Lorentz Transformation Question

[phisci]

Homework Statement

2 particles are created in a high-energy accelerator and move off in opposite directions. The speed of one particle, as measured in the laboratory is 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second particle, as measured in the laboratory.

Homework Equations

Lorentz Velocity Transformation Equation.

V_x= V_x^'+u/(1+(uV_x^'/C²))

The Attempt at a Solution

I let S be the frame of reference of the laboratory and S^, be the frame of reference of the first particle.

From the question, u=+0.650c and the velocity of the 2nd particle in the S' frame is
Vx'= -0.950c since it is approaching the first particle.

Hence when i plug in the values into the formula, I should get

Vx=[(-0.950+0.650)]c/(1+(-0.950x0.650)) = -0.784c

the speed is hence 0.784c.

However the solution given puts a negative sign to the 0.650 in the denominator. i.e.
Vx=[(-0.950+0.650)]c/(1+(0.950x-0.650))

Why did they negate the 0.650 and not the 0.950? Where have i gone wrong in my reasoning? Thanks!

 

[vela]

Probably just a typo. Your solution is fine.

 

[Rasalhague]

You can also use (0.950-0.650)c/1-0.650*0.950) = 0.784c. Since the question doesn't label one direction as positive, and only asks for speed (the magnitude of velocity), it doesn't really matter. I don't know why they chose to deploy the minus signs as they did; maybe it's a trivial difference in convention: which formula they started off with, and how they assigned the variables in this particular problem to those of the general formula etc., or maybe just a typo as vela suggests. I wouldn't lose too much sleep over it ;-)

 

[phisci]

okay noted. thanks alot! :)

 

[rwisz]

I would like to address your question by first clarifying that the Lorentz transformation equations are used to describe the relationships between space and time measurements in different frames of reference. In this case, we are dealing with the velocities of particles in different frames of reference.

Now, let's look at your attempt at a solution. You correctly identified the frames of reference and the velocities of the particles in those frames. However, the Lorentz velocity transformation equation that you used, Vx= Vx'+u/(1+(uVx'/C2)), is actually for the transformation of velocities from the S' frame to the S frame. In this case, we are interested in finding the velocity of the second particle in the S frame, so we need to use the inverse transformation, which is Vx'= (Vx-u)/(1-(uVx/C^2)).

Using this equation, we can calculate the velocity of the second particle in the S frame as Vx'= (0.650c-0.950c)/(1-(0.650c)(-0.950c)/c^2) = 0.784c. This is the same result that you obtained, but with the correct equation.

As for the solution given that puts a negative sign in front of the 0.650 in the denominator, this is because the velocity of the first particle in the S frame is positive (moving away from the observer), while the velocity of the second particle in the S' frame is negative (moving towards the first particle). In the equation, we use the velocities in the respective frames, so the negative sign is necessary to account for the direction of motion.

In summary, your reasoning was correct, but you used the wrong equation for the Lorentz velocity transformation. By using the correct equation, you will get the same answer as the given solution. I hope this helps clarify any confusion.