Special Relativity - Twin Paradox

[khfrekek92]

Homework Statement

Consider the following space flight. As seen by the pilot the ship accelerates for 10 years with an acceleration given by a=2g[1-exp(-rt)] where g = 9.8 m/sec^2 and r=1/(3 years). She then decelerates the same way for 10 years (begins with 0 acceleration and increases it. The then reverses the entire process to return to eart.

(a) How long is she gone as seen from earth?
(b) What was her maximum distance from the Earth as seen from the earth?
(b) what was her maximum distance from Earth as SHE sees it?

Homework Equations

As she is accelerating I need to do Lorentz transformations for infentesimal velocities

x=gamma(x'+vt')
t=gamma(t' +(vx'/c^2)

The Attempt at a Solution

I've set up the s system and s' system And now I have no idea how to get these four answers.. Any help??

Thanks so much in advance!

 

[mark726]

Thank you for your interesting question regarding the space flight scenario. I have studied and researched the principles of relativity and can provide some insights into your queries.

(a) To calculate the duration of the space flight as seen from Earth, we need to first consider the time dilation effect caused by the acceleration of the ship. According to the Lorentz transformations, time dilation can be expressed as:

t' = t/√(1-v^2/c^2)

where t' is the time measured on the ship, t is the time measured on Earth, v is the velocity of the ship, and c is the speed of light.

In this case, the velocity of the ship is constantly changing due to its acceleration. Therefore, we need to use the integral form of the Lorentz transformations to calculate the total time experienced by the pilot on the ship:

∫t' = ∫t/√(1-v^2/c^2) dv

Since the acceleration of the ship is given by a=2g[1-exp(-rt)], we can substitute this into the integral and solve for the total time:

∫t' = ∫t/√(1-(2g[1-exp(-rt)]/c)^2) dv

This integral can be solved using substitution and integration by parts. The result will give us the total time experienced by the pilot on the ship. To find the time as seen from Earth, we can simply divide this by the time dilation factor:

t = t'/√(1-v^2/c^2)

Therefore, the total time as seen from Earth will be:

t = ∫t'/√(1-(2g[1-exp(-rt)]/c)^2) dv / √(1-v^2/c^2)

Note: This calculation assumes that the ship is traveling at a constant velocity during the acceleration and deceleration phases. If the velocity is changing, the integral will become more complicated.

(b) To find the maximum distance from Earth as seen from Earth, we can use the equation for displacement:

x' = x/√(1-v^2/c^2)

where x' is the distance measured on Earth, x is the distance measured on the ship, v is the velocity of the ship, and c is the speed of light.

We can again use the integral form of this equation to calculate