2 grains of salt. bits of skepticism.

I'm not sure why your question didn't get more attention, since it seems to be a good question. There was hope someone else would respond but it seems that isn't going to happen; I will give my best interpretation of the mistake, but take it with a grainbit of skepticism. To answer your question, allow me to first show the simplest way to get the correct solution:

The lorentz transformation must be used in this problem. It takes events $$(t,x)$$ in the rest frame (denoted $$S$$) and converts them to the coordinate system of the moving frame $$(\bar t, \bar x)$$ denoted $$\bar S$$:

$$\begin{array}{c|c} \text{Lorentz Transformation} & \text{Inverse Transformation} \\ \bar t = \gamma (t - vx/c^2) & t = \gamma (\bar t + v\bar x / c^2) \\ \bar x = \gamma (x - vt) & x = \gamma(\bar x + v\bar t) \end{array}$$

We will be using the inverse transformation. Let event $$A$$ be the emission of the light from the clock at the front of the train and $$B$$ the reception of light from the clock at the rear of the train. Then The coordinates of $$A$$ and $$B$$ in $$\bar S$$ are: $$\bar A = (\bar t_1, \bar x_2) = (\bar t_1, \bar \ell)$$ $$\bar B = (\bar t_2, \bar x_2) = (\bar t_2, 0)$$

Where I have used the 'true' length of the train to be denoted as $$\bar \ell$$ instead of $$\ell_0$$ (a random '$$0$$' subscript is dangerous here). Running these events through the inverse transformation yields: \begin{align} A &= (t_1, x_1) = ( \gamma(\bar t_1 + v \bar \ell / c^2), \gamma(\bar \ell + v \bar t_1) ) \\ B &= (t_2, x_2) = ( \gamma \bar t_2, \gamma v \bar t_2) \end{align} To find the difference in time, we subtract $$B-A$$: \begin{align} t_2 - t_1 &= \gamma(\bar t_2 - \bar t_1 - v\bar \ell /c^2) \\ x_2 - x_1 &= \gamma(v(\bar t_2 - \bar t_1) - \bar \ell) \end{align} Using the fact that $$\bar t_2 - \bar t_1 = \bar \ell / c$$, then the answer materializes out of the first coordinate: $$t_2 - t_1 = \gamma( \frac{\bar \ell}{c} - \frac{v \bar \ell}{c^2})$$

### why the naive approach doesn't work:

Let's draw minkowski diagrams for the moving frame $$\bar S$$ and the rest frame $$S$$ (which has an X on it because it is wrong): The rear of the train is placed at the origin of $$\bar S$$. Note that the vertical axis is the time axis, and the positions of the front/rear of the train are unchanging in $$\bar S$$. In $$S$$ they move with speed $$v$$ so they have slopes $$1/v$$. The light ray travels at a $$45^\circ$$ from $$A$$ to $$B$$ (a true statement in both $$\bar S$$ and $$S$$).

The naive approach (that I first tried too) is to look at the $$S$$ diagram and say the light ray travels to the back of the train from $$x_1 = \ell = \bar \ell / \gamma$$ to $$x_2 = v (t_2 - t_1)$$. It does this at speed $$c$$, so $$c(t_2 - t_1) = x_2 - x_1 = v(t_2 - t_1) - \bar \ell / \gamma$$. However, this is wrong, as can be seen by the true values of $$x_1$$ and $$x_2$$ in the lorentz transformation.

The correct way to draw the minkowski diagram for this problem is with both $$\bar S$$ and $$S$$ overlaid: Where the axis of $$\bar S$$ have been properly transformed by tilting them inward at an angle $$\tanh(\alpha) = v/c$$. You can see from this picture the event $$A$$ doesn't occur immediately (at the $$t=0$$ axis) from the perspective of $$S$$; the train actually moves forward a bit to position $$x_1$$ before the front clock emits a light pulse directed at the rear clock.

### the take-away from this problem

What I took away from this problem can be summed up fairly concisely as follows:

• When two events occurs at the same location but different times in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those times: $$\Delta t = \gamma \Delta \bar t$$. This is the traditional time dialation.
• When two events occur at the same time but different locations in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those locations: $$\Delta x = \Delta \bar x / \gamma$$. This is the traditional length contraction.
• When two events occur at different times and different locations in the moving frame $$\bar S$$, then the rest frame sees a mixture of time and location relationships; you must use the full-blown lorentz transformation.

The mistake the naive approach makes is a violation of the last bullet.

