# Graphical Derivation of the CADO Equation

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

1. ### NotEinsteinValued Senior Member

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1,813
Then why use your own (sloppy) definition instead of the standard one? For example, you forgot to specify the medium this light pulse is traveling in. And why make it a light pulse, when just "light" is sufficient?

So... it's the proper frame? Why not just call it that then? Why invent confusing and non-standard terminology? Why go through all that trouble to, in a cumbersome way, define something that's already well established?

3. ### Neddy BateValued Senior Member

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1,767
Yes, I did think you meant they start moving simultaneously in the stay-home frame, but I thought perhaps it should be stated for clarity. The point being that if they start simultaneously in one frame, they do not start simultaneously in the other frame, so it is complicated.

I still have not had time to work out your triplet scenario using the inverse Lorentz transformations, so I still don't know for sure whether you have shown a flaw in the cado equation, but you might have.

But now I know that the cado equation is just the inverse Lorentz transformation with t substituted for γt' and with x substituted for -γx'. The 40 is not really t, it is γt'. And the 34.64 is not really x it is -γx'. So I would expect the cato equation to fail for cases where x is not equal to γx' and/or cases where t is not equal to γt'. And your triplet case might be one of those cases.

Last edited: Sep 6, 2018

5. ### Neddy BateValued Senior Member

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1,767
I think I know the answer now. The cado equation is derived from the inverse Lorentz transformation, under specific circumstances, as follows:

t = γ(t' + (vx' / c²))
where t is the time on some stay-home-frame clock, and t' is the time on some traveling-clock

We give a new name to t:
cado_t = γ(t' + (vx' / c²))

We distribute the gamma term:
cado_t = γt' + (vγx' / c²))

Now substituting t=γt' for such a case (i.e. in your website example where 40=2*20):
cado_t = t + (vγx' / c²))

Now substituting x=-γx' for such a case (i.e. in your website example where 34.64=-2*-17.32 which you don't really mention):
cado_t = t - (vx / c²))

And that is the cado equation.

7. ### NotEinsteinValued Senior Member

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1,813
True, so it's good that that's now made explicit.

At a glance, I agree with your assessment of my scenario. The issue I've highlighted is that the CADO-equation can't handle the case where the starting position of the two twins isn't equal when $v$ starts deviating from zero(, and that you're not allowed to vary $v$ back to zero and start moving in the same direction again later). In such cases there's an offset to x' (or x, whichever you fancy), and thus the correspondence you found between the inverse Lorentz transformation and the CADO-equation breaks, invalidating the use of the CADO-equation in such a case.

8. ### NotEinsteinValued Senior Member

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1,813
Our posts crossed; I see you've reached the same conclusion. The CADO-equation is only valid if:
t=γt'
and
x=-γx'
which means that t=t' and x=x' has to hold at the start of the motion.

(and perhaps $v$ is constant during the motion?)

9. ### Neddy BateValued Senior Member

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1,767
Yes, I'm glad you and I agree on that conclusion.

I'm not yet sure which cases violate t=γt' and x=-γx' but you definitely could be right about that.

10. ### Mike_FontenotRegistered Member

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81
That is incorrect. The CADO equation can handle arbitrary accelerations by the traveling twin, including two and three dimensional motions. I have used it to exactly solve the scenario that Brian Greene discusses in his NOVA program (about the alien creature riding a bicycle in a circle billions of light years from Earth, causing him to conclude that the current time on Earth is swinging forward and backward by centuries during each cycle of his circle). What the CADO equation can't handle is scenarios where the home twin also accelerates in arbitrary ways (which I make clear in the definition of the CADO equation in the webpage). But such scenarios CAN be handled by the CADO frame, it's just that the CADO equation can't be used. Those scenarios are just more difficult to handle. This is addressed in Section 14 of the webpage.

11. ### Mike_FontenotRegistered Member

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81
That's a very interesting example! I've never thought of a scenario like that. I haven't had time yet to work through your scenario, but I suspect that the CADO equation can handle it without any problems. I'll spend some time on it ASAP.

I CAN tell you that in general, the CADO FRAME CAN handle scenarios with overlapping FINITE (and spread-out) accelerations by both the "traveler" AND the "home" twin, but that the CADO EQUATION can't be used then ... other more difficult methods are required. But when the accelerations by both the "traveler" and the "home" twin consist ONLY of Dirac accelerations (i.e., when there are only instantaneous velocity changes involved), I suspect that the CADO equation does work without problems. But I'll need to try that case to find out for sure.

Last edited: Sep 7, 2018
12. ### Mike_FontenotRegistered Member

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81
You're right about my failure to specify the medium ... I should have said "in vacuum".

I intentionally specified it as a light PULSE, because I think that makes it clearer and less likely to be misunderstood.

Because I think my terminology is much clearer. "Proper frame" is not a term that is clearly understood by everyone, and it could easily result in confusion.

13. ### Q-reeusValued Senior Member

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3,172
Best thing to do is go study how an actual, generally applicable, CADO formula is derived and applied to a few particular cases:
https://en.wikipedia.org/wiki/Twin_...psed_times:_how_to_calculate_it_from_the_ship
As I wrote earlier, the correct expression looks nothing like your one.
In the case of constant circular motion about an axis intersecting the home twin, your formula predicts
since v*L (vector usage) is zero. Whereas the actual case has an accumulation of time differential
where γ = 1/√(1-v²/c²) has the usual meaning. Circular motion breaks the symmetry applying to linear relative motion - the traveling twin ages slower in an absolute sense.

