In a relativistic collision, if two particles come off at right angles, and each moves at 0.85c, what is their separation distance in a moving frame?

To determine the separation distance of two particles moving at right angles after a relativistic collision, where each moves at a speed of 0.85c, we can employ the principles of relativistic velocity addition and Pythagorean theorem in a modified way.

In relativistic mechanics, when two objects move perpendicularly to each other, their velocities do not combine in a straightforward manner as they would in classical mechanics. Instead, the effective relative velocity needs to account for the effects of special relativity.

Here, if we denote the speed of each particle as ( v = 0.85c ), we have to find the resultant velocity in the frame of reference that is moving relative to these particles. However, what is important here is that the distance traveled by each particle can be calculated individually and then combined geometrically.

For each particle, the distance it travels can be described as:

• Particle 1 moves along the x-axis,

• Particle 2 moves along the y-axis.

Using the time ( t ) they travel for our scenario, each particle covers a distance calculated by the formula ( d = vt ):

1. For particle 1 (along the x-axis), the distance is ( d_1 = 0.