# 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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1,498
And one of the things it tells us, as I pointed out, is that when large accelerations and distant objects are involved, you need GR to get a proper description of the situation.

3. ### NotEinsteinValued Senior Member

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1,498
Ah, I see you have lost track of what we are arguing about. You said in post #104:
Which you now agree with me was wrong, as I pointed out. It also depends on the acceleration involved.

No, I am saying that it's unphysical (as far as we know) for it to go backwards. If a theory predicts it to, the theory is wrong (at least in that case).

And yes; as I said, if the gravitational field into with the acceleration can be translated turn out to be strong, you need GR.

Then why did you feel the need to change what I said? (In which you actually failed: even at the larger distance there is an acceleration small enough to satisfy my argument.)

Look, if you think I'm wrong, please point it out to me, instead of dodging the point of my argument.

*sigh* Obviously I am aware of that, and I'm not arguing against that.

And I never claimed otherwise?

And then causality is allowed to get violated? Because that was the point of my criticism. You've once again missed it. Why do you keep doing that?

I have in the very same post you are quoted; read on.

But then again, everybody trained in the slightest bit of physics knows that when time starts moving backwards, causality is in trouble.

And even most ignoramuses know that traveling back in time isn't (currently) know to be possible. So why would it be here?

And here you go claiming a preferred frame again...

You know, I've never made the claim that's possible. You know why? Because I suspect it isn't.

See, I thought about this the first time it popped up. I think you'll find that in GR-terms, this "backwards in time travelling" happens in the "elsewhere"-part of the light-cone, so there can be no direct causal connection. In other words, the distance is most likely too large to send a message back in time through this effect.

Which is why I haven't made this argument. But thanks for bringing it up, I guess?

Although that would be a nice demonstration, please explain why you think that that is the only way to demonstrate it.

Again, with the preferred frame!

But it isn't from the alien's point of view, as you claim you understand the alien's point of view is just as valid. Do you see your preferred frame picking? Can you please learn to stop doing that?

What kind of a childish argument is that? "Watching a movie in reverse" == "time ticking backwards"?

And we're back to the preferred frame picking.

That argument isn't even remotely similar to mine, but let's see where you go with this...

It's not about the bar's proper frame. It never was. I don't understand why you keep bringing up arguments with that misconception.

And your argument is lacking. If it was that obvious your argument holds up, then why did it need to be introduced as a freaking postulate in SR?

In fact, here's an explicit case where your entire little story breaks down: non-inertial frames. Look up pseudo-forces, and be amazed at the forces that can be acting on that steel bar of yours!

5. ### NotEinsteinValued Senior Member

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Funnily enough, I agree.

Then please point them out to us. I mean, that should be simple for you, right? You've spend 20+ years sharpening your knowledge about all of this, so you should be an expert on this, right? With very clean arguments and clear terminology, you can describe all of our mistakes, right?

Aha, so you admit the mistakes are yours! You picked bad variable names.

But alas, that is not all. It's your definitions as well. Some are plain weird (the one you gave for the speed of light, for example), and some cannot be right (the one for CADO_H in conjunction with the CADO-equation).

But yes, please tell us how all your mistakes are actually ours, while you sneakily re-write your webpage in the background...

7. ### NotEinsteinValued Senior Member

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Perhaps you should spend those couple of minutes searching through my posts, so you can learn that you are, in fact, wrong in your conclusion.

But let me help you out a bit: you are right there is no gravity in SR. But then again, SR isn't the king of physics, GR is. And GR has (at least to first order) the Einstein Equivalence Principle, which says that acceleration and gravity are the same thing.

But wait! I hear you cry out, the twin paradox has acceleration!

And there we go: gravitational effects in a SR-scenario, which aren't modelled by it. So if these effects become significant, SR will fail to predict the right things.

But you'd know that if you took those few minutes searching through my posts, or spend any time in the last 20+ years learning about relativity beyond the CADO-equation.

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9. ### Mike_FontenotRegistered Member

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To be complete, the line in red should have said:
"and never any ambiguity about whose age CADO_H refers to, or who's opinion it is".

And to be complete, the line in blue should have said:
"And if there is ever any ambiguity in your mind about whose age the variable CADO_H refers to, or who's opinion it is,".

10. ### Neddy BateValued Senior Member

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1,674
Now you are saying SR is wrong, a least in these types of cases.

But we were talking about very small acceleration, remember?

