gravitational wavefunction

[chemguy]

An acceleration field (not necessarily gravitation) may be represented by two vectors.

 

A position vector identifies some point in space, and an acceleration vector will associate acceleration with the selected point. The acceleration vector must be “linked” to the position vector. This linkage is achieved by relating components of the acceleration vector to components of the position vector.

 

The two vector representation of a field leads to a gravitational wave function.

 

Please view;

http://doulting.shawwebspace.ca/asset/view/7853/gravitational_wave_function.pdf

 

[imatfaal]

Forgive me but I don't understand how the acceleration vector can be split into just two components. From my early reading you seem to split the acceleration into a radial/centripetal (for some reason also called normal but I am unsure what line it is normal to) and tangential components. I cannot see how you can use just two variables to describe the acceleration - three sure (either i,j,k or magnitude, phi, theta) but I don't see two doing the job

[studiot]

 

imatfaal

Forgive me but I don't understand how the acceleration vector can be split into just two components. From my early reading you seem to split the acceleration into a radial/centripetal (for some reason also called normal but I am unsure what line it is normal to) and tangential components. I cannot see how you can use just two variables to describe the acceleration - three sure (either i,j,k or magnitude, phi, theta) but I don't see two doing the job

 

 

Perhaps the assumption that the acceleration is always in a plane through the centre from which the position vector r is measured should have been stated explicity.?

 

Perhaps more interesting, is the assumption

 

chemguy : Link

One condition applies; X32 = ct

Giving; v32 = ∂X32/∂t = c

 

Since c is a constant we have

 

[math]\frac{{{\partial ^2}{X_{32}}}}{{\partial {t^2}}} = 0[/math]

 

 

In other words the acceleration component in the third dimension is zero.

 

But at the expense of a constant light speed in the third dimension.

 

Perhaps some additional words of explanation would not go amiss?

 

Perhaps also using g for the generalised acceleration rather than another symbol will lead to confusion with gravity?

 

Edited March 30, 2015 by studiot

[swansont]

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