Relative velocities at high speed

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SUMMARY

The discussion focuses on the relativistic velocity transformation equations used to calculate the speed of objects in different frames of reference, specifically in the context of two trains moving at 0.8c on perpendicular tracks. The equations provided are: v^'_x = (v_x - u)/(1 - uv_x/c^2), v^'_y = v_y/γ(1 - uv_x/c^2), and v^'_z = v_z/γ(1 - uv_x/c^2). The key takeaway is that the x-direction is treated differently due to the effects of length contraction along the direction of motion, necessitating a redefinition of axes when calculating relative velocities. This distinction is crucial for correctly applying the equations to find velocities in different directions.

PREREQUISITES
  • Understanding of special relativity concepts, particularly Lorentz transformations.
  • Familiarity with the concept of relativistic velocity addition.
  • Knowledge of the gamma factor (γ) in special relativity.
  • Basic understanding of vector components in physics.
NEXT STEPS
  • Study the derivation of Lorentz transformations in special relativity.
  • Learn about the implications of length contraction and time dilation in different frames.
  • Explore examples of relativistic velocity addition with multiple objects.
  • Investigate the role of reference frames in physics problems involving relative motion.
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Students of physics, particularly those studying special relativity, educators teaching advanced mechanics, and anyone interested in understanding relativistic effects on motion and velocity calculations.

Coldie
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Hi again,

Our textbook gives us equations to find the speed of objects in relation to others in the x, y, and z planes. These are:

[tex]v^'_x = \frac{v_x - u}{1 - uv_x/c^2}[/tex]
[tex]v^'_y = \frac{v_y}{\gamma(1 - uv_x/c^2)}[/tex]
[tex]v^'_z = \frac{v_z}{\gamma(1 - uv_x/c^2)}[/tex]

My first question is why is the equation for velocity in the x direction different from those in the y and z directions? Since all directions are relative, that would mean that simply turning 90º in any direction would mean that you'd have to use a different equation to find the velocity of something in a given axis?

My next question is from one of the homework problems, which says:

A and B are trains on perpendicular tracks. The velocities are in the station frame (S frame).
a) Find [tex]v_AB[/tex], the velocity of train B with respect to train A.
b) Find [tex]v_BA[/tex], the velocity of train A with respect to train B.

The picture shows that train A is going directly upwards from the train station at .8c, and train B is moving directly to the right from the station, also at .8c.

Now, just looking at the equations to finding the answer to part a you can tell that something's not right. The equation to find [tex]v^'_y[/tex] has [v_y] on the top, which in this case is 0 since train B is not moving vertically, so following the given equations, the vertical speed of train B relative to A is 0, which is not correct. The answer in the back is arrived at by using the equation for [tex]v^'_x[/tex] to find [tex]v^'_y[/tex] and using the equation for [tex]v^'_y[/tex] to find [tex]v^'_x[/tex]. Can someone tell me how you're supposed to know when to switch the x and y axes to use the proper equations to find the answer?

I hope I've made everything clear, thanks for the help.
 
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Um... bump?
 
Coldie said:
My first question is why is the equation for velocity in the x direction different from those in the y and z directions? Since all directions are relative, that would mean that simply turning 90º in any direction would mean that you'd have to use a different equation to find the velocity of something in a given axis?
But all directions are not equal!** These equations assume that the frame is moving with speed u along the +x axis. So, you must redefine your axes accordingly to make use of these formulas.

To find the velocity of train B wrt train A, train A becomes your primed frame so choose your axes to make its velocity in the x direction.

(**Recall that length contracts along the direction of motion but not perpendicular to it--thus the direction of motion selects one axis for special treatment.)
 

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