"Radar distance" on an accelerating rocketship

[jack476]

Homework Statement

This problem is from chapter 2 "Relativistic Kinematics", from Wolfgang Rindler's "Introduction to Special Relativity", second edition. The full statement is:

Consider a long uniformly accelerating rocketship. Prove that the "radar distance" (the proper time of a light echo multiplied by c/2) of a point on the rocket at parameter X₂, from an observer riding on the rocket at parameter X₁, is always X₁arcsinh((X₂² - X₁²)/2X₁X₂). Hint: Work in the rest frame of X₂.

Homework Equations

I assume this is going to involve the formula for hyperbolic motion, x² - (ct)² = X². I also think that the solution depends on the relationship αt/c = sinh(ατ/c). We also have X = c²/α

The Attempt at a Solution

The point at parameter X₂ appears to move to an observer along the trajectory x² - (ct)² = X₂² and the pulse of light along the trajectory x = X₁ + ct. By substituting x, we get ct = (X₂² - X₁²)/2X₁, and then with the third relation above we get αt/c = (X₂² - X₁²)/2X₁X₂. Then with the second relation, we get τ = (c/α)arcsinh((X₂² - X₁²)/2X₁X₂) and then multiplying both sides by c we get cτ/2 = (c²/2α)arcsinh((X₂² - X₁²)/2X₁X₂), which (I think) is giving the proper time that it takes for the pulse to travel from X₁ to X² But the problem is that c²/α gives X₂ and not X₁. The factor of one half is not an issue because presumably it goes away when the proper time for the pulse echoing back from X₂ to X₁ is added to get the total round trip time. I hope that made sense.

So what's the mistake I'm making here? Something's just not right and I'm just not clear what.

 

[TSny]

In this problem you have an observer in the rocket who is located at a point with parameter X₁. His wristwatch measures his proper time. You want to know the elapse of this observer's proper time (wristwatch time) between the event where the light pulse is sent out by this observer and the event where the pulse returns to this observer. If you could determine the coordinate time t of these two events, then you can use the equation α₁t/c = sinh(α₁τ/c) to determine the corresponding proper time for each event for this observer. The subscript 1 on the α emphasizes that this observer has acceleration associated with parameter X₁.

The Attempt at a Solution

The point at parameter X₂ appears to move to an observer along the trajectory x² - (ct)² = X₂² and the pulse of light along the trajectory x = X₁ + ct. By substituting x, we get ct = (X₂² - X₁²)/2X₁, and then with the third relation above we get αt/c = (X₂² - X₁²)/2X₁X₂. Then with the second relation, we get τ = (c/α)arcsinh((X₂² - X₁²)/2X₁X₂)

This gives the proper time of the arrival of the pulse at the reflection point on the worldline X₂ for an observer moving with the parameter X₂. But that's not what you need. You need the proper times for the emission and return events for the observer with parameter X₁.

Rindler's hint is to choose the emission and return events "symmetrically" on curve X₁ in the figure below such that the reflection event (on the curve X₂) occurs when the rocket is at rest in the figure.

 

[jack476]

Thanks for the help. Does this look right?

 

[TSny]

Looks essentially right. But it appears that you are saying that the outgoing pulse is represented by $x = X_2 - ct$. Is that correct, or does this equation represent the return pulse?

 

[jack476]

Looks essentially right. But it appears that you are saying that the outgoing pulse is represented by $x = X_2 - ct$. Is that correct, or does this equation represent the return pulse?

Yes, it seems I made a little writing error there. The outgoing pulse should be $x = ct - X_2$, and the returning pulse should be $x = X_2 - ct$.

But it doesn't matter in the end, because you end up squaring this expression when you plug it into the trajectory equation. Due to the symmetry about the x-axis it doesn't really matter.

 

[TSny]

Yes, it seems I made a little writing error there. The outgoing pulse should be $x = ct - X_2$, and the returning pulse should be $x = X_2 - ct$.

But it doesn't matter in the end, because you end up squaring this expression when you plug it into the trajectory equation. Due to the symmetry about the x-axis it doesn't really matter.

Yes. Good work.