Question

Standing beside railroad tracks, we are suddenly startled by a relativistic boxcar traveling past us as shown in the figure. Inside, a well-equipped hobo fires a laser pulse from the front of the boxcar to its rear. The laser pulse is reflected back to the hobo. The time interval measured by hobo for the roundtrip of the laser pulse is Δt0. Express the time interval measured by us standing beside the tracks, in terms of Δt0.

Answer 1

Let D the length of boxcar and c be the speed of light.
In the frame of hobo inside the boxcar, the laser pulse travels a total distance of 2D. The time taken for the round trip is then
\[\Delta t_0 = \dfrac{2D}{c}\]

Answer 2

In the frame of us beside the tracks, outside the boxcar :
Let \(v\) be the speed of boxcar relative to us.

Let \(\Delta t_1\) be the time taken for the laser pulse to go from front to rear. The boxcar travels a distance of \(v\Delta t_1\) towards the laser pulse. Therefore time taken by laser pulse to reach from front to rear end is
\[\Delta t_1 = \dfrac{D-v\Delta t_1}{c}\]

Answer 3

In the frame of us beside the tracks, outside the boxcar :

By similar reasoning, the time taken by laser pulse to go from rear end to front is
\[\Delta t_2 = \dfrac{D+v\Delta t_2}{c}\]

Answer 4

In the frame of us beside the tracks, outside the boxcar :

the time taken for the roundtrip of laserpulse is then
\[\Delta t = \Delta t_1+\Delta t_2 = \dfrac{2D}{c[1-(v/c)^2]}= \dfrac{\Delta t_0}{1-(v/c)^2}\]

Answer 5

But the time dilation equation from my textbook says
\[\Delta t = \dfrac{\Delta t_0}{\sqrt{1-(v/c)^2}}\]

what am i doing wrong ?

Answer 6

have you constracted the length of the car?

that extra square root would be a quick fix

Answer 7

hey i have just used the speed of light postulate as shown above
textbook hasn't covered length contraction yet...

Answer 8

that sqrt is really bothering me yeahh

Answer 9

as you observe it, the boxcar gets shorter in the direction of its travel. so you observe the hobo doing everything in slow motion and you observe the boxcar as shrunk. because c is constant, those two things have to happen to ensure that you observe the photon hitting the leading wall of the boxcar at the same time as the hobo. the formula is the same, the contraction factor is \(\gamma\), just as it is with time dilation.

which is why, without really looking at it, i think that's all you need to do to get square with the book.

Answer 10

Awesome! @baru have you understood yet why my equations are wrong ?

Answer 11

i think if you do this:

\(\Delta t_1 = \dfrac{D \gamma-v\Delta t_2}{c} \)
\(\Delta t_2 = \dfrac{D \gamma+v\Delta t_2}{c} \)

the it works

if that doesn't get you there, i have a good book that i can maybe dig out

Answer 12

@IrishBoy123 shouldn't the length in outisder frame be \(\dfrac{D}{\gamma}\) ?

My mistake is that I have wrongly assumed that \(D\) is same in both reference frames !

Answer 13

i understood that you didn't take into account length contraction while writing the equations from an outside perspective...

but qualitatively, the phenomenon still confuses me,
because even with the modified equations, light takes a shorter amount of time going from front to rear, than the other way, but we have only one time dilation formula
why is that?

Answer 14

you and me have the exact same confusion about the time dilation equation
i think i have figured it out this morning... il take explain what i think step by step

Answer 15

Lets define two things :
1) proper time
2) proper length

Answer 16

proper time :

When two events occur at the same location in an inertial reference frame, the time interval between them, measured in that frame, is called the proper time interval or the proper time.

Answer 17

Before even defining `proper length`, lets prove below fact

Measurements of the same time interval from any other inertial reference frame are always greater than the proper time.

Answer 18

since we don't know yet how the length changes with relative motion, it makes sense to put the length variable perpendicular to the relative motion while studying the relation between time in different frames

Answer 19

yes,i remember that
we fix two mirrors, one on the ceiling and one on the roof of the boxcar, and measure time for light to reflect between them right

Answer 20

one on the floor*

Answer 21

yes, lets do that again

Answer 22

http://i.stack.imgur.com/UdADX.png

Answer 23

in above link,
(a) is the frame inside the boxcar, sally is sitting inside the boxcar
(b) is the frame outside the boxcar, sam is standing on the platform

Answer 24

the boxcar is going toward right
and the light is dancing up and down

so this length D "must not" be affected by relative motion. there is no relative motion between sam and sally in the vertical axis

Answer 25

yep

Answer 26

so both sam and sally measure the same length D for the distance between mirrors

Answer 27

our focus in this experiment is to study the relation between time in both reference frames

Answer 28

with the technicalities aside, the relation is just algebra, lets do it quick

Answer 29

In sally's frame, inside the boxcar :
\[\Delta t_0 = \dfrac{2D}{c}\]

This is "proper time" because sally uses just "one clock" to measure this interval.

Answer 30

alright

Answer 31

In sam's frame, outside the boxcar, the light has to travel a distance of 2L:

\[\Delta t = \dfrac{2L}{c}\]

This is not "proper time" because sam must use two different clocks (one located at left end and other at right end) to measure this interval.

Answer 32

from the geometry it is easy to see that
\[L^2 = D^2 + (v\Delta t/2)^2\]

combining the 3 equations we end up with the time dilation equation
\[\Delta t = \dfrac{\Delta t_0}{\sqrt{1-(v/c)^2}}\]

This holds as long as \(\Delta t_0\) is the "proper time", measure using only one clock.

Answer 33

with me so far ?

Answer 34

http://i.stack.imgur.com/UdADX.png

im using this pic for the derivation of time dilation eqn

Answer 35

i didnt exactly get the we need to use 'two clocks' part

proper time: measure when we are at rest w.r.t the whole mirror apparatus

not proper time: apparatus is moving with respect to us.

right?

Answer 36

you're right, but i think it helps a lot to see it from the other angle too

Answer 37

sally requires just one clock to measure the time between the two events :

1) take the timestamp of the clock at emission of light
2) take the timestamp of the same clock at return of light

subtracrt the timestamps to get the proper time

Answer 38

sam cannot do the same using just one clock

Answer 39

he requires two different clocks because for him, both the events are spatially separated.

he has to take the timestamps from the two clocks located at the respective places.

Answer 40

proper time : timestamps from the same clock

not proper time : timestamps from different clocks

Answer 41

ohh, ok got it

Answer 42

forgetting length,
do we agree on when the time dilation equation holds ?

Answer 43

\[\Delta t = \dfrac{\Delta t_0}{\sqrt{1-(v/c)^2}}\]

holds only when \(\Delta t_0\) is "proper time" measured at the same location using one clock.

Answer 44

yep, agreed

Answer 45

also lets acknowledge again that \(\Delta t \gt \Delta t_0\) as \(\dfrac{1}{\sqrt{1-(v/c)^2}}\) for all \(v\lt c\)

Answer 46

yep, i see that

Answer 47

so we have this fact now :

`Measurements of the same time interval from any other inertial reference frame are always greater than the proper time.`

Answer 48

yep

Answer 49

lets look at length contraction
after that we can tackle sending laser pulse from left<->right

Answer 50

if i understand correctly, that sending laser pulse from left<->right is the source of our confustion right ?

Answer 51

yes!
our measurement for time taken for light should be greater than hobos regardless of direction

Answer 52

time dilation equation talks about "proper time" right ?

Answer 53

yes

Answer 54

so can i rephrase it like this :

our measurement for time taken for light should be greater than hobos "measurement for `proper` time" regardless of direction

Answer 55

yes

Answer 56

lets finish length contraction quick and get back to this..

Answer 57

alright

Answer 58

sam is on the platform

sally is in the boxcar

Answer 59

now sam and sally want to measure the length of the platform

Answer 60

sam's frame :

sam is stationary relative tot he platform, so sam's length measurements are "proper". Say the measured length of the platform by sam is \(L_0\). Also, sam notes that sally moves through this length in \(\Delta t\). Note that this time measurement requires two clocks. so it is not proper time.
\[L_0 = v\Delta t\]

Answer 61

Sally's frame :
For Sally, however, the platform is moving past her. She finds that the two events measured by Sam occur at the same place in her reference frame. She can time them with a single stationary clock, and so the interval Δt0 that she measures is a proper time interval. To her, the length L of the platform is given by

\[L = v\Delta t _0\]

Answer 62

divide both equations and get
\[\dfrac{L}{L_0} = \dfrac{\Delta t_0}{\Delta t}\]

Answer 63

using the earlier time dilation result, above equation becomes


\[\dfrac{L}{L_0} = \dfrac{\Delta t_0}{\Delta t} = \sqrt{1-(v/c)^2}\]

Answer 64

same as
\[L = L_0\sqrt{1-(v/c)^2}\]

since \(\sqrt{1-(v/c)^2} \) is less than 1 for all v < c, we have \(L\lt L_0\)

Answer 65

basically we have two results so far :

Answer 66

ok... i'm a bit confused about the 'proper' and two clocks thing again,

it the two events occur two different points in space w.r.t an observer, then the quantity is 'improper'?

Answer 67

yes

Answer 68

but there is nothing improper about it

Answer 69

alright

Answer 70

"proper time" is just a phrase used to refer to the time measurement made at a single location using a single clock

Answer 71

ok, got it

Answer 72

using those two results i hope we can handle our main problem of sending laser pulses left and right

Answer 73

me too :D

Answer 74

il rephrase the problem

Answer 75

Standing beside railroad tracks, we are suddenly startled by a relativistic boxcar traveling past us as shown in the figure. Inside, a well-equipped hobo fires a laser pulse from the front of the boxcar to its rear. (a) Is our measurement of the speed of the pulse greater than, less than, or the same as that measured by the hobo? (b) Is his measurement of the flight time of the pulse a proper time? c) which time measurement is greater, ours or hobos ?

Answer 76

http://assets.openstudy.com/updates/attachments/5703d696e4b03d42c803ca4f-ganeshie8-1459869390121-zzz.jpg

Answer 77

yep, looks good.

Answer 78

for part a, the speed of laser pulse must be same in all inertial reference frames according to speed of light postulate

Answer 79

for part b, what do u think

Answer 80

how many clocks does the hobo require to measure the time interval ?

Answer 81

i was about to say proper time,

but now i'm thinking, two events spatially seperated

Answer 82

yeah

The first timestamp is taken at the front of the car
the second timestamp is taken at the rear of the car

so he must use two different clocks

Answer 83

wow, that one completely caught me off guard...
go on

Answer 84

so clearly the time dilation equation that we have derived earlier does not apply here.

Answer 85

yep

Answer 86

However the length contraction equation still applies right ?

Answer 87

yep

Answer 88

Because the boxcar is moving left to right in the frame of outsider, the length of the boxcar must shrink by a factor of \(\gamma\)

\[L = \dfrac{L_0}{\gamma }\]

Answer 89

from the outsider frame, the time taken by the laser pulse to reach from front to rear is given by

\[\Delta t = \dfrac{L - v\Delta t}{c} = \dfrac{L_0/\gamma - v\Delta t}{ c}\]

Answer 90

isolating \(\Delta t\) do we get
\[\Delta t = \dfrac{L_0}{\gamma c(1+v/c)} \]

Answer 91

notice that \(\dfrac{L_0}{c}\) will be the time measured in the hobos frame inside the boxcar

Answer 92

clearly the time measured in the outside frame is "less" than the time measured in the inside frame

Answer 93

that is against what we expect

Answer 94

your saying that is ok,
because \(L_0/c\) is not the proper time

Answer 95

?

Answer 96

\(L_0/c\) is not proper time, so this is not violating time dilation equation. But I do see the issue here. If the clocks run slow inside the boxcar, they must run slow for all events. Not just the ones with proper time measurements. Hmm...