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my hypothesis: dark matter observations are relative


Maartenn100

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I have a hypothesis which I want to share here. It may be wrong, but I want to discuss about it. 

 What we see through our telescopes has also something to do with from which gravitational field we observe the celestial bodies. Our observation of dark matter in an observed galaxy f.e. is the result of the space-time curvature of our gravity field from where we observe the other field, and the curvature of the galaxy we observe.

It's a result of clocks and rulers. Of the influence of the field of gravity of the observers on their telescopes.

Because we consider our clock and our local ruler to be non-curved and always the standard for slower running time or curved spaces elsewhere in a relatively stronger (or relatively weaker field) of gravity somewhere else. The movements of the celestial bodies performed are therefore, in my opinion, partly relative. From another gravitational field, we see an object revolving around its star in a certain way which will be different for an observer from another field of gravity. That is the prediction from my hypothesis. From a different gravitational field, we will see stars moving differently in their galaxies. The dark matter that is assumed, because observed stars rotate too fast in their galaxy, is therefore, in my opinion, the result of the influence of our  gravitational field on our clock and ruler, on our telescope. On our observations. So what we call "dark matter" is, in my opinion, a kind of relativistic space-time perception. Depending on where we make the observation from. The result of the observer's gravitational field and the observed field provides the relativistic space-time curvature perception of the field concerned. Because an observer will always use his clock and ruler as the standard for normal ticking time and uncurved rulers.

Therefore, the amount of observed dark matter will differ for different observers in different gravitational fields.

 

 

 

Edited by Maartenn100
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15 minutes ago, Maartenn100 said:

Because we consider our clock and our local ruler to be non-curved and always the standard for slower running time or curved spaces elsewhere in a relatively stronger (or relatively weaker field) of gravity somewhere else. The movements of the celestial bodies performed are therefore, in my opinion, partly relative.  

Yes, but we can measure the local effect, and, if need be, correct for it. In any event, it's small.

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Yes, but we can measure the local effect, and, if need be, correct for it. In any event, it's small.

But small is relative, isn't it? The influence of the black hole in the centre of our galaxy and the whole system we are part of (the galaxy) and the cluster of galaxies on our clock and ruler is very big, isn't it? The influence of a system of clusters of galaxies on our clock and ruler is big, isn't it? 

But to us, our clock ticks 'normal'. We use our clock and ruler as a standard, to state: that galaxy we observe over there has this space-timeproperties.

It totally depends on our ruler and our clock. On our telescope in our field. 

In my theory there is no zero gravity condition thinkable. There is only curvature. There can be not no curvature of space and time.  And the amount of curvature always depends on the observer's standard for time and curvature of space.

 

Edited by Maartenn100
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20 minutes ago, Maartenn100 said:

From another gravitational field, we see an object revolving around its star in a certain way which will be different for an observer from another field of gravity. That is the prediction from my hypothesis. From a different gravitational field, we will see stars moving differently in their galaxies. The dark matter that is assumed, because observed stars rotate too fast in their galaxy, is therefore, in my opinion, the result of the influence of our  gravitational field on our clock and ruler, on our telescope.

It should be trivial for you to do the calculations to support this claim.

But it appears to be based on a misunderstanding. We don't just see stars orbit "too fast". As if the effective mass of the galaxy were different from what we estimated it be.

How would we estimate the mass of a galaxy? One way is by looking at the orbital speed of the stars. So you are suggesting that the orbital speed of the stars is not consistent with the mass we calculate by looking at the speed of the stars.

Or, to put it more simply, the orbital speed of the stars is not consistent with the orbital speed of the stars. That is obviously not the case.

The problem is the distribution of speeds with radius. We can estimate how the mass profile of a galaxy (ie. how much mass is within a given radius) by looking at the density of stars, amount of light emitted, etc. It is this observed mass profile that does match the mass distribution measured from orbital speeds.

So the problem is not that the stars are too fast, but that the distribution of speeds does not match the observed mass profile. Hence the need for extra mass, distributed throughout the galaxy.

For your idea to explain this, you would need to show that looking at another galaxy from our gravitational potential affects the observed mass differently in different parts of the observed galaxy.

