Graphical Derivation of the CADO Equation

[Neddy Bate]

That is in direct contradiction with:

If you want Charlie to pause, then Bob can wait until he sees Charlie stop. But Bob still has to leave before he sees Charlie start up again. Agreed? How do you propose he know when to leave?

That is in direct contradiction with:

Assuming Charlie only stops momentarily (because it is LESS complicated):

In the moment before Charlie stops, how old does he say Alice is? 10.
In the moment before Charlie stops, what time is displayed on a stay-home frame clock synched with Alice's, which is right next to him? 40.
So we know that Alice being 10 is simultaneous with a stay-home frame clock synched with Alice's right next to Charlie displaying 40.

In the moment after Charlie starts moving again, what time is displayed on a stay-home frame clock synched with Alice's, which is right next to him? 40.
Given that we know that Alice being 10 is simultaneous with a stay-home frame clock synched with Alice's right next to Charlie displaying 40, what time is Alice in the moment after Charlie starts moving again? 10.

 

[NotEinstein]

If you want Charlie to pause,

I don't "want Charlie to pause"; I've never said he can't/doesn't. That's something you came up with.

then Bob can wait until he sees Charlie stop. But Bob still has to leave before he sees Charlie start up again. Agreed? How do you propose he know when to leave?

Charlie and Bob should leave when all three parties involved agree that everyone is stationary with respect to each other. Only that way everybody agrees that $$\text{CADO}_H=\text{CADO}_T$$

Do we agree that Bob can (not necessarily "must", but "can") indeed wait until he sees Charlie stop?

Assuming Charlie only stops momentarily (because it is LESS complicated):

In the moment before Charlie stops, how old does he say Alice is? 10.

This is in direct contradiction with:

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

If Charlie calculated Alice to be 40, Alice is 40 according to Charlie. How can she also be 10 according to Charlie?

In the moment before Charlie stops, what time is displayed on a stay-home frame clock synched with Alice's, which is right next to him? 40.
So we know that Alice being 10 is simultaneous with a stay-home frame clock synched with Alice's right next to Charlie displaying 40.

In the moment after Charlie starts moving again, what time is displayed on a stay-home frame clock synched with Alice's, which is right next to him? 40.
Given that we know that Alice being 10 is simultaneous with a stay-home frame clock synched with Alice's right next to Charlie displaying 40, what time is Alice in the moment after Charlie starts moving again? 10.

(No comment.)

 

[Neddy Bate]

Charlie and Bob should leave when all three parties involved agree that everyone is stationary with respect to each other. Only that way everybody agrees that $$\text{CADO}_H=\text{CADO}_T$$

You came up with the NotEinstein scenario because you thought (CORRECTLY) that if Bob and Charlie are in the same frame, they should agree on the age of Alice. I showed that is the case. That will be the case no matter when anyone leaves, we just have to change the numbers accordingly.

Do we agree that Bob can (not necessarily "must", but "can") indeed wait until he sees Charlie stop?

Of course.

If Charlie calculated Alice to be 40, Alice is 40 according to Charlie. How can she also be 10 according to Charlie?

Seriously?

Before stopping when v = 0.866:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * -17.32))
t = 40 - (0.866 * 34.64))
t = 10

After stopping when v = 0.000:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.000 * -17.32))
t = 40 - (0.000 * 34.64))
t = 40

After starting up again when v = 0.866:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * -17.32))
t = 40 - (0.866 * 34.64))
t = 10

However much time he pauses just adds to this result, it doesn't change the underlying concept. This is SR.

 

[Neddy Bate]

So let Bob leave when he is 40 instead of 10, and let Charlie pause for 30 years after stopping, and then they both agree Alice is 40. Better?

 

[Neddy Bate]

I think I see the problem, NotEinstein expects Charlie to take his knowledge of Alice's age (from the stay-home frame) with him back into the travelling twin frame! It doesn't work that way.

Consider if Charlie does not stop. When he is t'=20 years old, he sees a stay-home-frame clock right near him, which he knows is synchronised with Alice's, and that clock displays t=40. That alone tells him that Alice is 40 in her own frame. He doesn't have to stop to know that.

