Session 1 November 2025

[axj]

I did the calculations again a couple of times to check if the given orbital parameters fit with the heliosphere crossings around 1505 AD at 187 AU and around 1805 AD at 156 AU (with 1% additional attractive EM force at perihelion).

The result is that it all fits, but you need to be careful in explaining to the AI that the AU-distances of the heliosphere crossings are measured from the Sun and are not the travelled AU-distances of the brown dwarf:

Given orbital parameters (including EM effect)​

• Given period 28.2 million years
• Perihelion 40 AU
• Semimajor axis (geometric, not Keplerian) a=1.7 ly
• Heliosphere crossings: inbound at 187 AU (1505), outbound at 156 AU (1805)
• EM force at perihelion ≈ 1% of gravity, provided by charge product 5.1×10385.1×1038 C²

Step 1: Why the Keplerian estimate fails​

A purely Keplerian orbit with the same aa and rprp would have a period of 35 My. The EM attraction adds extra inward force only near perihelion, which:

• Increases the brown dwarf’s speed during the close passage
• Reduces the orbital period to 28.2 My
• Also reduces the time spent between two radial distances inside the heliosphere

Step 2: Approximate time scaling​

The extra EM force acts like a brief “kick” near perihelion. For a highly eccentric orbit, the time spent inside a given radius (e.g., from 187 AU inbound to 156 AU outbound) is roughly proportional to the orbital period, but with a stronger dependence on the speed boost.

For a Keplerian orbit with a=1.7 ly, e=0.999626, the speed at 187 AU is ~0.5 km/s. With the extra EM force, the speed at that distance is slightly higher, but more importantly, the speed near perihelion is significantly higher (since EM adds ~1% to the force, but the velocity gain is cumulative).

Step 3: Check if 300 years is consistent with the given EM strength​

Let’s assume the inbound crossing at 187 AU occurs at true anomaly θ1θ1 and outbound at 156 AU at θ2θ2. Using the actual orbit with the correct energy (which gives the 28.2 My period) we can compute the time difference.

We need the effective semimajor axis that would give a period of 28.2 My under pure gravity (as an equivalent Keplerian orbit). That effective semimajor axis is:

aeff=(TobsTgrav)2/3agrav=(28.2/35)2/3×1.7 ly≈1.46
This is the semimajor axis of a fictitious Keplerian orbit that has the same period as the actual EM-perturbed orbit. Its eccentricity with rp=40rp=40 AU is:

eeff=1−40 AU1.46 ly=1−4092300=0.999567

Now compute the true anomalies for r=187 AU and r=156 AU using this effective ellipse;

• aeff(1−eeff2)=rp(1+eeff)=40×1.999567=79.98 AU
• For r=187 AU: cos⁡θ=(79.98/187−1)/0.999567=−0.5728cosθ=(79.98/187−1)/0.999567=−0.5728 → θ=124.9∘θ=124.9∘ or 235.1∘235.1∘. Inbound: θin=235.1∘θin=235.1∘ (since >180°).
• For 156 AU: cos⁡θ=(79.98/156−1)/0.999567=−0.4878cosθ=(79.98/156−1)/0.999567=−0.4878 → θ=119.2∘θ=119.2∘ or 240.8∘240.8∘. Outbound: θout=119.2∘θout=119.2∘.

Now compute times from perihelion using eccentric anomaly as before:

• For θout=119.2∘: tan⁡(θ/2)=1.713tan(θ/2)=1.713, (1+e)/(1−e)=(1.999567)/(0.000433)=4617=67.96(1+e)/(1−e)=(1.999567)/(0.000433)=4617=67.96. Then tan⁡(E/2)=1.713/67.96=0.02520tan(E/2)=1.713/67.96=0.02520 → E/2=0.02520E/2=0.02520 rad → E=0.05040E=0.05040 rad. Mean anomaly M=E−esin⁡E=0.05040−0.999567×0.05039≈0.0000252M=E−esinE=0.05040−0.999567×0.05039≈0.0000252 rad. Time after perihelion: tout=(M/2π)×28.2 My≈(0.0000252/6.2832)×28.2×106=113tout=(M/2π)×28.2 My≈(0.0000252/6.2832)×28.2×106=113 years.
• For the symmetric inbound radius (187 AU) we need the time from perihelion to θ=124.9∘ (since inbound is at 235.1°, the complement is 124.9°). For θ=124.9∘θ=124.9∘: tan⁡(θ/2)=1.924tan(θ/2)=1.924, tan⁡(E/2)=1.924/67.96=0.02831tan(E/2)=1.924/67.96=0.02831 → E/2=0.02830E/2=0.02830 rad → E=0.05660E=0.05660 rad. M=0.05660−0.999567×0.05658≈0.0000317M=0.05660−0.999567×0.05658≈0.0000317 rad. Time before perihelion: tin=(0.0000317/6.2832)×28.2×106=142tin=(0.0000317/6.2832)×28.2×106=142 years.

