The Man
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Again, without a calculation to compare it to, a measurement proves nothing about the utility of a theory.
The Theory of Relativity will begin to fall apart in 2016/2017 - Part IV
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Again, without a calculation to compare it to, a measurement proves nothing about the utility of a theory.
The Man
Unbanned zombie poster
No, that you construe any of this as a religion or dependent upon belief, is only your problem.
The Man
Unbanned zombie poster
No, while certain aspects of the math did exit already including the basic math of non-Euclidian space. Others developed specifically from this, namely Einstein notation and the field equations. Heck, I'm not sure if even a Pseudo Riemann Sphere was ever mathematically attempted before. Math doesn't need explaining, it derives from the relationships. What it does need is direct applicability to said relationships. In this case, mathematically, including time as a fourth dimension. This is just a simple matter of facts.
So when are you ever going to relate the mathematical relationships of your elastic space notion? Again, what you really need to start with is a modulus of elasticity or bulk modulus for space-time or even just space.
Bulk Modulus
The mathematics of elasticity have been around for centuries (1727 for the concept now named Young's modulus). It's applying them to space, space-time or even higher dimensions where the relevant mathematical developments must come for your notions to have any quantitatively verifiable meaning. Just like with curved space and GR, some math existed but it was making it so it was more applicable that needed to be developed.
ETA: Oh, and to inform you, again, TR has been around since Galilean invariance was proposed in 1632. Long before your assertion of some "very strange consequences that math had in the late eighteenth century". As you have been informed of this before, please stop trying to deliberately distort history to fit your liking.
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The Man
Unbanned zombie poster
Again, this is just numerology. You have no units in your calculation yet have units in your stated results. If the result is "1.000002117467 meter" and the "stretch inwards to the sun at 0.000002117467" then the upstretched value is 1 meter. What happens when it is feet or furlongs? Does not the amount of stretch vary over the amount of length considered? Again, why can't a " stretch of space" be curved?
But the gravitational effects "1 light year" are far less then at "1mm" so the ratio must change as the closest approach to the sun or whatever gets smaller and smaller.
What "missing deflection near the Sun"? Has "Edward" proposed current measurements of deflection were not accurate? If not then he is not asserting anything is "missing". Now, what he may be asserting is that other measurements closer to the sun may diverge from GR calculations but as none of that has been demonstrated yet, to me recollection, then still nothing is "missing".
W.D.Clinger
Philosopher
I could derive the mainstream result for gravitational lensing from the Schwarzschild metric, but my derivation would be indistinguishable from many published derivations so you still wouldn't believe I had done it myself. And it's true that I probably wouldn't have been able to do it myself if I hadn't put some effort into studying mainstream general relativity, in the process of which I checked several derivations of the mainstream result, including Eddington's Appendix Note 9.
In what follows, however, I will show my own work as I comment upon your formula for stretching of space.
Well, that's quite a coincidence!
So Bjarne's theory predicts a deflection (as viewed from earth) of about 0.002 arcseconds. Mainstream general relativity predicts a deflection of about 1.75 arcseconds at the radius of the sun, diminishing away from the sun. The mainstream predictions agree with observations. Bjarne's theory predicts a value far removed from what is actually observed.
Which is about 94 light-seconds, which implies a negligible deflection angle for galaxies outside the Milky Way as viewed from earth. I can't calculate a specific deflection angle without knowing the distance from earth, which I can't know because Bjarne hasn't mentioned a specific galaxy.
It is clear, however, that Bjarne's theory cannot explain spectacular examples of gravitational lensing such as those I cited earlier:
As I wrote
The mainstream predictions agree with observations. Bjarne's theory predicts a value far removed from what is actually observed.
Dowdye's theory, although it is so completely different from Bjarne's as to contradict Bjarne in almost every other respect, does have this in common with Bjarne's theory:Dowdye's and Bjarne's theories both fail to agree with observations made during solar eclipses, and cannot even begin to explain the many spectacular observations of gravitational lensing involving galaxies away from our own.
ETA: Let's not forget that Bjarne wrote:
Gravitational lensing often shows 10 times as much gravity / mass as there really is.
From the above, it is clear that Bjarne's theory predicts far less gravitational lensing than predicted by mainstream general relativity. So even if Bjarne were correct when he alleges the mainstream is predicting less gravitational lensing than is observed, Bjarne's and Dowdye's disparate theories would be predicting far less than observed, and would therefore be doing much worse than the mainstream theory.
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jonesdave116
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Dowdye was a religiously inspired crackpot. Who also lied about measurements at large distances from the Sun. His idiotic explanation for lensing was that it was caused by the plasma near to the Sun! Totally failing to realise, in his utter ignorance of the subject, that plasma produces refraction. It is therefore wavelength dependent. It affects longer wavelengths more than shorter ones. Gravitational lensing is wavelength independent. It affects all wavelengths to the same degree. And that is what we observe. Under certain circumstances, we can identify the plasma contribution in lensing observations. It is orders of magnitude less than the gravitational contribution;
Effects of plasma on gravitational lensing
Xinzhong Er, & Shude Mao (2014)
https://academic.oup.com/mnras/article/437/3/2180/1024645?view=extract&login=false (Free access)
You are doing yourself no favours by calling upon crackpots to back up your crackpot claims!
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Mojo
Mostly harmless
But apart from making predictions that are different from observed reality, what’s wrong with these theories?
The Man
Unbanned zombie poster
But apart from making predictions that are different from observed reality, what’s wrong with these theories?
Well, Bjarne's, once again, fails a basic dimensional analysis. The calculation given, "1/sqrt(1-2GM/(rc^2)", has a dimensionless or unitless result. The units of G, M, r & c2 all cancel each other in that relation. Yet Bjarne's result has units of meter with no other calculation, explanation or additional units in that calculation.
ETA: Further he asserts "a stretch inwards to the sun" with no basis in the calculation to ascribe any direction to the purported "stretch" nor an origin that such a direction would be based upon. Away from the sun or in other direction (dimension) fits the calculation as well.
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W.D.Clinger
Philosopher
In Bjarne's defense (yeah, I know), I believe he misspoke. Thrice. At least. He is, after all, good at getting things wrong.
With my highlighting:
The units in red are nonsense. I believe he really meant the words highlighted in light blue, and also meant the units highlighted in yellow.
ETA:
So long as we're discussing the errors in Bjarne's units, let's note that Bjarne's decision to express the deflection in meters rather than as an angle (such as arcseconds) was suspect as well. Given a distance from observer to lensing object (such as the sun or some galaxy outside the Milky Way), the angular deflection can be converted into a corresponding distance at the lensing object, and vice versa (a deflection expressed in meters at the lensing object can be converted into the corresponding angular deflection provided you know the distance to the lensing object).
Note, however, that Bjarne's first calculation allegedly yields a result in meters without mentioning the distance from earth to sun. That could not possibly be right unless Bjarne's formula was intended to hold only for the distance from earth to sun. We know Bjarne didn't intend that because his second calculation used exactly the same formula for an arbitrary galaxy the size of the Milky Way, mentioning "a 1000 LY path" (1000 LY = 9.5e18m) that had to be multiplied times the "ratio" 2.96e-9.
Question: Why did Bjarne's second calculation involve multiplying the formula-derived "ratio" by the distance from earth to lensing object, when Bjarne's first calculation involved no such multiplication?
Answer: Because Bjarne doesn't know what he's doing, and is just faking it.
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The Man
Unbanned zombie poster
OK, thanks for parsing that, it at least sounds likely.
ETA:
Wasn't the first one just multiplied by the r used for the ratio calculation?
Also, particularly if he is going to stick with elasticity and the calculation given for his ratio of such then he already has the spatial 'strain' and only needs the 'stress' (force per unit area) to get a Young's modulus equivalent for his 'elastic space'. Since he can't seem to get even there, it is doubtful to get anywhere from even there.
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Bjarne
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Correction
The last equation is obviously wrong. (because then the deflection angle is the same regardless if the travel distance near an objekt is 1 meter or 1 LY )
Hence the equation must not be DxDA
D = distance near the astronomic object
DA = deflection angle
But of course D2xDA
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Bjarne
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Read my previous post
The last equation DxDA is of course obvious wrong
D = distance near the astronomic object
DA = deflection angle
D must of course be: D2xDA
Otherwise the travel length near the astronomical object (sun galaxy etc) would give you the same angle, - regardless if the travel-path is 1mm - 1meter og 1 LY. Hence D2
Furthermore the travel distance is based on the whole path light travel near the sun (diameter of the sun) , if you want only the inwards deflection, for ex sample : towards earth-direction, - you must only use radius (which mean ) the half travel distance, - and hence not 7E8m but the half.
Now add just a little fade in / fade out effect, right before / after the sun, and you got it.
So near the Sun it seems TR and MTR is not much different...
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W.D.Clinger
Philosopher
Abstract: Bjarne doesn't know what he's doing, and is just faking it.
The fact that Bjarne's original formula/calculation yields a distance ratio or distance, both of which do depend on the distance between lensing object and observer, when Bjarne's formula/calculations fails to mention that distance, is what made Bjarne's original formula/calculation so obviously wrong.
With Bjarne's "correction" as quoted above, his revised formula is obviously wrong for an even more obvious reason: Multiplying the square of a distance times an angle is geometrically absurd.
Note also that Bjarne still hasn't explained how his theory would predict that deflection angle.
Proof that the deflection distance (and Bjarne's ratio of distances) depend upon the distance from lensing object to observer.
This is most easily explained with the aid of a picture. Here is some ASCII art:
For the moment, please ignore points V and W. I'll return to them later.
Please imagine a solid line drawn from A through Y and on to Z. Feel free to print the ASCII art above so you can draw that line on a piece of paper.
The picture is not drawn to scale. Angles are vastly exaggerated, and distances are vastly compressed. The sun is present just below point A, but is not drawn because circles are hard to draw in ASCII art, and the sun wouldn't be to scale anyway.
The solid line coming in from the left represents a ray of light from a faraway star. If the picture were drawn to scale, that star would be thousands of kilometers to the left of the picture. That ray of light encounters the edge of the sun at point A, where it bends downward and proceeds on to points Y and Z, following the solid line I asked you to imagine. The bending in that ray of light is so concentrated within an arc near the sun that, at this scale, the bend looks as though it occurs at a single point A. If the ray of light were not bent by the sun, it would continue along the dashed line to point X.
The dotted line coming in from the left and going to point Y represents the path a ray of light from the distant star to point Y would have taken if the sun hadn't obscured or bent that ray of light. Similarly for the dotted line coming in from the left and going to point Z. The lines going through AX, BY, and CZ are almost but not quite parallel; they converge at the distant star way off to the left of the picture.
The following angles are equal. Let's refer to that angle as θ.
The length of the line segment AB approximates the deflection distance as seen by an observer at point Y. The length of the line segment AC approximates the deflection distance as seen by an observer at point Z. The deflection distance seen by the observer at point Z is about twice the deflection distance seen by the observer at point Y, because the observer at point Z is twice as far from the sun as the observer at point Y.
Similarly, the ratio of those deflection distances to the radius of the sun depends upon the observer's distance from the sun.
Summary:
Bjarne's mental model. I believe Bjarne's mental model of gravitational lensing is slightly different from the picture above. I think Bjarne is imagining what an observer located at point X would see. Before the sun got in the way, an observer at X would see a ray of light from the distant star continuing along the path AX. Because rays of light from that star are bent by the sun's gravity, the observer at X perceives a ray of light that appears to be coming from along the path from V to X. I think Bjarne is defining his deflection distance as the distance from V to A.
Unfortunately for Bjarne's theory, Bjarne's definition of the deflection distance still depends upon the observer's distance from the lensing object. An observer at point W would perceive a ray of light that appear to be coming from a point located above V, so the deflection distance as conceived by Bjarne would be greater for an observer at W than for an observer at X.
From that, it follows that the ratio of deflection distance to the radius of the lensing object depends upon the observer's position. Bjarne's formula for a (closely related) ratio does not mention the observer's distance, so Bjarne's formula for his ratio can't possibly be right.
Note well that, even with Bjarne's allegedly corrected calculation for his notion of the deflection distance, Bjarne still hasn't corrected his formula for the ratio.
Converting between deflection angle and deflection distance.
From my picture, it looks as though the deflection angle θ, deflection distance x, and distance D from observer to lensing object would be related by this equation:
dotted imagined solid line AYZ. For the small deflection angles seen in gravitational lensing, the difference between arctan θ and arcsin θ is negligible.
For what I believe to be Bjarne's notion of deflection distance, the corresponding equations would be considerably more complex. They would have to account for the decrease in bending as the point V moves further from the sun. (It should go without sayng that Bjarne hasn't given us any quantitative predictions for that effect.)
Conclusion. Bjarne's theory of gravitational lensing is bunk, rubbish, flapdoodle.
tl/dr: Bjarne doesn't know what he's doing, and is just faking it.
Yes. I expected him to say the 1000 LY in his second calculation was really the radius of the lensing galaxy (instead of the "light path" as he wrote), which would have made his two examples consistent, albeit consistently wrong, but no. His "correction", quoted below, was even worse than I expected.
If the calculation is done correctly, as in mainstream general relativity, the deflection angle is independent of the travel path.
The fact that Bjarne's original formula/calculation yields a distance ratio or distance, both of which do depend on the distance between lensing object and observer, when Bjarne's formula/calculations fails to mention that distance, is what made Bjarne's original formula/calculation so obviously wrong.
With Bjarne's "correction" as quoted above, his revised formula is obviously wrong for an even more obvious reason: Multiplying the square of a distance times an angle is geometrically absurd.
Note also that Bjarne still hasn't explained how his theory would predict that deflection angle.
The radius of the sun is approximately 7e8m. Half of 7e8m is not the radius of the sun, but half the radius of the sun.
Proof that the deflection distance (and Bjarne's ratio of distances) depend upon the distance from lensing object to observer.
This is most easily explained with the aid of a picture. Here is some ASCII art:
Code:
V
——————————A-------------------------X-----------------------------W
|
..........B.........................Y
|
..........C..................................................ZPlease imagine a solid line drawn from A through Y and on to Z. Feel free to print the ASCII art above so you can draw that line on a piece of paper.
The picture is not drawn to scale. Angles are vastly exaggerated, and distances are vastly compressed. The sun is present just below point A, but is not drawn because circles are hard to draw in ASCII art, and the sun wouldn't be to scale anyway.
The solid line coming in from the left represents a ray of light from a faraway star. If the picture were drawn to scale, that star would be thousands of kilometers to the left of the picture. That ray of light encounters the edge of the sun at point A, where it bends downward and proceeds on to points Y and Z, following the solid line I asked you to imagine. The bending in that ray of light is so concentrated within an arc near the sun that, at this scale, the bend looks as though it occurs at a single point A. If the ray of light were not bent by the sun, it would continue along the dashed line to point X.
The dotted line coming in from the left and going to point Y represents the path a ray of light from the distant star to point Y would have taken if the sun hadn't obscured or bent that ray of light. Similarly for the dotted line coming in from the left and going to point Z. The lines going through AX, BY, and CZ are almost but not quite parallel; they converge at the distant star way off to the left of the picture.
The following angles are equal. Let's refer to that angle as θ.
θ = ∠XAY = ∠XAZ = ∠BYA =∠CZA
θ is the deflection angle. As can be seen from the fact that those angles are equal, the deflection angle is independent of the distance from lensing object to observer.The length of the line segment AB approximates the deflection distance as seen by an observer at point Y. The length of the line segment AC approximates the deflection distance as seen by an observer at point Z. The deflection distance seen by the observer at point Z is about twice the deflection distance seen by the observer at point Y, because the observer at point Z is twice as far from the sun as the observer at point Y.
Similarly, the ratio of those deflection distances to the radius of the sun depends upon the observer's distance from the sun.
Summary:
- The deflection angle seen by an observer does not depend upon an observer's distance from the lensing object.
- The deflection distance seen by an observer is (roughly) proportional to the observer's distance from the lensing object.
- The ratio of that deflection distance to the radius of the lensing object is (roughly) proportional to the observer's distance from the lensing object.
Bjarne's mental model. I believe Bjarne's mental model of gravitational lensing is slightly different from the picture above. I think Bjarne is imagining what an observer located at point X would see. Before the sun got in the way, an observer at X would see a ray of light from the distant star continuing along the path AX. Because rays of light from that star are bent by the sun's gravity, the observer at X perceives a ray of light that appears to be coming from along the path from V to X. I think Bjarne is defining his deflection distance as the distance from V to A.
Unfortunately for Bjarne's theory, Bjarne's definition of the deflection distance still depends upon the observer's distance from the lensing object. An observer at point W would perceive a ray of light that appear to be coming from a point located above V, so the deflection distance as conceived by Bjarne would be greater for an observer at W than for an observer at X.
From that, it follows that the ratio of deflection distance to the radius of the lensing object depends upon the observer's position. Bjarne's formula for a (closely related) ratio does not mention the observer's distance, so Bjarne's formula for his ratio can't possibly be right.
Note well that, even with Bjarne's allegedly corrected calculation for his notion of the deflection distance, Bjarne still hasn't corrected his formula for the ratio.
Converting between deflection angle and deflection distance.
