KJW

As I see it, the mathematics of general relativity doesn't describe the "underlying 4D manifold" itself. Even the overlayed coordinate system, though referenced by some of the equations, is not explicitly defined. I believe you see this as a bad thing... I do not. Over the coordinate system are two types of fields: (1), arbitrary fields of various types whose purpose is to provide a framework for the equations, for example the coordinate transformation law for tensors, where the tensor field is an arbitrary field; and (2), fields that implicitly define the coordinate system and the "underlying 4D manifold" itself, providing structure. However, particular metrics such as the Schwarzschild metric are expressed using coordinate variables that bear some resemblance to physical variables. In the Einstein equation, only the specification of the source term (the energy-momentum tensor field) provides any definition of the coordinate system specified by the metric tensor solution. So, if one has two distinct physical energy-momentum density distributions that have the same mathematical source term in terms of their respective coordinate system variables, then the metric tensor solutions in terms of their respective coordinate system variables will also be the same, differing only in the background metric specified by the auxiliary conditions. I regard this as a form of Mach's principle within general relativity. An example of this is frame-dragging, where the rotation of a star or planet tends to drag the surrounding spacetime around with it. Yes I agree geocentric coordinate system can be constructed within GR and will remain coherent. Similar way how Geocentric model was agreeing with observations thruogh epicycles adding new levels of mathematical complexity. This example clearly shows the value of ontological transparency and mathematical simplicity in physics. It leads directly to the "Operational differences" I pointed above.. While physicists will typically choose the simplest coordinate system in which to investigate a problem, being the simplest coordinate system does not in any way make this coordinate system any more valid than any other coordinate system. Simplicity or complexity are not criteria for determining the validity of a coordinate system. All coordinate systems are equally valid. As you know, reality doesn't come with a coordinate system. Coordinate systems are imposed by humans. But in fact, coordinate systems are also a theoretical notion and as such, we can speak of "all possible coordinate systems". And because reality doesn't come with a coordinate system, there is no basis for a preferred coordinate system. General relativity is about having equations that are the same in all coordinate systems. Thus, the equations of general relativity do not select a preferred coordinate system, or indeed any coordinate system. But it is not trivial that the equations of general relativity are the same in all coordinate systems. I mentioned earlier that tensor equations are the same in all coordinate systems. However, not all quantities are tensors (not all quantities obey the coordinate transformation law for tensors). That is no different from general relativity. However, when physicists refer to mass, they are referring to it in units of kilograms (or some other comparable units). And it is desirable to determine the spacetime curvature associated with mass specified in kilograms. To do this, it is necessary to use the measured value of G from the Cavendish experiment and apply Newtonian gravitation. Current technology is not sufficiently accurate to directly measure the spacetime curvature associated with kilogram masses. It is worth noting that for astronomical work, the quantity GM (m3 s–2) is far more accurately known than either G or M alone. And using the Schwarzschild radius to specify the mass of a star or planet does allow one to avoid G because the mass is no longer specified in kilograms. Time dilation relates to acceleration and therefore doesn't involve mass. This argument only proves my point. I don't see how. By comparing general relativity in the weak-field limit with Newtonian theory, I'm not in any way weakening general relativity or making it an approximation. You seem to be suggesting that general relativity is an approximation, and that your theory deviates from general relativity in a way that better agrees with empirical observation. This is a claim that needs justification.

In such a group, the strong can overpower the weak, but only in a society that has legal and other systems to protect wealth can someone be billions of times wealthier than someone else. Otherwise, a group can only possess the wealth that they can physically protect.

I didn't comment on the content of what you said. I just questioned why you included the link to the image when it served no purpose other than to induce the forum members to request that you post the image. But as for the content, it seems to lack the substance required for it to be posted as "Linear Algebra and Group Theory".

This seems to be nothing more than what you wrote in the opening post. So why include a link to this as if it were something more?