14. ### NotEinsteinValued Senior Member

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1,813
Well, I've already demonstrated that it can't, so good luck...

Sure, the proper frame can handle many such cases, because that's simply SR, and SR can handle those cases.

Like standard SR methods, yes.

As I've already demonstrated, no it can't. The accelerations aren't the problem, $L$ not being equal to zero when $v$ starts deviating from zero is.

OK, so your definition was sloppy, as I said. Glad we got that cleared up.

It's actually less clear: what's the profile of this pulse? I.e. is it a block pulse, a sine-pulse, or...? Because you've underspecified it, it's more likely to be misunderstood. If you explicitly say "light pulse", then that seemingly disqualifies light non-pulses, and that needs explanation.

It is not. It's very confusing, and your definitions are lacking, as I've demonstrated.

Except everyone that has studied the slightest amount of SR...

No, it doesn't. Your term does, as I've demonstrated.

Last edited: Sep 7, 2018
15. ### Mike_FontenotRegistered Member

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81
I've decided the way I am going to proceed in handling NotEinstein's very interesting scenario. Section 14 of the webpage (entitled "The CADO Frame When the Distant Person Is Also Accelerating") shows how to handle all cases where the overlapping accelerations are finite, and of finite duration. So it will also properly handle very short but finite acceleration profiles that are rectangular, with some proscribed area under the rectangle. The limit of such a rectangular acceleration profile, as the height of the rectangle goes to infinity, and the width goes to zero, is just the Dirac delta function. For very narrow and high (but finite) rectangles, it is well known that their effect becomes arbitrarily close to the effect of the Dirac delta function, as they get narrower and higher, as long as their area remains constant.. So that means that I can correctly solve the scenario for a very narrow and very high but finite rectangular acceleration profile. THEN, I'll use the CADO equation, and see if it agrees with the correct results for the very narrow and very high rectangular acceleration profile. If it agrees, then the CADO equation works for NotEinstein's scenario. If it doesn't agree, then the CADO equation doesn't work for his scenario. This analysis may take me a while ... it's been a long time since I wrote Section 14, and it won't be familiar to me. (It is already known, as stated in Section 14, that for overlapping elongated acceleration profiles for both twins, the CADO equation doesn't work, but that a more difficult procedure does work. What I don't yet know, is if the CADO equation works for simultaneous instantaneous velocity changes by both twins. The above plan should resolve that question.)

16. ### NotEinsteinValued Senior Member

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1,813
You have fun with that. Just know that it will utterly fail to address my scenario. You see, all instances of acceleration used in it are identical to those already present in the original twin paradox scenario. In other words, your claim that addressing the acceleration is the key is the same as an admittance that you haven't handled acceleration properly in the original twin paradox scenario, and thus currently all claims by you that the CADO-equation can handle it are unfounded.

On top of that, the acceleration isn't the problem anyway; perhaps you should try to understand the problem before addressing something arbitrarily in the hopes that will solve it.

17. ### Neddy BateValued Senior Member

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1,767
Small problem: If Bob and Charlie have the same acceleration profile as measured by Alice, then they will not remain in the same reference frame as each other, at least not during the acceleration itself. If they truly were in the same reference frame as each other, the distance between them, as measured by Alice, should become length contracted.

Could we simplify this by removing Bob altogether?

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

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1,767
Another small problem: Distance does matter in SR. See the x' variable in the inverse Lorentz transform equation for time: t = γ(t' + (vx' / c²))

I think the NotEinstein scenario (perhaps without Bob?) is still worth pursuing though, just out of curiosity.

19. ### NotEinsteinValued Senior Member

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1,813
It's not about the acceleration profile; as long as they end up going by the same speed $v$. That's all that's in the CADO-equation, so that's all that (apparently) matters.

Sure, but then it's more difficult to see there's a conflict with SR, which is why I left Bob in.

20. ### Neddy BateValued Senior Member

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1,767
Here is my 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 standing next to Alice the whole time, and he is 40 years old at the same time that Charlie started moving the second time. At that time, Bob starts moving at the same speed as Charlie. This time I will reset the clocks to 0 and then add the result to the known age of 40 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 40 at the end:
t = 0 + 40
t = 40

Yes indeed Bob and Charlie do calculate different ages for Alice. I think that is how it is in SR, and the CADO equation also, at least in this case, (unless I did something incorrectly, a distinct possibility!). The CADO equation would even work if the clocks were not reset to zero in the last case, as follows:
t = 40 - (0.866 * 0.00))
t = 40

Last edited: Sep 7, 2018
21. ### NotEinsteinValued Senior Member

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1,813
I think you've just violated causality. How is it possible for Alice to travel back in time? How can time dilation through the Lorentz transformation ever result in negative $\Delta t$?

22. ### Neddy BateValued Senior Member

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1,767
Wait, I think I did that wrong. Let's synchronize Alice's clock with one that is right next to Charlie at the time he starts moving the second time. That means we use x'=0.00 as follows:

Starting with the inverse Lorentz transformations:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * 0.00))
Distributing the gamma term results in the CADO equation:
t = 40 - (0.866 * 0.00))
t = 40

So Bob and Charlie do not calculate different ages for Alice after all! For that calculation, SR and the CADO equation both require adding a clock which is next to charlie and synchronized to Alice's.

23. ### Neddy BateValued Senior Member

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1,767
Well I think I may have made a mistake, see above.

But before Charlie stopped, he would have calculated Alice to be 10, even though a clock in the stay-home frame and synchronized with Alice's right next to him would have said t=40. So maybe he would say she was 10 again? My brain hurts.