But does that change the fact that she is only 10 at that time in his frame? No.

 

[Mike_Fontenot]

I've got another question. Imagine there is a triplet: Alice, Bob, and Charlie. Alice and Bob stay at home, but Charlie takes a rocket into space. But, at a distance D, Charlie comes to a complete stop (with respect to Alice and Bob). In other words, $$v=0$$. Afterwards, according to your formula, all will agree how old Alice is, because with $$v=0$$ your equation reduces to: $$\text{CADO}_T=\text{CADO}_H$$.
But now, both Bob and Charlie start moving (both in the same direction that Charlie was moving in earlier) with the same speed (using the same acceleration profile, if that's relevant). Even though they are in the same reference frame all the time, Bob and Charlie will now disagree on how old Alice is, due to Charlie starting with $$L=D$$ while Bob starts with $$L=0$$. That seems weird to me; I'm pretty sure SR predicts that Bob and Charlie would agree on Alice's age, because in SR time dilation is a function only of speed, not distance.

How do you resolve this conflict?

I just re-read your scenario description, and realized that it is not the interesting scenario that I thought it was. In fact, it's actually not interesting at all. The fact that you specify a velocity change by the traveler from +v1 to zero at some instant t in the traveler's life, immediately followed by a velocity change 0f 0 to +v1, means that the two changes are equivalent to no change at all in the traveler's velocity. What then remains to be calculated is nothing unusual.

I'm still going to pursue the question of what the general results of Section 14 say about the result of handling a very narrow and very high (but finite) rectangular acceleration profile, and whether the result is just equivalent to using the CADO equation.

 

[Neddy Bate]

In this post I will attempt to solve the NotEinstein scenario, with Bob and Charlie both taking off simultaneously in the stay-home frame:

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:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.000 * -17.32))
t = 40 - (0.000 * 34.64))
t = 40

So far so good.

Now let's say that Bob and Charlie have agreed to take off simultaneously in the stay-home frame, same velocity as before. Bob is right next to Alice, so if they both take off now, we expect him to calculate that Alice is 40.

For Bob's calculations, I will reset the clocks to 0 and then add the result to Alice's known age of 40 at the end:
t = γ(t' + (vx' / c²))
t = 2(0 + (0.866 * 0.00))
t = 0 - 0.00
t = 0
Finally, adding in the known age of 40 at the end:
t = 0 + 40
t = 40

So far so good.

Now, let's figure out what Charlie finds.
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * -17.32))
t = 40 - (0.866 * 34.64))
t = 10

But does that mean that Bob and Charlie are now in the same reference frame, and they calculate different ages for Alice? No.

According to Charlie now, Bob is still 10 years old and standing next to Alice. According to the moving frame, Bob will continue to stand there until Alice and Bob are both 40, which will not be until Charlie is 80.

In fact, when Charlie is 80, he can calculate the age of Alice and Bob as follows:
t = γ(t' + (vx' / c²))
t = 2(80 + (0.866 * -69.28))
t = 160 - (0.866 * 138.56)
t = 160 - 120
t = 40

And that is when Bob takes off and arrives in the moving frame. Because of relativity of simultaneity, their taking off simultaneously in the stay-home frame results in them arriving non-simultaneously in the moving frame. But they still agree with each other on Alice's age, because they are in the same frame.

 

[NotEinstein]

You came up with the NotEinstein scenario because you thought (CORRECTLY) that if Bob and Charlie are in the same frame, they should agree on the age of Alice.

Except that they don't: that's what the whole scenario is about! After Bob and Charlie start moving, they are still in the same reference frame, but they will disagree on the age of Alice according to the CADO-equation, because their $$L$$'s are different!

I showed that is the case.

If you showed it when Charlie is stationary again, sure, I never claimed it wouldn't be the case then.

That will be the case no matter when anyone leaves, we just have to change the numbers accordingly.

Please calculate the $$\text{CADO}_T$$ of both Bob and Charlie when they both are moving, and argue how those two terms are the same.

