[Dave Lev (OP)]

The total tidal force can be computed by multiplying the force per unit distance by the diameter of the Sun:
8.2 x 6.96 x 10^8= 5.7 x 10^9 newtons

There is a small error with the diameter of the Sun. You have used the radius instead of the diameter, but that isn't the key error.
Tidal heat is not just a direct outcome from the tidal force.
There are some other factors that we need to verify as rotational motion.
The Earth gets its magnetic fields mainly due to the Tidal force with the moon.
However, the moon itself is cold as ice.
The main difference is that the Earth rotates while the moon is face locked with the earth. Therefore, the earth moon tidal force doesn't contribute any tidal heat energy to the moon.
There are millions or even billions of stars that have a similar Sun mass.
If you would monitor them, you might find that each one of them has different energy.
Some of them might be considers as G stars and others as M stars.
The tidal heat energy is also based on other factors as the size and the conditions of the object. Is it a solid object, gas object or a liquid hot plasma object (As the sun).
If it was due to Fusion activity, then it is expected that all of them should have exactly the same energy.

Each one of the following factors could change the tidal heat due to the tidal force:
The Sun Equatorial circumference = 4.379 10^6 km
It sets one Equatorial rotation in 24.4 Days.
Therefore, the sun Equatorial rotation velocity is 7,477 Km/h
Io Equatorial rotation velocity is only 271 Km/h
Io Radius = 1,821 km
Sun Radius =  696,000 km
Io is a solid object while the sun is a hot liquid plasma

Therefore, the same tidal force would set at the Sun significantly more tidal heat energy than in IO.
If we could slow down the rotational motion of the Sun (or even stop it), we would reduce dramatically the solar energy.
The location in the spiral arm might also be a factor. if it is in the center of the arm then it should gain more tidal force from all the nearby stars. If it is at the edge of the arm or even outside the arm, its tidal heat energy would be reduced. 

The rotation of the Sun itself is what causes the magnetic pole orientation to be where it is, so of course it's going to be synchronized. All fusion does is make the Sun hot. It isn't directly responsible for the magnetic field, only indirectly.

The rotation of the dynamo in the core of the Sun is what causes the magnetic pole orientation to be where it is.
That core MUST be SOLID in order to gain the magnetic field.
On earth, the core temp is about 6500 c, never the less, it is stated that it is a solid plasma core due to the internal pressure.
On the Sun, 10 M c would be too high to maintain a solid plasma core.
Therefore, at this temp, a liquid dynamo won't generate any magnetic field.
Therefore, it is not just an issue of internal heat up to 10 M c, but about the correct heat that should maintain a proper rotational operation of the solid dynamo.

 

Here is a paper describing the math involved in calculating the upper mass limit on brown dwarf stars (equivalent to the minimum mass needed to start the fusion processes needed for main sequence stars). Temperature is discussed as well. https://arxiv.org/pdf/1909.08575.pdf

There is a fatal mistake in the first step of this explanation.
It is stated:
"1). As the protostar continues to contract, increasing its density, temperature rises reaching values of 10^3K. At this point, the hydrogen atoms begin to gradually ionize, until the protostar becomes a sphere composed mainly of plasma, which can be modeled as an ideal completely ionized hydrogen gas"
ionized hydrogen gas means a hydrogen atom that has lost its electron and is now positively charged.
There is no way for the Hydrogen to lose its electron without real source of EM power. Heat by itself isn't good enough.
They claim that:
"We know that the mass of a brown dwarf is less than ∼ 0.08M , and that its effective temperature is ∼ 10^3K"
They make some assumptions and some calculation and hope that it would jump from 10^3K to 10^7K:
"Stellar formation models agree on a value of Mmin ≈ 0.08M
 [2-5]. Therefore, if the mass reaches or exceeds this value, the process of contraction continues until the protostars center reaches the temperature required to ignite thermonuclear reactions, where hydrogen is fused into helium. This temperature is ∼ 10^7K and marks the birth of a star"

Hence, without real explanation how the temp is jumping from 10^3K to 10^7K, they now are using this high temp to generate an expansive pressure that can stop the gravitational collapse.
"These high temperatures generate an expansive pressure that can stop the gravitational collapse, allowing
the star to reach a state of equilibrium"
The assume as follow:
"If we assume that the protostar is made of an ideal hydrogen gas that is completely ionized,
then according to statistical mechanics EK ∼ N kT, where N is the total number of particles
(protons and electrons), k is the Boltzmann constant and T is the absolute temperature."
As we have assumed an electrically neutral hydrogen gas, then the number of electrons Ne is
equal to the number of protons Np
If we assume that T ≈ Tig, then M ≈ Mmin
Now it is clear to me that with all of those assumption, they can get any temp as they wish.

Sorry, I don't accept this kind of explanation as a real theory to gain the requested 10 million c and how the core could be solid at that temp for the proper operation of the dynamo.

[Kryptid]

There is a small error with the diameter of the Sun. You have used the radius instead of the diameter, but that isn't the key error.

Yes, you are correct, I did make an error there. So the tidal forces should be 50 million times weaker on the Sun from Alpha Centauri A than from Jupiter on Io instead of 100 million times. At least this shows that you looked at the math instead of skipping it.

Tidal heat is not just a direct outcome from the tidal force.
There are some other factors that we need to verify as rotational motion.

Yes, this is true.

The Earth gets its magnetic fields mainly due to the Tidal force with the moon.

Tidal action from the Moon does heat the Earth to a degree, but radioactive decay and residual heat from the Earth's formation also contribute to the Earth's interior being liquid.

However, the moon itself is cold as ice.

Only the night time side of the Moon is cold. During the day, it can become boiling hot.

The main difference is that the Earth rotates while the moon is face locked with the earth. Therefore, the earth moon tidal force doesn't contribute any tidal heat energy to the moon.

Not exactly. The Moon's orbit is elliptical. The change in distance over time will change the tidal forces over time and thus result in some degree of heating. This is what causes Io to heat up (but to a more extreme extent than our Moon).

There are millions or even billions of stars that have a similar Sun mass.
If you would monitor them, you might find that each one of them has different energy.
Some of them might be considers as G stars and others as M stars.

Yes, this is true.

The tidal heat energy is also based on other factors as the size and the conditions of the object. Is it a solid object, gas object or a liquid hot plasma object (As the sun).

The Sun is not liquid.

If it was due to Fusion activity, then it is expected that all of them should have exactly the same energy.

That is not what fusion theory predicts at all. Stars with more mass have higher pressure, hotter cores which burn hydrogen faster. If you are going to argue against fusion theory, please don't misrepresent it.

Io is a solid object while the sun is a hot liquid plasma

Again, the Sun is not a liquid.

Each one of the following factors could change the tidal heat due to the tidal force:
The Sun Equatorial circumference = 4.379 10^6 km
It sets one Equatorial rotation in 24.4 Days.
Therefore, the sun Equatorial rotation velocity is 7,477 Km/h
Io Equatorial rotation velocity is only 271 Km/h
Io Radius = 1,821 km
Sun Radius =  696,000 km

Since the Sun takes 24 to nearly 38 days to complete a rotation (depending on where you are doing the measuring), then that's also how long it takes to complete one tidal flexing cycle. It takes Io about 1.8 days to orbit Jupiter, so its tidal flexing period is much faster than the Sun's. The tidal flexing period matters more than rotation velocity when it comes to heating, since rapid changes in physical stresses generate more heat than slower changes. So it's not at all clear from those numbers you posted that the Sun has the advantage in terms of tidal heating over Io (especially in light of Io experiencing millions of times greater tidal forces than the Sun).

Therefore, the same tidal force would set at the Sun significantly more tidal heat energy than in IO.

Given that Io experiences millions of times the tidal force that the Sun does and experiences more rapid tidal flexes, that's awfully unlikely.

