Jump to content

Can the rotation of distant galaxies be explained without the use of Dark Matter and Energy?


Recommended Posts

Posted
8 minutes ago, tar said:

Genady,

We cannot know the position and velocity of a particle such as an electron of an atom with certainty.

what makes you think we can know the position and velocity of a star with certainty?

Do not try to drag quantum theory into a discussion about astronomical objects. It makes you look a fool.  Heisenberg’s uncertainty principle is not relevant at the macroscopic scale, because Planck’s constant is so small it only makes a difference at the atomic scale.

Posted

I think I will retire from the discussion for a while.   I am trying to offer a required adjustment to our understanding of galactic rotation and you are trying to teach me the standard model, which is find currently in trouble.    Dark Matter to me is much like the epicycles of the planets observed.  The standard model keeps requiring changes in the cosmological constant and the age of the universe and such and requires the place be made up 95% by dark matter and energy and never observed either one.   We infer the presence of such to true the model with observations.   Perhaps we can true the observations with our model by adjusting our model, not by proposing imaginary particles and forces.

Posted
1 minute ago, tar said:

I think I will retire from the discussion for a while.   I am trying to offer a required adjustment to our understanding of galactic rotation and you are trying to teach me the standard model, which is find currently in trouble.    Dark Matter to me is much like the epicycles of the planets observed.  The standard model keeps requiring changes in the cosmological constant and the age of the universe and such and requires the place be made up 95% by dark matter and energy and never observed either one.   We infer the presence of such to true the model with observations.   Perhaps we can true the observations with our model by adjusting our model, not by proposing imaginary particles and forces.

That is what MOND tries to do. But that is based on an actual model with detailed calculations to support it. What you seem to be doing is pulling random ideas out of your arse, with no attempt at quantifying their effect to show how they might explain what we observe. That’s no good.

Posted

I understand I am lacking in the math department exchemist, but I am not there yet with this idea.  I am trying to set the assumptions and groundwork for expecting the universe to be connected to itself and be composed of ordinary matter and energy that we have been studying for years and know a lot about.

If something only fits the math and has not correlation to reality and ordinary matter and energy, what is the use of it?

 

It is like people finding PI to the quadrillionth digit.   Why?  You cannot measure or fabricate to that precision so what is it for?

I think math is very useful, but it is a symbol system.  You have to know what is standing for what.  You have to be able to say it words to know what you are setting to what and what relationship you wish to illustrate.  

You cannot for instance figure that half a cow will give half as much milk.  Half a cow would give half as much hamburger.

Posted
3 hours ago, tar said:

My thought is that the distance and size of distant galaxies is not taken into consideration when applying rotational and gravitational equations.

 

To apply the equation the rotation of the galaxy in question is considered as one thing rotating at once.

This is not what we see.    We see the closer parts of the galaxy 10s of thousands of years before the further parts.   

 

That is we cannot use one equation that takes the whole galaxy at once because we are seeing the galaxy at different times and any motion needs to be thusly "played back" to arrive at a position of the various stars at one moment where they were all in the same moment and then derive their positions in the next moment.

Regards, TAR

like NGC 1068? Notice anything about it?

https://en.m.wikipedia.org/wiki/Spiral_galaxy

IMG_0922.jpeg

45 minutes ago, tar said:

Genady,

We cannot know the position and velocity of a particle such as an electron of an atom with certainty.

what makes you think we can know the position and velocity of a star with certainty?

There’s always uncertainty but we can quantify it.

The thing about spiral galaxies is that the motion is tangential, not radial, so it really doesn’t matter when the time-tag of the measurement is. I don’t see where you've made the case that it matters

Posted
30 minutes ago, exchemist said:

That is what MOND tries to do. But that is based on an actual model with detailed calculations to support it. What you seem to be doing is pulling random ideas out of your arse, with no attempt at quantifying their effect to show how they might explain what we observe. That’s no good.

yes i like the MOND solution.   I am maybe turning to looking for an explanation of the difference in gravity effects at large distances that would help base the MOND solutionk that involves the delay in gravity's effect on distant objects based on the "speed" of gravity. Thus timing and the consideration of which parts of the model exist "at the same time" becomes pertinent. 

