and $\sqrt{1-\frac{{\left(0.8c\right)}^2}{c^2}}$for $纬$into equation (2).

$$\begin{gathered} L\text{'}=\sqrt{1-\frac{{\left(0.8c\right)}^2}{c^2}}脳3000\mathrm{m} \\ =\sqrt{1-0.8}脳3000\mathrm{m} \\ =1800\mathrm{m} \end{gathered}$$

Therefore, the train don鈥檛 need to slow down because the initial reason of this issue is that train observers were unable to determine which command to follow. But when we return to the fixed frame of reference, the riddle has been solved (the ground).

In fact, the difficulty is that the two signs change simultaneously on the train's frame of reference. The two indicators don't necessarily agree with one another, though, in the context of the ground. The train's rear will depart from the first stop, and the distance to the second station will be

5000 m - 1800 =- 3200 m

Therefore, the time needed to arrive at the second station will be

$$\begin{gathered} 鈭�t=\frac{3200\mathrm{m}}{0.8脳3脳{10}^8\mathrm{m}/\mathrm{s}} \\ 鈮�13.3渭s \end{gathered}$$

This is effectively the interval between the two clocks, but we can get the same results by using the power of the Lorentz transformation. Please be aware that we will utilise the original length of for the distance between the two stations as we are attempting to determine the time difference between the two clocks on the ground frame.

Determine the time lag between the two events occurring.

Substitute 0 for $鈭�t$', $\left(0.8脳3脳{10}^8\right)$ for v and 5000 m for $鈭�x$and $\left(3脳{10}^8\mathrm{m}/\mathrm{s}\right)$for into equation (3).

$$\begin{gathered} 鈭�t=\frac{0.8脳3脳{10}^8\mathrm{m}/\mathrm{s}脳5000\mathrm{m}}{\left(3脳{10}^8\mathrm{m}/\mathrm{s}\right)} \\ =13.3渭s \end{gathered}$$

The result is that onlookers on the ground will see the two occurrences happening with a temporal lag of while the observer on the train sees the two signs changing simultaneously.