Question

A train travelling at a speed of 0.8 times the speed of light takes 8x10^-6 seconds to pass an observer on a nearby platform.
a) What is the time interval in the train's frame of reference?
? seconds.
What train length is measured by
a) observers on the train?
?m
b) the observer on the platform?
?m

Answer 1

this problem is based on Einstein's theory of relativity

Answer 2

length would seem to decrease to the observer on platform.

Answer 3

i think we need to see like whos reference frame has proper time. i think that it is the stationary object but i am not sure.

Answer 4

it could be equivalently said that they both are in motion with respect to each other.

Answer 5

however i have done these types of question as I described above ... i never understood these things properly.

Answer 6

lol me either :P

Answer 7

i will update when i find ... I've slacking too much.

Answer 8

LOL thx bro

Answer 9

wc

Answer 10

since they will be measuring according to their own frame of references, they both will measure length contraction (on the other frame of reference) <--- according to one of my senior.

Answer 11

You need to use the Lorentz transformations for solving problems about simultaneity. The Lorentz transformations relate the space and time intervals between two events observed in one inertial systems to the space and time intervals observed in another inertial system.

Take a look at:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html
and
http://en.wikipedia.org/wiki/Special_relativity

Using the equations on these sites, finding the answers to the problem are just about plugging in numbers and chugging out answers.

I need to go for now, but I will come back in a couple hours and show you how to solve step by step.

Answer 12

ok thx

Answer 13

Alrighty then. Let's get to it! (and you'll probably be happy to know there's a simpler way to solve than using Lorenntz that's much easier to understand)

a) What is the time interval in the train's frame of reference?
This is asking about time dilation.

Clocks in motion relative to an observer are measured to run slower by a factor
\[\gamma = (1 - v ^{2}/c ^{2})^{-1/2}\]
Where v is the velocity of the moving object and c is the speed of light.

What that means is that
\[t = t' \times \gamma\]
Where t is the time interval measured by the person on the platform, and t' is the time interval measured by someone on the train.

Answer 14

a) continued...

Now we plug in our givens
\[8 \times 10^{-6} = t' \times 1/\sqrt{1-0.8^{2} c ^{2}/c ^{2}}\]
and solve for t'

Notice that \[c ^{2}/c^{2} = 1\]

This means we can cancel that part out like so
\[8 \times 10^{-6} = t' \times 1/ \sqrt{1 - 0.8^{2}}\]

Which lets us solve the square root much easier
\[8 \times 10^{-6} = t' \times 1/0.6\]

Almost done now!
\[8 \times 10^{-6} = t' \times 1.667\]

Finally
\[t' = 4.8 \times 10^{-6}\]

Since \[4.8 \times 10^{-6} < 8.0 \times 10^{-6}\] we can feel confident that we're correct, because the time measured on the train should be less than the one measured by the person on the platform.

Answer 15

Do you have any concerns about what I did?

Answer 16

no its good thanks man i get it :)

Answer 17

Would you like me to walkthrough the length contraction questions?