Of course.

Good.

Seriously?

Before stopping when v = 0.866:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * -17.32))
t = 40 - (0.866 * 34.64))
t = 10

After stopping when v = 0.000:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.000 * -17.32))
t = 40 - (0.000 * 34.64))
t = 40

After starting up again when v = 0.866:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * -17.32))
t = 40 - (0.866 * 34.64))
t = 10

Alice goes from 40 to 10 between "After stopping" and "After starting up again". That's her traveling in time backwards; time dilation can only slow down time, not make it run backwards!

However much time he pauses just adds to this result, it doesn't change the underlying concept. This is SR.

Sure, and I never claimed otherwise.

So let Bob leave when he is 40 instead of 10, and let Charlie pause for 30 years after stopping, and then they both agree Alice is 40. Better?

As long as they all agree that $$\text{CADO}_H=\text{CADO}_T$$, you are complying with the scenario I described.

I think I see the problem, NotEinstein expects Charlie to take his knowledge of Alice's age (from the stay-home frame) with him back into the travelling twin frame! It doesn't work that way.

You are wrong: I require no such thing. All I require (as I've explained to you before) is that right before the moment both Bob and Charlie simultaneously start moving again with the same acceleration profile (in everybody's frame, possible since they are all stationary with respect to each other), that they all agree that $$\text{CADO}_H=\text{CADO}_T$$.

Consider if Charlie does not stop.

This is outside of my scenario, as it breaks the crucial point of construction that they must all agree that $$\text{CADO}_H=\text{CADO}_T$$, which I explicitly accomplish through $$v=0$$.

When he is t'=20 years old, he sees a stay-home-frame clock right near him, which he knows is synchronised with Alice's, and that clock displays t=40. That alone tells him that Alice is 40 in her own frame. He doesn't have to stop to know that.

But does that change the fact that she is only 10 at that time in his frame? No.

Perhaps you are right, and I need to think about this again. I'll go back to basics and start drawing Minkowski diagrams.

Anyway, none of that is actually relevant to my scenario, so I suggest we continue for now?

​

​

 

[NotEinstein]

I just re-read your scenario description, and realized that it is not the interesting scenario that I thought it was. In fact, it's actually not interesting at all. The fact that you specify a velocity change by the traveler from +v1 to zero at some instant t in the traveler's life, immediately

I never said "immediately"; it's only after everybody agrees that $$\text{CADO}_H=\text{CADO}_T$$.

followed by a velocity change 0f 0 to +v1, means that the two changes are equivalent to no change at all in the traveler's velocity. What then remains to be calculated is nothing unusual.

Then perhaps you could actually answer the question I asked you: "How do you resolve this conflict?"

I'm still going to pursue the question of what the general results of Section 14 say about the result of handling a very narrow and very high (but finite) rectangular acceleration profile, and whether the result is just equivalent to using the CADO equation.

(As I said, have fun!)

 

[NotEinstein]

In this post I will attempt to solve the NotEinstein scenario, with Bob and Charlie both taking off simultaneously in the stay-home frame:

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:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.000 * -17.32))
t = 40 - 0.000
t = 40

Actually, what $$v$$ are you using for $$\gamma$$, because you've set $$v=0$$ in the second term inside the parentheses?

 

[NotEinstein]

In this post I will attempt to solve the NotEinstein scenario, with Bob and Charlie both taking off simultaneously in the stay-home frame:

Now let's say that Bob and Charlie have agreed to take off simultaneously in the stay-home frame, same velocity as before.

According to Charlie now, Bob is still 10 years old and standing next to Alice. According to the moving frame, Bob will continue to stand there until Alice and Bob are both 40, which will not be until Charlie is 80.

And that is when Bob takes off and arrives in the moving frame. Because of relativity of simultaneity, their taking off simultaneously in the stay-home frame results in them arriving non-simultaneously in the moving frame. But they still agree with each other on Alice's age, because they are in the same frame.