3 minutes ago, Maartenn100 said:

The influence of the black hole in the centre of our galaxy and the whole system we are part of (the galaxy) and the cluster of galaxies on our clock and ruler is very big, isn't it?

No it is tiny. The mass of the Earth will have the largest effect. Followed by either the Moon or the Sun. Black holes or other galaxies will have pretty much zero effect.

4 minutes ago, Maartenn100 said:

But to us, our clock ticks 'normal'. We use our clock and ruler as a standard, to state: that galaxy we observe over there has this space-timeproperties.

1. The effect would be the same for all parts of that distant galaxy and so would not address the need for dark matter

2. The size of our local gravitation can be calculated and taken into account

3. There is almost certainly no point in doing (2) for most purposes because the effect would be smaller than measurement errors for distant galaxies. (It is required for some measurements, though.)

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I should say that the influence of curvature by masses on a clock is not knowable. Because there is no zeropoint to compare too. I mean, there is no zero gravity environment where we can put a clock or a ruler and compare it with our clock and ruler in our gravitational field. There is no objective standard for time and space. We can only compare the influence of Earth's gravity on our clock with another clock in another gravitational field.  

Small is therefore relative. Depending on with wich clock you compare it too.

All the environements are curved somehow. 

So, to us, observers, there is no curvature on our clocks. We have a normal idea of time.

To the observers in a spaceship above Earth, their idea of time is slowed down here on Earth. There is a small curvature. 

To an observer on Jupiter, time is stretched here on Earth. 

Edited by Maartenn100
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26 minutes ago, Maartenn100 said:

I should say that the influence of curvature by masses on a clock is not measurable. Because there is no zeropoint to compare too. I mean, there is no zero gravity environment where we can put a clock or a ruler and compare it with our clock and ruler in our gravitational field. We can only compare the influence of Earth's gravity on our clock with another clock in another gravitational field.  

We have equations that allow us to calculate the effect.

Clocks are affected by more than gravity, and we do the same thing in realizing the length of the second — e.g. you measure the temperature and the magnetic field present, and remove the shifts from the frequency (blackbody and Zeeman, respectively). You can do the same thing for gravity.

You could calculate that e.g. your clocks are running slow by a part in 10^10 relative to a point far away from a source of gravity and make that correction. Again, you would only do this if it mattered, and the shift is going to be small.

Quote

Small is therefore relative. Depending on with wich clock you compare it too.

Small is compared to the size of the effect you are measuring. If you are measuring the orbit of extrasolar some planet, and it's about one earth year, then the number I used would mean you are off by around a part in 10^17, which is almost certainly small relative to the precision of the measurement

 

Quote

All the environements are curved somehow. 

So, to us, observers, there is no curvature on our clocks. We have a normal idea of time.

To the observers in a spaceship above Earth, their idea of time is slowed down here on Earth. There is a small curvature. 

To an observer on Jupiter, time is stretched here on Earth. 

Again, these effects are quantifiable, and any discussion of this should include that.

 

44 minutes ago, Maartenn100 said:

 

But small is relative, isn't it? The influence of the black hole in the centre of our galaxy and the whole system we are part of (the galaxy) and the cluster of galaxies on our clock and ruler is very big, isn't it? The influence of a system of clusters of galaxies on our clock and ruler is big, isn't it? 

No, it's equal to the gravitational potential, and then divided by c^2. Near earth, the effect is around a part in 10^16 per meter of elevation. (GM/rc^2)

The black hole at the center of the galaxy has a big mass, yes, but r is also very large. If we were much closer to the galactic center, this would start to become a large effect.

Clocks are affected by our non-circular orbit about the sun, as they sample different gravitational potentials. You can leverage this to confirm aspects of general relativity. But you need good clocks to try, because the effect is small.

 

 

16 minutes ago, studiot said:

Except it's not zero gravity that's important. It's zero gravitational potential that would be the reference. But as I said before,w e can measure the effect, since we can measure gravity and distance, so we can correct for it, if we need to.

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12 minutes ago, swansont said:

 

 

Except it's not zero gravity that's important. It's zero gravitational potential that would be the reference. But as I said before,w e can measure the effect, since we can measure gravity and distance, so we can correct for it, if we need to.

 

I fully accept that this is a balance of forces, but does not the "equivalence principle" apply, since we are talking GR ?

We include the effect of rotation in the value of 'g' as measured on Earth, ie it is not solely gravitational.

Edited by studiot
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1 minute ago, studiot said:

 

I fully accept that this is a balance of forces, but does not the "equivalence principle" apply, since we are talking GR ?

I'm not sure how this matters. The clock shift depends on the potential. Zero force does not equate to zero potential.  

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2 minutes ago, swansont said:

 

I'm not sure how this matters. The clock shift depends on the potential. Zero force does not equate to zero potential.  

Have we not measured dilation effcts on clocks orbiting the Earth at (high) velocity?

Edited by studiot
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———

Another way of quantifying the effects would be gravitational redshift of light.

"As an example, take the white dwarf star Sirius B, with a gravitational field ~100,000 times as strong as the Earth’s." — the redshift is 3 x 10^-4

 

http://astronomy.swin.edu.au/cosmos/G/Gravitational+Redshift

1 minute ago, studiot said:

Have we not measured dilation effcts on clocks orbiting the Earth at (high) velocity?

Depends on what you mean by high, but yes. The kinematic effect depends on v, the gravitational effect depends on the change on gravitational potential.

If you put a clock in the center of a hollow planet, it would run slow, even though gravity would be zero in there. Because it would take energy to remove the clock and get it to an infinite distance away (i.e. the reference we use for gravitational PE)

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1 minute ago, swansont said:

———

Another way of quantifying the effects would be gravitational redshift of light.

"As an example, take the white dwarf star Sirius B, with a gravitational field ~100,000 times as strong as the Earth’s." — the redshift is 3 x 10^-4

 

http://astronomy.swin.edu.au/cosmos/G/Gravitational+Redshift

 

Good points.

You actually measure the corrections.

I have no 'feel' for the numbers involved or thus what it would take for one effect to balance another.

I am bearing in mind that the OP is claiming there are no points in space where the net effects are zero.

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1 hour ago, Maartenn100 said:

What we see through our telescopes has also something to do with from which gravitational field we observe the celestial bodies. Our observation of dark matter in an observed galaxy f.e. is the result of the space-time curvature of our gravity field from where we observe the other field, and the curvature of the galaxy we observe.

Some questions:
Does an observation made on earth differ* from an observation made with a space based telescope? Does differences match what your idea predict? 
 

*) Differ regarding the observation of dark matter as stated by OP. 

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25 minutes ago, studiot said:

I am bearing in mind that the OP is claiming there are no points in space where the net effects are zero.

I'm trying to point out that it doesn't matter.

The second in defined in terms of a Cesium-133 atom experiencing no external effects (no fields, at 0K) and yet we are still able to realize the second, because we make corrections, so we know what the frequency would be under the idealized conditions. Gravity can be corrected for as well. (We do it relative to the geoid already. It's just a matter of changing the reference for the correction) 

 

The gist of the OP would be similar to claiming that rulers are no good because the material they are made from has a temperature coefficient, and no ruler is going to be at the exact temperature where the length was precisely defined. And the answer would be the same: we can correct for that if we need to, and in any event the effect is small.

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2 hours ago, Maartenn100 said:

I should say that the influence of curvature by masses on a clock is not knowable.

Err, yes it is. We calculate and use it everyday.

2 hours ago, Maartenn100 said:

Because there is no zeropoint to compare too.

You don't need a "zero point". Think about why it is called the theory of relativity.

 

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3 hours ago, swansont said:

I'm trying to point out that it doesn't matter.

The second in defined in terms of a Cesium-133 atom experiencing no external effects (no fields, at 0K) and yet we are still able to realize the second, because we make corrections, so we know what the frequency would be under the idealized conditions. Gravity can be corrected for as well. (We do it relative to the geoid already. It's just a matter of changing the reference for the correction) 

 

I hope I have not offered the impression that we cannot correct for all the (known) effects.

I also find it disappointing that the OP has ignored my earlier comments trying to nudge his ideas in a more fruitful direction since others have had and investigated not dissimilar thoughts before.

