Light Speed In Forward and Backward Directions

[AlienGrey]

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.

 

[Ibix]

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...

 

[PeroK]

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!

 

[AlienGrey]

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.

 

[AlienGrey]

No you're not! You're only interested in trolling us. Admit it!

Just keep dancing.

 

[Ibix]

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.

 

[Vanadium 50]

so the onus is on you to show how I'm wrong.

I don't think that's how it works.

 

[AlienGrey]

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.

 

[weirdoguy]

I already did it

Then show it to us.

 

[Ibix]

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.

 

[Ibix]

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.

 

[AlienGrey]

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?

 

[AlienGrey]

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.

 

[Mark44]

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.

 

[Mark44]

Thread temporarily closed for moderation.

 

[berkeman]

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.

 

[Dale]

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