From my picture, it looks as though the deflection angle θ, deflection distance x, and distance D from observer to lensing object would be related by this equation:
x = D × arctan θ
The true equation isx = D × arcsin θ
because the line segments AB and AC would be perpendicular to the For what I believe to be Bjarne's notion of deflection distance, the corresponding equations would be considerably more complex. They would have to account for the decrease in bending as the point V moves further from the sun. (It should go without sayng that Bjarne hasn't given us any quantitative predictions for that effect.)
Conclusion. Bjarne's theory of gravitational lensing is bunk, rubbish, flapdoodle.
tl/dr: Bjarne doesn't know what he's doing, and is just faking it.
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The Man
Unbanned zombie poster
So the units of your result is something like Meter2 Degrees?
Reformed Offlian
Graduate Poster
Shouldn't that be radians? Not that it fixes anything, of course.
Ziggurat
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Radians are technically unitless, so it fixes something, but not enough.
The Man
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Certainly could be, as 360 degrees = 2pi radians, but depends on the calculation being used and the inputs. Heck, it could even be in Gradians (1/100th of a right angle).
Exactly.
The Man
Unbanned zombie poster
Must be a time dilation effect due to relativity falling apart so fast for so long.
Bjarne
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I off course mean DR ( Deflection Ratio) and this is also how my calculation is done.
I NEVER used the reflection angle in any calculation, and even never calculated that angle, and you KNOW that
Read the content of EQUATION above Kart Smart before bla bla bla bla too much
I used the deflection ratio in the equation, can you REALLY now see that ?
You are trying to get caught up in obvious typos
I am busy at work at do not use much time on this..
The last equation DxDR is of course obvious wrong
D = distance near the astronomic object
DR = deflection ratio, - must be ratio
D must of course be: D2xDR
Otherwise the travel length near the astronomical object (sun galaxy etc) would give you the same angle, - regardless if the travel-path is 1mm - 1meter og 1 LY. Hence D2
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smartcooky
Penultimate Amazing
W.D.Clinger
Philosopher
That explains why, in your "correction", you wrote "deflection angle" and abbreviated it as "DA".
With my highlighting:
Returning to your most recent "correction":
Bjarne, the few quantitative aspects of your theory that you have taken the time to present are pretty much indistinguishable from typos.
Okay. Let's go back and apply this most recent "correction" of yours to the first of your two original sample calculations.
Here is your calculation as you wrote it, with multiple "typos" in the units:
In that last multiplication, you multiplied the radius of the sun (7e8m) times
1/sqrt(1-2GM/(rc^2) - 1 ≈ GM/(rc^2)
where r is the radius of the sun. Although you haven't explicitly said what DR is supposed to be, it looks as though the "deflection ratio" of your most recent "correction" is that quantity, soDR = 1/sqrt(1-2GM/(rc^2) - 1 ≈ GM/(rc^2)
Your two most recent "corrections" say we're supposed to calculate the deflection distance asD2xDR
where D is the distance from observer to lensing object. For your example, that would be the distance from earth to sun, which is about 150e9m. Using the value you calculated for DR, that gives usD2xDR = (150e9m)2 x 0.00000211 = 4.7475e16 m2
I don't know how we're supposed to interpret 4.7475e16 m2 as a deflection distance, because a metric distance should be in meters, not square meters.I hesitate to regard your red superscript as an obvious typo, partly because you went to some trouble to show it in red, and also because it is one of the only two differences between your original calculation and your most recent "correction". (The other difference is taking D to be the distance from observer to lensing object, instead of the radius of the lensing object as in your original calculation. Even that is an assumption on my part, because you have never taken the time to state exactly what D is supposed to be.)
Because the units of your most recent "correction" are nonsensical, it is impossible to compare its 4.7475e16 m2 result to observations, but I will say this: I can't think of any units that would align the number 4.7475e16 with any number that has been reported or calculated from observations of light bending near the surface of the sun during a total eclipse.
In short, your most recent "correction" is a remarkable nonachievement, a debacle, a fiasco, a complete dud.
All of your attempts to present the quantitative predictions of your theory for gravitational lensing have been laughable. The reason for that is simple:
You don't know what you're doing, and are just faking it.
smartcooky
Penultimate Amazing
That explains why, in your "correction", you wrote "deflection angle" and abbreviated it as "DA".
