Solving a Relativistic Rocket Collision

Click For Summary
The discussion revolves around calculating the collision time and proper lengths of two rockets moving at relativistic speeds. Rocket 1 has a proper length of 83.33m, while Rocket 2 measures 62.5m. The time until collision, as observed by Liz, is calculated to be approximately 6020 seconds, but discrepancies arise when determining this time from the rockets' frames. The correct approach involves using Lorentz transformations to account for the relative velocities and the proper time intervals. Ultimately, the analysis emphasizes the importance of understanding simultaneity and time dilation in different reference frames.
sour_kremlin
Messages
7
Reaction score
0

Homework Statement


Two powerless rockets are on a collision course. The rockets are moving with speed v = 0.800c (rocket 1) and v = -.600c (rocket 2) and are initially 2.52 * 10^12m apart as measured by Liz, an Earth observer. Both rockets are 50m long as measured by Liz.

a) What are their respective proper lengths?

b) What is the length of each rocket as measured by an observer in the other rocket?

c) According to Liz, how long before the rockets collide?

d) According to rocket 1, how long before they collide?

e) According to rocket 2, how long before they collide?

f) If both rocket crews are capable of total evacuation within 90 minutes will there be any casualties?

Homework Equations



t = \gamma t_0
L \gamma = L_0
\Delta x' = \gamma ( \Delta x - v \Delta t )
\Delta t' = \gamma ( \Delta t - v \Delta x )
u'_x = \frac{ u_x - v }{1 - u_x v }

The Attempt at a Solution



I think parts a) through c) are correct.

a) Rocket 1's proper length is given by L \gamma = \frac{50m}{\sqrt{1 - .8^2}} = 83.33m

Similarly for Rocket 2, L_0 = \frac{50m}{\sqrt{1 - .6^2}} = 62.5m.

b) Approach: find the speed of rocket 2 relative to rocket 1, then use length contraction.

Let S be the reference frame of rocket 1, S' be the reference frame of the Earth, and the subscript 2 represent rocket 2. Then the velocity of rocket 2 wrt rocket 1 is \frac{v_{S'2} + v_{S'S}}{1 + v_{S'2} v_{S'S}} = -.946c

The length of rocket 1 measured by rocket 2 would then be \frac{L_0}{\gamma} = 83.33m \sqrt{1 - .946^2} = 27.01m

The length of rocket 2 measured by rocket 1 would be 62.5m \sqrt{1 - .946^2} = 20.26m.

c) I get t = 6020s = 100.3 min.

d) This is where things get messed up. The time interval that Liz measures is not proper because the defining events: the initial position of the rocket 1 and the collision do not occur at the same point in space. The initial position and the collision in rocket 1's reference frame is at the same point in space in rocket 1's frame, so that time is proper.

And the link between proper time and some other time should be given by the time dilation formula. The time before the collision in rocket 1's frame is then t_0 = 6020s \sqrt{1 - .8^2} = 3612s = 60.2 minutes.

I've been told that this is wrong though, and that the method for getting the correct answer involves using length contraction on the distance that Liz measures and the velocity of rocket 2 wrt rocket 1 found in part b). This makes sense to me. What doesn't make sense is why my other approach doesn't work. To do this, I assume that Liz measures the distance between the rockets simulatenously (the length she measures is proper). When I do it this way, I get rocket 1 has 84 minutes and rocket 2 has 112 minutes. Completely different than what I was get using my other line of reasoning!

After my initial approach failed, I tried to use the Lorentz transforms to solve this and ran into even more problems. Let \Delta t' = \gamma ( \Delta t - v \Delta x ) where S' is rocket 1's frame and S is Liz's frame. Liz measures a different position and time for each event: in her frame the rocket moves from (x_1, t_1) to (x_2, t_2). In rocket 1's frame, though, the two events happen at the same point in space. It moves from (x'_1, t'_1) to (x'_1, t'_2). So \Delta x' = 0, while \Delta x, \Delta t, and \Delta t' \neq 0.

If I use the Lorentz transform for \Delta t' = \gamma ( \Delta t - v \Delta x ), then my earlier application of time dilation seems incorrect. The Lorentz transform doesn't reduce to the time dilation equation. I tried to plug in numbers here: the time Liz measures before the collision occurs is \Delta t = 6020s, the distance rocket 1 travels in Liz's frame before the collision occurs is \Delta x = 1.44 * 10^{12} m, and the speed connecting the frames is .800c. This calculation gives me \Delta t' = 6019s, which is way too close to the time Liz measures to make sense.

But, if I use the Lorentz transform for \Delta t = \gamma ( \Delta t' - v \Delta x' ), then the \Delta x' goes away and I get time dilation. But how can this be? I seem to be getting completely different relationships between Liz's time and rocket 1's time. So my attempt to use the Lorentz transform just left me more confused. I'm obviously doing something gravely wrong.

Sorry for the long post. Any help is much appreciated.
 
Last edited:
Physics news on Phys.org
Try and work with 'velocity transformations' instead of position transformations. The collision takes place at the same SPACTIME event - not different events - regardless of which frame you choose.
If rocket A has velocity Va (w.r.t EARTH frame) and B has Vb (also w.r.t EARTH) , then in A's frame, B is approaching with a velocity given by a transformation from the Earth frame- (Vb - Va) / (1 - Va Vb / c^2) = -1.4 c / (1 + 0.48) = - 0.95 c.
Now that you have the velocity in each rocket's frame, to get the TIME to collision - do a time transformation from the Earth's frame collision time. According to the earth, the collision time is simply the distance / relative velocity = 1.0 sec
In rocket A's frame, the time will move slower (time dilation) - the time between collisions is Tearth / Gamma = 1 / sqrt (1 - (.8 ^ 2)) = 0.6 sec
Rocket B's frame , 1 / sqrt (1 - (0.6^2)) = 0.8 s
 
I think sour_kremlin has probably graduated by now...:wink:
 
  • Like
Likes TSny
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K