Mordred

Sigh no the SI units for G is [math] m^3\cdot kg^{-1}\cdot s^{-2}[/math] The SI unit for mass is kg. Radius obviously has units of length Those threeunits are on the RHS of the equal sign in formula one. On the left hand side you have the SI unit for time. Guess what it's seconds.( A single unit) So how can the left hand as side equal the right hand side when the units do not match regardless of value ? I have kg on the right hand side and units of length Yet on the left hand side I only have units of time. The right hand side does not cancel our the kg of meter units to leave only seconds. Please study the link I posted on dimensional analysis.

No how are you cancelling for example Kg in the first equation. You have no kg on the LHS. That's just one of the terms. Follow the procedures as per https://www.google.com/url?sa=t&source=web&rct=j&url=http://web.mit.edu/2.25/www/pdf/DA_unified.pdf&ved=2ahUKEwi6ybXastfpAhWTsJ4KHVrVCP8QFjAAegQIChAB&usg=AOvVaw3hTYBubBOwEQX7596gEKFR Easiest to convert to SI units.

These equations are invalid. If you perform a dimensional analysis you will find the LHS and RHS do not match.

Div operator is a short hand but you got the point. +1 An off topic side note the mass distribution is also how a matter only universe can expand. Which is a very tricky concept to understand. One would think a matter only universe would collapse. If you think about my last post one can see that as anistropies develop ie LSS and galaxy formation the density of the void regions decrease. In essence local gravitational anistropies aid expansion. However that's off topic...and involves the term and formula for critical density...

The Div operator is a vector. If you have a uniform matter field with no anistropies then you have no curvature and no gravity. Strange as this may be to understand but the term gravity is rather misleading. Marcus mentioned tidal forces. So let's add some detail. Take that uniform distribution. Now in that uniform distribution place two massive particles in free fall with an initial velocity or in this case momentum can be used ( momentum is the rest mass multiplied by the velocity) the two terms are not equal. The paths of those two particles do not change nor do they accelerate due to gravity. The field is uniform in distribution. Now take a region with spacetime curvature such as a planet. Drop the same two objects. The paths will converge toward the center of mass. So to understand the origin of gravity one can only answer the local anistropy regions of the mass distribution. The very term gravity is replaced by spacetime curvature. With EM the potential difference (voltage) induces current flow. The resistance restricts the current. Mass is resistance to inertia change or resistance to acceleration. So you can see the similarity. Photons do not couple with any field they interact with so have no invariant (rest mass). Though they do not couple they still interact with other particles in a region. We describe this interaction via the path of least action which a good study source would be Feymann path integrals. (The Feymann path integrals are also curved paths). Though gravity is not involved in the latter case. Another way to look at a uniform field of mass. (Which can be gained through all other fields in a multi particle system) Is take a multiparticle system uniformly distributed. You can arbitrarily choose any test particle in that field a the centre of mass then apply the shell theorem. However as any particle can be chosen with no difference you effectively have a scalar field. Gravity is at minimal treatment a vector field. (Attraction)

I think part of your confusion lies in that the Einstein field equations including the stress energy momentum tensor. Don't just describe spacetime curvature (gravity) but also describes how particles move in spacetime. The trick is the metric and Ricci tensor can both modify the stress energy tensor and vise versa. Remember the expression I gave mass tells spacetime how to curve spacetime tells matter/ particles how to move. The Einstein field equations cover both statements.

Since when provide a citation from. A peer reviewed source.

Equation 34 gives an example of the gravitational action. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.math.toronto.edu/mccann/assignments/426/DeGiuli.pdf&ved=2ahUKEwi_oLKjpcrpAhWEop4KHT1DAZMQFjACegQIAhAB&usg=AOvVaw15IQoP61wefGJ5yTd0oMlc I don't believe the OP is ready for the Langrangian at this time. However the article here provides some details.

A couple of points to add to the excellent post above. A scalar is rank 0. A vector is rank 1. A vector has both magnitude and direction. You need a higher rank when you require two vectors. Such as the example given by Markus. If I recall correctly the Kronecker delta function is also rank two. If I'm correct then hermitean groups would also be rank 2 but that's just a side note. The Poincare group is SO(3.1) which GR falls under. Which is a double cover [math]SU(2)\otimes SU(2)/\mathbb{Z}^2[/math] . So even in tensor ranks you require a minimal rank 2. Just to provide a tensor example. (Each of those groups is a tensor. The SO(3.1) is a 4×4 while each SU(2) is a 2×2 The Z parity operator is also 2×2. The proofs I have come across on rank 2 requirement were tensor related proofs. Which I looking for a more understandable example as they tend to be too complex for the average poster. Edit I did a quick search and I am correct the Kronecker delta function is a rank two tensor. https://mathworld.wolfram.com/KroneckerDelta.html So even a Cartesian space which uses the Kronecker delta function would be rank 2. The link above provides the differential geometry form using Euler angles. This link will show how Euler angles are employed and will understand their usage for different observers under different rotations. https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.weizmann.ac.il/sci-tea/benari/sites/sci-tea.benari/files/uploads/softwareAndLearningMaterials/quaternion-tutorial-2-0-1.pdf&ved=2ahUKEwid-fT1msrpAhU1KX0KHeAkDXQQFjABegQIAhAB&usg=AOvVaw07DuN5EIV2sfGXU19NMjbO It will also be a valuable tool to better understand rotations of the tensors. Such an example of tensor rotations is when you must rotate the Minkowskii or Lorentz tensor to describe acceleration (rapidity requires a rotation) or boost (A boost is also a type of rotation). Brian Crowell gave examples of each in that SR textbook I previously linked and provides some greater detail. A little side note the best tool to master GR is to study differential geometry. Once you understand differential geometry for Euclidean and curved geometries understanding GR8 becomes incredibly easy. (You won't even require Tensors ) they are a tool to handle multiple unknowns in essence an organization tool to keep track of multiple unknowns)

I know the OP probably won't understand this however it's informative. Here is how the metric tensor would look as a rotation of reference frames S and [math]\acute{S}[/math] around their common z axis in cylindrical coordinates. [latex]\eta=\begin{pmatrix}-(1-\frac{\omega^2r^2}{c^2})&0&\frac{\omega r^2}{c^2}&0\\0&1&0&0\\\frac{\omega r^2}{c^2}&0&r^2&0\\0&0&0&1\end{pmatrix}[/latex] Anyways this will give the time relation between proper time and coordinate time of [math]d\tau=\sqrt{1-\frac{\omega r^2}{c^2}dt}[/math] where [math]\tau[/math] is proper time. I too recall such a proof, if I can remember the source I will post it.

Not quite. You need to study the Principle of equivalence in regards to the time dilation of moving bodies. The shorthand for the principle is inertial mass is equivalent to gravitational mass. [math] m_i=m_g[/math] Relativistic moving bodies gain inertial mass (used to be called relativistic mass) however the current accepted term is the variant mass. The rest mass is now described as in invariant mass. (Mass all observers can agree on). The variant mass obviously depends on the observer. If you have any form of mass including gravitational mass then you will always have time dilation. However miniscule. To give you a clear example time dilation has been measured at a single meter difference in elevation. https://www.nist.gov/news-events/news/2010/09/nist-clock-experiment-demonstrates-your-head-older-your-feet For rotating bodies you need to study the Sagnac effect. https://en.m.wikipedia.org/wiki/Sagnac_effect Coincidently this is another test against the eather theories. PS glad to see your using the term fields as well as starting to look more into current studies of relativity. +1. This is an open source textbook on SR that the author wrote to cover many of the misconceptions common to science forums. The author is a PH.D who specializes in SR and GR. I have had numerous conversations with him over the last decade or so. http://www.lightandmatter.com/sr/

Correct now gravity is described under the Stress energy momentum tensor. This under GR is the portion that in essence curves the metric. So here is a simplified sample showing the Dust solution. Dust is matter only equation of state p=0. Dust solution no force acting upon particle. (No acceleration) [latex] T^{\mu\nu}=\rho_0\mu^\mu\nu^\mu[/latex] [latex]T^{\mu\nu}x=\rho_0(x)\mu^\mu(x)\mu^\nu(x)[/latex] Rho is proper matter density Four velocity [latex]\mu^\mu=\frac{1}{c}\frac{dx^\mu}{d\tau}[/latex] Leads to [latex]ds^2=-c^2d\tau^2=-c^2dt^2+dx^2+dy^2+dz^2=-c^2dt^2(1-\frac{v^2}{c^2})^\frac{1}{2}=\frac{1}{\gamma}[/latex] [latex]T^{00}=\rho_0(\frac{dt}{d\tau})^2=\gamma^2\rho_0=\rho[/latex] [latex]\rho[/latex] is mass density in moving frame. [latex]T^{0i}=\rho_0\mu^o\mu^i=\rho^o\frac{1}{c^2}\frac{dx^o}{d\tau}\frac{dx^2}{d\tau}=\gamma^2\rho_0\frac{\nu^i}{c}=\rho\frac{\nu^i}{c}[/latex] [latex]\nu^i=\frac{dx^i}{dt}[/latex] [latex]T^{ik}=\rho_0\frac{1}{c^2}\frac{dx^i}{d\tau}\frac{dx^k}{d\tau}=\gamma^2\rho\frac{\nu^i\nu^k}{c^2}=\rho\frac{\nu^i\nu^k}{c^2}[/latex] Thus [latex]T^{\mu\nu}=\begin{pmatrix}1 & \frac{\nu_x}{c}&\frac{\nu_y}{c} &\frac{\nu_z}{c} \\\frac{\nu_x}{c}& \frac{\nu_x^2}{c} & \frac{\nu_x\nu_y}{c^2}& \frac{\nu_x\nu_z}{c^2}\\ \frac{\nu_y}{c}& \frac{\nu_y\nu_z}{c^2} & \frac{\nu_y^2}{c^2}& \frac{\nu_y\nu_z}{c^2}\\ \frac{\nu_z}{c} &\frac{\nu_z\nu_x}{c^2}&\frac{\nu_z\nu_y}{c^2}&\frac{\nu_z}{c^2}\end{pmatrix}[/latex]

Here is a thought experiment. Let's start with an expression. "Matter tells spacetime how to curve, spacetime tells matter how to move" So let's look closer at the first part. First remove all matter (the full standard model of particles). End result curvature equals zero zip Nada. Now when you add other fields such as the Higgs field, weak field EM field, Strong field. Now you can get curvature bit only if the distribution of those fields is Non uniform. (Anistropic) I mentioned the path of least action previously in this thread. It is this principle that determines the path taken The descriptive spacetime is curved is actually a sloppy descriptive. What really is curved isn't the volume. What curves is the Geodesic paths for massless particles this is the null geodesic. The fields of the standard model and how they interact or couple to the particle in motion is what determines the amount of curvature of the geodesic path. Space is just volume. Spacetime is a metric that describes space with time as a dimension under a geometry basis. However spacetime without particles is simply a volume.

Energy doesn't exist on its own. It simply the ability to perform work. Space devoid of all particles is simply a volume. Light can transverse that volume without the need of a medium or eather. Space or spacetime isn't a medium to have permeability. It isn't some mysterious fabric which you often hear described in pop media or poorly written literature. It is far more than semantics with regards to the actual physics and the relevant applicable formulas. A term that no physicist would argue is to use the term field. This is a mathematical descriptive that entails any set of values or mathematical objects such as vectors, tensors, spinors etc under a coordinate treatment. In essence a field is an abstract mathematical object.

I suggest you look at velocity dispersion in astrophysical measurements. Quite frankly dispersion occurs in galaxy clusters etc. If you have a medium light no longer propogates at c.

That is an example for water which does show how refractive index does change for different wavelengths. Particularly with a material depending upon the materials properties. It is a good example of the dispersion effect So I will grant +1 for that graph and your effort in taking the effort based on our feedback. Now in gravitational lensing or gravitational redshift we do not have the frequency dependency. If I look with a spectrograph at a gravitational lens I do not see or measure a different ratio of change with a frequency dependancy

Good point particularly when it comes to GW waves. +1

As well as providing the same degree of accuracy and flexibility. The Einstein field equations can literally be used for any field theory. Which makes it incredibly useful as you can describe how all fields evolve with the energy momentum relations. At least until quantization becomes important.

Wrong the index of refraction depends upon the wavelength of light. You should really try learning a topic before making wrongful statements. Here is a brief article covering Snells law and dispersion https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.cis.rit.edu/class/simg232/lab2-dispersion.pdf&ved=2ahUKEwiKgobX7r7pAhVIj54KHTzECuAQFjAAegQIAxAB&usg=AOvVaw2BN4rxnCdl4Wu8zkJTi4sJ As previously stated any attempt to employ refractive index to spacetime will fail. Let's try the following thought experiment. The LIGO detector uses lasers in its 4 km arms. The location is fixed yet they calibrate the detector by placing a test weight near each detector. So the only thing that changes is the amount of mass near the detector. Yet the laser paths will experience a time delay. The elevation does not change nor does the local fields except for the addition of mass nearby. It is the mass term that causes the above delay the spacetime path becomes curved. Secondly a rocket ship travelling at near c will also experience time dilation So how does your model account for two rocket ships travelling at different speeds in space away from any gravitational bodies By the Principle of equivalence inertial mass and gravitational mass are indistinguishable from one another. [math] m_i =m_g [/math] nothing I have seen in your model accounts for this. I won't even get into your explanation of electron and photon behavior. You don't even require an EM field to have spacetime curvature or time dilation. All matter and force fields contribute to the mass term. Including mass less particles such as the Higgs boson. There is two main types of mass in GR. Invariant mass or rest mass and variant mass (relativistic or inertial mass}. Your theory doesn't have a chance of succeeding. For several of the reasons myself and others have mentioned.

Why would I ask to keep the frequency of light cobstant ? How would you get gamma rays, microwaves, x rays UV rays or even a color spectrum? I am asking you to learn how dispersion is frequency dependent. Then you would recognize a gravitational lens nor spacetime involves refraction. You have an emitter frequency (or more exact a range of frequencies from a distant star. ) Gravitational redshift affects that wide range of frequencies equally. Where a refractive index would not. Due to dispersion.

And yet I have zero problem understanding time dilation. The problem is explaining it to laymen. (Though it took understanding mean lifetimes and it's connection to the Langrangian) Refactive index is the wrong approach. Try gravitational redshift under refractive index.

Maybe you should look more closely at what causes a refractive index. Ie permittivity of a material for starters then step into how dispersions occur.

A refractive index requires scatterrings. Those scatterrings and the degree of change in angles will involve the wavelength. If I look at an Einstein Ring I will get distrortions that do not depend on wavelength. I have been involved in doing those tests studying the CMB as my speciality is early universe dynamics. In order to get a telescope even the Hubble telescope on must often use a gravitational lens to extend the distance to get deep field imaging. A simple analogy for time dilation would be signal propogation delay. Take a digital signal in electronics you can delay that signal through electrical cross talk. Now extend that analogy to the 18 coupling constants of the standard model of particle physics.

No I am a Professional Cosmologist with degrees in particle physics. I can prove any refractive index treatment of spacetime wrong. I can do the same with any treatment of spacetime as a medium wrong. Lol all I have to do is point out previous research papers. I have even been hired to research refractive indexes by a survey camera manufacturer. That grant paid my income for a year. I have also done spectronomy research on a couple of gravity wells.

No forget refractive index. The correct application is Principle of least action via the Langrangian. Which is completely different from refractive index. How can you have a refractive index when the mean average number density of particles amounts to 5 protons per cubic metre in interstellar space ?