# Graphical Derivation of the CADO Equation

Discussion in 'Physics & Math' started by Mike_Fontenot, Sep 5, 2018.

1. ### NotEinsteinValued Senior Member

Messages:
1,583
There's a reason why I didn't even bother addressing what effects Charlie's slowing down would have: it's not relevant to the scenario (that, and the headaches, yes

). According to the CADO-equation, after Charlie has come to his full stop, there's no age difference anymore (due to $v=0$), so what Charlie measures during the slowing down doesn't really matter. But it certainly is an interesting question nonetheless!

3. ### Neddy BateValued Senior Member

Messages:
1,727
I'm not talking about Charlie slowing down. Using instantaneous acceleration, there is a moment just before he stops when he would say Alice is 10 years old. Then when he stops a moment later, he says she is 40.

He could have known she was 40 before he stopped, because a clock at rest in the the stay-home frame right next to him, which is synchronized to Alice's would say 40. But before he stops, he would not agree that all the stay-home frame clocks are synchronized according to his own reference frame. They all display different times according to his frame.

So, if Charlie and Bob are truly in the same reference frame, and if Charlie says there is a stay-home-frame clock right next to him, which is synchronized to Alice's, which says 40, that means Bob should be moving past a 10 year old Alice.

5. ### Neddy BateValued Senior Member

Messages:
1,727
What do you mean there is no age difference? After he stops, Charlie is 20 and Alice is 40.

7. ### Neddy BateValued Senior Member

Messages:
1,727
I think the above comment is the key to the NotEinstein scenario. When Bob and Alice are both 10, that is the moment when Bob needs to join Charlie in his reference frame. That is the moment when he will find Charlie being 20 years old in his own frame. That is the only way they can both agree that Alice is 10, and Charlie will have already stopped, and started moving again.

8. ### NotEinsteinValued Senior Member

Messages:
1,583
Right, I should have been more clear. I was talking about there being no difference in CADO, $\text{CADO}_T=\text{CADO}_H$.

9. ### Neddy BateValued Senior Member

Messages:
1,727
In CADO, neither of those variables is the age of the traveling twin.

10. ### NotEinsteinValued Senior Member

Messages:
1,583
Right, and the age of the traveling twin has no relevance in my scenario, so that's fine.

11. ### Neddy BateValued Senior Member

Messages:
1,727
Okay, I thought you were saying there was no age difference between Charlie and Alice. As long as that is not the case, we are fine.

12. ### NotEinsteinValued Senior Member

Messages:
1,583
According to the CADO-equation, after Charlie has stopped and before he starts moving again, everybody must agree on the age of Alice ($\text{CADO}_T=\text{CADO}_H$ due to $v=0$). What is the age of Alice when this happens? Obviously, it can't be less than 40 years, because Charlie has already measured her being that age. Bob doesn't start moving before that's the case, so Bob cannot take of when Alice is 10.

13. ### NotEinsteinValued Senior Member

Messages:
1,583
Sorry for the confusion; I'll try to be more clear in the future and not refer to the CADO as age.

14. ### Neddy BateValued Senior Member

Messages:
1,727
With the above concept in mind, below is my REVISED attempt at the NotEintein scenario, using instantaneous acceleration, and the inverse Lorentz transformation.

I will start with the same basic twin scenario as described on the webpage, with v=0.866c so that gamma=2 and the time Charlie has been traveling is 40 years as measured by the stay-home twin Alice. So Alice says the distance traveled is 0.866*40=34.64 lightyears. Alice also says the age of the travelling twin is 40/2=20 years.

Then, at that moment, Charlie stops. The calculation that Charlie does should result in Alice being 40, if done correctly.

Using the inverse Lorentz transformations:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.000 * -17.32))
Distributing the gamma term results in the CADO equation:
t = 40 - (0.000 * 34.64))
t = 40

Looks like a correct result. And of course Charlie would be 20 years old. Now, in accordance with the NotEinstein scenario, Charlie will start moving again, at the same velocity as before.

Starting with the inverse Lorentz transformations:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * -17.32))
Distributing the gamma term results in the CADO equation:
t = 40 - (0.866 * 34.64))
t = 10

So now Charlie would say that Alice is only 10 years old again. Note that I left the clocks as they were, at 20 and 40. But I could have reset them to 0, calculated t=-30 and then added that to the known age of 40 and still arrived at t=10 that way.

Now let's see what Bob would say. He has been with Alice since they were born. When he is 10 years old he decides to start moving at the same speed as Charlie. This time I will reset the clocks to 0 and then add the result to Alice's known age of 10 at the end.

Starting with the inverse Lorentz transformations:
t = γ(t' + (vx' / c²))
t = 2(0 + (0.866 * 0.00))
Distributing the gamma term results in the CADO equation:
t = 0 - (0.866 * 0.00))
t = 0
Finally, adding in the known age of 10 at the end:
t = 0 + 10
t = 10

So Bob and Charlie are now in the same reference frame, and they calculate the same age for Alice, 10 year old.

Last edited: Sep 7, 2018
15. ### Neddy BateValued Senior Member

Messages:
1,727
Charlie is over 34 light years away from Alice when he stops momentarily, so Bob cannot wait until he actually sees Charlie stop. Bob must leave at age 10 for everything to work out.

16. ### NotEinsteinValued Senior Member

Messages:
1,583
You once again are violating causality; Alice cannot age backwards... See your own post #40.

17. ### NotEinsteinValued Senior Member

Messages:
1,583
I have nowhere in my scenario demanded Charlie only stops momentarily?

18. ### Neddy BateValued Senior Member

Messages:
1,727
Another interesting tidbit:

When Bob is 10 years old, Charlie is 8.66 light years away, as measured by the stay-home frame. Pretending for a moment that Charlie is at the front of a train, it would have to be twice that length in it's own frame, due to length contraction, in order for the rear of that train to reach Bob.

2*8.66=17.32 which is the value of the -x' in Charlie's calculation that determines Alice is 10 years old. So if Bob takes off when he is 10 years old, he will be just the right distance away from Charlie. Everything works out.

19. ### Neddy BateValued Senior Member

Messages:
1,727
I would ask you to show some example of how it could violate causality. Assume information can be sent at the speed of light. What information could ever be sent to the future? What effect could precede its cause? Show an example please.

20. ### Neddy BateValued Senior Member

Messages:
1,727
I did so for simplicity. However long he waits will just have to be added to the end results I have calculated.

21. ### Neddy BateValued Senior Member

Messages:
1,727
Alice does not age backwards, she is 10 when Bob leaves, and she is 40 when Charlie stops and starts again.

22. ### NotEinsteinValued Senior Member

Messages:
1,583
That is in direct contradiction with:

23. ### NotEinsteinValued Senior Member

Messages:
1,583
That is in direct contradiction with: