- #1

brotherbobby

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- Homework Statement
- (1) A 200 m long goods train is moving uniformly with a speed of 5 m/s when its tail end crosses the front of the platform chosen to be the origin O. 30 s after is has passed, its tain signal flashes on and off, as seen by the platform observer. A further 60 s later, the same tail signal is observed to flash again. Calculate the distance between the two events as observed by the platform observer (##\Delta x##)and an observer on the train (##\Delta x'##).

(2) Carry out the solution to the same problem above if the goods train starts moving from the origin O from rest with a uniform acceleration of ##0.1 ms^{-2}##.

- Relevant Equations
- (1) According to Galilean transformations, for two frames in standard configuration along the ##x## axis, an event ##E## occuring at spacetime position ##(x, t)## in the frame S is observed to occur in frame S' at a spacetime position ##(x', t')## where (a) ##x' = x- vt## and (b) ##t'=t##, where ##v## is the (uniform) velocity of frame S' with respect to frame S along the positive ##x## axis.

(2) In the event of an acceleration of S' relative to S, but zero initial velocity (##v_0 = 0##)at ##t = t' = 0##, we have (a) ##x' = x+\tfrac{1}{2}at^2## and ##t'=t##.

**
(1) Uniformly moving frames**

I begin with a drawing of the situation. The events are labelled as ##\color{red}{E_1}## and ##\color{red}{E_2}##. We note the time of those events : ##t_1 = t'_1 = 30s## and ##t_2 = t_2' = 30+60 = 90s##.

I attempt the problem in two different ways.

**(a) By calculating the spacetime coordinates of the events in the two frames**

Since those events happen at the same location (the tail end) for the train observer, the spatial separation between them is 0. Hence the distance between the two events in the moving frame : ##\boxed{\color{blue} {\Delta x' = 0\,\text{m}}}##.

The location of event ##E_1## in the Earth's frame : ##x_1 = vt_1 = 5\times 30 = 150## m, from the origin. The location of event ##E_2## in the Earth's frame : ##x_2 = vt_2 = 5 \times 90 = 450## m. The spatial separation between the two events : ##\Delta x = 450 - 150 = 300## m. Hence the distance between the two events in the ground frame : ##\boxed{\color{blue}{\Delta x = 300\,\text{m}}}##.

**(b) By direct use of the Galilean trasformations**

According to the Galilean transformations, the location of the two events are related as ##x' = x-vt\Rightarrow x = x' + vt'##. (This is the same equation which relates the event's position in the S frame as a function of the the events coordinates in the S' frame brought about by replacing the velocity ##v## by ##-v## since S is moving with a velocity in the negative ##x-x'## direction relative to S').

For a pair of events, the above transformation leads to their spatial separation as ##\color{red}{\Delta x = \Delta x' + v\Delta t'}##, taking "differences" and remembering that velocity ##v## is constant. Thus, for the pair of events in our problem, assuming ##\boxed{\color{blue} {\Delta x' = 0\,\text{m}}}## and temporal separation ##\Delta t' = 60\,\text{s}##, we have ##\boxed{\color{blue}{\Delta x}} = 0 + 5\times 60 = \boxed{\color{blue}{300\,\text{m}}}##.

**It looks good thus far, but this is where my confusions start, when the frame ##S'## accelerates.(2) Accelerating frames**

(a) By calculating the spacetime coordinates

Since those events happen at the same location (the tail end) for the train observer, the spatial separation between them is 0. Hence the distance between the two events in the moving frame : ##\boxed{\color{blue} {\Delta x' = 0\,\text{m}}}##.

The location of event ##E_1## in the Earth's frame : ##x_1 = \tfrac{1}{2}at_1^2 = \tfrac{1}{2}\times 0.1\times 30^2 = 45## m. The location of event ##E_2## in the Earth's frame : ##x_2 = \tfrac{1}{2}at_2^2 = \tfrac{1}{2}\times 0.1\times 90^2 = 405## m. The distance between the two events in the Earth's frame : ##\boxed{\color{blue}{\Delta x}} = 405 - 45 = \boxed{\color{blue}{360\,\text{m}}}##.

**(b) By using Galilean transformations for accelerated frames (my confusion scenario)**

The Galiean transformations link the spacetime coordinates for the same event in two different frames S and S' in standard configuration along the ##x## axis, viz. ##x' = x-\tfrac{1}{2}a^2\Rightarrow x= x'+\tfrac{1}{2}at'^2##, where we "invert" the original equations and write the ##x## coordinate of the event in terms of the ##x'## coordinates, remembering that in the process ##a\rightarrow -a## as the S frame moves with an acceleration ##-a## with respect to S'. For the pair of events, "differencing" the above equation leads to ##\color{red}{\Delta x = \Delta x'+at'\Delta t'}##. Question is - ##\color{red}{\text{what is}\, t'?}##. There are two events, so which t' to take? If we take ##\color{red}{t'}## to be the ##\color{red}{\text{average}}## of the two times, we get ##t' = \tfrac{t'_1+t'_2}{2} = \tfrac{30+90}{2} = 60## s. Using this, we get the spatial separation as : ##\boxed{\color{blue}{\Delta x}} = \Delta x'+at'\Delta t' = 0 + 0.1\times 60\times 60 = \boxed{\color{blue}{360\,\text{m}}}##, which yields the same answer as in part (a) above.

Question is - does it make sense to take ##\color{red}{t'\, \text{as the average of the times between those of the two events}}##?

There is a different way to do this problem which takes into account the

**velocity**of frame ##S'## at the time of event 1.

If we assume the spacetime location of event 1 to be the origin and try to find the location of event 2 relative to it, we observe that the frame ##S'## has a velocity relative to frame ##S## due to time of travel. Calling this velocity as ##v_0## and "starting" our clock with the timing of this event, the location of the second event is then given by : ##x_2 = x_2'+v_0 t_2' + \tfrac{1}{2} a{t'_{2}}^{2}##, where ##x'_2 = 0## since the 2nd event also happens to where we have shifted the origin and ##t' = 60##s, measured from the time of occurence of the first event. Hence, ##\boxed{\color{blue}{\Delta x}} = 0 + \underbrace{0.1\times 30}_{v_0} \times 60 + \tfrac{1}{2}\times 0.1\times 60^2 = 180 + 180 = \boxed{\color{blue}{360\,\text{m}}}##.

**Are the two methods above correct, or is it that my reasoning errors lead to the correct answers for these pair of events?**