Graphical Derivation of the CADO Equation

NotEinstein · Sep 8, 2018

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

As I said earlier, I have finished starting at Minkowski diagrams. Neddy, we're both right. You are right that Charlie can measure Alice aging backwards (i.e. to be travelling back in time) during periods of acceleration, and I'm right in that isn't what really happens. It's a known breakdown of SR. The acceleration Charlie undergoes is so large that you have to involve GR (this can be see through the Einstein Equivalency Principle; change the acceleration into a gravitational field, and calculate its strength; you'll find that GR is required to handle that acceleration). In GR, the entire concept of a plane of simultaneity is meaningless, and so the problem disappears there.

This also appears to resolve one other problem I had with the CADO-equation: if the distance $$L$$ and the acceleration (difference in $$v$$) are large enough, you run into exactly this issue. So, the CADO-equation could very well be correct (if all other problems can be resolved), where "correct" means "matching SR predictions". It's just that those predictions break down under certain conditions.

Neddy Bate · Sep 8, 2018

NotEinstein said:
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?

Sorry, t_cado should have been written cado_t. It is Alice's age as calculated by Charlie, in the moment he stops moving in your scenario:

For example, using CADO, at the time Charlie stops moving, he calculates Alice's current age in the home-frame:
cado_t = cado_h + (vL)
cado_t = cado_h + (0.000 * 34.64)
cado_t = 40

Neddy Bate · Sep 8, 2018

NotEinstein said:
This is not my argument, this is coming from the CADO-equation and the definition of $$\text{CADO}_H$$.

If that was not your argument, then I definitely misunderstood you. Sorry about that. I'm on an unfamiliar computer, so I don't really feel like going back to find whatever post gave me that impression.

But now you seem to be saying that you are ok with the idea that SR would say that, just after Charlie and Bob each accelerate, they each should calculate the age of Alice, and their results should not match, because they do not accelerate simultaneously according to the traveler-frame. Of course this follows from you having them accelerate simultaneously according to the home-frame. Agreed?

NotEinstein said:
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.

Whose ages will be the same, according to Alice? By my tally, Alice calculates Charlie's age to be 20 and Bob's age to be 40. Bob and Charlie both calculate Alice's age as 40, but only after Charlie has reached an age of 80 years old, because that is when Bob accelerates according to Charlie. At least that is what SR says for this case. Are you saying CADO fails this? Can you show the equations?

NotEinstein said:
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.

I don't see the variable t in any of the cado equations, though I see it mentioned on the website. Using only the CADO equations, can you show me where it goes wrong?

NotEinstein · Sep 8, 2018

Neddy Bate said:
Sorry, t_cado should have been written cado_t. It is Alice's age as calculated by Charlie, in the moment he stops moving in your scenario:

For example, using CADO, at the time Charlie stops moving, he calculates Alice's current age in the home-frame:
cado_t = cado_h + (vL)
cado_t = cado_h + (0.000 * 34.64)
cado_t = 40

In that case, you got the sign wrong (not that it matters in this particular case):

Mike_Fontenot said:
CADO_T = CADO_H - v * L

Neddy Bate · Sep 8, 2018

NotEinstein said:
As I said earlier, I have finished starting at Minkowski diagrams. Neddy, we're both right. You are right that Charlie can measure Alice aging backwards (i.e. to be travelling back in time) during periods of acceleration, and I'm right in that isn't what really happens. It's a known breakdown of SR. The acceleration Charlie undergoes is so large that you have to involve GR (this can be see through the Einstein Equivalency Principle; change the acceleration into a gravitational field, and calculate its strength; you'll find that GR is required to handle that acceleration). In GR, the entire concept of a plane of simultaneity is meaningless, and so the problem disappears there.

This also appears to resolve one other problem I had with the CADO-equation: if the distance $$L$$ and the acceleration (difference in $$v$$) are large enough, you run into exactly this issue. So, the CADO-equation could very well be correct (if all other problems can be resolved), where "correct" means "matching SR predictions". It's just that those predictions break down under certain conditions.

