Help with McMahon's Relativity

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In summary: When calculating the Cartan coordinate free geodesic acceleration between two geodesics when the Lie bracket is zero (flat) how do you get the second total derivative with respect to tau as the product of tangent vector grad (tangent vector grad displacement vector).There is no simple answer to this question. The most general approach would be to use the Lie bracket operator between the two geodesics, and then use the divergence theorem to calculate the second total derivative.
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mikah
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Please help a newbie? I am having trouble with the following:
1) Lamda tensor metric conversion between local non-holonomic basis and global holonomic basis, how can you just "read off" the diagonal from the line element say in spherical coordinates and when do you choose the inverse metric (just take the inverse of each component of the diagonal?) when converting with either one-forms or vectors? pp 100 to 103.

2) Why did they choose to make the components of the tangent vector the partial diff operator making the vector an operator?

3) When calculating the Cartan coordinate free geodesic acceleration between two geodesics when the Lie bracket is zero (flat) how do you get the second total derivative with respect to tau as the product of tangent vector grad (tangent vector grad displacement vector). Do you take double Lie derivative on operators? pp 135

Thank you in advance. Mikah
 
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More people will be able to help you if you put your questions in context; for example, quoting the relevant part of the text. Otherwise, you may be waiting a while for someone with that book to come along.
 
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mikah said:
2) Why did they choose to make the components of the tangent vector the partial diff operator making the vector an operator?
Think about how to calculate the rate of change of a function [itex]f:\mathbb R^n\rightarrow\mathbb R[/itex] along a curve in [itex]\mathbb R^n[/itex]. Suppose that [itex]C:(a,b)\rightarrow\mathbb R^n[/itex]. Then the rate of change of f at C(t) is [itex](f\circ C)'(t)[/itex]. The chain rule now tells us that

[tex](f\circ C)'(t)=f,_\mu(C(t)) C'^\mu(t)=C^\mu'(t) D_\mu|_{C(t)} f[/tex]

Note that [itex]C^\mu'(t)[/itex] are the components of the tangent vector of the curve.

Now suppose that [itex]C:(a,b)\rightarrow M[/itex] and [itex]f:M\rightarrow\mathbb R[/itex]. The rate of change of f at C(t) is still [itex](f\circ C)'(t)[/itex], but now the first equality above doesn't make sense as it stands. We can however make sense of it using a coordinate system x:

[tex](f\circ C)'(t)=(f\circ x^{-1}\circ x\circ C)=(x\circ C)^\mu'(t)\frac{\partial}{\partial x^\mu}\bigg|_{C(t)}f[/tex]

Note the similarities between this and the previous result. In both cases, the rate of change of f at C(t) came out equal to a linear combination of derivative operators acting on f, and in both cases the coefficients are equal to the components of a tangent vector of a curve in [itex]\mathbb R^n[/itex]. Also note that if we think of the identity map I (defined by I(x)=x for all x) on [itex]\mathbb R^n[/itex] as a coordinate system on [itex]\mathbb R^n[/itex], then we can put the first calculation into exactly the same form as the second (with I instead of x).
 

FAQ: Help with McMahon's Relativity

What is McMahon's Relativity?

McMahon's Relativity is a theory proposed by physicist Desmond McMahon, which suggests that the laws of physics remain the same in all inertial frames of reference. It is based on the concept of relativity proposed by Albert Einstein.

How is McMahon's Relativity different from Einstein's Theory of Relativity?

While Einstein's Theory of Relativity focuses on the relationship between space and time, McMahon's Relativity focuses on the constancy of the laws of physics in different reference frames. Additionally, McMahon's theory does not incorporate the concept of gravity as a curvature of space-time.

What are the key principles of McMahon's Relativity?

The key principles of McMahon's Relativity include the constancy of the laws of physics in all frames of reference, the equivalence of all inertial frames, and the absence of absolute space and time.

Is McMahon's Relativity widely accepted in the scientific community?

No, McMahon's Relativity is a relatively new theory and is not widely accepted in the scientific community. It has not been extensively tested or proven, and many physicists still hold Einstein's Theory of Relativity as the most accurate and widely accepted theory of relativity.

How can McMahon's Relativity be applied in practical situations?

McMahon's Relativity has not yet been applied in practical situations, as it is still a developing theory. However, it has the potential to offer new insights and perspectives in the field of physics and may have practical applications in the future.

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