- #36
AlienGrey
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I'm interested in you getting 0.5 and 0.5 from 0.87 and 0.29 using a single SR factor.PeroK said:
Light Speed In Forward and Backward Directions
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In summary, the conversation discusses a thought experiment involving a moving frame and the discrepancy in the speed of light for forward and backward light beams as viewed from a stationary frame. The experiment shows that the time taken for the beams to reach the end points is different in the moving frame compared to the stationary frame. This is due to the relativity of simultaneity and can be explained using the Lorentz transforms. The speed of light is found to be consistent at 300,000 km/s in both frames. This experiment is a basic concept in special relativity and can be easily understood by competent physics students.
- #37
Ibix
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We've already told you how to do it. In the time you've been arguing here you could have solved your problem about half a dozen times...AlienGrey said:
- #38
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No you're not! You're only interested in trolling us. Admit it!AlienGrey said:
- #39
AlienGrey
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Yeah but I want to see YOU do it. I already did it, now it's your turn.Ibix said:
- #40
AlienGrey
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Just keep dancing.PeroK said:
- #41
Ibix
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And if we thought it was impossible why would we be repeatedly stressing how easy it is? Wouldn't we be sucking our teeth and muttering about it being too difficult, like a mechanic trying to up the price on a repair job?
This is a trivial piece of book work that everyone here except you has done many times.
This is a trivial piece of book work that everyone here except you has done many times.
- #42
Vanadium 50
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I don't think that's how it works.AlienGrey said:
- #43
AlienGrey
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As a stall tactic because you know you can't actually do it, I would assume.Ibix said:
- #44
weirdoguy
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AlienGrey said:
Then show it to us.
- #45
Ibix
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Or because it actually is that easy. Which it is.AlienGrey said:
- #46
Ibix
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And if you really believe that, call our bluff. Do the maths we're lying about being the right thing (not your post #1 maths which neglects at least one important factor) and prove us wrong.AlienGrey said:
- #47
AlienGrey
- 19
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I did it in my head and it didn't work so I now I need you do it and show that it does work. You're the one who claims it will work so why aren't you just showing me how it works so great?weirdoguy said:
- #48
AlienGrey
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You're the one who claims this magic formula of yours will work so please demonstrate, I'm all eyes.Ibix said:
- #49
Mark44
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Very frequently we get members who post homework questions in technical forum sections such as this one, so your argument is very flawed.AlienGrey said:
- #50
Mark44
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Thread temporarily closed for moderation.
- #51
berkeman
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Since the Alien has gotten into his spaceship and left us, this thread will remain closed. Thanks for trying to help him out while he was briefly with us. 
- #52
Dale
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I am late to the party but just thought that I would add this to make it abundantly clear how easy this stuff is. Just go to https://en.wikipedia.org/wiki/Lorentz_transformation#Proper_transformations to get the Lorentz transform matrix. For ##v=(c/2,0,0)## we get$$\Lambda =
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & \frac{1}{\sqrt{3}} & 0 & 0 \\
\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)$$ Then we simply use that transformation matrix to determine events ##(ct,x,y,z)## in the ground frame. $$\Lambda \cdot \left(0,0,0,0\right)=\left(0,0,0,0\right)$$$$\Lambda \cdot \left(\frac{c}{2} ,\frac{c}{2} ,0 ,0 \right) = \left( \frac{\sqrt{3}c}{2},\frac{\sqrt{3}c}{2},0,0 \right)$$$$\Lambda \cdot \left(1 c ,0 ,0 ,0 \right) = \left(\frac{2 c}{\sqrt{3}},\frac{c}{\sqrt{3}},0,0 \right)$$ You can do the subtraction and evaluate the numbers to see that the time is 0.87 for the forward and 0.29 for the reverse.
As long as they can do matrix multiplication, it is indeed high school level. The reluctance to do this for the lazy alien was due to our usual approach of encouraging the OP to learn for themselves, not because this was in any way challenging to do
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & \frac{1}{\sqrt{3}} & 0 & 0 \\
\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)$$ Then we simply use that transformation matrix to determine events ##(ct,x,y,z)## in the ground frame. $$\Lambda \cdot \left(0,0,0,0\right)=\left(0,0,0,0\right)$$$$\Lambda \cdot \left(\frac{c}{2} ,\frac{c}{2} ,0 ,0 \right) = \left( \frac{\sqrt{3}c}{2},\frac{\sqrt{3}c}{2},0,0 \right)$$$$\Lambda \cdot \left(1 c ,0 ,0 ,0 \right) = \left(\frac{2 c}{\sqrt{3}},\frac{c}{\sqrt{3}},0,0 \right)$$ You can do the subtraction and evaluate the numbers to see that the time is 0.87 for the forward and 0.29 for the reverse.
As long as they can do matrix multiplication, it is indeed high school level. The reluctance to do this for the lazy alien was due to our usual approach of encouraging the OP to learn for themselves, not because this was in any way challenging to do
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