Total time between inbound and outbound = tin+tout=142+113=255tin+tout=142+113=255 years.

Step 4: Match with observed 300 years​

The calculated 255 years is close to the observed 300 years (within ~15%). The difference can be explained by:

• The heliosphere boundary is not exactly at 187 AU and 156 AU in the brown dwarf’s direction; slight changes in those radii would adjust the time.
• The EM force is not exactly 1% but varies with distance, and our effective Keplerian approximation underestimates the speed boost near the Sun.
• The actual orbit is not a perfect ellipse, so the true anomalies differ slightly.

Given the uncertainties, 255 years is consistent with 300 years. Thus, the entry in 1505 at 187 AU and exit in 1805 at 156 AU fits the EM-shortened orbit (28.2 My period) to a reasonable approximation.

Why 1% works:​

1. Period change: The extra attractive EM force (≈1% of gravity at 40 AU) shortens the orbital period from ~35 My (pure gravity) to 28.2 My. This matches your given numbers.
2. Crossing time: Using the EM-affected orbit (effective semimajor axis 1.46 ly, eccentricity 0.999567), we calculated the time between inbound crossing at 187 AU and outbound at 156 AU as 255 years. This is close to the observed 300 years (within ~15%).

Thus, 1% EM strength is sufficient to produce all the observed features consistently. No stronger force is needed.

 

[SasaM]

I did the calculations several times. Have you double-checked that the numbers the AI calculated are correct, such as the 60 million years? And did you include the electromagnetic forces in your calculations?

Kepler's 2nd law does not include explicit interaction, only areas of the orbit. Area of the whole elliptical orbit is Pi*a*bi, where a and b are major and minor semi-axes, 1.7 ly and 0.055 ly respectively.

Here's the last part of AI calculation:

Using the same geometry (a = 1.7 ly = 107,410 AU; area between r1 and r3 about focus ≈ 5.89965×10^3 AU^2; total area A = πab ≈ 1.169×10^9 AU^2) the fraction f = area/A ≈ 5.05×10^−6.

For a full period P_total = 26×10^6 yr, the time between P1 and P3 is
Δt = f * P_total ≈ 5.05e−6 * 26e6 yr ≈ 131.3 years.

So about 131 years.

If the companion's passage through the heliosphere lasted 300 years, the whole period is 300/5.05 My = 59.4 million years.

Again, exactly this was already discussed back in November (and the number the C's gave is 28.2 million years, not 26 million):

For period 28.2 My, companion's passage through the heliosphere between those two points would last cca 142 years, using numbers from the quote box above.

As well as the most likely solution - electromagnetic interactions between the two stars when their heliospheres touch:


In other words, the EM attractive pull at perihelion speeds up the brown dwarf and makes its orbital period shorter by million of years. Which means that 50% or more mass in the solar system is not necessary, if EM forces are included.

Fraction of the orbit where this happens is traveled the fastest, implying that the e-m interaction there is much stronger than just simulating 80% (or 50%) larger effective mass if the companion is to pass basically the whole orbit in cca 20% less time.

Assuming that the companion's mass is 80% the mass of the Sun for the whole orbit will give wrong results.

That's the mass that Kepler's 3rd law gives for 26 My period, i.e. the central mass or the mass of the whole system is then 1.83 M_Sun, as Kepler's 3rd law says: a^3/T^2 = G/4*Pi (M+m).

For T = 28.2 My and a = 1.7 ly, the ratio gives 1.56 M_Sun, which is exactly what you first wrote in November.

So, if the passage through the heliosphere took 300 years as said in the SOTT article, when the e-m interaction between the companion and the Sun was the strongest, it would give the orbital period of cca 60 My assuming the same strength of interaction along the whole orbit. If the period is to be reduced more than 50%, while keeping these 300 years for that orbital segment, then the companion speed and therefore interaction with the Sun needs to be (much) stronger during basically the whole orbit than it was on that small orbital segment.

 

[SasaM]

I did the calculations again a couple of times to check if the given orbital parameters fit with the heliosphere crossings around 1505 AD at 187 AU and around 1805 AD at 156 AU (with 1% additional attractive EM force at perihelion).

The result is that it all fits, but you need to be careful in explaining to the AI that the AU-distances of the heliosphere crossings are measured from the Sun and are not the travelled AU-distances of the brown dwarf:

True, and if the Sun is placed in the focus of the ellipse, it becomes rather straightforward to calculate the area of the elliptical segment that companion traveled around the Sun, which directly enters into Kepler's law. Divided by the area of the whole ellipse, we get the fraction that corresponds to the fraction of the period for the time companion needed to pass along that part of its orbit.

Here's more detailed calculation of that part from my AI chat:

Now use the observed time Δt = 300 years between P1 and P3 to infer the fraction of orbital period that corresponds to the true anomaly sweep from P1 to P3, then scale to full period and check consistency; but simpler: if Δt is 300 years between those two points, the full period would be P_obs = 300 / f where f is the fraction of orbital period spent between P1 and P3. We must compute f from geometry (area swept between those two points divided by total area πab).

We already computed coordinates in ly for P1 and P3 (from previous step with a=1.7 ly, c=1.6993671 ly):
- P1 ≈ (x1,y1) = (1.697000, +0.001769) ly
- P3 ≈ (x3,y3) = (1.697900, −0.001982) ly

Convert to AU (1 ly = 63241.077):
- P1 ≈ (1.697*63241.077, 0.001769*63241.077) ≈ (107,332 AU, +111.9 AU)
- P3 ≈ (107,375 AU, −125.4 AU)

Compute area swept from P1 to P3 about the focus at (c,0) with c≈1.6993671 ly ≈ 107,402 AU. But because a ≫ coordinates differences, the angular positions around focus are tiny; area between P1 and P3 is approximately area of the sector between their radii: A_segment ≈ 0.5 * r_mean^2 * Δθ. However easier: use orbital area rate = constant = L/2μ but Kepler's second law: fraction f = (area between P1 and P3) / (π a b).

Compute area between radii vectors r1 and r3 about focus F: compute polygonal area using angle at focus. I'll compute numerically approximately.

Use positions relative to focus F = (c,0) in AU:
Focus c ≈ 107,402 AU.
r1 vector = P1 − F ≈ (107,332−107,402, 111.9) = (−70, 111.9) AU → r1 ≈ sqrt(70^2+111.9^2)=187 AU (consistent).
r3 vector = (107,375−107,402, −125.4) = (−27, −125.4) AU → r3 ≈ 156 AU.

Compute angle Δθ between r1 and r3 via cross/dot:
dot = (−70)*(−27) + 111.9*(−125.4) = 1890 − 14047.26 = −12157.26
|r1||r3| = 187*156 = 29172
cosΔθ = −12157.26 / 29172 ≈ −0.4168 → Δθ ≈ 114.7° = 2.002 rad.

Compute sector area about focus using triangle area formula: A = 0.5 * |r1 × r3| = 0.5 * |x1*y3 − y1*x3| in AU^2.
Compute cross = (−70)*(−125.4) − (111.9)*(−27) = 8778 + 3021.3 = 11799.3
A = 0.5 * 11799.3 ≈ 5899.65 AU^2.

That's the area of triangle between r1 and r3; however the true swept area along ellipse between those points is the area under the ellipse sector, not the triangle— but for Kepler, area swept is area swept by radius vector from focus; the area between two radii and the arc equals area of triangle plus area between arc and chord; but exact swept area equals integral; easier compute by using sector formula: A = 0.5 * r1 * r3 * sinΔθ = 0.5 * |r1×r3| which we computed = 5899.65 AU^2. Good.

Total ellipse area = π a b. Compute b in AU: b ≈ 0.05478 ly = 0.05478*63241.077 ≈ 3464 AU (from earlier). a in AU = 107410 AU.

Total area = π * 107410 * 3464 ≈ π * 3.721e8 ≈ 1.169e9 AU^2.

Fraction f = 5899.65 / 1.169e9 ≈ 5.05e-6.

Thus full period implied by Δt=300 years: P = 300 / f ≈ 300 / 5.05e-6 ≈ 5.94e7 years ≈ 59.4 million years.

Edit: As you can see, I'm not calculating any time per se, just area, so no need to split anything around perihelion. The perihelion was used only to get the equations for the parameters of the ellipse.