If we could slow down the rotational motion of the Sun (or even stop it), we would reduce dramatically the solar energy.

Please demonstrate that this would be the case.

The location in the spiral arm might also be a factor. if it is in the center of the arm then it should gain more tidal force from all the nearby stars. If it is at the edge of the arm or even outside the arm, its tidal heat energy would be reduced.

Again, please demonstrate this.

The rotation of the dynamo in the core of the Sun is what causes the magnetic pole orientation to be where it is.
That core MUST be SOLID in order to gain the magnetic field.
On earth, the core temp is about 6500 c, never the less, it is stated that it is a solid plasma core due to the internal pressure.
On the Sun, 10 M c would be too high to maintain a solid plasma core.
Therefore, at this temp, a liquid dynamo won't generate any magnetic field.
Therefore, it is not just an issue of internal heat up to 10 M c, but about the correct heat that should maintain a proper rotational operation of the solid dynamo.

What is needed is a circulating, conductive fluid. Take this quote from the Wikipedia article on dynamo theory: https://en.wikipedia.org/wiki/Dynamo_theory

Dynamo theory describes the process through which a rotating, convecting, and electrically conducting fluid acts to maintain a magnetic field.

Both the Sun and the Earth meet this requirement. No solids are needed to create a magnetic field.

ionized hydrogen gas means a hydrogen atom that has lost its electron and is now positively charged.
There is no way for the Hydrogen to lose its electron without real source of EM power. Heat by itself isn't good enough.

Yes it is. Electrons are bound to atoms by a finite amount of force. If a material is sufficiently hot, then the atoms within it can collide with enough energy to dislodge the electrons. The only way that could be false would be if electrons were bound to atoms by infinite force. If that was the case, then direct stimulation by electromagnetic radiation wouldn't be sufficient to dislodge them either.

They make some assumptions and some calculation and hope that it would jump from 10^3K to 10^7K:

They are pointing out the 1,000 degree mark specifically because that's the order of magnitude where hydrogen becomes ionized. This is during a phase of the collapse and it does not stay at that temperature. As the protostar continues to collapse, it will continue to heat up in accordance with the gas laws.

Now it is clear to me that with all of those assumption, they can get any temp as they wish.

Not so. Scientists don't just make stuff up for the fun of it. If they did, then other scientists would call them out on it. Articles are peer-reviewed for a reason.

Sorry, I don't accept this kind of explanation as a real theory to gain the requested 10 million c

Because you misunderstood it.

and how the core could be solid at that temp for the proper operation of the dynamo.

The Sun doesn't have a solid core nor does it need one.

I'm still waiting for you to explain how the Sun generates neutrinos (and at the expected rate for fusion, at that).

[Bored chemist]

The Earth gets its magnetic fields mainly due to the Tidal force with the moon.

It is known that the earth' magnetic field has reversed many times in the past.
Are you saying that the Moon suddenly stopped and started orbiting backwards?

No, of course it didn't.

And that's how we know that, yet again, you are wrong.

There are millions or even billions of stars that have a similar Sun mass.
If you would monitor them, you might find that each one of them has different energy.

And, on average, they cancel out.
But that's not very important because, in reality, they are too far away to matter.
We already explained this to you.
But you do not pay attention to facts.

Why are you on a science page if you do not want to pay attention to facts?

[Bored chemist]

There is a fatal mistake in the first step of this explanation.

Which of these is more likely?
(1) there is a mistake and you are the only one to notice, even though you don't understand science or
(2) there is no mistake there, but you think there is because don't understand it.

[Origin]

If it was due to Fusion activity, then it is expected that all of them should have exactly the same energy.

Wrong.  Your statement doesn't even make sense.

The rotation of the dynamo in the core of the Sun is what causes the magnetic pole orientation to be where it is.
That core MUST be SOLID in order to gain the magnetic field.

Wrong.

There is no way for the Hydrogen to lose its electron without real source of EM power. Heat by itself isn't good enough.

Wrong.

