So how can we say or prove time dilation in a moving frame or disprove simultaneity using this thought experiment? Am I wrong somewhere?

Let’s step away from thought experiments for a moment and look directly at the Lorentz transform. In the usual configuration the x direction is the direction of relative movement and the x’ axis is parallel to the x axis and similarly for the other axes. So the Lorentz transform is:

$$\Delta t’ = \gamma \left(\Delta t -\frac{v \Delta x}{c^2}\right)$$ $$\Delta x’ = \gamma ( \Delta x - v \Delta t)$$ $$\Delta y’ = \Delta y$$ $$\Delta z’ = \Delta z$$

In your scenario x is the direction along the road and y is the direction across the road. Since in your scenario $\Delta t=0$ and $\Delta x=0$ we can plug into the first formula and find $\Delta t’=0$.

So it is expected that both frames agree that they happened simultaneously. This thought experiment is not capable of studying the relativity of simultaneity.

Similarly, for time dilation since $\Delta t’$, $\Delta t$, and $\Delta x$ are all zero then the first equation reduces to $0=\gamma 0$ which is true regardless of $\gamma$ and therefore provides no information about time dilation.

This scenario is set up in a way that it is not sensitive to either time dilation or the relativity of simultaneity

So how can we say or prove time dilation in a moving frame or disprove simultaneity using this thought experiment? Am I wrong somewhere?

Let’s step away from thought experiments for a moment and look directly at the Lorentz transform. In the usual configuration the x direction is the direction of relative movement and the x’ axis is parallel to the x axis and similarly for the other axes. So the Lorentz transform is:

$$\Delta t’ = \gamma \left(\Delta t -\frac{v \Delta x}{c^2}\right)$$ $$\Delta x’ = \gamma ( \Delta x - v \Delta t)$$ $$\Delta y’ = \Delta y$$ $$\Delta z’ = \Delta z$$

In your scenario x is the direction along the road and y is the direction across the road. Since in your scenario $\Delta t=0$ and $\Delta x=0$ we can plug into the first formula and find $\Delta t’=0$.

So it is expected that both frames agree that they happened simultaneously. This thought experiment is not capable of studying the relativity of simultaneity.

Similarly, for time dilation since $\Delta t’$, $\Delta t$, and $\Delta x$ are all zero then the first equation reduces to $0=\gamma 0$ which is true regardless of $\gamma$ and therefore provides no information about time dilation.

This scenario is set up in a way that it is not sensitive to either time dilation or the relativity of simultaneity. $\Delta x=0$ makes the relativity of simultaneity irrelevant and $\Delta t =0$ makes time dilation irrelevant

So how can we say or prove time dilation in a moving frame or disprove simultaneity using this thought experiment? Am I wrong somewhere?

Let’s step away from thought experiments for a moment and look directly at the Lorentz transform. In the usual configuration the x direction is the direction of relative movement and the x’ axis is parallel to the x axis and similarly for the other axes. So the Lorentz transform is:

$$\Delta t’ = \gamma \left(\Delta t -\frac{v \Delta x}{c^2}\right)$$ $$\Delta x’ = \gamma ( \Delta x - v \Delta t)$$ $$\Delta y’ = \Delta y$$ $$\Delta z’ = \Delta z$$

In your scenario x is the direction along the road and y is the direction across the road. Since in your scenario $\Delta t=0$ and $\Delta x=0$ we can plug into the first formula and find $\Delta t’=0$.

So it is expected that both frames agree that they happened simultaneously. This thought experiment is not capable of studying the relativity of simultaneity

So how can we say or prove time dilation in a moving frame or disprove simultaneity using this thought experiment? Am I wrong somewhere?

Let’s step away from thought experiments for a moment and look directly at the Lorentz transform. In the usual configuration the x direction is the direction of relative movement and the x’ axis is parallel to the x axis and similarly for the other axes. So the Lorentz transform is:

$$\Delta t’ = \gamma \left(\Delta t -\frac{v \Delta x}{c^2}\right)$$ $$\Delta x’ = \gamma ( \Delta x - v \Delta t)$$ $$\Delta y’ = \Delta y$$ $$\Delta z’ = \Delta z$$

In your scenario x is the direction along the road and y is the direction across the road. Since in your scenario $\Delta t=0$ and $\Delta x=0$ we can plug into the first formula and find $\Delta t’=0$.

So it is expected that both frames agree that they happened simultaneously. This thought experiment is not capable of studying the relativity of simultaneity.

Similarly, for time dilation since $\Delta t’$, $\Delta t$, and $\Delta x$ are all zero then the first equation reduces to $0=\gamma 0$ which is true regardless of $\gamma$ and therefore provides no information about time dilation.

This scenario is set up in a way that it is not sensitive to either time dilation or the relativity of simultaneity