No, your argument was supposed to be that time going backwards is unphysical and therefore cannot happen at any acceleration. Arguing that you can choose a small acceleration which is small enough for it to "not happen" is equivalent to arguing that it "does happen" whenever your small acceleration is exceeded.

Therefore this is a sidetrack which does not support what you were supposed to be arguing in the first place.

YES I THINK YOU ARE WRONG. I think, in the case of small accelerations, GR should reduce to SR. In SR, the current "now time" of a distant point depends on the location (x) and the relative velocity (v) between the two frames as per t'=γ(t-(vx/c²)). I don't know how I can make it any clearer.

I do not see it as a violation of causality to have a glass hit the floor in the earth frame, followed by the glass breaking in the earth frame. If you think it is a violation of causality for an alien to say the events are reversed, then you think SR violates causality.

You have said as much anyway, when saying that GR would have to supersede SR in every case where this type of thing happens, otherwise causality would be violated. I think you are wrong about that, so I will not be agreeing with any of this nonsense, sorry.

If everybody agrees with you, you should be able to find lots of knowledgeable people stating that SR predicts things which are unphysical, and which violate causality. That would be an argument from authority, but if you find enough of them, maybe I will agree. No, actually I still won't, sorry.

You must be very peeved that the NOVA tv show explained the exact effect that I have been talking about, and never once mentioned that the effect couldn't really happen, because causality would be violated. Maybe you should write to them and tell them how they are in disagreement with "everybody trained in the slightest bit of physics."

Here is the transcription from the video which Mike_Fontenot posted in post #90:

"To get a feel for the bizarre effect this can have,
imagine an alien in a galaxy 10 billion light years from earth.
And way over there on earth, a guy at a gas station.

"Now if the two are sitting still,
not moving in relation to one another,
their clocks tick off time at the same rate,
and so they share the same now slices,
which cut straight across the loaf.

"But watch what happens if the alien hops on his bike,
and rides directly away from earth.
Since motion slows the passage of time,
their clocks will no longer tick off time at the same rate,
and if their clocks no longer agree,
their now slices no longer agree either.

"The alien's now slice cuts through the loaf differently,
it's angled toward the past.

"Since the alien is biking at a leisurely pace,
his slice is angled to the past by only a minuscule amount,
but across such a vast distance,
that tiny angle results in a huge difference in time.

"So, what the alien would find on its angled now slice,
what he considers is happening right now on earth,
no longer includes our friend at the gas station,
or even 40 years earlier when our friend was a baby.

"Amazingly, the alien's now slice swept back
through 200 years of history,
and now includes events that we consider part of the distant past,
like Beethoven finishing the 5th symphony.

"And if that's not strange enough,
the direction you move makes a difference too.

"Watch what happens if the alien turns around
and bikes toward earth.
The alien's new now slice is angled toward the future,
and so it includes events that won't happen on earth
for 200 years.

"Perhaps our friend's great great great granddaughter
teleporting from Paris to New York."

Last edited: Sep 12, 2018
11. ### Neddy BateValued Senior Member

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1,674

It was not one moment of time in the traveler's life. It was two different moments of time in the traveling frame.

The reason is because NotEinstein demanded that Bob and Charlie accelerate simultaneously according to Alice's frame. This turns out to mean that Bob and Charlie accelerate at two different times in the traveling frame.

According to Alice's frame, Bob's coordinates when he accelerates are x_bob=0.00 and t_bob=0
Here are the pertinent Lorentz equations for that event:
t' = γ(t - (vx / c²))
t'_bob = γ(t_bob - (vx_bob / c²))
t'_bob = 2(0 - (0.866 * 0.00)
t'_bob = 0

According to Alice's frame, Charlie's coordinates when he accelerates are x_charlie=34.64 and t_charlie=0
Here are the pertinent Lorentz equations for that event:
t' = γ(t - (vx / c²))
t'_charlie = γ(t_charlie - (vx_charlie / c²))
t'_charlie = 2(0 - (0.866 * 34.64)
t'_charlie = -60

So Charlie accelerates 60 years before Bob accelerates, (according to the traveling frame).

Last edited: Sep 12, 2018
12. ### phytiRegistered Senior Member

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287
The list compares A-events with B-events.

A B B
----------------------

1 1.3 1.2

2 2.6 2.5

3 3.9 3.0

4 ___ 3.5

5 ___ 4.0

6___ 4.5

7 4.0 5.0

8 5.3 5.5

9 6.7 6.8

10 8.0 8.0

Column 2 is from fig.1, and column 3 is from fig.2. The 2 columns differ.

