Science > Physics > Could someone check this is right - SR length and time
| Topic: |
Science > Physics |
| User: |
"blackboab" |
| Date: |
02 Oct 2005 09:48:57 AM |
| Object: |
Could someone check this is right - SR length and time |
A train is travelling at 0.8 times the speed of light.
an observer on board measures the speed of light and gets c and a meter
stick measures a meter as before . there is no time dilation on length
contraction as far as the observer on board the train is concerned. in
this inertial reference frame all is as it should be.
an observer looking at the train sees it differently. from his
perspective the clock on the train is running slower than his own
clock. the meter rod on board the train also looks shorter than his
meter rod.
the correction factor for the time is 1/ sqrt (1 - v*v/c*c) which is
1/sqrt(1 - 0.8c*0.8c / c*c) = 1/sqrt( 1-0.64) = 1/sqrt(0.36) = 1/0.6 =
1.66
therefore it takes 1.66 seconds by his clock before the clock on the
train moves 1 second.
similarly the meter rod on the train appears to him to be only 0.6
meters long
(the correction factor is sqrt (1 - v*v/c*c)= 0.6 )
of course the observer on the train notices no such time dilation or
length contraction of his clock and rod.
but if he were to observe the clock of the stationary observer it would
appear to him to take 1.66 seconds to move one second and the
stationary observers meter rod would appear to be only 0.6 meters long.
assuming the train starts and stops without acceleration then the two
cloks would have been found to have beaten out an exact same number of
seconds during the time of the journey.
the clock on the moving train would only run slowly in an absolute
sense if it underwent acceleration.
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 10:17:02 AM |
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"blackboab" <blackboab@gmail.com> wrote in message
news:1128264537.482150.205090@o13g2000cwo.googlegroups.com...
| A train is travelling at 0.8 times the speed of light.
Nah... 0.6c, you have hard numbers.
The train is 32 cars long. Each car is 60 meters long.
The speed of light is 5 car lengths per microsecond.
5 * 60 / 1,000,000 = 300,000,000 meters/second.
Speed of train, 3 car lengths per microsecond, 0.6c.
Light leaves the station as the caboose leaves the station, catches up
with the engine at 80 car lengths down the track, taking 16 microseconds
to get there.
The light traveled 80 car lengths at speed 5 , 80/5 = 16
The engine started the race at 32, (80-32)/3 = 16
Then the light reflects and meets the caboose 4 microseconds later,
which is now
60 car lengths down the track.
The trackside clock records 20 usec.
The train clock records 16 usec.
Why?
Hint:
[quote]
we establish by definition that the "time" required by light to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
[quote]
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
,
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Over to you.
Androcles.
| an observer on board measures the speed of light and gets c and a
meter
| stick measures a meter as before . there is no time dilation on length
| contraction as far as the observer on board the train is concerned. in
| this inertial reference frame all is as it should be.
|
| an observer looking at the train sees it differently. from his
| perspective the clock on the train is running slower than his own
| clock. the meter rod on board the train also looks shorter than his
| meter rod.
|
| the correction factor for the time is 1/ sqrt (1 - v*v/c*c) which is
| 1/sqrt(1 - 0.8c*0.8c / c*c) = 1/sqrt( 1-0.64) = 1/sqrt(0.36) = 1/0.6 =
| 1.66
|
| therefore it takes 1.66 seconds by his clock before the clock on the
| train moves 1 second.
|
| similarly the meter rod on the train appears to him to be only 0.6
| meters long
| (the correction factor is sqrt (1 - v*v/c*c)= 0.6 )
|
| of course the observer on the train notices no such time dilation or
| length contraction of his clock and rod.
|
| but if he were to observe the clock of the stationary observer it
would
| appear to him to take 1.66 seconds to move one second and the
| stationary observers meter rod would appear to be only 0.6 meters
long.
|
| assuming the train starts and stops without acceleration then the two
| cloks would have been found to have beaten out an exact same number of
| seconds during the time of the journey.
|
| the clock on the moving train would only run slowly in an absolute
| sense if it underwent acceleration.
|
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 12:07:58 PM |
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if the observer O at the front of the train fires a beam of light at a
mirror M at the end of the train (train is of any length you desire )
are you saying the light takes longer to get from O to M than from M to
O ?