I'm not sure why your question didn't get more attention, since it seems to be a good question. There was hope someone else would respond but it seems that isn't going to happen; I will give my best interpretation of the mistake, but take it with a grain of skepticism. To answer your question, allow me to first show the simplest way to get the correct solution:

The lorentz transformation must be used in this problem. It takes events $$(t,x)$$ in the rest frame (denoted $$S$$) and converts them to the coordinate system of the moving frame $$(\bar t, \bar x)$$ denoted $$\bar S$$:

$$\begin{array}{c|c} \text{Lorentz Transformation} & \text{Inverse Transformation} \\ \bar t = \gamma (t - vx/c^2) & t = \gamma (\bar t + v\bar x / c^2) \\ \bar x = \gamma (x - vt) & x = \gamma(\bar x + v\bar t) \end{array}$$

We will be using the inverse transformation. Let event $$A$$ be the emission of the light from the clock at the front of the train and $$B$$ the reception of light from the clock at the rear of the train. Then The coordinates of $$A$$ and $$B$$ in $$\bar S$$ are: $$\bar A = (\bar t_1, \bar x_2) = (\bar t_1, \bar \ell)$$ $$\bar B = (\bar t_2, \bar x_2) = (\bar t_2, 0)$$

Where I have used the 'true' length of the train to be denoted as $$\bar \ell$$ instead of $$\ell_0$$ (a random '$$0$$' subscript is dangerous here). Running these events through the inverse transformation yields: \begin{align} A &= (t_1, x_1) = ( \gamma(\bar t_1 + v \bar \ell / c^2), \gamma(\bar \ell + v \bar t_1) ) \\ B &= (t_2, x_2) = ( \gamma \bar t_2, \gamma v \bar t_2) \end{align} To find the difference in time, we subtract $$B-A$$: \begin{align} t_2 - t_1 &= \gamma(\bar t_2 - \bar t_1 - v\bar \ell /c^2) \\ x_2 - x_1 &= \gamma(v(\bar t_2 - \bar t_1) - \bar \ell) \end{align} Using the fact that $$\bar t_2 - \bar t_1 = \bar \ell / c$$, then the answer materializes out of the first coordinate: $$t_2 - t_1 = \gamma( \frac{\bar \ell}{c} - \frac{v \bar \ell}{c^2})$$

### why the naive approach doesn't work:

Let's draw minkowski diagrams for the moving frame $$\bar S$$ and the rest frame $$S$$ (which has an X on it because it is wrong): The rear of the train is placed at the origin of $$\bar S$$. Note that the vertical axis is the time axis, and the positions of the front/rear of the train are unchanging in $$\bar S$$. In $$S$$ they move with speed $$v$$ so they have slopes $$1/v$$. The light ray travels at a $$45^\circ$$ from $$A$$ to $$B$$ (a true statement in both $$\bar S$$ and $$S$$).

The naive approach (that I first tried too) is to look at the $$S$$ diagram and say the light ray travels to the back of the train from $$x_1 = \ell = \bar \ell / \gamma$$ to $$x_2 = v (t_2 - t_1)$$. It does this at speed $$c$$, so $$c(t_2 - t_1) = x_2 - x_1 = v(t_2 - t_1) - \bar \ell / \gamma$$. However, this is wrong, as can be seen by the true values of $$x_1$$ and $$x_2$$ in the lorentz transformation.

The correct way to draw the minkowski diagram for this problem is with both $$\bar S$$ and $$S$$ overlaid: Where the axis of $$\bar S$$ have been properly transformed by tilting them inward at an angle $$\tanh(\alpha) = v/c$$. You can see from this picture the event $$A$$ doesn't occur immediately (at the $$t=0$$ axis) from the perspective of $$S$$; the train actually moves forward a bit to position $$x_1$$ before the front clock emits a light pulse directed at the rear clock.

### the take-away from this problem

What I took away from this problem can be summed up fairly concisely as follows:

• When two events occurs at the same location but different times in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those times: $$\Delta t = \gamma \Delta \bar t$$. This is the traditional time dialation.
• When two events occur at the same time but different locations in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those locations: $$\Delta x = \Delta \bar x / \gamma$$. This is the traditional length contraction.
• When two events occur at different times and different locations in the moving frame $$\bar S$$, then the rest frame sees a mixture of time and location relationships; you must use the full-blown lorentz transformation.

The mistake the naive approach makes is a violation of the last bullet.