Fig.2 uses the SR convention of sending frequent signals to A to get the uniformly spaced clock events, and assigning the times to half the round trip time. In fig.1 events 4, 5, 6 are missing. In fig.2, the A profile is discontinuous in 2 positions.

What does your CADO theory say about the missing events and the odd shaped profile in fig.2?

Included a drawing to provide a reality check for this case.

The x axis of A and B are parallel, B is moving to the right at v, the series of vertical lines are images from A moving at c toward B. Why should any images be missing from reversing course?

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13. ### Mike_FontenotRegistered Member

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In my post, you added underlining in this phrase, as shown:

"or are using two different values for CADO_H for the same instant t in the traveler's life,"

The home twin will never say that a given traveler has two different ages for a single instant in her own life ... that's an impossibility. The two twins start out the same age, and thereafter, he either ages at her same rate of ageing, or slower (according to her). For her, his age never goes backwards. So if that is happening in your's and NotEinstein's scenario, then the scenario is impossible.

Then, later in your post, you say

"This turns out to mean that Bob and Charlie accelerate at two different times in the traveling frame."

Why is that a problem? Are you sometimes calling Charlie the "traveler", and sometimes calling Bob the "traveler"? I.e., are you wanting to determine the current age of the home twin according to two different accelerating observers, at a different time in the lives of those two observers? If so, that is two separate and unrelated calculations. Each of those two observers uses a separate CADO equation, with different values of the observer's age "t", and generally different values of the three variables on the right-hand-side of the CADO equation. They are completely separate and unrelated operations.

14. ### Neddy BateValued Senior Member

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1,674
Correct. There is nothing in the NotEinstein scenario that makes Alice say that either Bob's or Charlie's ages go backwards.

I am calling the "travelling frame" the frame that both Bob and Charlie are in after they instantaneously accelerate to a constant speed of 0.866c.

Yes, we want to do two separate calculations. One is Bob's calculation of Alice's age right after Bob accelerates, and the other is Charlie's calculation of Alice's age right after Charlie accelerates.

According to "the traveling frame" Bob's coordinates when he accelerates are x'_bob=0.00 and t'_bob=0
Here are the pertinent inverse Lorentz equations for that event:
t = γ(t' + (vx' / c²))
t_alice_by_bob = γ(t'_bob + (vx'_bob / c²))
t_alice_by_bob = 2(0 + (0.866 * 0.00))
t_alice_by_bob = 0

According to "the traveling frame" Charlie's coordinates when he accelerates are x'_charlie=69.28 and t'_charlie=-60
Here are the pertinent inverse Lorentz equations for that event:
t = γ(t' + (vx' / c²))
t_alice_by_charlie = γ(t'_charlie + (vx'_charlie / c²))
t_alice_by_charlie = 2(-60 + (0.866 * 69.28))
t_alice_by_charlie = 0

How are you going to do the Charlie calculation using CADO? Your CADO_H is defined as Alice's clock time, which is CADO_H=0 at the time Charlie accelerates. Your L is defined as the distance between Alice and Charlie as measured by Alice which is L=34.64. So you're going to get something like:
CADO_T = 0 - (0.866 * 34.64)
Aren't you? But that is not right, because we know the time was t_alice=0 when Charlie accelerated.

What you really want to do is something like this:
CADO_T = -120 - (0.866 * -138.56)
But how do you justify those values for CADO_H and L?

Last edited: Sep 12, 2018
15. ### NotEinsteinValued Senior Member

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1,498
I don't think I've ever claimed otherwise? But it's more correct to say: I realized that SR predicts the wrong things in these cases.

I see you're confusing things again. No, we were not talking about specifically-small accelerations there; that was in another part of the discussion. Also, that's no retort to what I said, so I don't see what your point is anyway?

Yep, as I thought: you've confused two different arguments. Please re-read my posts and try to not confuse them. One argument is (as you just said) that time going backwards is unphysical (according to our current knowledge). Another is that small-enough accelerations don't cause backwards flowing time (as you claimed) at a set distance, but merely a slowdown.

True, if by "happening" you mean "SR predicts it".

I didn't realize I was supposed to be arguing something? Tell me, oh great master, what am I supposed to argue next?

Depending on your definition of "small", I could probably agree with this.

You are talking about accelerations, and then you give an equation not containing it... How are the two related?

Well, try harder anyway, because you're not making much sense.

