Why am I getting this relativity velocity addition problem wrong?

In summary, the spacecraft S2 is escaping from S1 at a speed of 0.95c and is capable of firing a missile with a speed of 0.98c. In part A, where the question is asked from the perspective of S2, the velocity addition formula is not needed as the missile is simply approaching S1 at a speed of 0.03c. However, in part B where the question is asked from the perspective of S1, the velocity addition formula is needed as you are switching frames of reference to calculate the speed of the missile approaching S1, which is 0.9994c.
  • #1
DunWorry
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Homework Statement


A spacecraft S2 is capable of firing a missile which can travel 0.98c. S2 is escaping from S1 at a speed of 0.95c when it fires a missile towards S1.

part A) According to the pilot of S2, what speed does the missile approach S1?
Part B) according to pilot of S1, what speed does the missile approach it?


Homework Equations


Call the S1 frame x and S2 frame y and speed of missile U

Velocity addition V[itex]_{x}[/itex] = [itex]\frac{v_{y} + U}{1 + \frac{v_{y} U}{C^{2}}}[/itex]

The Attempt at a Solution



My problem lies with part A. The answer is just a simple 0.98c - 0.95c = 0.03c. However I can't get this result with the velocity addition formula, why is it in this case the velocity addition formula does not work/ does not apply?

I tried imagining S2 moving to right (positive) and firing the missile backwards towards S1 (left direction which is negative). Taking the frame of reference of S2, the spaceship S2 is stationary and S1 is moving to the left at a velocity of -0.95c, the missile is also moving to left with speed -0.98c

if I try use the velocity addition formula Velocity addition V[itex]_{x}[/itex] = [itex]\frac{-0.98 - 0.95}{1 + \frac{0.98 x 0.95}{C^{2}}}[/itex] I get -0.9994C, which is wrong. The answer is just 0.98c - 0.95C but I cannot see what I am doing wrong with the velocity addition formula or why it is not needed in this case.

I solved part B) using the formula V[itex]_{x}[/itex] = [itex]\frac{0.98 - 0.95}{1 - \frac{0.98 x 0.95}{C^{2}}}[/itex]. The signs are as they are as in the frame of S1, the ship S2 is moving in + direction with speed 0.98C and the missile is moving with -0.95C. It seems to work for part B but not for part A and I cannot see why.

Thanks
 
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  • #2
For part (A) you don't need to use the velocity addition formula. You already know how fast the missile travels relative to S2 and also how fast S1 moves relative to S2. You just want to know how fast the missile is "closing" on S1 as measured by S2.

It's the same as asking if S2 rolls a ball at 5 m/s along her x-axis and then rolls a ball at 7 m/s along her x-axis, how fast is the second ball closing on the first ball according to S2? No relativity needed since all measurements are in one inertial frame. You are not switching frames of reference.
 
  • #3
You use the relativistic formula when you are working between two frames.

For instance, in part B) you add adding two velocities from S2's frame but want an answer for S1 frame.

When you are working just in one frame, it is is not used. For example in part A) you are adding two velocities according to S2 and want an answer for that frame.

Think of it is this way. in part (A, you have a missile traveling away from S2 at .98c

After 1 sec, the missile will be 0.98 light secs further away.

S1 is traveling away a 0.95c, so after 1 sec, it will be 0.98 light sec away.

This means that, according to S2, after 1 sec the missile will be 0.03 light sec closer to each other. which works out to a difference of 0.03 c between S1 and the missile according to S2.
 
  • #4
Ahhh I see that's much clearer now. So its basically because for part A the measurements are given from the frame of S2 and you are working in the same frame of reference because you want an answer for S2 so it is not needed. However in part B you are trying to take the position from S1 and so you are using measurements which were given from the frame in S2 and therefore need to use the velocity addition because you are switching frames of reference
 
  • #5
for your question! The answer to part A is not just a simple subtraction because you are dealing with velocities in different frames of reference. When using the velocity addition formula, you need to make sure that the velocities are in the same frame of reference. In this case, the velocity of the missile is given in the frame of S2, while the velocity of S1 is given in the frame of S1. To apply the velocity addition formula, you need to convert the velocity of S1 into the frame of S2. This can be done using the Lorentz transformation equations. Once both velocities are in the same frame of reference, you can use the velocity addition formula to find the resulting velocity.

For part A, the correct approach would be to use the Lorentz transformation to convert the velocity of S1 into the frame of S2. This would give you a velocity of -0.9987c for S1 in the frame of S2. Then, you can use the velocity addition formula to find the resulting velocity of the missile in the frame of S2, which would be 0.9994c. This is the correct answer, as the missile is approaching S1 with a velocity of 0.9994c in the frame of S2.

For part B, the Lorentz transformation is not needed because both velocities are already in the frame of S1. Therefore, you can just use the velocity addition formula to find the resulting velocity of the missile in the frame of S1, which would be 0.03c.

It is important to always make sure that all velocities are in the same frame of reference when using the velocity addition formula. I hope this helps clarify the issue and allows you to solve similar problems in the future.
 

Related to Why am I getting this relativity velocity addition problem wrong?

1. Why is the velocity addition problem in relativity so difficult to understand?

The concept of velocity addition in relativity can be difficult to understand because it goes against our intuition about how velocities combine. In classical mechanics, velocities simply add together, but in relativity, they do not. This is due to the fact that the speed of light is constant and the laws of physics are the same in all inertial reference frames.

2. How can I improve my understanding of velocity addition in relativity?

One way to improve your understanding of velocity addition in relativity is to practice solving problems and familiarizing yourself with the equations involved. You can also try to visualize the concepts by using thought experiments or diagrams to help you understand the relative motion of objects in different reference frames.

3. Why do I keep getting the wrong answer when solving velocity addition problems in relativity?

There could be several reasons why you are getting the wrong answer. One possibility is that you are not using the correct equations or not applying them correctly. Another possibility is that you are not taking into account the different reference frames and their relative velocities. It is important to carefully analyze the problem and use the correct equations for each situation.

4. Is there an easier way to solve velocity addition problems in relativity?

There is no easy way to solve velocity addition problems in relativity, but there are some tips that can make it easier. One tip is to always use the correct equations and to carefully consider the relative motion of objects in different reference frames. Another tip is to practice and familiarize yourself with the concepts and equations involved.

5. How does the concept of velocity addition in relativity relate to the theory of relativity?

The concept of velocity addition in relativity is a direct consequence of Einstein's theory of special relativity. This theory states that the laws of physics are the same in all inertial reference frames and that the speed of light is constant for all observers, regardless of their relative motion. Therefore, the way velocities add together must also be consistent in all reference frames, leading to the concept of velocity addition in relativity.

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