Yeah, I saw a documentary a while ago in which it said that the so-called "land of the free" has 5% of the world's population but 25% of the world's prisoners.

It’s also worth noting that the “proportionality constant” between geometry and energy, which GR must import from the Newtonian limit, arises automatically in RG from internal normalization. In GR this constant is written as: [math]\alpha = \frac{8\pi G}{c^2}.[/math] In RG the same quantity follows algebraically from the geometric normalization of energy density: [math]\rho_{\max} = \frac{c^2}{8\pi G r^2},[/math] hence [math]\alpha = \frac{1}{\rho_{\max}\,r^2}.[/math] This shows that the coupling between curvature and energy is not something imposed externally to fit Newtonian gravity - it’s already embedded in the relational structure itself. The gravitational constant G comes from Newtonian gravitation theory, so if this constant is anywhere in your theory, it came from Newtonian gravitation theory, even if indirectly. And if you avoid using this constant, then any mass or energy that is related to geometry can't be expressed in terms of kilograms or other familiar units of mass or energy. As I said, general relativity must reduce to Newtonian theory in the weak-field limit. So, comparing general relativity in the domain where the relationship between geometry and energy has already been empirically established allows one to extend this relationship to the domain where general relativity is on its own. It is not about accepting Newtonian theory as an ontologically correct theory but accepting Newtonian theory as an empirically accurate theory within its domain of applicability.

I understand your point. And its seems reasonable at first. But that line of reasoning is structurally identical to early geocentric arguments: “Look around you - it is obvious that the Sun goes around the Earth.” Both rely on perceptual immediacy rather than logical derivation. The fact that perception appears spatial does not logically justify the postulate of a background. It only proves that our measurements project into four observable parameters - not that the universe is a 4-dimensional container. You say this like you believe the geocentric viewpoint is invalid. In general relativity, one can construct a geocentric coordinate system, and such a coordinate system is just as valid as any other coordinate system, in accordance with the principle of general relativity. What you are essentially saying is that some observations are invalid because they don't conform to some theoretical framework. However, it's important to note that any geocentric coordinate system needs to be constructed in accordance with general relativity, and in such a case will never disagree with observation in any way that a heliocentric coordinate system agrees with observation. So, it isn't a case of heliocentrism verses geocentrism because general relativity isn't choosing one over the other. Thus, I put it back to you: How can the observation of a space in which we exist be invalid?

I don't. I do visit RationalWiki often, though it's been a while since I last visited it. Thanks for letting me know, I'll keep an eye on it.

This assumes that the spammer is a persistent person who will give their info/email, wait for a response, then flood us with spam, only to get immediately banned. A bot, or a 'casual' spammer, will not; they just visit random sites and spam at will. I am a user of two other forums, which require an admin to email you back before you are allowed to be a posting member ( or sometimes even see additional content ) There is a compromise (assuming it is available in the forum software): Simply require a new member to wait say 24 hours after joining before they can post. In that case, the moderators don't have to do anything because there wouldn't be a basis to reject membership anyway.