Neither do I, and I've never claimed otherwise.

Yes, that is indeed a logical conclusion. Luckily, I also think SR isn't the correct theory to use in those cases.

Well, if you are unwilling to even give a single (coherent) argument for your position, I'll not be taking you seriously anymore.

If you are that close-minded, I'll not even try. Thanks for saving me some time!

Not really.

That you can't handle thinking somebody else being wrong doesn't mean I can't either.

That contains no explanation to the backwards-in-time-travelling whatsoever, so I don't know why you felt the need to post it?

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

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54
I think I see the source of the confusion here. Here's a "very quick and dirty" explanation of the problem:

When both twins accelerate, and all accelerations are finite, the CADO equation can't be used. Those problems can always be solved, but never with the CADO equation. A different procedure must be used.

When all accelerations are Dirac delta functions (so that only instantaneous velocity changes are involved), the problem can always be solved by the same procedure used for finite accelerations. The CADO equation can sometimes be used for some of the velocity changes, but sometimes it can't.

I think the contradiction you are seeing in your example means that for at least one of your velocity changes, the CADO equation can't be used.

Last edited: Sep 13, 2018 at 2:56 PM
17. ### phytiRegistered Senior Member

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287
Neddy Bate #107;

Interesting, but the 'jump' is not supposed to be about what anyone sees with their eyes. It is about the definition of the current "now" in a given reference frame. Also, instead of a jump, it can shift gradually with gradual acceleration. The 'jump' is caused by the idealized instantaneous acceleration which is not realistic, and only serves to simplify the exercises.
-----
But if B sees them, they can't be missing!
Yes, it does simplify to a 1st approximation. The curved portion of a profile can always be approximated with a few straight line segments
If the 'jump' is the result of an unrealistic speed profile, then it's irrelevant. It's simpler to calculate time dilation for each segment of the profile.
The 1st drawing shows the C-frame on the opposite side of B, with a fictitious 'jump' backward, which is more nonsense. Note when B pings C (B0,E,B12), he assigns it (green) to B6, whereas the red aos would have assigned it to B7.5. The aos is only applicable to inertial motion.
The 2nd drawing hopefully makes the point. What time does B assign for a light pulse from E?

18. ### Neddy BateValued Senior Member

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Imagine two aliens standing next to each other in a galaxy 10 billion light years away from earth. One is named Amy and the other is named Bikey because he likes to ride his bike.

The Scenario According to Special Relativity (SR):

1. Amy and Bikey are both stationary with respect to earth, so their "now slice" includes the guy at the gas station on earth.

2. Bikey starts to ride his bike directly toward earth, while Amy remains stationary. So Amy's 'now slice' still includes our friend at the gas station, but Bikey's 'now slice' includes our friend's great great great great granddaughter teleporting from Paris to New York some 200 years in the future of the guy at the gas station.

3. Bikey stops his bike, and this makes him stationary with respect to Amy. Since Amy's 'now slice' always included our friend at the gas station, we know that Bikey's 'now slice', also must include our friend at the gas station, now that Bikey has stopped his bike.

4. Bikey starts to ride his bike directly away from earth, while Amy remains stationary. So Amy's 'now slice' still includes our friend at the gas station, but Bikey's 'now slice' includes Beethoven finishing his fifth symphony, some 200 years in the past of the guy at the gas station.

5. Bikey stops his bike, and this makes him stationary with respect to Amy. Since Amy's 'now slice' always included our friend at the gas station, we know that Bikey's 'now slice', also must include our friend at the gas station, now that Bikey has stopped his bike.

.....................................

The Scenario According to NotEinstein (NE):

1. Amy and Bikey are both stationary with respect to earth, so their "now slice" includes the guy at the gas station on earth.

2. Bikey starts to ride his bike directly toward earth, while Amy remains stationary. So Amy's 'now slice' still includes our friend at the gas station, but Bikey's 'now slice' includes our friend's great great great great granddaughter teleporting from Paris to New York some 200 years in the future of the guy at the gas station. THIS IS ALLOWED BECAUSE IN THE NE SCENARIO, CLOCKS CAN ALWAYS MOVE FORWARD, THEY JUST CAN'T MOVE BACKWARDS!

