Special Relativity with 2 light beams moving towards each other

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SUMMARY

The discussion centers on calculating the relative speed of two light beams moving towards each other using the formula (v+w) / (1 + cw/c²). The user attempts to apply this equation with both velocities set to the speed of light (c), resulting in an incorrect conclusion of 2c. The correct interpretation confirms that no object can exceed the speed of light, leading to the conclusion that the left-moving beam approaches the right-moving beam at speed c from the perspective of the right-moving beam.

PREREQUISITES
  • Understanding of Special Relativity principles
  • Familiarity with the speed of light (c) as a universal constant
  • Knowledge of relativistic velocity addition formula
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the relativistic velocity addition formula
  • Explore implications of Special Relativity on simultaneity and time dilation
  • Learn about Lorentz transformations in detail
  • Investigate experimental evidence supporting Special Relativity
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Students of physics, educators teaching Special Relativity, and anyone interested in understanding the behavior of light and relativistic speeds.

Felipe Doria
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Homework Statement



If one light beam is heading toward you from the right, and another is heading toward you from the left, how quickly from the perspective of the right-moving beam is the left-moving beam approaching it?

1) 0
2) c
3) 1.5c
4) 2c

Homework Equations



(v+w) / (1 + cw/c2)

The Attempt at a Solution



I tried plugging in c into the equation for both v and w but it gives me (c + c) / (1 + c2/c2) which equals 2c. It is one of the possible answers but I know nothing can go faster than c so I'm thinking the answer is probably c. Am I right? If so, why? If not, what am I doing wrong?
 
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Felipe Doria said:

Homework Statement



If one light beam is heading toward you from the right, and another is heading toward you from the left, how quickly from the perspective of the right-moving beam is the left-moving beam approaching it?

1) 0
2) c
3) 1.5c
4) 2c

Homework Equations



(v+w) / (1 + cw/c2)

The Attempt at a Solution



I tried plugging in c into the equation for both v and w but it gives me (c + c) / (1 + c2/c2) which equals 2c. It is one of the possible answers but I know nothing can go faster than c so I'm thinking the answer is probably c. Am I right? If so, why? If not, what am I doing wrong?

(c+c)/(1+c^2/c^2) isn't equal to 2c.
 

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