With my highlighting:
Returning to your most recent "correction":
Bjarne, the few quantitative aspects of your theory that you have taken the time to present are pretty much indistinguishable from typos.
Okay. Let's go back and apply this most recent "correction" of yours to the first of your two original sample calculations.
Here is your calculation as you wrote it, with multiple "typos" in the units:
In that last multiplication, you multiplied the radius of the sun (7e8m) times
1/sqrt(1-2GM/(rc^2) - 1 ≈ GM/(rc^2)where r is the radius of the sun. Although you haven't explicitly said what DR is supposed to be, it looks as though the "deflection ratio" of your most recent "correction" is that quantity, so
DR = 1/sqrt(1-2GM/(rc^2) - 1 ≈ GM/(rc^2)Your two most recent "corrections" say we're supposed to calculate the deflection distance as
D2xDRwhere D is the distance from observer to lensing object. For your example, that would be the distance from earth to sun, which is about 150e9m. Using the value you calculated for DR, that gives us
D2xDR = (150e9m)2 x 0.00000211 = 4.7475e16 m2I don't know how we're supposed to interpret 4.7475e16 m2 as a deflection distance, because a metric distance should be in meters, not square meters.
I hesitate to regard your red superscript as an obvious typo, partly because you went to some trouble to show it in red, and also because it is one of the only two differences between your original calculation and your most recent "correction". (The other difference is taking D to be the distance from observer to lensing object, instead of the radius of the lensing object as in your original calculation. Even that is an assumption on my part, because you have never taken the time to state exactly what D is supposed to be.)
Because the units of your most recent "correction" are nonsensical, it is impossible to compare its 4.7475e16 m2 result to observations, but I will say this: I can't think of any units that would align the number 4.7475e16 with any number that has been reported or calculated from observations of light bending near the surface of the sun during a total eclipse.
In short, your most recent "correction" is a remarkable nonachievement, a debacle, a fiasco, a complete dud.
All of your attempts to present the quantitative predictions of your theory for gravitational lensing have been laughable. The reason for that is simple:
You don't know what you're doing, and are just faking it.
So, I was right! Its jibber-jabber!
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MRC_Hans
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Highligted: Especially since none of the attempts at formulas seem to contain ... well, gravity, or mass of the deflecting object. And since Bjarne claims that his elastic space is stretched by gravity ....
Hans
The Man
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Highligted: Especially since none of the attempts at formulas seem to contain ... well, gravity, or mass of the deflecting object. And since Bjarne claims that his elastic space is stretched by gravity ....
Hans
Well technically the equation he uses for his "relativistic stretch of space", DA or now DR is "1/sqrt(1-2GM/(rc^2)" having both mass and the gravitational constant. However, also technically, as asserted and demonstrated by W.D.Clinger, it is an approximation of the Schwarzschild metric. So he is, as usual, just absconding with the math of GR while not understanding it and claiming GR is invalid in general. The thing is if he gave up all his other nonsense that just exacerbates the errors he'd find that the parts that work is just GR.
MRC_Hans
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But he can't even approximate the mass of a galaxy without using TR (and even then, not really, since he does not accept dark matter).
But that's how it is when you make stuff up as you go.
Hans
The Man
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Which of course would put his 'calculation' (provided he could get one with reasonable units) of the lensing by such a galaxy way off from the observational evidence. As such and other observational evidence is part of the evidence for dark matter.
Fact of the matter is, that he's using TR though poorly while bemoaning it in general. It's like riding around town, backwards, in a stolen BMW while shouting out the window that BMW's don't run.
Bjarne
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I did not wrote that, you did.
I wrote quote.
The following equation will shows you the relativistic stretch of space: 1/sqrt(1-2GM/(rc^2)
At the surface of the Syn u get 1/sqrt(1-2*6.67e-11*2e30/(7e8*3e8^2) = 1.000002117467 meter ( a stretch inwards to the sun at 0.000002117467 meter )
This is a ratio that shows how much space is stretched a given place.
*****************************************************************************************************************************************
The next step is now trying to understand how much that stretch of space, included photons "embedded" in that space, - will be reflected inward to a heavy astronomic object.
So fare we are not implying any observer at all, unless the observer is a man living at the surface of the sun - or in other words only an observer that can observe the distortion of space.
D is as mentioned the travel-distance light travels close by a heavy astronomic object .
Which is relevant if we assumes that a photon will be pulled inwards to the sun, due to the inwards stretch of space (calculated above) .