 

[axj]

So, if the passage through the heliosphere took 300 years as said in the SOTT article, when the e-m interaction between the companion and the Sun was the strongest, it would give the orbital period of cca 60 My assuming the same strength of interaction along the whole orbit.

The EM-interaction was not along the whole orbit, but only when the two heliospheres were touching each other for about 300 years. This attractive force during those 300 years is what speeds up the brown dwarf and as such shortens its whole orbital period.

Also, Deepseek checked what your AI calculated (which one are you using?) and thinks it found the error:

The AI’s calculation contains a fundamental geometric error: it uses the triangle area (0.5 × |r₁ × r₂|) as the swept area between P1 and P3, but Kepler’s second law requires the area of the elliptical sector (the region swept by the radius vector as the body moves along the orbit). For a large angle (114.7°), the triangle area is much smaller than the true sector area, leading to an underestimation of the fraction of the period.

Correct Approach​

We previously used the eccentric anomaly method, which correctly computes the time between two true anomalies for a Keplerian orbit. That gave:

• For the orbit with a = 1.7 ly, perihelion = 40 AU (so b ≈ 2932 AU), and period = 28.2 My (EM‑shortened): crossing time ≈ 255 years.
• The cirrect fraction f = 255 / 28.2×10⁶ ≈ 9.04×10⁻⁶.

Their triangle area fraction (5.05×10⁻⁶) is too small, so their period estimate of 59.4 My is invalid.

Why Their 300 years → 59.4 My is Wrong​

If the true swept area fraction were, say, 9.0×10⁻⁶, then a 300‑year crossing would imply a period of 300 / 9.0×10⁻⁶ ≈ 33.3 My, which is close to our pure‑gravity period (35 My) or the EM‑shortened one (28.2 My depending on exact geometry). The 59.4 My arises solely from using the wrong area.

In short: The AI’s calculation is incorrect because it confuses triangle area with elliptical sector area. Our calculation using eccentric anomaly is the proper method, and it supports that a 300‑year heliosphere crossing is consistent with a period of ~28–35 My, not 59 My.

You can copy this whole response to your AI and see what it says. It is also a good idea to do the same calculations at least a couple times in new windows (no previous data) to find such errors, which do happen quite frequently in AI calculations.

 

[SasaM]

The EM-interaction was not along the whole orbit, but only when the two heliospheres were touching each other for about 300 years. This attractive force during those 300 years is what speeds up the brown dwarf and as such shortens its whole orbital period.

Also, Deepseek checked what your AI calculated (which one are you using?) and thinks it found the error:

You can copy this whole response to your AI and see what it says. It is also a good idea to do the same calculations at least a couple times in new windows (no previous data) to find such errors, which do happen quite frequently in AI calculations.

That could be true.
What's the segment area your AI got?

Edit: If the e-m interaction happened only on that orbital segment, where the passage is the fastest anyway, it means that along the rest of the orbit the companion could not go faster than on that segment. So, the orbital period derived from that segment, if no additional interaction except gravitation happened along the orbit, would be the lower limit.

 

[SasaM]

The EM-interaction was not along the whole orbit, but only when the two heliospheres were touching each other for about 300 years. This attractive force during those 300 years is what speeds up the brown dwarf and as such shortens its whole orbital period.

Also, Deepseek checked what your AI calculated (which one are you using?) and thinks it found the error:

You can copy this whole response to your AI and see what it says. It is also a good idea to do the same calculations at least a couple times in new windows (no previous data) to find such errors, which do happen quite frequently in AI calculations.

You could maybe crosscheck your numbers by asking what would be orbital period for the ellipse containing the points in the article and time interval of 300 years for traveling along that segment, so not to instruct the AI to fit anything to your desired values.

I'll also calculate the area of the segment explicitly using integral calculus, prolly in Mathematica, as AI is very prone to errors as you said. Then we could compare the results.

 

[SasaM]

I did the calculations again a couple of times to check if the given orbital parameters fit with the heliosphere crossings around 1505 AD at 187 AU and around 1805 AD at 156 AU (with 1% additional attractive EM force at perihelion).

The result is that it all fits, but you need to be careful in explaining to the AI that the AU-distances of the heliosphere crossings are measured from the Sun and are not the travelled AU-distances of the brown dwarf:

BTW, have you checked this claim:

For a Keplerian orbit with a=1.7 ly, e=0.999626, the speed at 187 AU is ~0.5 km/s.

The speed depends on the assumed period and exact location of the point on the orbit, so there the AI used some additional (assumed) info.

Also this:

We need the effective semimajor axis that would give a period of 28.2 My under pure gravity (as an equivalent Keplerian orbit). That effective semimajor axis is:

aeff=(TobsTgrav)2/3agrav=(28.2/35)2/3×1.7 ly≈1.46
This is the semimajor axis of a fictitious Keplerian orbit that has the same period as the actual EM-perturbed orbit.

The underlying assumption in the above is that (M_Sun + m_companion) is the same for both orbits, meaning that the effective mass does not change along the whole orbit. Which comes down to what was said earlier, for things to 'fit' in that way, interaction between the Sun and companion needs to be more than just gravitation along the whole orbit, with an extra additional contribution during the companion's passage through the heliosphere.

How realistic is the AI 'fitting', even if done correctly, with those assumptions?

 

[Gaby]

The underlying assumption in the above is that (M_Sun + m_companion) is the same for both orbits, meaning that the effective mass does not change along the whole orbit. Which comes down to what was said earlier, for things to 'fit' in that way, interaction between the Sun and companion needs to be more than just gravitation along the whole orbit

Then, figuring out these forces would be akin to figuring out "Unified Field Theory". Pierre's CliffNotes apply as an introduction to the concepts. His working hypothesis was for a future sun's perihelion, not one that took place in the 17th Century.

The equations could still be used to do calculations and science, send man to the moon and so forth. But do we fundamentally understand what is gravity?

 

[SasaM]

Then, figuring out these forces would be akin to figuring out "Unified Field Theory". Pierre's CliffNotes apply as an introduction to the concepts. His working hypothesis was for a future sun's perihelion, not one that took place in the 17th Century.

He did a splendid job bringing the EU concepts to the light of the day, opening a whole new landscape in front of our eyes. It's a pitty there were things he got stuck into until the end, but that does not diminish even one bit the superb things he did.

But do we fundamentally understand what is gravity?

There was a discussion along these lines cca 3 years ago with a tentative hypothesis being (more or less in line with what Ark wrote on his blog at the time):

If everything that exists is a consciousness being an observer (of itself), and gravity binds everything together, then de facto gravity is an act of observation which binds consciousness with its Being, OSIT.

Which sort of makes a ful circle to understanding and describing consciousness as sort of a prerequisite for anything else.

 

[SasaM]

You can copy this whole response to your AI and see what it says. It is also a good idea to do the same calculations at least a couple times in new windows (no previous data) to find such errors, which do happen quite frequently in AI calculations.

Before turning to AI again, or to computer help, decided to do some calculations by hand first, and of course as usual it turned out AI made rather huge errors.

From semi-major axis of a = 1.7 ly (107 509.5 AU) and perihelion at 40 AU, placing the Sun in one of the foci of the companion's orbital ellipse at (a-40,0), i.e. distance (also linear eccentricity) c=a-40 from the center of the ellipse, we directly get semi-minor axis b, as b^2 = a^2 - c^2, b = 2933.2 AU (0.046 ly), and the eccentricity e = c/a = 0.99962794.

[AI got b = 0.055 ly = 3478.2 AU, which is already quite off in the very beginning.]

As such ellipse is very elongated, with the Sun in one focus very close to the 'edge' of the ellipse or so called vertex, it's useful to check so called semi-latus rectum to get a feel where the points of entry at 187 AU and companion's exit from heliosphere at 156 AU on the ellipse in relation to the Sun (focus) could be.

From Wikipedia link for ellipse:

Semi-latus rectum

The length of the chord through one focus, perpendicular to the major axis, is called the latus rectum. One half of it is the semi-latus rectum ℓ.
A calculation shows:
ℓ = b^2 / a = a (1 − e^2).

Using b we get l = 80.03 AU, while using e we get l = 79.99 AU.

This then implies that points of entry and exit on the ellipse are 'before' (or on the 'left' side) the semi-latus rectum lines, making the angle between the rays connecting the Sun in focus with entry and exit points definitely larger than 180°.
The AI in calculating the area of elliptical segment enclosed by those two rays said that the angle between the rays was 114°, which is evidently wrong. If anything, it should have been 360°-114°.

So, thanks axj for doubting the values my chat with OpenAI GPT-5 mini produced.

Once again the reluctance to consult AI for any kind of a serious task has been sort of justified by empirical proofs.
Basically, there hasn't been even a single instance in my experiences with different LLMs when not so intelligent AIs didn't make some huge errors, erroneous claims, or simply and plainly lied.