text: # Measurement of the speed of light form different perspectives
  
  I've been showing a special interest in Einstein's theory of relativity and how he proved the speed of light to be always the same. At first it was a bit hard for me to understand, but now I THINK I understand what this means.
  
  So basically Einstein's theory claims that the speed of light is always the same, even if supposedly someone (say, person A) would move towards a light beam, he would measure the same speed in respect to a person (say, person B) measuring the speed of the light beam from a relative rigid position. Furthermore, the theory states that the time for person A is moving slower in respect to person B, because person A is moving in respect to person B (the time of a moving object runs slower in respects to a rigid object).
  
  Keeping these features in mind, I've stated an example to try and explain why person A and B are measuring the same speed of light, even though person B is moving towards the light (you would think he'd measure a greater speed), which is as follows:
  
  We have person A, whose moving with v=100,000 km/s straight towards a light beam with c=300,000 km/s. Then there's person B, who is standing at a rigid position in respect to the earth, so we can state that person B's velocity is 0 km/h. A "logical" measurement of the speed of the light beam for person A would be 100,000 + 300,000 = 400,000 km/s and 300,000 + 0 = 300,000 km/s for person B. This, however, is not what we observe. We observe the measurements of the speed of light of the two persons to be exactly the same (c = about 300,000 km/s). So I personally thought that the measurements of the speed of the light beams are exactly the same because time is slowing down for person A in respect to person B. So, person A WOULD measure 400,000 km/s, were it not that 1 second for him is not the same second for person B. A simple calculation would then conclude that in 1 second of person B's time 0.75 seconds passes in person A's time (400,000 times 0.75 equals 300,000 km/s equals the speed of light).
  
  Can somebody tell me if my way of thinking is legit? Thanks in advance.
  
  • A measures 300,000 km/s due to (special relativistic) velocity addition. – Kyle Kanos Jul 19 '15 at 14:46
  • Also, asking "Is this correct" type questions are not a good fit for this Q&A site because the answer (yes or no) has too few characters to make a complete answer (minimum 15 chars). – Kyle Kanos Jul 19 '15 at 14:47
  • If you think about light beams moving parallel to the direction of B's travel, and light beams moving perpendicular to the direction of B's travel, you'll discover that you cannot account for the constancy of light speed with time dilation alone, because you'll need different dilation factors to account for the two kinds of beams. – WillO Jul 19 '15 at 15:12
  • @WillO I see your point. Person A would still measure 300,000 km/s for light beams perpendicular to the path he's traveling. I almost thought I had it figured out.. So what is it then that makes A measure parallel and perpendicular light beams at the same speed? Is there something (obvious) I'm missing here? – AMG Jul 19 '15 at 15:21
  • @user86273: You really should read an elementary book about relativity. – WillO Jul 19 '15 at 15:27
  
  The problem is that special relativity uses Lorentz transformation and not Galilean transformation. If we consider an inertial reference frames $S$ it's perfectly fine to just use the vector sum to add two velocities. If we consider velocity $v_1$ and $v_2$, the velocity summation $s$ is so that $s = v_1 + v_2$.
  However, in special relativity $s$ takes the form of $\frac{v_1+v_2}{1+(\frac{v_1 v_2}{c^2})}$. If we set $c = 1$ we can simplify it to $\frac{v_1+v_2}{1+v_1 v_2}$.
  Using your example of $v = 100 \text{ km/s} = 0.3336c$, we find that $s = \frac{1 + 0.3336}{1 + 1 \cdot 0.3336} = 1.$
  In fact, for any velocity $v$ we find that $s = \frac{1 + v}{1 + 1 \cdot v} = \frac{1 + v}{1 + v} = 1.$ So the speed of light is constant in every reference frame, even though two reference frames moves relative to each other.