Sorry, I don't accept this kind of explanation as a real theory

Who cares what you think?  You make up absurd claims that are disproven in almost every one of you posts.  I think it is safe to say that you put out more pseudoscience than anyone else on this site.  It is really a shame because you clearly aren't dumb, just incredibly confused, stubborn or a troll

[Dave Lev (OP)]

Both the Sun and the Earth meet this requirement. No solids are needed to create a magnetic field.

Solid inner core is a mandatory request:
https://www.ipgp.fr/~aubert/the-geodynamo.html
At the heart of our planet, the Earth?s core is mainly composed of liquid (outer core) and solid (inner core) iron. Its formation and differentiation is one of the significant events in the early history of our planet.
https://iopscience.iop.org/article/10.1088/1361-6404/ab8780
The inner core with a radius of 1210 km is solid, the outer core is liquid.

The Sun doesn't have a solid core nor does it need one.

Sorry, inner core must be solid. Without it there is no dynamo.

Here's what you should have said.
"However, the tidal impact of millions/billions stars around the Sun is  severe averages to  zero because, for every star pulling in one direction there is (on average) another star pulling in the opposite direction."

You have a severe mistake
https://exploringthecosmos.tumblr.com/image/35967694059
https://exploringthecosmos.tumblr.com/post/35967694059/tides-contrary-to-what-one-may-think-tides-are
"Contrary to what one may think, tides are caused by a differential force, not simply the mere force of gravity. Consider the fact that there are two tidal bulges on Earth; if tides were simply caused by the gravitational pull of our Moon, there would only be one tidal bulge on the side of Earth closest to the Moon when in fact, there is one tidal bulge on each side"
Therefore, if we could place one more moon exactly at the other side of the earth, then we would get two identical bulges
Each bulge would be:
GM/(r+R)^2 + GM/(r-R)^2

I'm still waiting for you to explain how the Sun generates neutrinos (and at the expected rate for fusion, at that).

I need few more days.

[Kryptid]

Yes, the Earth has a solid inner core. No, the Earth's inner core doesn't need to be solid in order for there to be a dynamo. If you disagree, then give us the exact quote from your source which states that it is necessary.

[Bored chemist]

Without it there is no dynamo.

You keep saying that.
It is wrong.
Why do you keep saying it?

You have a severe mistake

No, I have not.
Your picture does not show anything like the scenario I was talking about. It's the same picture I have known about for 40 years or more. Posting a copy of it here is stupid because all the real scientists already know about it.
And we know that it does NOT prove your point.
You need to stop assuming that everyone else is wrong.


I was pointing out that, if you have two distant stars in opposite directions, the pull from them cancels out and you end up with a tiny effect on the Sun or Earth.
Not the "severe" effect that you somehow seem to need to believe in.

two stars.png (2.52 kB . 755x186 - viewed 888 times)

[paul cotter]

Do you even read the articles you post? The first one you quoted says that the inner core is solid, correct, but that the magnetic field arises in the liquid outer core, same as the solar plasma. Nowhere is a solid core "mandatory".

[Dave Lev (OP)]

Yes, the Earth has a solid inner core.

Well, don't you agree that the earth core should be liquid?
It is stated:
https://iopscience.iop.org/article/10.1088/1361-6404/ab8780
" In the center of the Earth, it is about 6000 K, at the surface of the core it is slightly more than 4000 K."
It is made out of iron, so why they insist for a solid core?
The answer is very simple:
Based on their simulation it had been discovered that only if the inner solid conductive core rotates in a liquid conductive outer core the geodynamo can work!
https://cfn-live-content-bucket-iop-org.s3.amazonaws.com/journals/0143-0807/41/4/045803/revision2/ejpab8780f1_online.jpg?AWSAccessKeyId=AKIAYDKQL6LTV7YY2HIK&Expires=1688973293&Signature=ORMCwMbQAxSGmue%2FiOs4qQqMO%2Fo%3D

The geodynamo has some specific characteristics:
"The self-excited dynamo has some characteristics that we also find in the geodynamo, and which we would like to highlight here.
In order for the self-excited dynamo to start, the stator must first be supplied from another source, because as long as no current flows in the stator coils, no current is induced in the rotor coils.
If the dynamo runs too slowly, it can 'extinguish'.
If the rotor runs sufficiently fast, an arbitrarily small magnetic field is sufficient to start the dynamo effect. The faster the dynamo runs, the more unstable becomes the currentless state."