I was focusing on the mathematics. You said general relativity is an incredibly complex theory and that you're not sure you'll ever have a proper understanding of it. I don't think it is an especially difficult theory, but it does require a level of mathematical skill that I cannot assume you possess. General relativity is based on calculus, so it is necessary to have skills in that mathematical subject. Also, because spacetime is four-dimensional, the calculus one needs is multivariable calculus. The two chain rules of multivariable calculus is at the bare minimum of the skills required as these deal with the notion of coordinate transformations, a key concept in general relativity. My post so far has only dealt with the concept of tensors. Although modern mathematics seeks to remove the notion of coordinates, they are actually key to the concept of tensors, which is not about going beyond scalars, vectors, and matrices, but about the notion of covariance and how the coordinate transformation law manifests that. I touched upon the machinery of tensors when I mentioned dummy indices and the Einstein summation convention, as well as the distinction between superscript indices and subscript indices with regards to the coordinate transformation law. But I have yet to even start upon general relativity itself. I would say that general relativity doesn't really start until the concept of "covariant derivatives" are introduced. This leads to the notion of a "connection field", and subsequently to the notion of a "curvature field". Getting back to calculus, suppose one is given a vector field [math]V_j(x^1,...,x^n)[/math], and we enquire: Does there exist a scalar field [math]\phi(x^1,...,x^n)[/math] such that: [math]V_j = \dfrac{\partial \phi}{\partial x^j}[/math] is satisfied? How would one determine if [math]V_j(x^1,...,x^n)[/math] is the partial derivative of a scalar field [math]\phi(x^1,...,x^n)[/math]? We know that the partial derivative of [math]\phi(x^1,...,x^n)[/math] satisfies: [math]\dfrac{\partial^2 \phi}{\partial x^k \partial x^j} - \dfrac{\partial^2 \phi}{\partial x^j \partial x^k} = 0[/math] and therefore it is necessary for [math]V_j(x^1,...,x^n)[/math] to satisfy: [math]\dfrac{\partial V_j}{\partial x^k} - \dfrac{\partial V_k}{\partial x^j} = 0[/math] If [math]\dfrac{\partial V_j}{\partial x^k} - \dfrac{\partial V_k}{\partial x^j} = F_{jk} \ne 0[/math], then this is an obstruction to the existence of scalar field [math]\phi(x^1,...,x^n)[/math]. The non-zero Riemann curvature tensor field is also an obstruction to the existence of a solution to a particular system of partial differential equations corresponding to a coordinate transformation. The necessary condition is straightforward; the sufficient condition is less straightforward. Look around you. It is clear that we exist in a space that is at least four-dimensional. Even if we can't derive this space logically from first principles, we can assume its existence on the basis of empirical observation. Although it is assumed that a differentiable manifold exists and that one can overlay a coordinate grid onto it, the mathematics itself does not indicate a priori the physical nature of the differentiable manifold nor provide any physical interpretation of the coordinate variables. In the mathematics I presented in this and my previous post, I made no mention of a metric, nor did I even specify the dimensionality of the space, simply indicating it as [math]n[/math]. In the mathematics of tensor calculus, most formulae are independent of the dimensionality of the space, but some formulae do depend on the dimensionality of the space, so it's a good idea to work in [math]n[/math]-dimensional space, even if general relativity itself is four-dimensional. And why spacetime is four-dimensional is an interesting question. It turns out that four-dimensional spaces have special properties not possessed by other-dimensional spaces, though whether these answer the question of why spacetime is four-dimensional is debatable. The way I personally see it, the whole of physics is based on the question of whether or not one description of reality is describing the same reality as another description of reality. This involves two ideas: (1), that there are many different ways to describe the same reality; and (2), that there are different possible realities which can be distinguished by their descriptions even though different descriptions are not necessarily describing different realities. Special relativity deals with the notion that the measured values of various quantities depend on the frame of reference from which it is measured. So, the notion of coordinate transformations does have a physically measurable consequence, although not all coordinate systems correspond to frames of reference. However, the set of all possible coordinate systems is a theoretical notion that identifies a particular reality and distinguishes it from a different reality.

It should be noted that a mixture of solid potassium permanganate and glycerol will burst into flames, so potassium permanganate can be considered a fire hazard with regards to contact with organic substances and other readily oxidisable materials.

\mathfrak{e}^{\alpha_1 ... \alpha_n} = \mathfrak{e}_{\alpha_1 ... \alpha_n} = \left\{\begin{array}{cl} 1 & \text{if}\ \ \alpha_1 ... \alpha_n\ \ \text{is an EVEN permutation of}\ \ 1 ... n\\-1 & \text{if}\ \ \alpha_1 ... \alpha_n\ \ \text{is an ODD permutation of}\ \ 1 ... n\\0 & \text{if}\ \ \alpha_1 ... \alpha_n\ \ \text{is NOT a permutation of}\ \ 1 ... n\end{array}\right.

