Particle hitting an inclined wall viewed in different frames

In summary: I'm not sure if my argument for the incident and reflected angle is correct, can anyone help check my work? Thanks.
  • #1
Whitehole
132
4

Homework Statement


Consider an inertial frame S with coordinates ##x^μ = (t, x, y ,z)##, and a frame S' with coordinates ##x^{μ'}## related to S by a boost with velocity parameter v along the y-axis. Imagine we have a wall at rest in S', lying along the line x' = -y'. From the point of view of S, what is the relationship between the incident angle of a particle hitting the wall (traveling in the x-y plane) and the reflected angle? What about the velocity before and after?

Homework Equations


Lorentz Transformations
##θ_i = incident~ angle, ~~~θ_f = reflected~ angle##
##γ = Lorentz~ factor##

The Attempt at a Solution


We know that ##θ_i' = θ_f'##. Since S' is moving along the y-axis, the wall would appear to be contracted only in the y direction. Suppose the wall has length L, then the y component of the wall as viewed in S is
##y = \frac{L~sin(φ')}{γ}##
The x component of the wall is the same as in the S' frame, so the angle ##φ## that the wall makes with respect to the x-axis is
##φ = tan^{-1}(\frac{\frac{L~sin(φ')}{γ}}{L~cos(φ')}) = tan^{-1}(\frac{tan(φ')}{γ})##
This angle ##φ## will be smaller compared to ##φ'##, their difference ##φ' - φ## is the increase in angle the observer in S will see for the incident angle (decrease for the reflected angle), so the incident and reflected angles as seen in S are
##θ_i = θ_i' + (φ' - φ). ~~~ θ_f = θ_f' - (φ' - φ)##
The velocity before and after in frame S are just related by Lorentz transformations to the velocity before and after in the frame S'.

I'm not sure if my argument for the incident and reflected angle is correct, can anyone help check my work? Thanks.

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  • #2
##θ_i = θ_i' + (φ' - φ). ~~~ θ_f = θ_f' - (φ' - φ)##
Why ?
First, you try find ##θ_i&θ_f## in a simple case if the wall only moves along in x-axis, the face of wall parallel with y-axis. Then φ' - φ=0 so ##θ_i = θ_i'## but in this case ##θ_i = θ_i'## seems incorect.
3219865203_1690627977_574_574.jpg

View attachment 102709[/QUOTE]
 
Last edited:
  • #3
Hamal_Arietis said:
##θ_i = θ_i' + (φ' - φ). ~~~ θ_f = θ_f' - (φ' - φ)##
Why ?
First, you try find ##θ_i&θ_f## in a simple case if the wall only moves along in x-axis, the face of wall parallel with y-axis. Then φ' - φ=0 so ##θ_i = θ_i'## but in this case ##θ_i = θ_i'## seems incorect.
3219865203_1690627977_574_574.jpg

View attachment 102709
[/QUOTE]
Because the length contraction is along the y' axis and by taking the angle as viewed in S, that angle would be smaller compared to the angle that the wall makes with the x-axis in S. But then the decrease in angle for the wall is the increase for the incident angle (with respect to the normal of the wall), vice versa for the reflected angle.
 
  • #4
Hamal_Arietis said:
##θ_i = θ_i' + (φ' - φ). ~~~ θ_f = θ_f' - (φ' - φ)##
Why ?
First, you try find ##θ_i&θ_f## in a simple case if the wall only moves along in x-axis, the face of wall parallel with y-axis. Then φ' - φ=0 so ##θ_i = θ_i'## but in this case ##θ_i = θ_i'## seems incorect.
3219865203_1690627977_574_574.jpg

View attachment 102709
[/QUOTE]
By assuming that the wall is parallel to the y-axis then I think the incident and reflected angles shouldn't change, but in this case of course the wall is inclined.
 
  • #5
incident and reflected angles must change because for incident angel,we see in the S frame, incident angle moves with velocity v along Ox , but for reflected angle it moves contrariwise with velocity -v. So the Lorentz Transformations will be different.
 
  • #6
Hamal_Arietis said:
##θ_i = θ_i' + (φ' - φ). ~~~ θ_f = θ_f' - (φ' - φ)##
Why ?
First, you try find ##θ_i&θ_f## in a simple case if the wall only moves along in x-axis, the face of wall parallel with y-axis. Then φ' - φ=0 so ##θ_i = θ_i'## but in this case ##θ_i = θ_i'## seems incorect.
3219865203_1690627977_574_574.jpg

View attachment 102709
[/QUOTE]
Sorry for the long gap before the reply. So in this case, I can suppose that the incident velocity is u' and decompose its components by the velocity transformation law

##u_x = \frac{ u'cosθ'_i + v }{ 1 + \frac{u'vcosθ'_i}{c^2} } ~~~, ~~~u_y = \frac{ u'sinθ'_i }{ γ(1 + \frac{u'vsinθ'_i}{c^2}) } ##

The components of the reflected velocity would just be the same except for a negative sign in the x component based on your diagram.

##u_x = \frac{ -u'cosθ'_f + v }{ 1 - \frac{u'vcosθ'_f}{c^2} } ~~~, ~~~u_y = \frac{ u'sinθ'_f }{ γ(1 + \frac{u'vsinθ'_f}{c^2}) } ##

But how do I relate the incident and reflected angle as seen in S?
 

FAQ: Particle hitting an inclined wall viewed in different frames

What is the concept of a particle hitting an inclined wall viewed in different frames?

The concept refers to the scenario where a particle is moving towards an inclined wall, and its motion is observed from different reference frames. This means that the position, velocity, and acceleration of the particle can be perceived differently depending on the observer's frame of reference.

How does the angle of inclination affect the particle's motion?

The angle of inclination affects the component of the particle's velocity that is perpendicular to the wall. This component determines the direction of the particle's motion after hitting the wall. A steeper angle of inclination will result in a larger change in direction for the particle, while a smaller angle will result in a more gradual change.

What is the role of the frame of reference in this scenario?

The frame of reference is crucial in understanding the particle's motion when hitting an inclined wall. It helps determine the velocity and acceleration of the particle, as well as the angle at which it bounces off the wall. Without a frame of reference, it would be difficult to accurately describe the particle's motion.

How does the particle's initial velocity affect its motion?

The particle's initial velocity plays a significant role in its motion. It determines the speed at which the particle approaches the wall and the angle at which it bounces off. A higher initial velocity will result in a faster and more significant change in direction after hitting the wall, while a lower initial velocity will result in a slower and less significant change.

What are some real-world applications of this concept?

This concept has various real-world applications, such as understanding the movement of a ball bouncing off a wall or a car colliding with a barrier. It also has applications in fields such as physics, engineering, and sports, where the motion of objects or bodies needs to be analyzed from different reference frames.

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