Relativistic Spaceship moving past Earth

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Homework Help Overview

The discussion revolves around a relativistic scenario involving a spaceship moving at 99% the speed of light past Earth, with a specific focus on the behavior of water in a swimming pool aboard the ship. The original poster seeks to determine the tilt of the water surface as observed from Earth, given that the water rises at a rate of 5 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the velocities of the spaceship and the water, with some suggesting the use of relativistic velocity addition. Others question the assumptions about the water's behavior and its relation to gravitational effects.

Discussion Status

The discussion is active, with participants providing various insights and suggestions for approaching the problem. There is a mix of interpretations regarding the water's behavior and the necessary equations to describe it in different frames of reference.

Contextual Notes

Some participants express confusion over the use of different symbols in the equations, indicating a potential misunderstanding of the relativistic transformations involved. The original poster's assumptions about the water's behavior are also questioned, highlighting the need for clarity in the problem setup.

PermaTrashed
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Homework Statement


A relativistic spaceship is moving "horizontally" past Earth at 99% speed of light (c), and the water in the ship's swimming pool rises at 5 m/s. What is the tilt of the water surface with respect to the horizontal, AS OBSERVED FROM EARTH?

Homework Equations


The Attempt at a Solution


I tried using simple trigonometry to find the angle between the velocity of the ship and the velocity of the water, but cannot translate this to find an angle between the acceleration of the Earth vector and the velocity of the water.
 
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Welcome to PF!

Hi PermaTrashed! Welcome to PF! :smile:

I'm not sure that you're seeing the issue here.

The astronauts are pouring water into the pool, and in their frame the water level is a horizontal line moving vertically at 5 m/s.

So write the equation for the water surface in x,y,z,t then convert that into x',y',z',t' and put t' = 0 to find what the water looks like at any fixed time on Earth. :wink:
 
Wow I do have it wrong then, I totally thought the water rising was due to the gravitational field of Earth, it doesn't say anything about astronauts pouring water into it, thanks a lot!
 
if the direction of the velocity of water is parallel to the ships velocity, you will need relativistic velocity addition, which is by no means easy to derive:
http://en.wikipedia.org/wiki/Velocity-addition_formula
otherwise if the water moves perpendicular to the spaceship, the answer is simply 5 m/s
 
ardie said:
if the direction of the velocity of water is parallel to the ships velocity, you will need relativistic velocity addition, which is by no means easy to derive:
http://en.wikipedia.org/wiki/Velocity-addition_formula
otherwise if the water moves perpendicular to the spaceship, the answer is simply 5 m/s

But it's not asking for the velocity, it's asking for the tilt of the water WRT the horizontal axis in Earth's frame
 
yup! :smile: use the method i suggested …
tiny-tim said:
So write the equation for the water surface in x,y,z,t then convert that into x',y',z',t' and put t' = 0 to find what the water looks like at any fixed time on Earth. :wink:
 
What I don't understand is that from reference of Earth, the tilt of the surface of water won't change, it still will be horizontal, just moving in the diagonal direction. Is this angle (the angle at which it travels diagonally) what I'm looking for?
 
PermaTrashed said:
… from reference of Earth, the tilt of the surface of water won't change, it still will be horizontal, just moving in the diagonal direction.

No.

y' = y, but t' ≠ t.

What is the equation (in x y z and t) for the water surface in the spaceship frame?
 
in the spaceship frame, y'=5.0t where t=ζ(t'+vx'/c^2)?? Then in the Earth's frame x=ζ(x'+vt), then its just the angle between x and y?
 
  • #10
PermaTrashed said:
in the spaceship frame, y'=5.0t

Nooo, you're already confusing me :confused:

in any frame, either use y and t, or use y' and t'.

Start again.​
 
  • #11
Right, so y'=5.0m/s(t') in the spaceship frame, and then x=ζ(x'+vt)?
 
  • #12
PermaTrashed said:
Right, so y'=5.0m/s(t') in the spaceship frame, and then x=ζ(x'+vt)?

just write it y'=5t' …

now translate y' and 5t' into x y and t :smile:
 
  • #13
Word, that makes more sense, so t'=ζ(t-vx/c2) and then I find the angle between y=5t' and x=ζ(x'+vt) then plug in t' and x'=0?
 
  • #14
before you can use the Lorentz transformations to find out the contraction of the length, you will need the relative velocity of the two objects, and for that you need the relativistic velocity addition i just wrote, although moving at 99% of the speed of light, you probably get something very similar as a result.
 
  • #15
(just got up :zzz: …)
PermaTrashed said:
Word, that makes more sense, so t'=ζ(t-vx/c2) and then I find the angle between y=5t' and x=ζ(x'+vt) then plug in t' and x'=0?

(why are you using zeta (ζ ) instead of the usual gamma (γ)? :confused:)

that's very confusing :redface:

just write y' = 5t' as y = 5γ(t-vx/c2),

then put t = constant (eg 0) :smile:
 
  • #16
Sorry I couldn't find the gamma key, so that all seems correct then? Thanks for all the help!
 

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