GR is incredibly complex theory and there's so much remains for me to learn. I don't claim that I have a "proper" understanding of it. In fact Im not even sure that I ever will. Are you aware of the two chain rules in multivariable calculus? [math]\dfrac{d\bar{x}^\mu}{d\xi} = \displaystyle \sum_{\nu=1}^{n} \dfrac{dx^\nu}{d\xi} \dfrac{\partial \bar{x}^\mu}{\partial x^\nu}\ \ \equiv\ \ \dfrac{d\bar{x}^\mu}{d\xi} = \dfrac{dx^\nu}{d\xi} \dfrac{\partial \bar{x}^\mu}{\partial x^\nu}[/math] [math]\dfrac{\partial \psi}{\partial \bar{x}^\mu} = \displaystyle \sum_{\nu=1}^{n} \dfrac{\partial \psi }{\partial x^\nu} \dfrac{\partial x^\nu}{\partial \bar{x}^\mu}\ \ \equiv\ \ \dfrac{\partial \psi}{\partial \bar{x}^\mu} = \dfrac{\partial \psi }{\partial x^\nu} \dfrac{\partial x^\nu}{\partial \bar{x}^\mu}[/math] where the second expression on each line uses the "Einstein summation convention" (repeated indices, called "dummy indices", [math]\nu[/math] in this case, indicate an implicit summation over those indices). Note that: [math]\dfrac{\partial \bar{x}^\sigma}{\partial x^\nu} \dfrac{\partial x^\nu}{\partial \bar{x}^\tau} = \dfrac{\partial \bar{x}^\sigma}{\partial \bar{x}^\tau} = \delta^\sigma_\tau[/math] [math]\dfrac{\partial x^\varrho}{\partial \bar{x}^\mu} \dfrac{\partial \bar{x}^\mu}{\partial x^\gamma} = \dfrac{\partial x^\varrho}{\partial x^\gamma} = \delta^\varrho_\gamma[/math] where [math]\delta^\sigma_\tau[/math] is called the "Kronecker delta" and is the identity matrix, and therefore [math]\dfrac{\partial \bar{x}^\sigma}{\partial x^\nu}[/math] and [math]\dfrac{\partial x^\nu}{\partial \bar{x}^\tau}[/math] are the inverse matrix of each other. So, what is called a "coordinate transformation" in physics is a "change of variable" in calculus. The chain rule for [math]\dfrac{d\bar{x}^\mu}{d\xi}[/math] is the archetype for the coordinate transformation law for contravariant vectors in general: [math]\bar{V}^\mu = V^\nu \dfrac{\partial \bar{x}^\mu}{\partial x^\nu}[/math] The chain rule for [math]\dfrac{\partial \psi}{\partial \bar{x}^\mu}[/math] is the archetype for the coordinate transformation law for covariant vectors in general: [math]\bar{V}_\mu = V_\nu \dfrac{\partial x^\nu}{\partial \bar{x}^\mu}[/math] This leads to the coordinate transformation law for the general absolute¹ tensor: [math]\bar{T}^{\sigma...\varrho}_{\tau...\gamma} = T^{\alpha...\beta}_{\mu...\nu}\ \dfrac{\partial \bar{x}^\sigma}{\partial x^\alpha}...\dfrac{\partial \bar{x}^\varrho}{\partial x^\beta}\ \dfrac{\partial x^\mu}{\partial \bar{x}^\tau}...\dfrac{\partial x^\nu}{\partial \bar{x}^\gamma}[/math] The coordinate transformation law for tensors guarantee that tensors have the following two important properties: 1: A tensor that is zero in any coordinate system is zero in every coordinate system. 2: A tensor equation that is true in any coordinate system is true in every coordinate system. ¹ There are also relative tensors of various weights for which the coordinate transformation law has an extra factor, emerging naturally as a result of the nature of determinants.