The inward path of a photon follows will of course continue to change direction / angle, - (notice in this context angle comes in) (relative to a straight line it would follow if there were no object nearby) - depending on how long a path it follows, in such stretches environment.
similar to a stone accelerering down on the Sun or inwards in a galaxy.
You brag about how good a mathematician you are and how bad I am.
Don't you not think that you just have to go ahead and prove how smart you are, and thus calculate all these angles you serveral times derailed to, - and observers on Earth of course ?
No, you will and can not.
You do not know how to do it.
All you can do is criticize everything you want to misunderstand, right?
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The Man
Unbanned zombie poster
Fantastic, so how exactly does one determine the value of "D"? Which in at least on example of yours was the radius of the sun. Then how does one get the observed deflection from that to compare to observational results?
How do you determine what area constitutes "close by a heavy astronomic object"?
As your space around such a "heavy astronomic object" is 'stretched' would that make you "D" length of space 'stretched' as well? Further if your "D" length of space is 'stretched' towards the sun wouldn't that make it curved?
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W.D.Clinger
Philosopher
True. I did write that. It seemed to me to be the most reasonable guess as to what you meant by D. In the post that began with the sentence quoted above, you went on to write
If the phrase I highlighted does not mean the distance from lensing object to observer, then I still don't know what you mean by "D is as mentioned the travel-distance light travels close by a heavy astronomic object". I hope you will eventually find a way to communicate what you mean by that sentence.
Yes, you did write that, complete with misspellings and your incorrect assertion that the result (1.000002117467) was in units of meters. I assumed the misspellings and incorrect units were typos/mistakes. If you insist we read exactly what you write without trying to guess what you meant when you wrote it, then there won't be much for us to say beyond the fact that most of what you write is jibber-jabber.
Actually, the deflection angle is essentially independent of the distance from source to lensing object and from lensing object to observer. That's because the deflection angle is almost completely determined by the part of the photon's path in which it is quite close to the lensing object, and that part of the path is very short compared to the two distances I mentioned in the previous sentence.
The deflection angle is determined by how closely the photon passes by the lensing object, by the mass of the lensing object, and by the shape of the lensing object. (In many cases, as with the sun, the shape of the lensing object is so close to being a ball that we can approximate it as such with negligible error.)
From your statements in multiple posts, it is apparent you do not understand that the deflection angle is the same regardless of the distance from lensing object to observer. You also do not understand that the deflection distance and ratio depend upon the distance from lensing object to observer. Your lack of understanding of those facts remains apparent even if we accept that you have never considered that distance in your calculations; indeed, never taking account of that distance in your calculations of deflection distance is what made it so obvious that you do not understand that the deflection distance and ratio depend upon that distance.
You don't have to be a good mathematician or a smart person to accept that the deflection angle is given by the formula derived over a century ago using the theory of general relativity.
Actually, Bjarne, I have cited and stated the correct formula for deflection angle several times within this thread. I have directed your attention to a book, written over a century ago, that contains that formula along with calculations using that formula. I have also given you a link to a historical/tutorial paper written at a level suitable for high school students, which contains several modern observational tests of deflection angles predicted by that formula.
I have explained how to calculate the deflection angle from the deflection distance and distance from lensing object to observer, and I have performed that calculation using your numbers. I have also explained how to calculate the deflection distance from the deflection angle and distance from lensing object to observer.
Question: Why are you not aware that I and quite a few other participants in this thread know how to compute the deflection angle for light bending around the sun, and for many other examples of gravitational lensing?
Answer: Because you don't know what you're doing, and are just faking it.
Bjarne
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Try this 90*60*60/(r/dd) ---- dd = deflection distance near the object, what do you think you get then ?
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The Man
Unbanned zombie poster
Looks like what you get is just more numerology.
r would seem to be some Radius, but radius of what and why?
How is dd "deflection distance near the object" measured or represented. If it is just radially from the center of the object with the origin at the closest approach then a +dd just means it is dd further away from the origin while a -dd would be closer. Naturally ones coordinate system can be set up so that -dd is further, + is closer and with the origin anywhere. However, your dd still has to be a displacement from a specific position in some direction. So where specifically is your dd displaced from.
90, 60 & 60 what and why?
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W.D.Clinger
Philosopher
What I think I'd get is a meaningless number.
If both r and dd are measured in meters, or any other units of length, then any number I'd get would be dimensionless.