3. Bikey stops his bike, and this makes him stationary with respect to Amy. Since Amy's 'now slice' always included our friend at the gas station, we might think that Bikey's 'now slice', also must include our friend at the gas station, but NotEinstein tells us that is not allowed, because clocks can only move forward. SO THE NE SCENARIO SAYS BIKEY'S NOW SLICE STILL INCLUDES THE GREAT GRAND DAUGHTER TELEPORTING EVEN THOUGH HE IS IN THE SAME FRAME AS AMY! WAIT THAT CAN'T BE RIGHT, SO THE NE SCENARIO SAYS THAT BIKEY CAN ONLY DECELERATE HIS BIKE AT A SUPER SLOW RATE, SO THAT IT TAKES HIM 200 YEARS TO STOP THE BIKE, SO THAT THE CLOCKS DON'T RUN BACKWARDS! WAIT THAT CAN'T BE RIGHT, SO THE NE SCENARIO SAYS THAT BIKEY'S DECELERATION CREATES A HUGE GRAVITATIONAL FIELD THAT PULLS THE TIME ON EARTH FORWARD 200 YEARS, AND IT PULLS AMY'S CLOCK RIGHT ALONG WITH IT. YEAH, YEAH, THAT'S THE TICKET!

4. Bikey starts to ride his bike directly away from earth, while Amy remains stationary. So Amy's 'now slice' still includes our friend at the gas station, but Bikey's 'now slice' includes Beethoven. No wait, NE tells us clocks can't run backwards! SO THE NE SCENARIO IS THAT AMY'S CLOCK HAS BEEN YANKED FORWARD 200 YEARS BY BIKEYS'S HUGE GRAVITATIONAL FIELD, AND NOW BIKEY AND AMY BOTH AGREE THE TIME ON EARTH IS WHEN THE GREAT GRAND DAUGHTER IS TELEPORTING, EVEN THOUGH BIKEY IS MOVING AND AMY IS STATIONARY! WHOOPS! I GUESS THEY'LL JUST AGREE ON THAT THE WHOLE TIME BIKEY IS PEDALING, EVEN THOUGH THEY ARE IN DIFFERENT FRAMES! OH WELL, JUST BLAME IT ON THAT HUGE GRAVITY FIELD IN THE NE SCENARIO!

5. Oh forget it! The NE scenario says the acceleration of pedaling a bike creates a huge HUGE HUGE GRAVITY FIELD! IT'S JUST LIKE WHEN BIKEY FIRST STARTED PEDALING TOWARD EARTH AND THE HUGE GRAVITY FIELD CAUSED THE CLOCKS TO ALL... OH WAIT! THERE WAS NO HUGE GRAVITY FIELD IN THAT CASE, AND THE CLOCKS BEHAVED JUST LIKE SR. BUT THAT WAS BECAUSE IT'S ALL FINE AND DANDY WHEN CLOCKS RUN FORWARD. SO SOMETIMES THE ACCELERATION OF PEDALING A BIKE CREATES A HUGE GRAVITY FIELD AND SOMETIMES IT DOESN'T! GET OVER IT! GR IS COMPLICATED! ALL WE KNOW IS CLOCKS DON'T RUN BACKWARDS!! EVER!!! PERIOD. STOP QUESTIONING THE GREAT GR EXPERT!

Last edited: Sep 13, 2018 at 6:13 PM
19. ### Neddy BateValued Senior Member

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1,674
Yes, that is pretty much the case. You could say that the CADO equation works in that case only if you use completely different CADO_H and L values:

CADO_T = -120 - (0.866 * -138.56)

The reasoning there would be that we need to look at a time when your t variable was -60.

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Seek help.

21. ### NotEinsteinValued Senior Member

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Mike_Fontenot: My initial "conflict" was due to me not realizing certain scenario's have to be treated with GR, even though they don't explicitly involve gravity. Please note that I never said the CADO-equation was wrong; I just asked how the conflict could be resolved. I think that if you sharpen the language you use (mainly your definitions, and make it more explicit in which situations the CADO-equation holds), you'll have a very nice derivation of a surprisingly simple result. I didn't know my inquiries would lead to somebody blowing a fuse in this way. Please accept my apology; I will now bow out of the thread.

22. ### Neddy BateValued Senior Member

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I was just having a bit of fun.

I do think you should explain how you think the alien scenario goes, with clocks only moving forward. That is, if you have any idea how that would work at all.

23. ### phytiRegistered Senior Member

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It’s simpler to assign t and x to the rest frame and t’ and x’ to the moving frame.

At t, she perceives him at x with a clock reading of t’.

His t’ relates to the variables as the LT

t’ = γ(t-vx)