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Ewww! That's not something I would want to watch. 🤮

@Anton Rize, it seems to me that you do not have a proper understanding of general relativity. For example, you talk about spacetime being an independent background that you reject, whereas in general relativity, spacetime is not an independent background. For the Einstein equation, given the energy-momentum density distribution, one obtains the metric tensor field as a solution. The metric tensor field contains the information that determines both the energy-momentum density field and the gravitational field. Yes, one often considers trajectories of test masses in spacetime. That's because the definition of a test mass is a point-like particle whose mass is small enough to have negligible effect on the spacetime. This allows the trajectory of the test mass to be determined against the background spacetime. But that is a mere simplification of the mathematics. It works adequately for planets orbiting a star. But the mutual orbits of black holes around each other cannot be treated in this manner and can only be determined as the dynamics of the spacetime itself. Don't confuse the mathematical simplifications that physicists may make with the inadequacy of the theory. You also mentioned asymptotically flat spacetime. It should be noted that general relativity must reduce to Newtonian theory in the weak-field limit as indeed it does. Also, the Einstein equation is a relationship between the mathematical realm of Ricci calculus and the physical realm. In order to establish how much spacetime is curved by energy-momentum density, one needs to consider Newtonian theory in the weak-field limit. This provides the proportionality constant between the Einstein tensor and the energy-momentum density tensor in the Einstein equation. Also, the foundations of general relativity are more fundamental than you seem to think, and indeed more fundamental than you seem to indicate concerning your theory. For example, you treat energy as foundational, whereas the Einstein tensor field that is equivalent to the energy-momentum density field is derived.

That's not necessarily true. Bear in mind that Einstein did not have the Schwarzschild metric when he published general relativity and the anomalous precession of the perihelion of Mercury. I have read that the value of the precession was obtained as the non-conservation of the Laplace-Runge-Lenz vector, although it is not clear to me how the non-Newtonian potential ¹ was obtained without the Schwarzschild metric. I'm currently working on a problem that involves the Schwarzschild metric, and that problem turned out to be mathematically rather complicated rather quickly. It seems to me that the mathematics could be simplified by using a weak-field approximation. I wasn't suggesting that the Schwarzschild metric was an approximation of the physical solar system, even though the anomalous precession of the perihelion of Mercury is only a part (about 7.5%) of the total precession of the perihelion of Mercury. ¹ The Laplace-Runge-Lenz vector is a constant of motion under an exact inverse-square force law. That not a big achievement. The curvature field of the solar solar system is quite small and only a first-order GR effect would be significant. The approximation I was suggesting would be fine for Mercury around the Sun but may be inadequate for a close orbit around a black hole or neutron star.

Thanks. However, you didn't mention the physics that is described by the formula: [math]\text{Perihelion shift:}\ \ \ \sigma = \dfrac{2\pi Q^2_{\text{merc}}}{1 - e^2_{\text{merc}}}[/math] In particular, in what way does the derivation of this formula indicate that it is describing a perihelion shift? I'm still not convinced about your overall theory. I would like to explore how it is that these two quantities: [math]\beta^2 = \dfrac{v^2}{c^2}\\\kappa^2 = \dfrac{2GM}{c^2 r}[/math] lead to the known formula for the perihelion shift. One thing to note is that: [math]\sqrt{1 - \beta^2}[/math] is the time dilation associated with speed, and that: [math]\sqrt{1 - \kappa^2}[/math] is the time dilation associated with gravitation. The total time dilation for both speed and gravitation: [math]\sqrt{1 - \beta^2} \sqrt{1 - \kappa^2}[/math] [math]= \sqrt{(1 - \beta^2) (1 - \kappa^2)}[/math] [math]= \sqrt{1 - (\beta^2 + \kappa^2) + \beta^2 \kappa^2}[/math] [math]\approx \sqrt{1 - (\beta^2 + \kappa^2)}\ \ \ \text{ for } \beta \approx 0 \text{ and } \kappa \approx 0[/math] [math]= \sqrt{1 - Q^2}[/math] If [math]v[/math] is the orbital speed, then: [math]\beta^2 \approx \dfrac{GM}{c^2 r} = \dfrac{1}{2} \kappa^2[/math] and therefore: [math]Q^2 = \beta^2 + \kappa^2 = \dfrac{3}{2} \kappa^2 = \dfrac{3GM}{c^2 r}[/math] I note that [math]\beta^2 + \kappa^2[/math] is a weak field approximation. However, it is not clear to me the extent to which the standard formula for the perihelion shift is an approximation. I find it interesting that the standard formula for the perihelion shift gives a non-zero value for circular orbits. Although I expect this value to be a limit value for infinitesimal eccentricity, it is not clear to me if it is the same value as given by the deviation from Newtonian theory.