Edited by studiot
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There is another factor involved that isn't mentioned. The mass of a galaxy can be determined by the mass to luminosity relation. However when you compare that mass to the galaxy rotation curve you should get a Kepler decline. This is due to the baryonic mass distribution. 

 However we don't see this, instead we get a rotation curve that can only be gained through a uniform mass distribution. 

We can account for gravitational redshift in our measurements so relativity is accounted for where applicable. It does have effects on sprectrographic measurements in particular the 21 cm hydrogen line.

 It was the mass/luminosity measurements that led Zwicky to first realize that something was missing in terms of mass.

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On 12/19/2019 at 1:43 PM, Maartenn100 said:

And the amount of curvature always depends on the observer's standard for time and curvature of space.

The curvature of spacetime is a tensorial quantity, so it is invariant covariant. All observers agree on it, regardless of what coordinate systems they use.

Edited by Strange
Correct typo
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On 12/19/2019 at 5:01 AM, Maartenn100 said:

I should say that the influence of curvature by masses on a clock is not knowable. Because there is no zeropoint to compare too. I mean, there is no zero gravity environment where we can put a clock or a ruler and compare it with our clock and ruler in our gravitational field. There is no objective standard for time and space. We can only compare the influence of Earth's gravity on our clock with another clock in another gravitational field.  

Small is therefore relative. Depending on with wich clock you compare it too.

All the environements are curved somehow. 

So, to us, observers, there is no curvature on our clocks. We have a normal idea of time.

To the observers in a spaceship above Earth, their idea of time is slowed down here on Earth. There is a small curvature. 

To an observer on Jupiter, time is stretched here on Earth. 

The problem is that gravitational time dilation is not related to some "zero point", but merely the difference in potential.

An analogy:   You and someone else are on a ladder.   Gravitational time dilation is related to how far apart you are on the ladder ( say 10 rungs apart), it doesn't depend on how far you are from the "bottom" of the ladder.   If you are on rung 20 and he is on rung 10 you would measure the same time dilation between you and him as you would if you were on rung 120 and he was on rung 110.  So where it is, or even if there is a "zero point" for gravity has no bearing.

Also, the difference in potential and subsequent measure of time dilation also effects the light we see coming from those distant galaxies. The gravitational Doppler shift exactly follows the time dilation.  So if there were a large enough relative time difference between us and the distant galaxy to account for the higher perceived star velocities, it would show up as an equivalent shift in the spectral lines.

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23 hours ago, Markus Hanke said:

The curvature of spacetime is a tensorial quantity, so it is invariant. All observers agree on it, regardless of what coordinate systems they use.

Apologies, I just noticed a mistype here - I meant to say covariant, not invariant. That’s an important difference, though in both cases all observers will agree on it.

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Please, explain this in terms of observers and their clocks and rulers. An observer near a black hole will have a different idea of 'the stretch of time' here on Earth. To the observer near the black hole, time is not curved here on Earth, but stretched. (according to his idea of time near that black hole).

To the observer here above Earth, time is contracted on Earth. (very small, but contracted).

Both observers (near the black hole and an observer above Earth) have their own clock and their own ruler to determine the 'curvature of spacetime' on Earth. They have their own idea of standard timeflow and their own idea of an uncurved ruler. So they will have a particular idea of how much time is contracted here on Earth. They use their own stick to determine that. Their own reference frame.

It depends on their standard for normal time flow (and their idea of a straight uncurved line). To an observer near a black hole, he has a different idea of a normal timeflow then an observer above Earth. 

(please try to explain this in terms of observers and their clocks and rulers).

So, my point is: wherever you are as an observer, you use your standard idea of normal timeflow to determine the 'stetch or contraction' of time elsewhere by relatively more/less heavy massive bodies and relatively higher or lower speeds. The amount of 'curvature of spacetime' (the stretch or contraction of time f.e.) depends on your standard clock and standard ruler, wherever you are.

So, whatever you observe outthere about space and time, has something to do with your own clock and ruler, locally.