Regards, TAR

22 minutes ago, swansont said:

like NGC 1068? Notice anything about it?

https://en.m.wikipedia.org/wiki/Spiral_galaxy

IMG_0922.jpeg

There’s always uncertainty but we can quantify it.

The thing about spiral galaxies is that the motion is tangential, not radial, so it really doesn’t matter when the time-tag of the measurement is. I don’t see where you've made the case that it matters

I notice that we are seeing it from its axis of rotation.  That is, if we look at a galaxy edge on the closer part of the galaxy is on the edge closest to us but here the fringes of the galaxy are all the same distance, hence happening at the same time as the other portions of the fringe.   Although I would still imagine that the center of the pictured galaxy is closer to us then are its fringes.    Thus implying that we see the rotation near the center before we see the rotation at the fringes that was occurring simultaneously

Regards, TAR

Posted
20 minutes ago, tar said:

I notice that we are seeing it from its axis of rotation.  That is, if we look at a galaxy edge on the closer part of the galaxy is on the edge closest to us but here the fringes of the galaxy are all the same distance, hence happening at the same time as the other portions of the fringe.   Although I would still imagine that the center of the pictured galaxy is closer to us then are its fringes.    Thus implying that we see the rotation near the center before we see the rotation at the fringes that was occurring simultaneously

Regards, TAR

The center isn’t rotating, and I addressed the issue of timing, though you didn’t quote that part.

Posted

Thread.

Lets say mass accretes and tiny pieces of matter gather in one location to make a massive object.   This mass does not instantly affect all the rest of the universe, does it?

Objects close to it I would imagine are effected first and objects further away later.   After a time X after accretion  a close object has been subjected to its gravity for nearly all of time X whereas an object  far away from the accretion is yet to fell its effects.   

Posted
1 minute ago, tar said:

Thread.

Lets say mass accretes and tiny pieces of matter gather in one location to make a massive object.   This mass does not instantly affect all the rest of the universe, does it?

The effect of the mass doesn’t “turn on” when it forms a clump with another mass. It’s always on.

 

Posted
18 minutes ago, tar said:

yes i like the MOND solution.   I am maybe turning to looking for an explanation of the difference in gravity effects at large distances that would help base the MOND solutionk that involves the delay in gravity's effect on distant objects based on the "speed" of gravity. Thus timing and the consideration of which parts of the model exist "at the same time" becomes pertinent. 

Regards, TAR

 

 While changes in a gravitational field will propagate at c, a static field, like that exerted by the centre of mass of a galaxy, has no “speed”. And this is just another vague idea, thrown out by you seemingly at random, with no attempt to work it out to show  how it might produce the effect on rotation rate that we observe.

If you want to make a serious scientific suggestion, you have to show how your idea might produce the observed effects. 

Posted
5 minutes ago, swansont said:

The center isn’t rotating, and I addressed the issue of timing, though you didn’t quote that part.

I am not sure why you know the center is not rotating.  Don't you have to watch a galaxy for a thousand years before you notice a change in position of a unique star?   You can only use red shift and blue shift if the galaxy is edge on.  Looking at it form the axis  you don't know any of it is rotating.

1 minute ago, swansont said:

The effect of the mass doesn’t “turn on” when it forms a clump with another mass. It’s always on.

 

SwansonT, long time no argue with.  Good to see you.

Well Jupiter did not have a strong gravitational pull until it formed out of the accretion disc of the solar system.

You cannot measure Jupiter's gravitational effect on Sirius unless there is a Jupiter.

'

 

Posted
4 minutes ago, tar said:

I am not sure why you know the center is not rotating. 

Because that’s how rotation works.

4 minutes ago, tar said:

Don't you have to watch a galaxy for a thousand years before you notice a change in position of a unique star?   You can only use red shift and blue shift if the galaxy is edge on.  Looking at it form the axis  you don't know any of it is rotating.

You are missing information here. You can get doppler shift information even if it’s only a component of the velocity.

If the galaxy was edge-on, you can get the information from the opposite edges, which would in fact be the same distance away, but wouldn’t get the information from the interior part, since you can’t see it.

 

4 minutes ago, tar said:

SwansonT, long time no argue with.  Good to see you.

Well Jupiter did not have a strong gravitational pull until it formed out of the accretion disc of the solar system.