This is not my scenario. In my scenario, when Bob and Charlie take off together, all three agree on Alice's age, by construction. And since they are stationary with respect to each other at that time, by construction, they can all synchronize clocks, and thus Bob and Charlie will take off at the same time according to each other, by construction. I don't know what scenario you've "solved", but it isn't mine.

 

[Neddy Bate]

Actually, what $$v$$ are you using for $$\gamma$$, because you've set $$v=0$$ in the second term inside the parentheses?

Yes, there is an error there, thanks for the correction. But somewhere earlier I calculated it correctly, I think. I will fix it or post a new one.

 

[Neddy Bate]

This is not my scenario. In my scenario, when Bob and Charlie take off together, all three agree on Alice's age, by construction. And since they are stationary with respect to each other at that time, by construction, they can all synchronize clocks, and thus Bob and Charlie will take off at the same time according to each other, by construction. I don't know what scenario you've "solved", but it isn't mine.

This is exactly your scenario. Because you demand that Bob and Charlie leave simultaneously in the home frame, they must arrive non-simultaneously in the traveling frame. That is relativity of simultaneity. It can't be simultaneous in both frames because they are spacially separated along the axis of motion (x axis).

Once they both arrive in the same frame, they agree Alice is 40, in this case. Just as they both agreed Alice was 10 in the other case. No contradictions in SR!

 

[Mike_Fontenot]

[...]

Just for fun, in your scenario I inserted the velocity change by the new guy, in between the two (zero-sum) velocity changes by the traveler, and redid the analysis. I found that the CADO equation worked just fine. Basically, when the traveler instantaneously changes velocity, his conclusion about the effect of the resulting sudden age change of a distant person doesn't end up depending at all on whether the distant person is instantaneously changing velocity at that instant or not! All an instantaneous velocity change by the distant person then does is put a "kink" in the ongoing world line of the distant person at the point of it's intersection with the line of simultaneity, which doesn't affect the point of intersection of the traveler's line of simultaneity with the distant person's world line at all, and thus doesn't affect the traveler's conclusion about the distant person's age. This can be easily seen from the Minkowski diagram.

 

[NotEinstein]

This is exactly your scenario. Because you demand that Bob and Charlie leave simultaneously in the home frame, they must arrive non-simultaneously in the traveling frame.

Arrive where? Bob never arrives anywhere? If you meant that they leave at different times in the moving frame; that's fine, I never claimed otherwise. I also don't see how that's an issue?

That is relativity of simultaneity. It can't be simultaneous in both frames because they are spacially separated along the axis of motion (x axis).

Sure, but the moving frame isn't of importance, unless it can make the distance between Bob and Charlie zero in their frame. Which seems impossible to me: Bob and Charlie leave at the same time in their proper frames, with the same acceleration profile and velocity. How can Bob ever catch up to Charlie?

Once they both arrive in the same frame,

What does that mean, "arrive in the same frame"? They never leave the same frame! At every point in time, their clocks are synced, and they perform the same actions. How can that desync?

they agree Alice is 40, in this case.

But also according to the CADO-equation? Note that while $$v$$ is the same for both Bob and Charlie, $$L$$ is vastly different.

Just as they both agreed Alice was 10 in the other case. No contradictions in SR!

Of course this doesn't show a contradiction in SR, and I never claimed otherwise.

 

[NotEinstein]

Just for fun, in your scenario I inserted the velocity change by the new guy, in between the two (zero-sum) velocity changes by the traveler, and redid the analysis.

Who is "the new guy", and who is "the traveler"?

I found that the CADO equation worked just fine. Basically, when the traveler instantaneously changes velocity,

I never said "instantaneously".

his conclusion about the effect of the resulting sudden age change of a distant person doesn't end up depending at all on whether the distant person is instantaneously changing velocity at that instant or not!

I never mentioned any "sudden age change of a distance person" where "the distant person is instantaneously changing velocity"? Or are you talking about Alice, in which case you are trivially wrong because that's the $$v$$ in the CADO-equation, which obviously has an effect: that's what the entire equation is about!

