Solving for c: Speed of Q2 in Q1 Reference Frame

AI Thread Summary
The discussion focuses on calculating the speed of Q2 as observed from the reference frame of Q1 using the relativistic velocity addition formula. The user initially misapplies the formula, mistakenly calculating the speed of Q2 as 0.922c instead of correctly determining v_{BC}. Clarifications are provided on the definitions of the variables and the need to consider the direction of motion, particularly noting that if Q1 is moving at 0.860c relative to the observer, the observer's speed relative to Q1 must be negative. The suggestion is made to either rearrange the variables or apply classical velocity addition for verification. Understanding the correct application of the formula is crucial for accurate results in relativistic contexts.
Pochen Liu
Messages
52
Reaction score
2
Homework Statement
Stellar system Q1 moves away from us at a speed of 0.860c. Stellar system Q2, which lies in the same direction in space but is closer to us, is moving away from us at a speed 0.560c. What multiple of c gives the speed of Q2 as measured by an observer in the reference frame of Q1?
Relevant Equations
*attached
I've tried using this equation:

243343

Where:
u' = 0.86c
v = 0.3c
u = Is the speed of Q2 as measured by an observer in the reference frame of Q1 = 0.922c

Where have I gone wrong? Have I missed a negative symbol?
 
Physics news on Phys.org
Before plugging the numbers in, make sure you understand the variables in velocity addition formula.

If you have an observer ##A##, a second observer ##B##, and an object ##C##, then let:

  • ##v_{BA}## = the velocity of ##B## as measured by ##A##.
  • ##v_{CB}## = the velocity of ##C## as measured by ##B##.
  • ##v_{CA}## = the velocity of ##C## as measured by ##A##.
If all three velocities are in a straight line, then according to relativity,

##v_{CA} = \dfrac{v_{BA} + v_{CB}}{1+ \frac{v_{BA} v_{CB}}{c^2}}##

In your case, the three objects are:
  1. You = ##A##
  2. Q1 = ##B##
  3. Q2 = ##C##

In this case, what they are asking is not ##v_{AC}##. They tell you that that is equal to ##0.860c##. They are asking for ##v_{BC}##.

So you have two options, that work the same:
  • Write down the formula for ##v_{CA}, v_{BA}, v_{CB}## and solve, algebraically for ##v_{CB}## in terms of the other two, or
  • Switch the labels around. Let ##A## be Q1, ##B## be you, and ##C## be Q2.
If you do the second one, you have to remember that if Q1 is moving at 0.860c relative to you, then you are moving at -0.850c relative to Q1.
 
Pochen Liu said:
Problem Statement: Stellar system Q1 moves away from us at a speed of 0.860c. Stellar system Q2, which lies in the same direction in space but is closer to us, is moving away from us at a speed 0.560c. What multiple of c gives the speed of Q2 as measured by an observer in the reference frame of Q1?
Relevant Equations: *attached

I've tried using this equation:

View attachment 243343
Where:
u' = 0.86c
v = 0.3c
u = Is the speed of Q2 as measured by an observer in the reference frame of Q1 = 0.922c

Where have I gone wrong? Have I missed a negative symbol?
Try the calculation using the classical velocity addition and see what you get.
 
  • Like
Likes Pochen Liu
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 '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