The individual parts still had gravitational pull. And at distances far enough away, it won’t matter if the mass is concentrated or not (it’s why a black hole has the same gravitational attraction as a normal star of equal mass, as long as you’re further away than the radius of the star. (i.e. the earth wouldn’t notice a gravity difference if the sun were a 1 solar mass BH)

Posted
9 minutes ago, exchemist said:

 While changes in a gravitational field will propagate at c, a static field, like that exerted by the centre of mass of a galaxy, has no “speed”. And this is just another vague idea, thrown out by you seemingly at random, with no attempt to work it out to show  how it might produce the effect on rotation rate that we observe.

If you want to make a serious scientific suggestion, you have to show how your idea might produce the observed effects. 

If the spiral arm of a galaxy contains a lot of stars and mass and you pick a point on the fringes of the galaxy the gravitational pull of the arm will speed up a mass that is "behind" it, rotationally and retard the rotational movement if it is"ahead" of the arm and the arm would add to the gravitation mass of the center of the galaxy if center of mass of the arm was between the fringe point and the center of mass of the galaxy.    Thus "when" the arm is in the various positions relative to the fringe point matters in terms of the acceleration  the arm's mass will impart on the fringe point.

swansont, the galaxy pictured was seen from its axis of rotation.  All the fringe was equal distant more or less.   An edge on galaxy the far side would be further away and hence what you saw of it was its position a long time before what you see on the closer edge.

 

Regards, TAR

9 minutes ago, swansont said:

Because that’s how rotation works.

You are missing information here. You can get doppler shift information even if it’s only a component of the velocity.

If the galaxy was edge-on, you can get the information from the opposite edges, which would in fact be the same distance away, but wouldn’t get the information from the interior part, since you can’t see it.

 

The individual parts still had gravitational pull. And at distances far enough away, it won’t matter if the mass is concentrated or not (it’s why a black hole has the same gravitational attraction as a normal star of equal mass, as long as you’re further away than the radius of the star. (i.e. the earth wouldn’t notice a gravity difference if the sun were a 1 solar mass BH)

edge on the close edge is more recent than the far edge.   Looking from the axis of rotation the edges are all approximately equal in age.

If my theory is correct the rotational speeds of galaxies viewed from the axis of rotation would be more consistent with Kepler's law than galaxies viewed edge on.

I do not know if there are  observations categorized in this manner.  Edge on or from the axis of rotation.

not theory, not even hypothesis.  If my speculation is correct the edge on galaxy would be figured to require more dark matter to explain its rotation than the axis viewed galaxy because the time component is not throwing off the observation as much in the axis view

Posted
33 minutes ago, tar said:

edge on the close edge is more recent than the far edge.   Looking from the axis of rotation the edges are all approximately equal in age.

But not left or right. If it’s edge-on, you can’t see the far edge anyway, so that’s moot.

Quote

 

If my theory is correct the rotational speeds of galaxies viewed from the axis of rotation would be more consistent with Kepler's law than galaxies viewed edge on.

I do not know if there are  observations categorized in this manner.  Edge on or from the axis of rotation.

not theory, not even hypothesis.  If my speculation is correct the edge on galaxy would be figured to require more dark matter to explain its rotation than the axis viewed galaxy because the time component is not throwing off the observation as much in the axis view

 

You’ve provided no reason why this timing matters. The stars would have to speed up or slow down, which requires the distance from the center change, which requires a radial velocity component. Is there evidence of this?

Posted
2 hours ago, tar said:

If the spiral arm of a galaxy contains a lot of stars and mass and you pick a point on the fringes of the galaxy the gravitational pull of the arm will speed up a mass that is "behind" it, rotationally and retard the rotational movement if it is"ahead" of the arm and the arm would add to the gravitation mass of the center of the galaxy if center of mass of the arm was between the fringe point and the center of mass of the galaxy.    Thus "when" the arm is in the various positions relative to the fringe point matters in terms of the acceleration  the arm's mass will impart on the fringe point.

 

So what? How does this do away with the need for extra, invisible mass to account for the observed rotation rates? 

Posted
10 hours ago, tar said:

what you saw of it was its position a long time before what you see on the closer edge.

