# Light Speed In Forward and Backward Directions

• B
• AlienGrey
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.
PeroK said:
It is impossible to teach anyone anything they do not want to learn. If you are not interested in learning SR, then there is no point to the discussion.
I'm interested in you getting 0.5 and 0.5 from 0.87 and 0.29 using a single SR factor.

AlienGrey said:
Show how to get 0.5 and 0.5 from 0.87 and 0.29 using the same SR factor. We both know you can't, isn't that right?
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:
I'm interested in you getting 0.5 and 0.5 from 0.87 and 0.29 using a single SR factor.
No you're not! You're only interested in trolling us. Admit it!

Ibix said:
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...
Yeah but I want to see YOU do it. I already did it, now it's your turn.

malawi_glenn
PeroK said:
No you're not! You're only interested in trolling us. Admit it!
Just keep dancing.

malawi_glenn
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.

AlienGrey said:
so the onus is on you to show how I'm wrong.
I don't think that's how it works.

russ_watters
Ibix said:
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.
As a stall tactic because you know you can't actually do it, I would assume.

malawi_glenn and Motore
AlienGrey said:

Then show it to us.

russ_watters
AlienGrey said:
As a stall tactic because you know you can't actually do it, I would assume.
Or because it actually is that easy. Which it is.

AlienGrey said:
As a stall tactic because you know you can't actually do it, I would assume.
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.

weirdoguy said:
Then show it to us.
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?

Dale, malawi_glenn and weirdoguy
Ibix said:
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.
You're the one who claims this magic formula of yours will work so please demonstrate, I'm all eyes.

malawi_glenn and weirdoguy
AlienGrey said:
Is this the homework section? No, so you doubting it being homework seems irrelevant.
Very frequently we get members who post homework questions in technical forum sections such as this one, so your argument is very flawed.

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.