Question

http://puu.sh/o92q9/fd27ddb5f2.png

Answer 1

@Astrophysics

Answer 2

\(\gamma = \dfrac{1}{\sqrt{1-0.75^2}}=1.5\)

Answer 3

omg I hate velocity addition

Answer 4

One thing that never goes away: Draw a diagram!

Answer 5

Actually velocity addition is needed if you solve it in Earths frame which is a bit tricky

Answer 6

I can show you that after, but keep going you got gamma correct!

Answer 7

So what you're doing is solving it in the spaceships frame (label it S') and then transforming it to earths' frame (label it S)

Answer 8

since \(9\times 10^9m\) is the distance measured in A's frame,
the signal must take a time of \(9\times 10^9/c = 30s\) to reach A, in A's frame.

Answer 9

In A's frame, suppose it takes \(a\) seconds for the reply to reach from A to B.
Then the total time taken for the round trip in A's frame is given by
\[\Delta t = a+30\]

Answer 10

In B's frame, the measured time for the roundtrip is "proper time". so we can use time dilation equation to relate the two measurements :

\[\Delta t_0 = \dfrac{\Delta t}{\gamma} =\dfrac{a+30}{\gamma}\]

It remains to find \(a\) hmm

Answer 11

does that look good so far ?
im gonna see if finding \(a\) is easy...

Answer 12

Hey sorry, I think what you wrote looks ok, but I'm not entirely sure how you will find a, this one is kind of hard I can't find what I did but I'll rework it out, the answer however is \[\Delta t = 22.7s\]

Answer 13

im getting \(a = 4.29\)
http://www.wolframalpha.com/input/?i=solve+a+%3D+%289*10^9+-+%28a%2B9*10^9%2F299792458+%29*0.75*299792458%29%2F%28299792458%29


and \(\Delta t_0 = \dfrac{4.29+30}{1.5} = 22.87\)

Answer 14

That should do it! Maybe I rounded something wrong or something, but what you did looks good to me

Answer 15

Nice! happy studying qm :)

Answer 16

Lol thanks, it's a pain! Once I'm done in about 2 weeks lets do more problems, I got quite a bit that are fun, and some that are very tricky and frustrating xD. And if you want we can derive some of the equations to like the Lorentz transformations (and some theory) if you're interested, have fun!

Answer 17

Oh one last question I think you might like say we have a space that is an isosceles triangle of length x0 and width y0. It speeds past an observer at relativistic velocity, v so if we adjust the velocity v lets the say, how would the observer measure it? Like it y0 too small and x0 is correct, is the ships length longer x0 etc, what do you think will happen?
I think you like a qualitative questions right haha?

Answer 18

spaceship*

Answer 19

nothing relativistic happens in vertical direction, so y0 remains same in both frames

Answer 20

x0 must shrink in the observer frame as this length is in the direction of relative velocity and thus contracts

Answer 21

Bravo! So our spaceships turns into an equilateral triangle, as the contraction is only in the x - direction, ganeshie you're too good!

Answer 22

if you watch food factory in discovery channel, you might like this idea :

Answer 23

there is a method for separating large potatoes from the small potatoes
im not finding the right picture but it looks something like this
http://media.syracuse.com/news/photo/2010/09/8929800-large.jpg

Answer 24

The small potatoes slip through the holes and the large potatoes go to the next stage.

if the potatoes are moving at great enough speed relative to the belt, all the potatoes will shrink and must slip through the holes !

Answer 25

None will go to the next stage... its not weird as we know relativity now :)

Answer 26

Hahaha, nice!

Answer 27

hopefully the earth's gravitational force is strong enough to attract the potato down