Special Relativity of rocket relative to you

[tombarrtt]

Homework Statement

Studying Special Relativity at the moment, and having a little trouble getting to grips with it. I've got stuck on this question, and was wondering if anyone could clear it up for me?

Here's the question:

"A rocket is traveling at 0.6c, v, along the x-axis relative to you. It fires a missile in the y-axis (perpendicular to the rocket in the rocket's reference frame) at 0.7c relative to you.

What speed and at what angle to the x-axis do you see the missile travelling?"

I think you have to use the Lorentz transformations for velocity, but I'm not sure.

Homework Equations

I think: ux' = (ux-v)/(1-uxv/c^2)

and: uy' = uy/(gamma*(1-uxv/c^2))

The Attempt at a Solution

I think you have to use the Lorentz transformations for velocity, but I'm not sure. I'm just getting confused with which reference frames to choose for the rocket, and what the variables "ux", "ux'" and "v" are and how to put them in.

 

[tiny-tim]

welcome to pf!

hi tombarrtt! welcome to pf!

"A rocket is traveling at 0.6c, v, along the x-axis relative to you. It fires a missile in the y-axis (perpendicular to the rocket in the rocket's reference frame) at 0.7c relative to you.

What speed and at what angle to the x-axis do you see the missile travelling?"
…
I think you have to use the Lorentz transformations for velocity, but I'm not sure. I'm just getting confused with which reference frames to choose for the rocket, and what the variables "ux", "ux'" and "v" are and how to put them in.

i think the Lorentz equations for velocity are too difficult to remember :redfae:

start with the equation of the missile in the rocket's frame …

that's x' = 0, y' = u't'

(you don't know yet what u' is, but you do know that the equivalent speed u in your frame is 0.7c)

now use the Lorentz transformation to convert that to your frame, find u, and put it equal to 0.7c …

what do you get? ​

 

[tombarrtt]

Sorry I'm just still really confused. "v" is the x-velocity of the rocket right, so 0.6c? Then what are Ux and Ux'? Is Ux' the transformed velocity into our frame? But then what is Ux, surely that's the same as v?

I'm just so confused by this :/

 

[tiny-tim]

hi tombarrtt!

solving maths problems often just involves giving everything sensible names, so that you can clearly see what you're doing

in this case, you need to differentiate clearly between the speed of the rocket and of the missile

so I'm using v for the rocket, and u for the missile ​

 

[Nickie]

Hello,

I can understand the confusion you may be experiencing with the concept of Special Relativity. Let me try to explain it in a simple way.

Firstly, let's define the variables in this problem. "ux" and "uy" represent the velocity components of the missile in the x and y directions, respectively, in the rocket's reference frame. "v" represents the velocity of the rocket in your reference frame. And "ux'" and "uy'" represent the velocity components of the missile in the x and y directions, respectively, in your reference frame.

Now, to solve this problem, we need to use the Lorentz transformations for velocity, which you have correctly identified. These transformations allow us to calculate the velocity of an object in one reference frame, as observed from another reference frame.

In this case, we are observing the missile from your reference frame, so we will use the transformations for velocity in the y direction:

uy' = uy/(gamma*(1-uxv/c^2))

First, we need to calculate the value of "gamma", which is the Lorentz factor and is given by:

gamma = 1/(sqrt(1-(v/c)^2))

Here, v represents the velocity of the rocket in your reference frame, which is 0.6c. So, gamma = 1/(sqrt(1-(0.6c/c)^2)) = 1.25.

Now, we can plug in the values for "uy" and "v" in the above equation and solve for "uy'":

uy' = (0.7c)/(1.25*(1-(0.6c/c)^2)) = 0.84c

So, the missile will appear to be moving at a velocity of 0.84c in the y direction, as observed from your reference frame.

To calculate the angle at which the missile is traveling, we can use basic trigonometry. The tangent of an angle is equal to the opposite side (uy') divided by the adjacent side (0.6c). So, tan(theta) = uy'/0.6c. Solving for theta, we get:

theta = tan^-1(0.84c/0.6c) = 53.1 degrees

Therefore, the missile will appear to be traveling at a velocity of 0.84c at an angle of 53.1 degrees to the x