But this is entirely irrelevant to Dark Matter. The mathematical relationship we are talking about here (Tully-Fisher relation and Faber-Jackson relation) are statistical statements; they relate the average rotational velocity of a large collection of stars in a galaxy to their total combined mass. Notions of simultaneity for an arbitrary observer as to the position of a single star at any given time never come into this at all, so this entire discussion is pretty much mood.

The other thing of course is that DM is needed for a lot more than just galaxies’ rotation curves; you cannot just ignore all the other evidence we have for its existence when discussing this subject.

Posted
18 hours ago, tar said:

My thought is that the distance and size of distant galaxies is not taken into consideration when applying rotational and gravitational equations.

 

To apply the equation the rotation of the galaxy in question is considered as one thing rotating at once.

This is not what we see.    We see the closer parts of the galaxy 10s of thousands of years before the further parts.   

 

That is we cannot use one equation that takes the whole galaxy at once because we are seeing the galaxy at different times and any motion needs to be thusly "played back" to arrive at a position of the various stars at one moment where they were all in the same moment and then derive their positions in the next moment.

Regards, TAR

I think you essentially want to point out that in current theories, dark matter is only used to explain the problem of stars moving too quickly in the outer region of galaxies. You offer an alternative explanation that the gravitational pull that galaxies exert on each other is the cause and that dark matter is not necessary. As an explanation for the previous calculation results that require dark matter, you say that the error lies in the fact that, due to the fact that light only moves at the speed of light, the current position of the galaxies deviates from the positions at the time they emitted the light and you assume that this is why the calculations are wrong because gravity would emanate from the current position of the galaxies. But this is exactly where the error in your assumption lies.

Gravity also spreads at the speed of light.

This means that all objects that we can observe in the universe always exert their gravitational attraction from the point at which we observe them, regardless of where they actually are at that time. Because gravity and light move at the same speed.

However, it becomes much more difficult when you consider that we as observer A are at a different distance from a gravitational source B than this gravitational source B is from a gravitational source C. Let's assume the distance from A to B is 10 light years and the distance from A to C is 20 light years. Theoretically, the distance from B to C could be, for example, 30 light years (these are more like distances between stars, but it is easier to use small numbers as an example). This would give the observer who is in position A a false impression of the gravitational attraction that B and C exert on each other. However, I assume that such things are taken into account in the scientists' calculations.

Posted (edited)
On 1/18/2025 at 8:12 PM, tar said:

We see the closer parts of the galaxy 10s of thousands of years before the further parts.

IMHO it doesn't matter for for such scales of galaxies' sizes and their rotation speed.

But I agree with following:

Quote

My thought that the calculations are missing something is that they need dark matter to provide the extra mass to arrive at the observed rotation.
...
it is easier in my thinking to discard imaginary dark matter and look for the manner in which the equations are misapplied.
...
Dark Matter is a made up thing.  Easier to discard it and bold an image that matches our senses.
...
dark matter is by definition not observed
...
Bottom line, they seem to not have "found" the particle required to account for the dark matter they need in their velocity of rotation calculations.

As I know, physicists use incorrect conclusion of Inertia Law. There is no movements without forces! Therefore, inertial motion is accelerated. It means that stars in the galaxies should continuously accelerate and go away from galaxy center. I said and showed it in my closed (blocked) topic, here.

Astronomers at early 20th century lived without dark matter, and they was sure that everything was taken in account and correctly calculated. Now your opponents assure you that dark matter is real and "everything was taken in account and correctly calculated". 😄

Next step is a correction of basic principles of Physics to explain astrophysical phenomenas, again without dark matter. 😉 After it everybody will say that "everything was taken in account and correctly calculated" :) 

 

Edited by kba
Posted
2 hours ago, kba said:

Therefore, inertial motion is accelerated.

If you put an accelerometer in free fall (inertial motion), it reads exactly zero at all times. This is an experimental fact. Thus, due to F=ma with a=0, no force acts on freely falling test particles.

Posted (edited)
3 hours ago, Markus Hanke said:

If you put an accelerometer in free fall (inertial motion), it reads exactly zero at all times. This is an experimental fact. Thus, due to F=ma with a=0, no force acts on freely falling test particles.