All an instantaneous velocity change by the distant person then does is put a "kink" in the ongoing world line of the distant person at the point of it's intersection with the line of simultaneity, which doesn't affect the point of intersection of the traveler's line of simultaneity with the distant person's world line at all, and thus doesn't affect the traveler's conclusion about the distant person's age. This can be easily seen from the Minkowski diagram.

Let's try this again:
$$\text{CADO}_T=\text{CADO}_H-\frac{v*L}{c^2}$$
Right before Bob and Charlie take off:
v=0 (they are all stationary with respect to each other);
L=0 for Bob;
L=large distance for Charlie.
Due to $$v=0$$, we have: $$\text{CADO}_T=\text{CADO}_H$$
Everybody agrees on this. Alice has the same age (CADO) for everybody.

Bob and Charlie start moving with respect to Alice, with the same velocity. So for them now:
v=large velocity for both Bob and Charlie;
L: steadily increasing with the same increments for both Bob and Charlie (due to their velocities).

Then, because $$v*L$$ is larger for Charlie (due to his larger $$L$$), don't we have:
$$\text{CADO}_{T\text{,Bob}}>\text{CADO}_{T\text{,Charlie}}$$ ?

There's your age (CADO) difference due to distance.

 

[Neddy Bate]

If you meant that they leave at different times in the moving frame; that's fine, I never claimed otherwise. I also don't see how that's an issue?

Yes, I meant that they accelerate at two different times according to the moving frame, and I'm glad you agree. This means that one accelerates before the other, according to the moving frame.

The importance of this issue is as follows: Whatever age Alice might be (according to the moving frame) when one of them accelerates, she will not be the same age (according to the moving frame) when the other one accelerates. "Different times" means that time must elapse between the time the two guys accelerate, and Alice must age during that elapsed time.

Note that this fully invalidates your argument that Bob's calculated result for Alice's age, immediately after he accelerates into the moving frame, should be equal to Charlie's calculated result for Alice's age, immediately after he accelerates into the moving frame.

Since that is not the case in SR, there is no reason to expect it to be the case in CADO either.

Let's look at the moment right after CADO_T=CADO_H. All three people agree that Alice is 40 years old. Then the two guys accelerate.

You and I seem to have no problem agreeing that Bob, immediately after he accelerates, would calculate that Alice is 40 years old. Of course, immediately after he accelerates, he is in the moving frame, not the home frame anymore.

But you and I do seem to have a little problem agreeing that Charlie, immediately after he accelerates, could possibly calculate that Alice is some age other than 40 years old. Of course, immediately after he accelerates, he is in the moving frame, not the home frame anymore.

Your requirement that both Charlie and Bob should calculate the SAME age for Alice, immediately after each one has accelerated into the moving frame, even though you agree these events happen at different times according to moving frame, would result in them perpetually being in disagreement about her age, even though they are both in the same frame. In case you still don't see the reason why, please consider your earlier statement:

"If you meant that they leave at different times in the moving frame; that's fine, I never claimed otherwise. I also don't see how that's an issue?"

 

[Neddy Bate]

I see now that the CADO equation does indeed make some of these calculations easier.

For example, using CADO, at the time Charlie stops moving, he calculates Alice's current age in the home-frame:
t_cado = t + (vx)
t_cado = 40 + (0.000 * 34.64)
t_cado = 40
PRETTY EASY.

Using the inverse Lorentz transform, for the same calculation, we must consider the following things:

Unlike the CADO equation, simply changing the velocity to v=0.000 will not work.
t = γ(t' + (vx' / c²))
t = 1(20 + (0.000 * -17.32))
t = 20
WRONG! DOES NOT EQUAL 40!
(I made that error earlier in the thread, and I thank NotEinstein for pointing it out to me.)

Instead, we need to consider the moment before Charlie stops moving, and a home-frame clock which is adjacent to Charlie, and synched to Alice's:
t = γ(t' + (vx' / c²))
t = 2(20 + (0.866 * 0.000))
t = 40

......


Here is another example of where the CADO equation makes some of these calculations easier.