Thanks. However, you didn't mention the physics that is described by the formula: [math]\text{Perihelion shift:}\ \ \ \sigma = \dfrac{2\pi Q^2_{\text{merc}}}{1 - e^2_{\text{merc}}}[/math] In particular, in what way does the derivation of this formula indicate that it is describing a perihelion shift? I'm still not convinced about your overall theory. I would like to explore how it is that these two quantities: [math]\beta^2 = v^2/c^2\\\kappa^2 = 2GM/(c^2 r)[/math] lead to the known formula for the perihelion shift. One thing to note is that: [math]\sqrt{1 - \beta^2}[/math] is the time dilation associated with speed, and that: [math]\sqrt{1 - \kappa^2}[/math] is the time dilation associated with gravitation. The total time dilation for both speed and gravitation: [math]\sqrt{1 - \beta^2} \sqrt{1 - \kappa^2}[/math] [math]= \sqrt{(1 - \beta^2) (1 - \kappa^2)}[/math] [math]= \sqrt{1 - (\beta^2 + \kappa^2) + \beta^2 \kappa^2}[/math] [math]\approx \sqrt{1 - (\beta^2 + \kappa^2)}\ \ \ \text{ for } \beta \approx 0 \text{ and } \kappa \approx 0[/math] If [math]v[/math] is the orbital speed, then: [math]\beta^2 \approx \dfrac{GM}{c^2 r} = \dfrac{1}{2} \kappa^2[/math] and therefore: [math]\beta^2 + \kappa^2 = \dfrac{3}{2} \kappa^2 = \dfrac{3GM}{c^2 r}[/math] I note that [math]\beta^2 + \kappa^2[/math] is a weak field approximation. However, it is not clear to me the extent to which the standard formula for the perihelion shift is an approximation. I find it interesting that the standard formula for the perihelion shift gives a non-zero value for circular orbits. Although I expect this value to be a limit value for infinitesimal eccentricity, it is not clear to me if it is the same value as given by the deviation from Newtonian theory.

For quite some time, I have been wondering what would happen if the blue states seceded from the US to Canada.

@Anton Rize , the "math" tags and everything between them needs to be on the same line, though the line is allowed to be long enough for the text to wrap.