Therefore, the dynamo must be solid and it can't run to fast or too slow.

Did you ever try to run one kind of a liquid in other liquid without mixing them all together?
Therefore, if the inner core is liquid it would mix up with the outer liquid core and shut down the geodynamo proper operation.
However, if you still think that there is a possibility for a liquid dynamo to rotate in a liquid outer core and maintain its proper functionality, then please show the article to support this imagination.

[Bored chemist]

Well, don't you agree that the earth core should be liquid?

No.
We agree with reality which tells us that it happens to have a solid core.
You should try it some time.

Based on their simulation it had been discovered that only if the inner solid conductive core rotates in a liquid conductive outer core the geodynamo can work!

Based on reality, we know that you do not need a solid core; we checked.

Why do you keep ignoring reality and trying to push your weird impossible ideas?

[Dave Lev (OP)]

Based on reality, we know that you do not need a solid core; we checked.
Why do you keep ignoring reality and trying to push your weird impossible ideas?

What did they really check in this toy demo?
Did they check the core conditions in the sun?
Did the calculate the impact of the 10 M c imagination temp on the Sun' metal core?
Could it be that the Metal core at the Sun had been transformed to A Gas?
https://www.scienceabc.com/pure-sciences/can-a-metal-be-gas.html
Can A Metal Be A Gas?
"Yes, absolutely! Although metals usually occur in a solid state at room temperature (which is probably why we associate the word ?metal? with solid objects), metals can be in a gaseous state. "

"Does A Metal Remain A Metal When It Turns Into Gas?
We've established that metals can turn into gases if they're heated to their boiling points, but once a metal is heated to its boiling point and becomes gas, is it still metal? In other words, can a metal be in a gaseous state and still be considered a metal?
The short answer? No.

Gaseous metals don't retain the properties of their solid counterparts, including the metallic bonds, metallic conductivity, ductility, luster or other metallic properties. This is why metals are no longer considered metals when they assume a gaseous state, they're just gas with certain characteristic properties of the parent element, i.e. mercury gas"

So, if the metal core in the sun is transformed to gas at 10 M c and it doesn't retain the properties its solid counterparts, including the metallic bonds, metallic conductivity, ductility, luster or other metallic properties, then how it could generate any sort of magnetic field?
Do you have better answer than this toy demo?

[Kryptid]

The Sun's core isn't made of metal anyway. It's made of hydrogen plasma (which is electrically conductive).

[Bored chemist]

What did they really check in this toy demo?

That you do not need a solid core to male a magnetic field in a conductive liquid.

In doing so they proved that you are wrong to say "

he dynamo must be solid

Did you ever try to run one kind of a liquid in other liquid without mixing them all together?

Yes, it's a very common thing to do in (old school) analytical chemistry.
https://en.wikipedia.org/wiki/Liquid%E2%80%93liquid_extraction

Why did you ask?

Gaseous metals don't retain the properties of their solid counterparts, including the metallic bonds, metallic conductivity, ductility, luster or other metallic properties. This is why metals are no longer considered metals when they assume a gaseous state, they're just gas with certain characteristic properties of the parent element, i.e. mercury gas"

Here is a picture of mercury, as a gas, carrying an electric current.
https://edisontechcenter.org/MercuryVaporLamps.html
It forms a plasma- just like the hydrogen in the sun.

I can't see why you think the question is relevant.

Do you have better answer than this toy demo?

We don't need one.
They didn't have a solid core and they got a magnetic firld.
So we know that you do not need a solid core to get a magnetic field.
Even this toy is good enough to prove that you are wrong.

