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2 grains of salt. bits of skepticism.
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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): minkowski diagrams 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: proper minkowski diagram 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): minkowski diagrams 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: proper minkowski diagram 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): minkowski diagrams 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: proper minkowski diagram 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
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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): minkowski diagrams 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: proper minkowski diagram 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.