There is an old example given where all some has to do is walk in one direction on earth, and that makes them calculate the time on an extremely distant planet to change drastically. Then they walk the opposite direction, and the time on the extremely distant planet changes drastically in the opposite way. I don't think GR is required due to the gravitational field generated by walking!

NotEinstein · Sep 8, 2018

Neddy Bate said:
If that was not your argument, then I definitely misunderstood you. Sorry about that. I'm on an unfamiliar computer, so I don't really feel like going back to find whatever post gave me that impression.

But now you seem to be saying that you are ok with the idea that SR would say that, just after Charlie and Bob each accelerate, they each should calculate the age of Alice, and their results should not match, because they do not accelerate simultaneously according to the traveler-frame. Of course this follows from you having them accelerate simultaneously according to the home-frame. Agreed?

Agreed, and it's quite possible this solves my conflict completely, but only through calculations can prove that.

Neddy Bate said:
Whose ages will be the same, according to Alice?

Yes, according to Alice, as I said in that very same sentence.

Neddy Bate said:
By my tally, Alice calculates Charlie's age to be 20 and Bob's age to be 40.

That's before the clock sync. After the clock sync, everybody is back to $$t=0$$.

Neddy Bate said:
Bob and Charlie both calculate Alice's age as 40, but only after Charlie has reached an age of 80 years old, because that is when Bob accelerates according to Charlie. At least that is what SR says for this case. Are you saying CADO fails this? Can you show the equations?

You're the one using biological age; I'm merely using clocks. I sync the clocks after the initial, set-up phase, as anything happening before that isn't actually relevant to my scenario.

Neddy Bate said:
I don't see the variable t in any of the cado equations, though I see it mentioned on the website. Using only the CADO equations, can you show me where it goes wrong?

Re-read post #11, followed by post #76.

NotEinstein · Sep 8, 2018

Neddy Bate said:
There is an old example given where all some has to do is walk in one direction on earth, and that makes them calculate the time on an extremely distant planet to change drastically. Then they walk the opposite direction, and the time on the extremely distant planet changes drastically in the opposite way.

I've never heard of this particular example before; can you please give some more details, because it's too vague for me to understand what exactly you're talking about?

Neddy Bate said:
I don't think GR is required due to the gravitational field generated by walking!

I can't comment on this, as you haven't given enough details.

Mike_Fontenot · Sep 8, 2018

Neddy Bate said:
I don't see the variable t in any of the cado equations, though I see it mentioned on the website. Using only the CADO equations, can you show me where it goes wrong?

Just to clarify my notation, I reserve the lower-case variable "t" to always denote the current age of the traveler. It normally serves as the independent variable ... i.e., you are normally free to choose it to have any value you want ... it just specifies the instant in the traveler's life that you want to focus on at the moment. ALL of the other variables in the CADO equation are functions of the variable "t", even though occasionally I don't show that explicitly, for simplicity. Since I don't want the variable "t" to be confused with anything else, I don't use the letter "t" as part of the two CADO variables: I write CADO_T(t) to denote the age of the home twin, according to the Traveler, when the traveler is age "t". And I write CADO_H(t) to denote the age of the home twin, according to the Home twin, when the traveler is age "t". Hope that helps.

Neddy Bate · Sep 8, 2018

NotEinstein said:
I've never heard of this particular example before; can you please give some more details, because it's too vague for me to understand what exactly you're talking about?

I can't comment on this, as you haven't given enough details.

Mike_Fontenot gave an example earlier, in post #27, " 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). "

Haven't you ever noticed that in the standard twin scenario, during the entire journey (except for the turnaround acceleration), the traveling twin calculates that the stay-home twin's clock is ticking at a slower rate than his own? Yet, when the traveling twin arrives back home, the stay-home twin is the one who is older? It was considered the 'twin paradox' by folks who did not realize that during the turnaround acceleration, the stay-home twin's clock raced forward according to the traveling twin?

Oh but wait. You are worried about time racing backward, not forward. So, just let the traveling twin do a 360 degree turn at the turnaround, instead of a 180 degree turn, and he must find the stay-home twin's clock race forward and then backward. Did you not realize this is part of SR?

Mike_Fontenot · Sep 8, 2018

Neddy Bate said:
There is an old example given where all some has to do is walk in one direction on earth, and that makes them calculate the time on an extremely distant planet to change drastically. Then they walk the opposite direction, and the time on the extremely distant planet changes drastically in the opposite way. I don't think GR is required due to the gravitational field generated by walking!

The best version was part of a NOVA TV show, and the alien was riding a bicycle around in a small circle. The physicist who talks about it is Brian Greene, and he's the one that came up with it.

Here's a link to one of the YouTubes that has that example:

Scan forward to 6:00 ... that's where the alien example starts.

Q-reeus · Sep 9, 2018

Mike_Fontenot said:
The best version was part of a NOVA TV show, and the alien was riding a bicycle around in a small circle. The physicist who talks about it is Brian Greene, and he's the one that came up with it.

Here's a link to one of the YouTubes that has that example:

Scan forward to 6:00 ... that's where the alien example starts.

Apart from that 'heretical' block spacetime viewpoint agitating one member here who is prone to frequently SHOUT that TIME DOES NOT EXIST, Brian Greene's alien on a bicycle scenario is misleading. It's based on a projection/extrapolation. Whose conclusion would only be 'real' if the alien kept cycling at the same rate and direction for the entire 3+ billion years it would take light to travel between the alien's 'then' location in the Andromeda galaxy and 'here and now' on Earth.
Not the only thing Brian Greene got wrong in that for-tv piece - see my comments here:
http://www.sciforums.com/posts/3496506/
http://www.sciforums.com/posts/3496738/

And btw you ignored my circular motion example in #30 refuting general validity of your CADO formula. The formulae given in the Wiki reference there are btw not totally general and only covers rectilinear motion howbeit allowing arbitrary speed and acceleration profiles within that (1+1)D restriction.

NotEinstein · Sep 9, 2018

Neddy Bate said:
Mike_Fontenot gave an example earlier, in post #27, " 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). "

Ah, I misread what you said (didn't catch the "distant planet" part). Sorry, my mistake.

In that case, I invite you to calculate the acceleration that the walking Earth-person measured to be applied to the distant planet, and translate that through the Einstein Equivalency Principle to a gravitational field. Then check whether that gravitational field is strong enough to warrant the usage of GR. Hint: it's making an entire planet change movement direction (assuming the distant planet is stationary with respect to Earth) within seconds.

Neddy Bate said:
Haven't you ever noticed that in the standard twin scenario, during the entire journey (except for the turnaround acceleration), the traveling twin calculates that the stay-home twin's clock is ticking at a slower rate than his own?

Obviously, yes.

Neddy Bate said:
Yet, when the traveling twin arrives back home, the stay-home twin is the one who is older? It was considered the 'twin paradox' by folks who did not realize that during the turnaround acceleration, the stay-home twin's clock raced forward according to the traveling twin?

Clocks racing forwards are not the same as the clock racing backwards.

Neddy Bate said:
Oh but wait. You are worried about time racing backward, not forward.

(Exactly.)

Neddy Bate said:
So, just let the traveling twin do a 360 degree turn at the turnaround, instead of a 180 degree turn, and he must find the stay-home twin's clock race forward and then backward. Did you not realize this is part of SR?

Nope, as I never encountered that scenario before, or at least, I don't remember it. It's also superseded by GR, so this effect is not that important either. But yeah, I'm glad you pointed it out to me! Learning something every day.

phyti · Sep 9, 2018

Mlke;
I don't know how you derived the expression, but here is one method from a spacetime graphic. The axis of simultaneity is red and light path is blue. The angle theta provides similar triangles because the t and x axes are symmetrical relative to light path.

j=time jump

j/x = x/t = v, thus j=xv

since x is the same for outbound and inbound,

Neddy Bate · Sep 10, 2018

NotEinstein said:
That's before the clock sync. After the clock sync, everybody is back to $$t=0$$.

You're the one using biological age; I'm merely using clocks. I sync the clocks after the initial, set-up phase, as anything happening before that isn't actually relevant to my scenario.

Oh okay, I missed the part where you wanted every clock to be reset to 0 & synched just before Bob & Charlie accelerate simultaneously (according to the stay-home frame).

All of these calculations take place just after Bob & Charlie accelerate:

ALICE:

ALICE's calculation of Bob's time:
Using the Lorentz transformation equation:
t' = γ(t - (vx / c²))
t' = 2(0 - (0.866 * 0.00))
t' = 0 - (0.866 * 0.00)
t' = 0 - 0.00
t' = 0
Using the CADO equation:
CADO_T' = CADO_H' - (v * L')
CADO_T' = 0 - (0.866 * 0.00)
CADO_T' = 0 - 0.00
CADO_T' = 0
(Note: To find Bob's biological age according to Alice, we can add the time that was displayed on his clock just before it was reset to 0.)
(CADO_T' = 0 + 40)
(CADO_T' = 40)

BOB:

BOB's calculation of Alice's time:
Using the inverse Lorentz transformation equation:
t = γ(t' + (vx' / c²))
t = 2(0 + (0.866 * 0.00))
t = 0 + (0.866 * 0.00)
t = 0 + 0.00
t = 0
Using the CADO equation:
CADO_T = CADO_H - (v * L')
CADO_T = 0 + (0.866 * 0.00)
CADO_T = 0 + 0.00
CADO_T = 0
(Note: To find Alice's biological age according to Bob, we can add the time that was displayed on her clock just before it was reset to 0.)
(CADO_T = 0 + 40)
(CADO_T = 40)

CHARLIE:

CHARLIE's calculation of Alice's time:
Using the inverse Lorentz transformation equation:
t = γ(t' + (vx' / c²))
t = 2(60 + (0.866 * -69.28))
t = 120 - (0.866 * 138.56)
t = 120 - 120.00
t = 0
Using the CADO equation:
CADO_T = CADO_H - (v * L')
CADO_T = 120 - (0.866 * 138.56)
CADO_T = 120 - 120.00
CADO_T = 0
(Note: To find Alice's biological age according to Charlie, we can add the time that was displayed on her clock just before it was reset to 0.)
(CADO_T = 0 + 40)
(CADO_T = 40)

(Also note: Charlie's own time is t'=60 so to find his biological age, we can add the time that was displayed on his clock just before it was reset to 0.)
(t' = 60 + 20)
(t' = 80)

In summary, there does not appear to be anything in the NotEinstein scenario that causes the CADO equation to perform any differently than the Lorentz equations.

Neddy Bate · Sep 10, 2018

NotEinstein said:
It's also superseded by GR, so this effect is not that important either.

This is the first and only time I've ever heard that GR 'supersedes' certain effects in SR.

So, let's go back to the original twin scenario. On the webpage, the basic twin scenario is set up with v=0.866c so that gamma=2 and the time the traveling twin has been traveling is 40 years as measured by the stay-home twin. So the stay-home twin says the distance traveled is 0.866*40=34.64 lightyears. The stay-home twin also says the age of the travelling twin is 40/2=20 years.

Before the turnaround, the traveling twin calculates the age of the stay-home twin to be:
CADO_T = CADO_H - (v * L)
CADO_T = 40 - (0.866 * 34.64))
CADO_T = 40 - 30
CADO_T = 10
After the turnaround, the traveling twin calculates the age of the stay-home twin to be:
CADO_T = 40 - (-0.866 * 34.64))
CADO_T = 40 + 30
CADO_T = 70

So what do you supposed would happen if the traveling twin turned around again immediately after that?
A. The traveling twin calculates the age of the stay-home twin to be 10 again.
B. The traveling twin calculates the age of the stay-home twin to still be 70.
C. Something else?

Neddy Bate · Sep 10, 2018

Here's a strange calculation which I left out of post #94 because it didn't make much sense:

ALICE's calculation of Charlie's time:
Using the Lorentz transformation equation:
t' = γ(t - (vx / c²))
t' = 2(0 - (0.866 * 34.64))
t' = 0 - (0.866 * 69.28)
t' = 0 - 60.00
t' = -60

To understand this negative result we have to consider that the Lorentz transforms are based on the idea that all clocks are synched to each other in their own frames. So, when Bob's clock says t'=0 after the acceleration, there is an assumption that there is a whole "moving frame" filled with clocks which all say t'=0 in that frame. But Alice is not in that frame, she is in the "stay-home frame." So, according to her, Bob's clock says t'=0 but a moving-frame-synched-clock out by Charlie would say t'=-60.

It does not mean that Charlie becomes a negative age when he accelerates. He simply accelerates and finds the moving clock in his place already there, displaying t'=-60. This means that he will have to wait 60 more years before Bob accelerates, because that happens at t'=0.

Charlie's own clock can still say 0 right after the acceleration, from having been reset to 0 right before the acceleration, per NotEinstein's request. And his biological age is still 20 right after the acceleration. So we see that if Charlie has to wait 60 more years for Bob to accelerate, then Charlie's biological age will be 80 years at that time.

NotEinstein · Sep 10, 2018

Neddy Bate said:
Oh okay, I missed the part where you wanted every clock to be reset to 0 & synched just before Bob & Charlie accelerate simultaneously (according to the stay-home frame).

I've only mentioned it a couple of times before... :rolleye:

Neddy Bate said:
All of these calculations take place just after Bob & Charlie accelerate:

ALICE:

ALICE's calculation of Bob's time:
Using the Lorentz transformation equation:
t' = γ(t - (vx / c²))
t' = 2(0 - (0.866 * 0.00))
t' = 0 - (0.866 * 0.00)
t' = 0 - 0.00
t' = 0
Using the CADO equation:
CADO_T' = CADO_H' - (v * L')
CADO_T' = 0 - (0.866 * 0.00)
CADO_T' = 0 - 0.00
CADO_T' = 0
(Note: To find Bob's biological age according to Alice, we can add the time that was displayed on his clock just before it was reset to 0.)
(CADO_T' = 0 + 40)
(CADO_T' = 40)

BOB:

BOB's calculation of Alice's time:
Using the inverse Lorentz transformation equation:
t = γ(t' + (vx' / c²))
t = 2(0 + (0.866 * 0.00))
t = 0 + (0.866 * 0.00)
t = 0 + 0.00
t = 0
Using the CADO equation:
CADO_T = CADO_H - (v * L')
CADO_T = 0 + (0.866 * 0.00)
CADO_T = 0 + 0.00
CADO_T = 0
(Note: To find Alice's biological age according to Bob, we can add the time that was displayed on her clock just before it was reset to 0.)
(CADO_T = 0 + 40)
(CADO_T = 40)

CHARLIE:

CHARLIE's calculation of Alice's time:
Using the inverse Lorentz transformation equation:
t = γ(t' + (vx' / c²))
t = 2(60 + (0.866 * -69.28))
t = 120 - (0.866 * 138.56)
t = 120 - 120.00
t = 0
Using the CADO equation:
CADO_T = CADO_H - (v * L')
CADO_T = 120 - (0.866 * 138.56)
CADO_T = 120 - 120.00
CADO_T = 0
(Note: To find Alice's biological age according to Charlie, we can add the time that was displayed on her clock just before it was reset to 0.)
(CADO_T = 0 + 40)
(CADO_T = 40)

(Also note: Charlie's own time is t'=60 so to find his biological age, we can add the time that was displayed on his clock just before it was reset to 0.)
(t' = 60 + 20)
(t' = 80)

In summary, there does not appear to be anything in the NotEinstein scenario that causes the CADO equation to perform any differently than the Lorentz equations.

You are using two different values of CADO_H for Bob and Charlie; this is obviously wrong. CADO_H is defined as the age Alice measures herself to be, in this case right after Bob and Charlie accelerate. Alice cannot measure herself to be two different ages; that's nonsense.