You can measure acceleration value a=1E-9 m/s²? Such acceleration we have from mass of half Universe on the distance equal to approximately half size of Universe. It is because inertial acceleration is caused by gravity of half Universe on the movement direction.

Edited by kba
Posted
42 minutes ago, kba said:

You can measure acceleration value a=1E-9 m/s²?

No. The most accurate accelerometers we have are around ~10E-7m/s^2.

44 minutes ago, kba said:

Such acceleration we have from mass of half Universe on the distance equal to approximately half size of Universe.

Please show in detail how you arrived at that number.

51 minutes ago, kba said:

It is because inertial acceleration is caused by gravity of half Universe on the movement direction.

There is no such thing as “inertial acceleration”. An inertial frame is defined to be one where there is zero proper acceleration; if you measure acceleration, no matter how small, then you’re not in an inertial frame. So even if the universe were somehow more gravitationally attractive in one particular direction (which it isn’t), you’d still be locally in free fall in that direction, and your accelerometer reads zero.

 

Posted

Also, this here:

1 hour ago, kba said:

Such acceleration we have from mass of half Universe on the distance equal to approximately half size of Universe.

is of course incorrect, irrespective of numbers. If you are surrounded by a spherical shell of matter, assuming an approximately uniform distribution, the net Newtonian gravitational force acting on you is exactly zero, irrespective of your state of motion, and irrespective of where within the shell you are located.

Posted (edited)
5 hours ago, Markus Hanke said:

Please show in detail how you arrived at that number

All what you need are known mass of Universe, Universe's size, and Newton's laws. My previous post need the correction. I said "half of Universe' size. Actually, the distance is about 1/2-1/3 of half Universe's size - if we consider Universe's volume as ball.

Approximately half of Universe's size we can consider with full mass of Universe.

I already wrote formula to calculate number in my blocked topic:

a = MG/R^2 = 0.00000000868 m/s².

It is nearly to 1e-8 m/s², but anyway it is smaller than 1e-7 m/s².

M, G, and R value you can take from Wikipedia.

4 hours ago, Markus Hanke said:

If you are surrounded by a spherical shell of matter, assuming an approximately uniform distribution, the net Newtonian gravitational force acting on you is exactly zero

This is correct if you consider gravity as static force. But I declare that gravity is dynamic one.

Quote

 you’d still be locally in free fall in that direction, and your accelerometer reads zero.

This is correct for any frame which is under only gravitational force. According to equvalence principle, bodies, that falling in the uniform gravitational field together, wouldn't register relative acceleration of each others. Look, even GR predicts that bodies inside uniform gravitational field will accelerate. You can consider any inertial frame as accelerated by means of gravity of half mass of Universe (as in the uniform gravitational field) in any direction. Only relativelly moving observer could register its acceleration.

And any gravitational frame you can consider similarly.

Edited by kba
Posted
10 hours ago, kba said:

All what you need are known mass of Universe

We don’t know the mass of the universe. And what about other sources of gravitation, like radiation? You haven’t accounted for that at all.

10 hours ago, kba said:

Universe's size

We don’t know this either. We know only the size of the observable part of it.

10 hours ago, kba said:

if we consider Universe's volume as ball.

Right now the evidence is more consistent with the universe being spatially flat on cosmological scales.

10 hours ago, kba said:

Approximately half of Universe's size we can consider with full mass of Universe.

No. You can’t.

10 hours ago, kba said:

This is correct if you consider gravity as static force. But I declare that gravity is dynamic one.

Gravity isn’t a force at all, except as an approximation in the Newtonian limit.

10 hours ago, kba said:

Look, even GR predicts that bodies inside uniform gravitational field will accelerate.

It says no such thing. Free fall is always along geodesics, irrespective of spacetime geometry, meaning a=0 at all times.

10 hours ago, kba said:

You can consider any inertial frame as accelerated

No. You can’t.

10 hours ago, kba said:

I already wrote formula to calculate number in my blocked topic:

If the topic was locked, then you shouldn’t bring this up again. 

We’re done here.

Posted (edited)
46 minutes ago, Markus Hanke said:

you shouldn’t bring this up again. 

I saw the question in the title of topic and answered to its author. Not to you.

You asked other questions, and I answered to you too.

 

Edited by kba

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.