|
| If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. |
|
|||||||
| Register | FAQ | Members List | Calendar | Mark Forums Read |
![]() |
|
|
LinkBack | Thread Tools | Display Modes |
|
|||
|
Quote:
|
|
|||
|
Not true. It is quite consistent with the FLRW solution, though there now appears to be evidence of a non-zero cosmological constant.
|
|
|||
|
Last August there was a thread by Doug Sweetser; "Causality and the Quaternion Derivative":
Causality and the Quaternion Derivative It was a fairly lively thread and worth reading. Jim |
|
||||
|
Quote:
|
|
||||
|
Yup, I saw that in yawyaw's OP here:
Quote:
not to mention space-time curvature of GR.I don't think distant red shift at cosmological distances can be explained this easily, though I am personally of the opinion such red shift may be from other than Doppler space-expansion, so am open to suggestion. ![]() |
|
|||
|
Quote:
And now for an obvious question. It is not sufficient for a new theory to explain a single phenomena, such as the cosmological red-shift. If you have a new theory of gravitation then it should also explain: 1) The behaviour of the perihelion of Mercury, 2) Gravitational red-shift 3) Gravitational lensing, and the gravitational deflection of light by the Sun, as well as the more usual gravitational effects. Does your model do this? |
|
||||||||||||
|
Quote:
Quote:
Quote:
Quote:
Quote:
Quote:
Quote:
Quote:
Quote:
Quote:
Quote:
Quote:
Considering that you see an imaginary in the very real Einstein interval you quoted, I won't hold my breath.
__________________
papageno "Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes) "It's all about context!" - Vince Noir (The Mighty Boosh) "I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama) |
|
||||
|
Quote:
1) I have developed the orbit equation and get r=P(1 +ct/R)/(1 + ecos v) where P=R(sin(rv))^2. This may or may not answer to Mercury's perihelion. The vector equilibrium equation is 0=dcv/cdt + cDelxv t + ur/R3. 2 and 3) I think the Quaternion Theory handles the gravitational redshift and I am uninformed about gravitational lensing. On gravitational deflection by the sun, this Theory challenges Einstein's deflection. This theory clearly gives the deflection as cos(rv) = v/c=sqrt(GM/c^2R) = 1.457859E-3 giving a angle of 89degrees 54' 59.93". This gives a deflection angle of 5 arc minutes and 0.07 arc seconds. This is much larger than Einstein's 1.75' arc seconds. This difference raises issues. The computation here is the redshift deflection caused by gravity changing the angle of the velocity away from the sun. Is Einstein's deflection changing the angle away or towards the sun, a blue shift? Einstein computes the angle by the tangent. The cosine is the redshift angle, so is Einstein's angle a redshift deflection? Looking at Einstein's computation, the tangent is computed using the time across the diameter of the sun and the ratio of the deflection to the radius or half the denominator: Tan v =y/R=1/2gt^2/R should be the Tan. Einstein drops the 1/2 and doubles the time t=2R/c giving him a factor of 4 over what I think it should be and this accounts for the 1.75 instead of 0.44'. This gives Einstein a Tan= y/R=4GM/Rc^2=8.4656E-6. I think a case can be made for Tan=y/2R=GM/Rc^2=2.1164E-6 and angle 0.44'. There also is the question of the direction of the deflection is it a redshift or blueshift, away from the sun or towards the sun. I welcome your comments about the tangent and whether it is a red or blue shift. My inclination is to go with Quaternion Gravity. Could there also be a defect in the observation of Einstein's deflection and does the observation show redshift or blueshift away or towards the sun. Thanks again for your comments. |
|
||||
|
Quote:
I am not familiar with 4 vectors. From wiki 4-vector: inner product of u v=u0v0 -u1v1 -u2v2-u3v3 seems to be the same as the scalar product of quaternions. In quaternions uv=u0v0 -u1v1 -u2v2 -u3v3 +u0(v1 + v2 + v3) + v0(u1 + u2 =u3) + (u1 + u2 +u3)x(v1 +v2 +v3) . The rule for multiplying quaternions is AB=(Ar +Av)(Br + Bv)=(ArBr -Av.Bv) + (ArBv + AvBr + AvxBv) Where AvxBv = -BvxAv For spacetime position p=r + ix + jy + kz where r=ct p^2 = (r^2 -(x^2 + y^2 +z^2)) + 2r(ix + jy +kz) pp* =r^2 + x^2 + y^2 + z^2 where p* = r-(ix + jy + kz) Quaternions and 4-vectors seem similar but do 4-vector use Hamilton's rules for i^2=j^2=k^2=ijk=-1? |
|
|||
|
Quote:
Current measurements have verified GR to within about 2 percent of the 1.75 arc seconds. Your theory's prediction is outside of that uncertainty and hence, does not make an accurate prediction of the bending of light around the sun. I was going to ask how exactly does your theory tie into SR and how exactly does your theory allow covariance in physical laws. But, since it's prediction is so far outside the actual observations, on the bending of light around the sun, it's pretty much useless anyway.
__________________
Some try to tell me, thoughts they cannot defend,... - Moody Blues. |
|
||||
|
Quote:
This is not a gravity problem. Gravity has two aspects the real and the vector. The real is attractive and the vector is 'repulsive". Thanks for your comment. Last edited by yawyaw : 27-April-2008 at 04:51 PM. Reason: Forgot comment |
|
|||
|
Quote:
Quote:
![]() |
|
|||
|
Quote:
|