I'm not sure why your question didn't get more attention, since it seems to be a good question. There was hope someone else would respond but it seems that isn't going to happen; I will give my best interpretation of the mistake, but take it with a bit of skepticism. To answer your question, allow me to first show the simplest way to get the correct solution:

The lorentz transformation must be used in this problem. It takes events $$(t,x)$$ in the rest frame (denoted $$S$$) and converts them to the coordinate system of the moving frame $$(\bar t, \bar x)$$ denoted $$\bar S$$:

$$\begin{array}{c|c} \text{Lorentz Transformation} & \text{Inverse Transformation} \\ \bar t = \gamma (t - vx/c^2) & t = \gamma (\bar t + v\bar x / c^2) \\ \bar x = \gamma (x - vt) & x = \gamma(\bar x + v\bar t) \end{array}$$

We will be using the inverse transformation. Let event $$A$$ be the emission of the light from the clock at the front of the train and $$B$$ the reception of light from the clock at the rear of the train. Then The coordinates of $$A$$ and $$B$$ in $$\bar S$$ are: $$\bar A = (\bar t_1, \bar x_2) = (\bar t_1, \bar \ell)$$ $$\bar B = (\bar t_2, \bar x_2) = (\bar t_2, 0)$$

Where I have used the 'true' length of the train to be denoted as $$\bar \ell$$ instead of $$\ell_0$$ (a random '$$0$$' subscript is dangerous here). Running these events through the inverse transformation yields: \begin{align} A &= (t_1, x_1) = ( \gamma(\bar t_1 + v \bar \ell / c^2), \gamma(\bar \ell + v \bar t_1) ) \\ B &= (t_2, x_2) = ( \gamma \bar t_2, \gamma v \bar t_2) \end{align} To find the difference in time, we subtract $$B-A$$: \begin{align} t_2 - t_1 &= \gamma(\bar t_2 - \bar t_1 - v\bar \ell /c^2) \\ x_2 - x_1 &= \gamma(v(\bar t_2 - \bar t_1) - \bar \ell) \end{align} Using the fact that $$\bar t_2 - \bar t_1 = \bar \ell / c$$, then the answer materializes out of the first coordinate: $$t_2 - t_1 = \gamma( \frac{\bar \ell}{c} - \frac{v \bar \ell}{c^2})$$

### why the naive approach doesn't work:

Let's draw minkowski diagrams for the moving frame $$\bar S$$ and the rest frame $$S$$ (which has an X on it because it is wrong): The rear of the train is placed at the origin of $$\bar S$$. Note that the vertical axis is the time axis, and the positions of the front/rear of the train are unchanging in $$\bar S$$. In $$S$$ they move with speed $$v$$ so they have slopes $$1/v$$. The light ray travels at a $$45^\circ$$ from $$A$$ to $$B$$ (a true statement in both $$\bar S$$ and $$S$$).

The naive approach (that I first tried too) is to look at the $$S$$ diagram and say the light ray travels to the back of the train from $$x_1 = \ell = \bar \ell / \gamma$$ to $$x_2 = v (t_2 - t_1)$$. It does this at speed $$c$$, so $$c(t_2 - t_1) = x_2 - x_1 = v(t_2 - t_1) - \bar \ell / \gamma$$. However, this is wrong, as can be seen by the true values of $$x_1$$ and $$x_2$$ in the lorentz transformation.

The correct way to draw the minkowski diagram for this problem is with both $$\bar S$$ and $$S$$ overlaid: Where the axis of $$\bar S$$ have been properly transformed by tilting them inward at an angle $$\tanh(\alpha) = v/c$$. You can see from this picture the event $$A$$ doesn't occur immediately (at the $$t=0$$ axis) from the perspective of $$S$$; the train actually moves forward a bit to position $$x_1$$ before the front clock emits a light pulse directed at the rear clock.

### the take-away from this problem

What I took away from this problem can be summed up fairly concisely as follows:

• When two events occurs at the same location but different times in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those times: $$\Delta t = \gamma \Delta \bar t$$. This is the traditional time dialation.
• When two events occur at the same time but different locations in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those locations: $$\Delta x = \Delta \bar x / \gamma$$. This is the traditional length contraction.
• When two events occur at different times and different locations in the moving frame $$\bar S$$, then the rest frame sees a mixture of time and location relationships; you must use the full-blown lorentz transformation.

The mistake the naive approach makes is a violation of the last bullet.

1

I'm not sure why your question didn't get more attention, since it seems to be a good question. There was hope someone else would respond but it seems that isn't going to happen; I will give my best interpretation of the mistake, but take it with a grain of skepticism. To answer your question, allow me to first show the simplest way to get the correct solution:

The lorentz transformation must be used in this problem. It takes events $$(t,x)$$ in the rest frame (denoted $$S$$) and converts them to the coordinate system of the moving frame $$(\bar t, \bar x)$$ denoted $$\bar S$$:

$$\begin{array}{c|c} \text{Lorentz Transformation} & \text{Inverse Transformation} \\ \bar t = \gamma (t - vx/c^2) & t = \gamma (\bar t + v\bar x / c^2) \\ \bar x = \gamma (x - vt) & x = \gamma(\bar x + v\bar t) \end{array}$$

We will be using the inverse transformation. Let event $$A$$ be the emission of the light from the clock at the front of the train and $$B$$ the reception of light from the clock at the rear of the train. Then The coordinates of $$A$$ and $$B$$ in $$\bar S$$ are: $$\bar A = (\bar t_1, \bar x_2) = (\bar t_1, \bar \ell)$$ $$\bar B = (\bar t_2, \bar x_2) = (\bar t_2, 0)$$

Where I have used the 'true' length of the train to be denoted as $$\bar \ell$$ instead of $$\ell_0$$ (a random '$$0$$' subscript is dangerous here). Running these events through the inverse transformation yields: \begin{align} A &= (t_1, x_1) = ( \gamma(\bar t_1 + v \bar \ell / c^2), \gamma(\bar \ell + v \bar t_1) ) \\ B &= (t_2, x_2) = ( \gamma \bar t_2, \gamma v \bar t_2) \end{align} To find the difference in time, we subtract $$B-A$$: \begin{align} t_2 - t_1 &= \gamma(\bar t_2 - \bar t_1 - v\bar \ell /c^2) \\ x_2 - x_1 &= \gamma(v(\bar t_2 - \bar t_1) - \bar \ell) \end{align} Using the fact that $$\bar t_2 - \bar t_1 = \bar \ell / c$$, then the answer materializes out of the first coordinate: $$t_2 - t_1 = \gamma( \frac{\bar \ell}{c} - \frac{v \bar \ell}{c^2})$$

### why the naive approach doesn't work:

Let's draw minkowski diagrams for the moving frame $$\bar S$$ and the rest frame $$S$$ (which has an X on it because it is wrong): The rear of the train is placed at the origin of $$\bar S$$. Note that the vertical axis is the time axis, and the positions of the front/rear of the train are unchanging in $$\bar S$$. In $$S$$ they move with speed $$v$$ so they have slopes $$1/v$$. The light ray travels at a $$45^\circ$$ from $$A$$ to $$B$$ (a true statement in both $$\bar S$$ and $$S$$).

The naive approach (that I first tried too) is to look at the $$S$$ diagram and say the light ray travels to the back of the train from $$x_1 = \ell = \bar \ell / \gamma$$ to $$x_2 = v (t_2 - t_1)$$. It does this at speed $$c$$, so $$c(t_2 - t_1) = x_2 - x_1 = v(t_2 - t_1) - \bar \ell / \gamma$$. However, this is wrong, as can be seen by the true values of $$x_1$$ and $$x_2$$ in the lorentz transformation.

The correct way to draw the minkowski diagram for this problem is with both $$\bar S$$ and $$S$$ overlaid: Where the axis of $$\bar S$$ have been properly transformed by tilting them inward at an angle $$\tanh(\alpha) = v/c$$. You can see from this picture the event $$A$$ doesn't occur immediately (at the $$t=0$$ axis) from the perspective of $$S$$; the train actually moves forward a bit to position $$x_1$$ before the front clock emits a light pulse directed at the rear clock.

### the take-away from this problem

What I took away from this problem can be summed up fairly concisely as follows:

• When two events occurs at the same location but different times in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those times: $$\Delta t = \gamma \Delta \bar t$$. This is the traditional time dialation.
• When two events occur at the same time but different locations in the moving frame $$\bar S$$, then the rest frame sees a direct relationship between those locations: $$\Delta x = \Delta \bar x / \gamma$$. This is the traditional length contraction.
• When two events occur at different times and different locations in the moving frame $$\bar S$$, then the rest frame sees a mixture of time and location relationships; you must use the full-blown lorentz transformation.

The mistake the naive approach makes is a violation of the last bullet.