[Bored chemist]

Could it be that the Metal core at the Sun had been transformed to A Gas?

No. Because the sun doesn't have a metal core.

Did the calculate the impact of the 10 M c imagination temp on the Sun' metal core?

That does not parse as English.

[Kryptid]

Let's refer back to the link calculating tidal forces I posted earlier: https://www.astro.uvic.ca/~jwillis/teaching/astr201/maths.7.tidal_heating.pdf

According to the equation calculating the power of the tidal forces affecting Io, the total power is 6 x 10¹⁷ watts. The power is inversely proportional to the tidal cycle period, so if we reduce the period from the 1.769 days of Io to the 25.05 days of the Sun, that reduces the total power to 4.24 x 10¹⁷ watts. The power is also linearly proportional to the tidal force. Since the tidal force on the Sun from Alpha Centauri A is 50 million times lower than that of Io, then the power also reduces 50 million-fold to 8.47 x 10⁸ watts.

The Sun's total power output is 3.828 x 10²⁶ watts. If we assume (very, very generously) that tidal power is converted into heat with 100% efficiency, then the heights of the tides on the Sun would need to be 4.519 x 10¹⁷ times higher than they are on Io (according to the equation). Since Io's tidal heights are about 50 meters, that corresponds to tidal heights on the Sun of 2.26 x 10¹⁶ kilometers. That's over 16 billion times larger than the Sun's own diameter. So even if every star in the Milky Way (400 billion of them) could exert the same amount of tidal force on the Sun that Alpha Centauri A does (which, of course, is impossible because they are much too far away), that would still require tides nearly 112,000 kilometers high (about the size of the planet Saturn) on the Sun.

Since the Sun does not have such extreme tides (despite extremely unrealistic assumptions that would help your model), then we know that your model cannot possibly be correct.

[Dave Lev (OP)]

That's over 16 billion times larger than the Sun's own diameter. So even if every star in the Milky Way (400 billion of them) could exert the same amount of tidal force on the Sun that Alpha Centauri A does (which, of course, is impossible), that would still require tides nearly 56,000 kilometers high (larger than the diameter of Uranus) on the Sun.

Let's assume that you are fully correct with your calculation.
How do we know that there are no other invisible dark stars or even blackholes?
Don't you agree that we can only see those stars that emits light.
However, there is always a possibility that some proto stars have not been converted into Sun like star.
It is stated: "10,000 black holes discovered at the center of Milky Way"
https://www.indiatoday.in/education-today/gk-current-affairs/story/10-000-black-holes-discovered-at-the-center-of-milky-way-columbia-university-of-new-york-1205215-2018-04-05
If there are so many BHs just in the core, why there can't be millions or even billions of invisible BHs in the galaxy?

Astronomers Find a Sun-like Star Orbiting a Nearby Black Hole
https://www.universetoday.com/157655/astronomers-find-a-sun-like-star-orbiting-a-nearby-black-hole/
In this case we see the BH due to its accretion disc.
However, without the accretion disc, that BH would be invisible.

Astronomers Uncover Black Hole Closer to Earth Than Ever Before
https://scitechdaily.com/astronomers-uncover-black-hole-closer-to-earth-than-ever-before/#google_vignette

Do we also have an idea about the total tidal heat impact of the solar system itself based on planets, Kepler belt and Oort cloud?
Could it be that the total mass there is several times the Sun mass?

Could There Be a Black Hole Lurking on the Edge of the Solar System?
https://www.newsweek.com/black-hole-hidden-edge-solar-sysem-1741515
"There also could be a black hole on the edge of the solar system," Smethurst said. "Like there's this idea that there's another planet out there that we haven't discovered yet, that could be shepherding some of the objects out there into these weird orbits that we've seen, and we haven't found it."

Therefore, you can't just eliminate the idea that somehow there is enough matter around the Sun and in the galaxy that could set the requested tidal heat on the Sun.

[Kryptid]

Could it be that the total mass there is several times the Sun mass?