So either you are using a different definition of CADO_H than the one Mike_Fontenot is using, or you've just proven the CADO-equation to be lacking, if not downright wrong.

NotEinstein · Sep 10, 2018

Neddy Bate said:
This is the first and only time I've ever heard that GR 'supersedes' certain effects in SR.

Maybe I've expressed it not very clearly? The mainstream consensus is that GR is more general that SR. In other words, GR is always equal or better than SR. Whenever they disagree, it's SR that's inaccurate.

SR predicts some traveling-back-in-time, GR says nothing like that happens. GR wins; it's a case SR can't handle correctly.

Neddy Bate said:
So, let's go back to the original twin scenario. On the webpage, the basic twin scenario is set up with v=0.866c so that gamma=2 and the time the traveling twin has been traveling is 40 years as measured by the stay-home twin. So the stay-home twin says the distance traveled is 0.866*40=34.64 lightyears. The stay-home twin also says the age of the travelling twin is 40/2=20 years.

Before the turnaround, the traveling twin calculates the age of the stay-home twin to be:
CADO_T = CADO_H - (v * L)
CADO_T = 40 - (0.866 * 34.64))
CADO_T = 40 - 30
CADO_T = 10
After the turnaround, the traveling twin calculates the age of the stay-home twin to be:
CADO_T = 40 - (-0.866 * 34.64))
CADO_T = 40 + 30
CADO_T = 70

So what do you supposed would happen if the traveling twin turned around again immediately after that?
A. The traveling twin calculates the age of the stay-home twin to be 10 again.
B. The traveling twin calculates the age of the stay-home twin to still be 70.
C. Something else?

I think that the traveling twin will once again see something that looks just like a huge gravitational field, which is something that requires GR to handle properly. Now that I know this is a weak spot of SR, I'm not comfortable using it to make predictions in such cases anymore.

Neddy Bate · Sep 11, 2018

NotEinstein said:
I've only mentioned it a couple of times before...

And now that I've done the calculations for it, we see that it makes no difference to the end result. Because when the previous times on the clocks (which were reset to 0) are added back in, we get the biological ages again, as expected.

NotEinstein said:
You are using two different values of CADO_H for Bob and Charlie; this is obviously wrong. CADO_H is defined as the age Alice measures herself to be, in this case right after Bob and Charlie accelerate. Alice cannot measure herself to be two different ages; that's nonsense.

So either you are using a different definition of CADO_H than the one Mike_Fontenot is using, or you've just proven the CADO-equation to be lacking, if not downright wrong.

Yes, I do have two different values of CADO_H and that may well show that the CADO equations are lacking.

But I also have three different values of t' for Charlie, one value in post #94 which you are quoting (t'=60), and two different values in a post just a little further down which was supposed to pertain to the same scenario, post #96 (t'=-60 and t'=0).

But what I am doing with the different values of t' is required in order for the Lorentz equations to give the correct results. And I give the same flexibility to the CADO equations by just copying & pasting the resulting values of γt' and -γx' into the CADO equations. So if CADO is lacking in that sense, then so is SR.

Neddy Bate · Sep 11, 2018

NotEinstein said:
Maybe I've expressed it not very clearly? The mainstream consensus is that GR is more general that SR. In other words, GR is always equal or better than SR. Whenever they disagree, it's SR that's inaccurate.

SR predicts some traveling-back-in-time, GR says nothing like that happens. GR wins; it's a case SR can't handle correctly.

I think that the traveling twin will once again see something that looks just like a huge gravitational field, which is something that requires GR to handle properly. Now that I know this is a weak spot of SR, I'm not comfortable using it to make predictions in such cases anymore.

The twins do not have to have any mass, and the accelerations do not have to be instantaneous. Shouldn't GR reduce to SR in that case? I assume you would answer yes, because you call this a "weak spot" in SR, so it must predict the result that you think is wrong, namely time going backwards for very distant clocks (not local).

What does GR say happens with extremely small masses? Is this SR effect always superseded regardless of mass?

Graphical Derivation of the CADO Equation

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