 

Edited by Maartenn100
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So, you see the universe through the lens of your clock. An observer near a black hole, who will think that time on Earth is stretched (according to his idea of time), will observe Earth differently then we (here on Earth) do. We have the same clock then Earth has. (same time rate passage, same timeflow). So Earth stands still, relative to us here on Earth. Because we have the same clocks and rulers.

(please reply with the analogy of observers and their clocks and rulers)

 

Edited by Maartenn100
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Ask yourself these questions:

What is a straight line for observers in different, curved spacetime environements? Is it the same straight line for all observers? Or do we disagree about our ruler? 

For every observer, his particular idea of straight uncurved ruler (a straight line) is an individual idea of space, like he has an individual clock (time rate passage). 

Whatever an observer will see as a straight line, in his own referenceframe, will be curved somewhere else by heavier masses.

Whatever an observer will see as a straight line in his own reference frame, will be seen as an expanded space somewhere else in less heavy environements (in intergalactic space f.e.).

We have individual clocks, but we also have individual rulers. We all have a particular idea of straight uncurved and unexpanded spaces in our own reference frames.

 

Edited by Maartenn100
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4 hours ago, Maartenn100 said:

Please, explain this in terms of observers and their clocks and rulers. An observer near a black hole will have a different idea of 'the stretch of time' here on Earth. To the observer near the black hole, time is not curved here on Earth, but stretched. (according to his idea of time near that black hole).

To the observer here above Earth, time is contracted on Earth. (very small, but contracted).

Both observers (near the black hole and an observer above Earth) have their own clock and their own ruler to determine the 'curvature of spacetime' on Earth. They have their own idea of standard timeflow and their own idea of an uncurved ruler. So they will have a particular idea of how much time is contracted here on Earth. They use their own stick to determine that. Their own reference frame.

It depends on their standard for normal time flow (and their idea of a straight uncurved line). To an observer near a black hole, he has a different idea of a normal timeflow then an observer above Earth. 

(please try to explain this in terms of observers and their clocks and rulers).

So, my point is: wherever you are as an observer, you use your standard idea of normal timeflow to determine the 'stetch or contraction' of time elsewhere by relatively more/less heavy massive bodies and relatively higher or lower speeds. The amount of 'curvature of spacetime' (the stretch or contraction of time f.e.) depends on your standard clock and standard ruler, wherever you are.

So, whatever you observe outthere about space and time, has something to do with your own clock and ruler, locally.

 

But you need one of the observers to be close to something like a black hole in order for this to be significant.  So for example, lets say we have a star orbiting a galaxy at the same distance from the center of the galaxy as our Sun is from ours( _26,000 ly).  We we also assume that the mass of that galaxy is 150 billion solar masses.    We can plug these numbers in and at least get a ball park figure of how much time dilation you could expect.

if you do that, you get an answer of 0.999999548 for the time dilation factor.  This is totally insignificant and many, many many factors too small to account for the rotation curves we see.

The only way time dilation could be a significant factor is if the galaxy contained a great deal more mass than what we see, much more "extra mass" than what we presently calculate for Dark matter.   You would have made the problem worse in terms of extra mass needed rather than better.

And, as I already mentioned, if time dilation were the culprit, it would betray itself as a shift in the light spectrum.  In a typical galaxy rotation curve, the star' orbital speeds remain fairly constant as you move out from the center.  According to the amount of visible mass in the galaxy and how it is distributed, We would expect the star's orbital speeds to decrease with distance.   If star B was twice as far out as star A,  you would expect it to have around 70% the orbital speed of star A.   If the reason we saw Star A with the same speed as star B was time dilation, this is a large time dilation factor, one that would also effect the light we see coming from these stars.   The spectral lines produced by star A would be greatly shifted compared to those coming from Star B.  This is not something we see in the spectrum from these stars.

And while we use a shift in the spectrum due to Doppler shift to measure these differences in star orbital speed, the shift we measure is very small compared to what would be needed if time dilation were the cause.  A star orbiting at 200 km/sec has a Doppler shift factor of 0.999335, compared to a factor of 0.999533 for one traveling at 70% of that speed. This only something in the order of a 0.01% difference, compared to the 70% difference needed for the measured speed to be due to a time dilation effect.

 

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