𝐃𝐢𝐦𝐞𝐧𝐬𝐢𝐨𝐧𝐥𝐞𝐬𝐬 𝐩𝐫𝐨𝐣𝐞𝐜𝐭𝐢𝐨𝐧 𝐩𝐚𝐫𝐚𝐦𝐞𝐭𝐞𝐫𝐬 𝐟𝐨𝐫 𝐌𝐞𝐫𝐜𝐮𝐫𝐲: κₘₑᵣ꜀ = √(Rₛₛᵤₙ / aₘₑᵣ꜀) βₘₑᵣ꜀ = √(Rₛₛᵤₙ / 2aₘₑᵣ꜀) 𝐂𝐨𝐦𝐛𝐢𝐧𝐞𝐝 𝐞𝐧𝐞𝐫𝐠𝐲 𝐩𝐫𝐨𝐣𝐞𝐜𝐭𝐢𝐨𝐧 𝐩𝐚𝐫𝐚𝐦𝐞𝐭𝐞𝐫: Qₘₑᵣ꜀ = √(κₘₑᵣ꜀² + βₘₑᵣ꜀²) 𝐂𝐨𝐫𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐟𝐚𝐜𝐭𝐨𝐫 𝐟𝐨𝐫 𝐭𝐡𝐞 𝐞𝐥𝐥𝐢𝐩𝐭𝐢𝐜 𝐨𝐫𝐛𝐢𝐭 𝐝𝐢𝐯𝐢𝐝𝐞𝐝 𝐛𝐲 𝐨𝐧𝐞 𝐨𝐫𝐛𝐢𝐭𝐚𝐥 𝐩𝐞𝐫𝐢𝐨𝐝: (1 − eₘₑᵣ꜀²) / 2π 𝐅𝐢𝐧𝐚𝐥 𝐖𝐈𝐋𝐋 RG 𝐩𝐫𝐞𝐜𝐞𝐬𝐬𝐢𝐨𝐧 𝐫𝐞𝐬𝐮𝐥𝐭: Δφ₍WILL₎ = (Qₘₑᵣ꜀²) / [(1 − eₘₑᵣ꜀²) / 2π] = (2πQₘₑᵣ꜀²) / (1 − eₘₑᵣ꜀²) I didn't look at the link because I assume that (as required by the forum rules) you've posted all that I requested. I don't see any of the physics that your formula is describing. Nevertheless, I am quite intrigued by the maths. I checked your maths to see if your claim is valid: [math]\text{From above:}\ \ \ Q^2_{\text{merc}} = \dfrac{3r_{\text{s sun}}}{2a_{\text{merc}}} = \dfrac{3GM_{\text{sun}}}{c^2 a_{\text{merc}}}[/math] [math]\text{From Kepler's third law:}\ \ \ GM_{\text{sun}} = \dfrac{4\pi^2 a^3_{\text{merc}}}{T^2_{\text{merc}}}[/math] [math]Q^2_{\text{merc}} = \dfrac{12\pi^2 a^2_{\text{merc}}}{c^2 T^2_{\text{merc}}}[/math] [math]\text{Perihelion shift:}\ \ \ \sigma = \dfrac{2\pi Q^2_{\text{merc}}}{1 - e^2_{\text{merc}}} = \dfrac{24\pi^3 a^2_{\text{merc}}}{c^2 T^2_{\text{merc}}(1 - e^2_{\text{merc}})}[/math] which agrees with the known formula. However, how do you justify the "correction factor for the elliptic orbit" [math](1 - e^2_{\text{merc}})/2\pi[/math]? [If the above LaTeX doesn't render, please refresh the webpage.]

No. What matters more than calculated results is how one obtains one's formulae. Please explain how you obtained this formula: [math]\Delta_{WILL}=\dfrac{2\pi Q_{Merc}^{2}}{\left(1-e_{Merc}^{2}\right)}[/math] [If the above LaTeX doesn't render, please refresh the webpage.]

Actually, the 2,2-dimethylpropyl radical is also known as the neopentyl radical. Attached to some leaving group, it is notorious for its unreactivity towards nucleophilic substitution reactions. The steric hinderance of the three methyl groups on the beta carbon atom strongly inhibits SN2 reactions, and because the leaving group is attached to a primary carbon atom, SN1 reactions are also very slow (and if forced will lead to rearrangement to the more stable tertiary carbocation).

The formulae refer to open-chain unbranched or branched hydrocarbons. Cyclic hydrocarbons have different formulae. Each cycle (the number of cycles is defined as the number of bonds that need to be cut to form an acyclic structure¹) removes two hydrogen atoms. However, even in the case of cyclic hydrocarbons, each double bond removes two hydrogen atoms, and each triple bond removes four hydrogen atoms. ¹ Thus norbornane C7H12 is a bicyclic hydrocarbon, even though one might think it has three cycles: