What is the speed of each particle relative to the other?

Click For Summary
Two particles shot out simultaneously at right angles with velocity v perceive each other's speed as v√2 due to their relative motion. The relative velocity can be calculated using vector analysis, where the relative speed is the hypotenuse of a right triangle formed by their velocities. For example, if one ship travels at 4c/5 and another at 3c/5, their relative velocity is c/5. When dealing with relativistic speeds in different directions, the relativistic velocity addition formula is necessary for accurate calculations. Understanding these principles is crucial for analyzing relative motion in physics.
pivoxa15
Messages
2,250
Reaction score
1
Two particles are shot out simultaneously from a given point with velocity v. They start at right angles to each other. What is the speed of each particle relative to the other?

I got one particle thinks the other one is traveling at v\sqrt2 whereas they are traveling at v and vice versa. But I think I calculated it from a frame that is at rest relative to both particles and imagined what each particle would think if they had information in my frame.
 
Last edited:
Physics news on Phys.org
pivoxa15 said:
Two particles are shot out simultaneously from a given point with velocity v. They start at right angles to each other. What is the speed of each particle relative to the other?

I got one particle thinks the other one is traveling at v\sqrt2 whereas they are traveling at v and vice versa. But I think I calculated it from a frame that is at rest relative to both particles and imagined what each particle would think if they had information in my frame.
If you stick to a vector analysis, it is not so confusing. The relative velocity of a to b is the velocity of a relative to the origin - velocity of b relative to the origin (or + the velocity of the origin relative to b). So just subtract the velocity vectors.

The relative velocity, therefore is \vec v_a + (-\vec v_b). Relative speed is the length of that vector, which is the hypotenuse of a right triangle with sides of length v: v_{rel}\sqrt{2v^2} = \sqrt{2}v.

AM
 
Last edited:
So veclocity of a to b is usually just \vec v_a -\vec v_b.

So if we look at another example, ship a is traveling at 4c/5. ship b is traveling at 3c/5 in 1D motion. The relative velocity of ship a to b or what ship b thinks a is traveling is v_a -v_b = c/5.
 
Last edited:
If the speeds are relativistic and in different directions, you need to use the vector form of the relativistic "velocity addition" formula. See the section "The velocity addition formula for non-parallel velocities" on the following page from the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html
 
Last edited:
But if I am just doing velocity addition for different frames (not events in frame), i.e know what the other frame is traveling relative to me, do I still need the relativistic addition formula? If I do apply it than I could pretend the 'event' is a person standing still in the other frame. So the velocity of that person is 0 in their frame S'. I get the speed I see them is v the relative velocity of the frames. This value can be calculated with the method I suggested in #3?
 
pivoxa15 said:
But if I am just doing velocity addition for different frames (not events in frame), i.e know what the other frame is traveling relative to me, do I still need the relativistic addition formula?
Yes. The relativistic addition formula tells you how velocities transform from one frame to another, which is what you want.
 
(a) The polarisation pattern is elliptical with maximum (1,1) and minimum (-1,-1), and anticlockwise in direction. (b) I know the solution is a quarter-wave plate oriented π/4, and half-wave plate at π/16, but don't understand how to reach there. I've obtained the polarisation vector (cos π/8, isin π/8) so far. I can't find much online guidance or textbook material working through this topic, so I'd appreciate any help I can get. Also, if anyone could let me know where I can get more...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 29 ·
Replies
29
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K