Longitude contraction / time dilatation

[pepediaz]

Homework Statement

An inertial observer O ' in a spacecraft, moving with four-speed U = 2 (c, u) seen by a
Earth observer O, passes by Earth and synchronizes his clock with O's clock. This
observer turns to Titan, a moon of Saturn that is (at that time) at a distance
Δx = 1.2 billion km from Earth.
How long does that observer take to reach Titan as seen from Earth? And measured by
the same? How far is the land of Titan seen by O '?

Relevant Equations

Lorentz transformation

I get the same values for both observers, which is not nice I think.

 

[PeroK]

Homework Statement:: An inertial observer O ' in a spacecraft , moving with four-speed U = 2 (c, u) seen by a
Earth observer O, passes by Earth and synchronizes his clock with O's clock. This
observer turns to Titan, a moon of Saturn that is (at that time) at a distance
Δx = 1.2 billion km from Earth.
How long does that observer take to reach Titan as seen from Earth? And measured by
the same? How far is the land of Titan seen by O '?
Relevant Equations:: Lorentz transformation

I get the same values for both observers, which is not nice I think.

Yes, that doesn't sound right. Please post what you did.

 

[pepediaz]

Be ϒ the Lorentz transformation:

Δx/U = Δx /(2cc-2uu) = t
Δx'/U' = ϒΔx /ϒ(2cc-2uu) = t'

Thus, t=t'

 

[PeroK]

Be ϒ the Lorentz transformation:

Δx/U = Δx /(2cc-2uu) = t
Δx'/U' = ϒΔx /ϒ(2cc-2uu) = t'

Thus, t=t'

That's hard to read and difficult to understand. Can you explain what you're doing?

First, what does $U = 2(c, u)$ mean?

 

[pepediaz]

I'm calculating the time for each observer. U is the four-velocity, and I have expressed it in form of invariant product (as an absolute value, but with a "-" between squared components).

For the observer inside the spaceship, I have considered that their distance and velocity is that of the inertial observer, but multiplied by the Lorentz factor, I think that's wrong, but I don't know how to do it different.

 

[PeroK]

I'm calculating the time for each observer. U is the four-velocity, and I have expressed it in form of invariant product (as an absolute value, but with a "-" between squared components).

For the observer inside the spaceship, I have considered that their distance and velocity is that of the inertial observer, but multiplied by the Lorentz factor, I think that's wrong, but I don't know how to do it different.

Sorry, none of this makes much sense. Let me repeat the question you didn't answer:

First, what does $U = 2(c, u)$ mean?

 

[pepediaz]

U is the four-velocity, and it has a temporal component (c) and three space-coordinate components(u, as a vector).

 

[PeroK]

U is the four-velocity, and it has a temporal component (c) and three space-coordinate components(u, as a vector).

What is the $2$?

 

[pepediaz]

I guess it is a coefficient, but I'm not sure.

 

[PeroK]

I guess it is a coefficient, but I'm not sure.

Isn't that a problem?

What is the general form of a four-velocity? Hint: think about the gamma factor $\gamma$.

 

[pepediaz]

So, I can obtain inertial velocity equating γ=2.

Once done this, I continue having the same problem:

t = Δx /v, but t' = Δx '/v' = γ(Δx)/γ(v)

 

[PeroK]

So, I can obtain inertial velocity equating γ=2.

Once done this, I continue having the same problem:

t = Δx /v, but t' = Δx '/v' = γ(Δx)/γ(v)

Okay, so you have $\gamma = 2$. That gives you $u$, if you need it. And in the Earth frame, we have $t = \frac{\Delta x}{u}$.

In the spacecraft frame, what do we have?

 

[pepediaz]

I guess t' = Δx '/v' = γ(Δx)/v

 

[PeroK]

I guess t' = Δx '/v' = γ(Δx)/v

How far is Titan from Earth in the spacecraft frame?

How fast is Titan moving in the spacecraft frame?

 

[pepediaz]

In the spacecraft frame, the distance would be γ(Δx) and the velocity would be v, or γ(v), this one I don't know.

 

[PeroK]

In the spacecraft frame, the distance would be γ(Δx) and the velocity would be v, or γ(v), this one I don't know.

That's not correct. The distance to Titan is length contracted. And, velocities must be equal and opposite between frames. If the rocket is moving at speed $u$ in the Earth frame, then the Earth and Titan must be moving at speed $u$ in the spacecraft frame.

 

[pepediaz]

I understand it now! Thanks!

The time in spacecraft frame is a half of the inertial frame time.

 

[pepediaz]

And another thing, be the red point this spacecraft , the purple line the world line with slope = 2 and the green point a photon thrown towards the Earth from the spacecraft , would this spacetime diagram ok? (distance would be 1.2 billion km, and the time, 1000 times of shown, but scale is badly displayed, even the red point wouldn't be at that position, but I did it like this just for showing it).