Using CADO, at the time Bob starts moving, he calculates Alice's current age in the home-frame:
t_cado = t + (vx)
t_cado = 40 + (0.866 * 0.00)
t_cado = 40
PRETTY EASY.

Using the inverse Lorentz transform, for the same calculation, we need to
reset the clocks to 0, and remember to add Alices formerly known age of 40 to the end result.
t = [ γ (t' + (vx' / c²)) ] + 40
t = [ 2(0 + (0.866 * 0.000)) ] + 40
t = 0 + 40
t = 40

......

The only problem I see is that the differences between the two methods can cause confusion.
(But I do want to thank Mike_Fontenot for sharing this.)

 

[NotEinstein]

Yes, I meant that they accelerate at two different times according to the moving frame, and I'm glad you agree. This means that one accelerates before the other, according to the moving frame.

I've finished staring at Minkowski diagrams, and I agree with this. I'm not sure I ever disagreed, but let's make my agreement explicit.

The importance of this issue is as follows: Whatever age Alice might be (according to the moving frame) when one of them accelerates, she will not be the same age (according to the moving frame) when the other one accelerates. "Different times" means that time must elapse between the time the two guys accelerate, and Alice must age during that elapsed time.

Agreed.

Note that this fully invalidates your argument that Bob's calculated result for Alice's age, immediately after he accelerates into the moving frame, should be equal to Charlie's calculated result for Alice's age, immediately after he accelerates into the moving frame.

This is not my argument, this is coming from the CADO-equation and the definition of $$\text{CADO}_H$$. It's defined as:

the quantity CADO_H denotes the home-twin's conclusion about the home-twin's age, when the traveler's age is t.

Since, according to Alice, both travelers start moving at the same instant, and we synced clocks before, right after they undergo the same acceleration, their ages will still be the same. So the instant after an instantaneous acceleration, the two travelers are still age $$t$$, and Alice measures her age to be $$t$$ as well, so $$\text{CADO}_{H\text{,Bob}}=\text{CADO}_{H\text{,Charlie}}$$. But $$\text{CADO}_{T\text{,Bob}}\neq\text{CADO}_{T\text{,Charlie}}$$, which is the conflict I'm talking about.

Since that is not the case in SR, there is no reason to expect it to be the case in CADO either.

True.

Let's look at the moment right after CADO_T=CADO_H. All three people agree that Alice is 40 years old. Then the two guys accelerate.

You and I seem to have no problem agreeing that Bob, immediately after he accelerates, would calculate that Alice is 40 years old. Of course, immediately after he accelerates, he is in the moving frame, not the home frame anymore.

But you and I do seem to have a little problem agreeing that Charlie, immediately after he accelerates, could possibly calculate that Alice is some age other than 40 years old. Of course, immediately after he accelerates, he is in the moving frame, not the home frame anymore.

I agree with this, and the CADO-equation indeed shows this: Bob's and Charlie's $$\text{CADO}_T$$ are different after acceleration.

Your requirement that both Charlie and Bob should calculate the SAME age for Alice, immediately after each one has accelerated into the moving frame, even though you agree these events happen at different times according to moving frame, would result in them perpetually being in disagreement about her age, even though they are both in the same frame.

Exactly, and that's the conflict I pointed out.

In case you still don't see the reason why, please consider your earlier statement:

"If you meant that they leave at different times in the moving frame; that's fine, I never claimed otherwise. I also don't see how that's an issue?"

You've only went through half the job: you're now worked through the $$\text{CADO}_T$$-side of the CADO-equation. Please consider the $$\text{CADO}_H$$-side next before drawing any conclusions about whether my conflict is real or not.

 

[NotEinstein]

I see now that the CADO equation does indeed make some of these calculations easier.

For example, using CADO, at the time Charlie stops moving, he calculates Alice's current age in the home-frame:
t_cado = t + (vx)
t_cado = 40 + (0.000 * 34.64)
t_cado = 40
PRETTY EASY.

I'm not sure what $$t_\text{cado}$$ denotes, but it's doesn't appear to be $$\text{CADO}_H$$, because that's Alice's age according to Alice, by definition. Could you please elaborate?