No, the Sun is the most massive object in the Solar System by a wide margin (99.8% of the Solar System's mass is in the Sun): https://solarsystem.nasa.gov/solar-system/sun/in-depth/#:~:text=Most%20of%20the%20nebula's%20material,of%20our%20solar%20system's%20mass.

If there are so many BHs just in the core, why there can't be millions or even billions of invisible BHs in the galaxy?

Do you have evidence that there are that many? Let's also not forget that the vast majority of them would be much too far away to help with your tidal energy problem. Sagittarius A*, which is far and away the most massive black hole in the galaxy, still produces only 9.48 x 10⁴ newtons of tidal force on the Sun (over 120,000 times weaker than Alpha Centauri A's pull) despite weighing over 4 million solar masses. It's just too far away to matter. And if something that massive doesn't matter, then any black holes on the other side of the galaxy from us with more typical masses (like 10 solar masses) are going to be much less relevant.

So if you want your idea to be taken seriously, you need to show us that the needed number of black holes at the needed distances are really out there. Otherwise, you are just speculating.

Remember, my calculation was immensely generous in your favor. I assumed that every single star in the galaxy was only 4.3 light-years away from the Sun all at the same time. That's equivalent to having a 400 billion solar mass black hole (one weighing around 100,000 times as much as Sagittarius A*) sitting only 4.3 light-years from us. If something even that unrealistically monstrous at such a very close distance to us can't give the Sun its needed tidal energy output, then how do you expect your idea to work?

[Bored chemist]

How do we know that there are no other invisible dark stars or even blackholes?

Because the effects of their gravity would be apparent.
That's how we know about dark matter.

It's important to remember that tidal forces drop of as an inverse cube, but attraction drops as an inverse square.
So, if the tidal forces were detectable, the attractive forces would be much much bigger and we would notice them.

Could it be that the total mass there is several times the Sun mass?

No.
That much mass would affect the orbits of the planets.

Therefore, you can't just eliminate the idea that somehow there is enough matter around the Sun and in the galaxy that could set the requested tidal heat on the Sun.

Yes we can.
Kryptid came up with a clever way of doing it.

You are saying that the tidal forces of distant objects provide (significant) heat in the sun.
To provide heat they have to do work.
Work is done when a mass moves through a distance.
In this case, the mass would be the part of the sun subject to tides and the distance would be the difference between "high tide" and "low tide" on the sun.
And as was pointed out, that would require the sun to shrink and grow in response to the tide effects.
But the diameter of the sun is essentially constant.

So we know it is not being stretched and squashed by tidal effects.
So we know that practically no work is being done on it.
So we know that can't be the source of the sun's heat.

[Kryptid]

I wanted to see what kind of data existed on physical distortions of the Sun's shape (tidal effects from other stars would definitely affect this if your model is correct). It seems that the Sun is very close to being a perfect sphere with little in the way of distortions: https://science.nasa.gov/science-news/science-at-nasa/2008/02oct_oblatesun

It seems that, on average, the Sun varies from a perfect sphere by only about 6 kilometers (8 milliarcseconds). Based on the measurements, there seems to be times where the distortion is more extreme than that: https://www.researchgate.net/figure/Left-solar-figure-as-from-the-RHESSI-measurements-Right-solar-oblateness-measured-by_fig4_51910936

The distortion appears to reach a maximum of about 24 milliarcseconds (which would be equivalent to 18 kilometers). If we assume (again, generously) that this area of maximum distortion is caused by the tidal effects of other stars, then I can calculate the needed tidal forces for your model to work. Since I calculated that a 400 billion solar mass object sitting 4.3 light-years away would need to produce 112,000 kilometer tides on the Sun in order to produce the needed power to explain the Sun's luminosity, then the required tidal forces would need to be be (112,000/18) = 6,222 times larger than what I originally calculated. This would be equivalent to a 2.489 quadrillion solar mass black hole sitting 4.3 light-years from the Sun. That is over 2,000 times the mass of all the stars in the entire Milky Way galaxy.