Lorentz Transformation of y-velocity

AI Thread Summary
The discussion centers on transforming the y-velocity of a particle moving along the y'-axis in a rocket frame to the laboratory frame using Lorentz transformation equations. Participants note that the textbook lacks clarity and examples, leading to confusion in applying the transformations. The correct approach involves using the relativistic Velocity Addition Theorem, particularly since the rocket moves along the x-axis in the laboratory frame. Key equations include Vx = V rel and Vy = Vy'(1-Vrel^2)^0.5, which are derived under the assumption of the rocket's motion. Overall, the conversation highlights the need for precise definitions in physics problems to avoid misinterpretation.
muffinbottoms
Messages
2
Reaction score
0

Homework Statement



A Particle moves with uniform speed V'y = Δy'/Δt' along the y'-axis of the rocket frame. Transform Δy' and Δt' to laboratory displacements Δx, Δy, and Δt using the Lorentz transformation equations. Show that the x-component and the y-component of the velocity of this particle in the laboratory frame are given by the expressions ... (under relevant equations)

Homework Equations



Vx = V rel
Vy = Vy'(1-Vrel^2)^.5


The Attempt at a Solution



Okay so the textbook i got this problem from is lacking in both directions and example problems. This is what I have so far..

x = x' because the particle is moving along the y-axis
z=z'

Δt = vγy' + γt
Δx = x'
Δy= γy' + Vγt'
 
Physics news on Phys.org
or is it that
t' = -Vrelγy+γt = γ(-Vrel(y)+t)
y' = γy-Vrelγt = γ(y-Vrel(t))

how should i continue?
 
muffinbottoms said:

Homework Statement



A Particle moves with uniform speed V'y = Δy'/Δt' along the y'-axis of the rocket frame. Transform Δy' and Δt' to laboratory displacements Δx, Δy, and Δt using the Lorentz transformation equations. Show that the x-component and the y-component of the velocity of this particle in the laboratory frame are given by the expressions ... (under relevant equations)

Homework Equations



Vx = V rel
Vy = Vy'(1-Vrel^2)^.5


The Attempt at a Solution



Okay so the textbook i got this problem from is lacking in both directions and example problems. This is what I have so far..

x = x' because the particle is moving along the y-axis
z=z'

Δt = vγy' + γt
Δx = x'
Δy= γy' + Vγt'

The relevant equations are not correct. If the particle is moving relative to the rocket, and the rocket is moving relative to the laboratory, then you have to use the relativistic Velocity Addition Theorem to get the velocity of the particle relative to the laboratory.
 
I was able to get the given expressions by assuming that the rocket moves only along the x-axis in the laboratory frame with relative speed v_rel, as suggested by v_x = v_rel, though the fault is on the book for not mentioning that specifically, thus forcing you to assume that the relative velocity could be in any direction.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top