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 02:38:50 PM |
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"blackboab" <blackboab@gmail.com> wrote in message
news:1128272878.684110.179890@g44g2000cwa.googlegroups.com...
| if the observer O at the front of the train fires a beam of light at a
| mirror M at the end of the train (train is of any length you desire )
| are you saying the light takes longer to get from O to M than from M
to
| O ?
Nope.
Why don't you work the problem yourself, I gave you easy numbers?
It does you no good if you look to me for answers.
I'll help, but you can't learn without trying.
Androcles.
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| User: "Dirk Van de moortel" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 04:11:59 PM |
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"blackboab" <blackboab@gmail.com> wrote in message news:1128264537.482150.205090@o13g2000cwo.googlegroups.com...
A train is travelling at 0.8 times the speed of light.
With respect to something.
Let's say, with rspect to the station.
an observer on board measures the speed of light and gets c and a meter
stick measures a meter as before . there is no time dilation on length
contraction as far as the observer on board the train is concerned. in
this inertial reference frame all is as it should be.
an observer looking at the train sees it differently. from his
perspective the clock on the train is running slower than his own
clock. the meter rod on board the train also looks shorter than his
meter rod.
the correction factor for the time is 1/ sqrt (1 - v*v/c*c) which is
1/sqrt(1 - 0.8c*0.8c / c*c) = 1/sqrt( 1-0.64) = 1/sqrt(0.36) = 1/0.6 =
1.66
therefore it takes 1.66 seconds by his clock before the clock on the
train moves 1 second.
similarly the meter rod on the train appears to him to be only 0.6
meters long
(the correction factor is sqrt (1 - v*v/c*c)= 0.6 )
of course the observer on the train notices no such time dilation or
length contraction of his clock and rod.
but if he were to observe the clock of the stationary observer it would
appear to him to take 1.66 seconds to move one second and the
stationary observers meter rod would appear to be only 0.6 meters long.
assuming the train starts and stops without acceleration then the two
cloks would have been found to have beaten out an exact same number of
seconds during the time of the journey.
No, the situation is not symmetric.
The train must jump twice into another inertial frame:
1) from the station frame into the moving frame
2) from the moving frame into the station frame.
The station does not make jumps like that.
The observer in the station will feel nothing.
The observer on the train will get killed twice.
Suppose the train starts at the event where both clocks
read t = 0 and t' = 0. Assume the train immediately gets
speed v w.r.t. the station and keeps this speed during a
time T, as seen on the station clock. At that moment the
train will be at distance vT, according to the station.
You can verify that the train clock will read T/gamma
at this event. If the train then suddenly stops and the
clock is not destroyed, it will still read T/gamma. Since
the train remains at rest now, its clock will continue
running at the same rate as the station clock, but it will
remain ahead. At station time T+k, the train clock will
read T/gamma + k.
the clock on the moving train would only run slowly in an absolute
sense if it underwent acceleration.
Not necessarily - jumping frames (which can be seen
as limits of infinite acceleration during zero time) does the
job as well.
Dirk Vdm
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 02:02:50 AM |
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Dirk
thanks for that.
so during a time t when the train is moving with a constant velocity
its clock will beat out the same number of seconds as the station (ary)
clock.
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| User: "Dirk Van de moortel" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 11:19:51 AM |
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"blackboab" <blackboab@gmail.com> wrote in message news:1128322970.559049.61080@o13g2000cwo.googlegroups.com...
Dirk
thanks for that.
so during a time t
On who's clock?
when the train is moving with a constant velocity
its clock will beat out the same number of seconds as the station (ary)
clock.
beat out between which two events?
Dirk Vdm
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 01:07:17 PM |
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Dirk Van de moortel wrote:
"blackboab" <blackboab@gmail.com> wrote in message news:1128322970.559049.61080@o13g2000cwo.googlegroups.com...
Dirk
thanks for that.
so during a time t
On who's clock?
when the train is moving with a constant velocity
its clock will beat out the same number of seconds as the station (ary)
clock.
beat out between which two events?
could they not use some solar event such as todays eclipse ?
they could agree to start their clocks when the eclipse started and
stop their clocks when the eclipse ended.
when they compared their clocks they would both have recorded the same
duration for the eclipse ?
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| User: "Dirk Van de moortel" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 01:41:21 PM |
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"blackboab" <blackboab@gmail.com> wrote in message news:1128362837.482404.20030@f14g2000cwb.googlegroups.com...
Dirk Van de moortel wrote:
"blackboab" <blackboab@gmail.com> wrote in message news:1128322970.559049.61080@o13g2000cwo.googlegroups.com...
Dirk
thanks for that.
so during a time t
On who's clock?
when the train is moving with a constant velocity
its clock will beat out the same number of seconds as the station (ary)
clock.
beat out between which two events?
could they not use some solar event such as todays eclipse ?
they could agree to start their clocks when the eclipse started and
stop their clocks when the eclipse ended.
when they compared their clocks they would both have recorded the same
duration for the eclipse ?
That would be a very bad example, since the times of first and
second exterior contact (i.e. start and end of eclipse), and even
the difference between these, very much depend on where you
are when looking at them. For some people on Earth there
probably was no eclipse at all today.
One should take events on which everyone unambiguously agrees.
For instance the event where *I see* first contact and the event
where *I see* last contact. I could raise flags on those occasions.
Different observers measure different times between those fixed
and universally agreed upon events. For instance, if I measure a
time T between these events, then you, moving w.r.t. me, will
measure a time T*gamma between them.
By the way, in my first reply (on 2-Oct) I made a little mistake.
As an exercise, try to find it.
Dirk Vdm
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 02:00:58 PM |
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ok. how about observing a binary star system and using its period as
the start and end point of the observation ?
re deliberate mistake : you can only die once ?
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 02:51:42 PM |
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"blackboab" <blackboab@gmail.com> wrote in message
news:1128366058.523776.217450@o13g2000cwo.googlegroups.com...
| ok. how about observing a binary star system and using its period as
| the start and end point of the observation ?
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| re deliberate mistake : you can only die once ?
Dunno who you are writing to...
Androcles.
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| User: "Dirk Van de moortel" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 02:58:11 PM |
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"Androcles" <Androcles@ MyPlace.org> wrote in message news:iZf0f.53966$iW5.19682@fe3.news.blueyonder.co.uk...
"blackboab" <blackboab@gmail.com> wrote in message
news:1128366058.523776.217450@o13g2000cwo.googlegroups.com...
| ok. how about observing a binary star system and using its period as
| the start and end point of the observation ?
|
| re deliberate mistake : you can only die once ?
Dunno who you are writing to...
Of course you know, little liar ;-)
Dirk Vdm
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
04 Oct 2005 01:20:06 AM |
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dirk
was the observation of the binary star a reasonable event they could
set their clocks by ?
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| User: "Dirk Van de moortel" |
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| Title: Re: Could someone check this is right - SR length and time |
04 Oct 2005 02:18:35 AM |
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"blackboab" <blackboab@gmail.com> wrote in message news:1128406806.649584.239840@g47g2000cwa.googlegroups.com...
dirk
was the observation of the binary star a reasonable event they could
set their clocks by ?
Before I answer this, do the exercise I gave you.
When you have done that, I might help you find
the answer to this question (and formulate it more
properly) as another exercise.
Dirk Vdm
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
05 Oct 2005 11:25:05 AM |
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you dont know do you, fair enough
i think i am on the right lines
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| User: "Dirk Van de moortel" |
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| Title: Re: Could someone check this is right - SR length and time |
05 Oct 2005 11:34:20 AM |
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"blackboab" <blackboab@gmail.com> wrote in message news:1128529505.376009.244210@g44g2000cwa.googlegroups.com...
you dont know do you, fair enough
Oh Yes, I perfectly know.
You are just not ready for it yet.
i think i am on the right lines
I think you can't do the exercise. Have you tried? It's not difficult.
If you need help, say the word, I can give a few hints.
Dirk Vdm
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| User: "Dirk Van de moortel" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 02:04:11 PM |
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"blackboab" <blackboab@gmail.com> wrote in message news:1128366058.523776.217450@o13g2000cwo.googlegroups.com...
ok. how about observing a binary star system and using its period as
the start and end point of the observation ?
re deliberate mistake : you can only die once ?
No deliberate mistake.
Try to find it first.
Dirk Vdm
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| User: "Sam Wormley" |
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| Title: Re: Could someone check this is right - SR length and time |
05 Oct 2005 11:31:56 AM |
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blackboab wrote:
ok. how about observing a binary star system and using its period as
the start and end point of the observation ?
Those times are different for every observer!
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| User: "Sam Wormley" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 02:05:43 PM |
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blackboab wrote:
ok. how about observing a binary star system and using its period as
the start and end point of the observation ?
re deliberate mistake : you can only die once ?
Those times are different for every observer!
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| User: "Sam Wormley" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 09:39:28 AM |
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blackboab wrote:
so during a time t when the train is moving with a constant velocity
its clock will beat out the same number of seconds as the station (ary)
clock.
The train will see the station clock run slow.
The station will see the train clock run slow.
Such is the way the world works.
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| User: "Steven Gray" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 12:38:20 PM |
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"blackboab" <blackboab@gmail.com> wrote in news:1128264537.482150.205090
@o13g2000cwo.googlegroups.com:
but if he were to observe the clock of the stationary observer it would
appear to him to take 1.66 seconds to move one second and the
stationary observers meter rod would appear to be only 0.6 meters long.
assuming the train starts and stops without acceleration then the two
cloks would have been found to have beaten out an exact same number of
seconds during the time of the journey.
If it starts and stops, then by definition it has accelerated.
the clock on the moving train would only run slowly in an absolute
sense if it underwent acceleration.
You're asking what a theory says about a situation that the theory doesn't
allow. The only sensible answer that you're going to get is that there are
no sensible answers to nonsense questions.
--
Steve Gray
sgray2@cfl.rr.com
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 02:39:46 PM |
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"Steven Gray" <sgray2@NOcfl.rr.comSPAM> wrote in message
news:Xns96E38B0326245sgray2cflrrcom@65.32.5.122...
| "blackboab" <blackboab@gmail.com> wrote in
news:1128264537.482150.205090
| @o13g2000cwo.googlegroups.com:
|
| > but if he were to observe the clock of the stationary observer it
would
| > appear to him to take 1.66 seconds to move one second and the
| > stationary observers meter rod would appear to be only 0.6 meters
long.
| >
| > assuming the train starts and stops without acceleration then the
two
| > cloks would have been found to have beaten out an exact same number
of
| > seconds during the time of the journey.
|
| If it starts and stops, then by definition it has accelerated.
|
| > the clock on the moving train would only run slowly in an absolute
| > sense if it underwent acceleration.
| >
| You're asking what a theory says about a situation that the theory
doesn't
| allow. The only sensible answer that you're going to get is that
there are
| no sensible answers to nonsense questions.
Don't be silly.
Androcles.
| --
| Steve Gray
|
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 02:37:48 PM |
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it is a thought experiment and in the thought experiment we are only
concerned with a portion of the trains motion when it is travelling at
a constant velocity.
so for that period of motion the clock on board the train will beat out
the exact same number of seconds as the clock of the observer on the
ground,
there is no acceleration in this thought experiment.
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| User: "Androcles Androcles@ MyPlace.org" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 02:51:35 PM |
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"blackboab" <blackboab@gmail.com> wrote in message
news:1128281868.046310.107930@g47g2000cwa.googlegroups.com...
| it is a thought experiment and in the thought experiment we are only
| concerned with a portion of the trains motion when it is travelling at
| a constant velocity.
Yep. Stick with it, work out why Einstein thinks he can fool you
into thinking clocks slow down.
Androcles.
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| so for that period of motion the clock on board the train will beat
out
| the exact same number of seconds as the clock of the observer on the
| ground,
|
| there is no acceleration in this thought experiment.
|
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| User: "Steven Gray" |
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| Title: Re: Could someone check this is right - SR length and time |
02 Oct 2005 03:47:48 PM |
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"blackboab" <blackboab@gmail.com> wrote in news:1128281868.046310.107930
@g47g2000cwa.googlegroups.com:
it is a thought experiment and in the thought experiment we are only
concerned with a portion of the trains motion when it is travelling at
a constant velocity.
so for that period of motion the clock on board the train will beat out
the exact same number of seconds as the clock of the observer on the
ground,
there is no acceleration in this thought experiment.
If it's a thought experiment, then think about how you're going to compare
the clocks. If you're going to do it by bringing them together in the same
time and place, then something has to accelerate.
--
Steve Gray
sgray2@cfl.rr.com
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| User: "blackboab" |
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| Title: Re: Could someone check this is right - SR length and time |
03 Oct 2005 03:22:55 AM |
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a powerful telescope ?
distant galaxies cannot be brought here for observation yet physicists
seem to know an awful lot about the speed of light coming from them.
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