It seems increasingly likely that your notion of "deflection distance" is neither of the two possible notions I described in a previous post. Fair enough. Guessing at what you might mean seems to irritate you, as though you are offended when someone actually tries to make sense of what you write, so why don't you just explain exactly what you mean by "deflection distance"?
In other words: what The Man said.
Bjarne
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The following equation will shows you the relativistic stretch of space: 1/sqrt(1-2GM/(rc^2)
At the surface of the Syn u get 1/sqrt(1-2*6.67e-11*2e30/(7e8*3e8^2) = 1.000002117467 meter
This reveals how much at such position is a stretching inwards to the sun
The distortion is as you (hopefully) can understand 0.000002117467 meter
It is very likely that when matter pulls / stretches space, also light that moves in the space, - will follow.
Therefore, regardless of whether the path where the light travels will be 1mm, 1 meter or 1 light year, - then the ratio of the deflection will be the same, (assuming the data of equations are the same all the way.)
Therefore just past the sun + a little extra for fade in / out the total reflection will be based on the equation:
td*dr
- td=travel distance near the astronomic object
- dr=deflection ration (in this case of the surface of the sun)
And now the equation to figure out how that will affect the angle of the path, relative to an expected straight path near the sun, Ok ?
This is the question you got yesterday (and as expected not understood) :
90*60*60/(r/dtd)
- 90*60*60 = degree*minutes*arcsecond
- r = radius from path of the photon to center of the Sun
- dfd = deflected travel distance
Which mean
90*60*90/(7e8/2500)= 1.735 arc seconds
So what is your problem ?, - and what is way off?
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Ziggurat
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Well, one problem is that there's only 60 arcseconds in an arcminute. If you don't magically switch from 60 to 90 for no reason, you don't get 1.735, you get 1.157.
I'm impressed that you're actually doing math. I'm not surprised you're doing it wrong, but I didn't even think you would try. So I was wrong about that.
MRC_Hans
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W.D.Clinger
Philosopher
After Bjarne has had an opportunity to answer the following questions, I will list some of Bjarne's more fundamental errors in a later post.
The relatively superficial questions below involve what appears to be Bjarne's slapdash presentation and numerology. (Bjarne's numerology appears to be a dishonest attempt to arrive at numbers close to those observed, but that presumption could be overcome if Bjarne gives us a plausible account of how Bjarne arrived at his various numbers.)
Please explain where you got the number 1.185e9. I don't recall seeing that number in any of your previous calculations, and I can't think of any physical constant or relevant measurement that yields that number.
I assume you meant "ratio" when you wrote "ration". Please confirm. (If you really meant "ration", then you are introducing yet another novel and as yet unexplained quantity into your calculations.)
Why, Bjarne, did your calculation deviate from your own formula by 50%?
(I assume you meant "dtd" when you wrote "dfd". Please confirm. If you really meant "dfd", and are using "dfd" to stand for "deflected travel distance", then what does the "dtd" in your formula mean?)
The relatively superficial questions below involve what appears to be Bjarne's slapdash presentation and numerology. (Bjarne's numerology appears to be a dishonest attempt to arrive at numbers close to those observed, but that presumption could be overcome if Bjarne gives us a plausible account of how Bjarne arrived at his various numbers.)
Bjarne, you persist in writing the highlighted words in this calculation. How did you come up with those units?
I'm sorry, Bjarne, but I still don't understand what you mean by "travel distance near the astronomic object". Please explain what that means.
Please explain where you got the number 1.185e9. I don't recall seeing that number in any of your previous calculations, and I can't think of any physical constant or relevant measurement that yields that number.
I assume you meant "ratio" when you wrote "ration". Please confirm. (If you really meant "ration", then you are introducing yet another novel and as yet unexplained quantity into your calculations.)
As Ziggurat noted, your formula says "60" twice where your calculation says "90":
Why, Bjarne, did your calculation deviate from your own formula by 50%?
(I assume you meant "dtd" when you wrote "dfd". Please confirm. If you really meant "dfd", and are using "dfd" to stand for "deflected travel distance", then what does the "dtd" in your formula mean?)
acbytesla
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I can't believe this conversation is still going on. Bjarne clearly hasn't presented the math to disprove Relativity. That he is trying to disprove Einstein by arguing in a Skeptics forum as opposed to publishing in a Physics or Astronomy Science Journal is telling.
This is such a waste of time.
This is such a waste of time.
Bjarne
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None og the formulars are perfect, just rough approximations.
acbytesla
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