How Do You Calculate Spacetime Intervals for a Merry-Go-Round Ride?

[ssmfour]

Homework Statement

Two friends with super-synchronized clocks. One goes on merry-go-round the other stands outside at the same point. The merry-go-round has a constant v of 45 m/s and takes 300s to go around. Find dt, ds, and dτ.

Homework Equations

v=dx/dt
v=2πR/p using p=period
ds^2=dt^2-dx^2
dτab=(1-v^2)^1/2 x dtab
dt≥ds≥dτ

The Attempt at a Solution

Put into SR Units first:
45 m/s (1 s/3x10^8m)=1.5 x 10^-7
300 s (1 x 10^9 ns/1 s)=3 x 10^11 ns

v=dx/dt
1.5 x 10^-7=(3.0 x 10^11 ns)/dt
dx= (1.5 x 10^-7)(3.0 x 10^11 ns)
dx=45000 ns

ds=√(3.0 x 10^11)^2 - (45000 ns)^2
ds=√9.0 x 10^22
ds= 3.0 x 10^11

v=2πR/p
1.5 x 10^-7= 6.28R/(3.0 x 10^11 ns)
R=7165 ns

dtab=2πR/v
=2π(7165 ns)/(1.5 x 10^-7)
=3.0 x 10^11


dτab=(1+(-v^2))^1/2 x dtab
= (1-1/2(v^2)) x dtab
= (1-1/2(2.25 x 10^-14)) x dtab
because v ≪ 1:
dτab=dtab

My answer fits the equation dt≥ds≥dτ but it seems wrong that they're all the same. Never had any examples before where this has happened. Was hoping someone can just check to see if I made any mistakes.

Thanks

 

[anathema]

!

Thank you for your post. I am a scientist and I would be happy to help you with your question.

First, let's start by writing out the given information in a more organized manner:

Given:
- Two friends with super-synchronized clocks
- One friend is on a merry-go-round with a constant velocity of 45 m/s
- The other friend is standing outside at the same point
- The merry-go-round takes 300 seconds to go around

Unknowns:
- dt (change in time)
- ds (change in space)
- dτ (proper time)

From the given information, we can deduce that the change in time (dt) and the change in space (ds) will be the same for both friends since they are traveling at the same point and at the same speed. This means that dt = ds.

To find dt, we can use the equation v = dx/dt, where v is the velocity (45 m/s) and dx is the change in space (ds). Solving for dt, we get:

dt = dx/v
dt = ds/45 m/s

To find ds, we can use the equation ds^2 = dt^2 - dx^2. Since we know that dt = ds, we can substitute that in and get:

ds^2 = ds^2 - dx^2
dx^2 = 0
dx = 0

This means that the change in space (ds) is equal to 0, which makes sense because the friend standing outside is not moving, so there is no change in space for them.

To find dτ, we can use the equation dτ = dt (1 - v^2)^1/2. Since we know that dt = 0, we can substitute that in and get:

dτ = 0 (1 - v^2)^1/2
dτ = 0

This means that the proper time (dτ) is also equal to 0. This may seem counterintuitive, but it makes sense because proper time is the time experienced by an observer in their own frame of reference. Since the friend standing outside is not moving, they are not experiencing any change in time.

So, to summarize, the change in time (dt) and the change in space (ds) are both equal to 0, and the proper time (dτ) is also equal to 0