Question

how to do part b

Answer 1

OF WHAT

Answer 2

part a)

\(\Delta x = x_2-x_1 = 480-1200=-720\)
\(\Delta t = 5\mu s\)

\(\Delta x' = \gamma(\Delta x - v\Delta t) = 0 \\\implies v = \frac{\Delta x}{\Delta t} = \dfrac{-720}{5 \mu} = -1.44\times 10^8 = -0.48c\)

Answer 3

\(v = -0.48c\)

negative sign tells me that \(S'\) is going in the negative x direction, but whats the intuition behind this ? how to make sense of the fact that two events happened at two different locations in S CAN happen in S' at the same location ?

Answer 4

physics>maths any day of the week for me!!

Answer 5

math is more awesome
physics = applied math

:)

Answer 6

exactly :") XD

Answer 7

familiar with special relativity ?

Answer 8

am a high school student.. but i luv what your doing!!!

Answer 9

I don't know either... but try playing around with the coordinates between (x1,0) and (x2,5\(\mu\)) and see what happens? Perhaps learning what happens to other related points may help your intuition and see what's going on generally? If you do explore and find stuff out, please tell me too, I'm curious!

Answer 10

https://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals_in_flat_space

I guess it has something to do with the timelike interval section in the link above? If the order of events can be changed why not the spatial location?

Answer 11

Oh right, it happens when the events are spatially separated by a distance of \(v\Delta t\). Then the moving object travels this separation in time \(\Delta t\) such that both the coordinates of the align exactly in the stationary frame. Nothing to do with special relativity. Thanks :)

Answer 12

But when the align they could potentially have exactly the same spatial coordinates (if their y and z coodinates are the same)? I don't see the intuition of how they could be in the same spot?

Answer 13

Suppose you're in a train is \(100\) meter long and let

The first event happens at \(x=100, t=0\)
The second event happens at \(x=0, t=1\)

In your frame the events are spatially separated by 100 meter

Answer 14

I am on the platform watching you go at speed \(100 m/s\) to the right.

w/o using special relativity, newton would say that the train would move exactly 100m in 1 second.
so both the events will happen at x=100 in my frame.

Answer 15

Oh, i c :P Basically, it's because the thing actually moves to that point from my frame. Thx!

Answer 16

yeah in platform frame, the front of the train at t = 0 and the back of the train at t = 1 sec will be at the same point

Answer 17

yeah in platform frame, the front of the train at t = 0 and the back of the train at t = 1 sec will have same x coordinate

Answer 18

Do you have to move even faster when the speed is high to compensate for relativistic effects?

Answer 19

if the spatial separation of the events is \(v\Delta t\) in the train frame, then the two events will have same x coordinate in the platform frame.

shouldn't matter what \(v\) is right ?

Answer 20

Classical relativity :
\(\Delta x' = \Delta x - v\Delta t\)

special relativity
\(\Delta x' = \gamma(\Delta x - v\Delta t)\)

Answer 21

either way, letting \(\Delta x'=0\) gives \(\Delta x = v\Delta t\)