Mordred

Your understanding is wrong. At sufficient distance from mass. The geodesic is a straight line. Read your post again and find the line in error.

Nice link A guy can have fun playing around with that calculator.

One of the key aspects to understand expansion is to understand the potential and kinetic energy aspects with conservation of energy. For this we can detail using Newtons laws. [latex] F=\frac{GMm}{r^2}[/latex] Mass density we will use [latex]\rho[/latex] which is the mass per unit volume. Now assume a field of test particles. Motion and mass currently unimportant. One of the aspects of the shell theorem in Newtons laws is the test particle will only notice a force from the center of mass. In a homogeneous and isotropic distribution any test particle or CoM can be used. As we're dealing with test particles we just need the mass relation. [latex]M=\frac{4\pi\rho^3}{3}[/latex] So [latex]E_p=-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}[/latex] Kinetic energy is [latex]E_k=1/2m\dot{r}^2[/latex] [latex]U=E_k+E_p[/latex] U is just a dimensionless constant to equate total energy must be set as a constant value. So the above translates to [latex]U=\frac{1}{2}m\dot{r}^2-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}[/latex] Now with the vector relation of the radius to length we can denote the scale factor. [latex]\overrightarrow{r}=a(t)\overrightarrow{x}[/latex] Where a is a function of time. This leads to [latex]U=\frac{1}{2}m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2 m[/latex] Multiply each side by [latex]2/ma^2x^2[/latex] Leads to [latex](\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}[/latex] [latex]kc^2=-2U/mx^2[/latex] [latex]k=-2U/mc^2x^2[/latex] K is the curvature constant. Please note the above doesn't care about what mass or momentum the test particles have except as a consequence of the above relations. [/quote[

The evaperation time is roughly proportional the cube of its mass. [latex]T_ev=\frac{c^2M_o^3}{3}[/latex]. Assuming the blackbody temperature of the BH is greater than the blackbody temperature of its surroundings. Further details here. https://en.m.wikipedia.org/wiki/Hawking_radiation One of the examples they give is in order for a BH to evaperate with the blackbody temperature of the Universe it must have mass less than 0.8% the mass of the Earth

So which is it. A) you will do the work and show the math presenting your idea.? B) you will present the idea without understanding the needed math and expect others to do the math for you.? Quite frankly I probably know the answer. Neither here nor there, you have difficulty accepting an expanding universe. Where as I can't see how a universe will neither expand nor contract. Which breaks down to a beginning and ending. It's virtually impossible for a thermodynamic system as complex and diverse as our universe standard model has in the sheer number of degrees of freedom to be stable over the course of 13 billion years. I've spent a lot of years studying the particle physics aspects including the QM aspects in Cosmology. Over the years I've come to realize a steady state is highly unrealistic This is strictly an analogy.... Take several thousand magnets in the shape of metal balls. -assume you manage to place these magnets in a perfect stable placement so that you have a stable grid. Currently no movement. this system can be defined as having strictly potential energy. Move 1 of those magnets the entire system becomes unbalanced. You will get random movements with every magnet. The system converts potential energy to kinetic energy. That's just due to a miniscule change in an electromagnetic field analogy. ( You can apply that to a homogeneous and isotropic beginning to gravity. All it takes is a miniscule overdensity to start a change in the overall distribution) this is accounted for in LCDM. Lots of cosmology models use potential and kinetic energy metrics.

Roflmao, you admit zero knowledge of the math involved in both relativity and QM yet believe you found a link ? Very appropriate response

The question really breaks down to a matter of choice on the accuracy. For example between your feet and your head there is an extremely miniscule time dilation. The effect is so miniscule it would take several thousand years to become noticeable. In our everyday applications Newtonian physics is usually accurate enough for everyday needs. Mainly because our everyday needs don't involve any significant time scale. An exception to this is measurement data such as redshift over large distance or change in gravity well, velocity etc. In truth it depends on the needs behind the application. For example it's important for GPS satellites due to a need for higher precision. Yet using Newtonian physics works quite well for calculating falling objects including asteroids and meteors. As you can see the meteor gains a far greater velocity than the satellite. Yet we choose a less accurate metric. Were not particularly concerned with time dilation for the meteor as the event is extremely short lived. Einsteins equations are more accurate anytime you involve motion and a difference in gravity no matter how miniscule. Yet Newtons equations on many applications give a good enough approximation.

On that note, an old friend of mine was doing a study on kinetic energy in falling objects. He had a plate he cooled to the same baseline temperature then dropped different mass objects onto the plate and recorded the temperature via a highly sensitive infrared camera. Obviously there being lots of room for systematic errors this took a while. However it was fun looking through that camera. You could see every single footprint long after people left the room lol.

Man this thread has gotten infinetely foolish. Cladking your assertions make absolutely no sense. In point of detail great care is taken to avoid problems that lead to infinite values. In QFT for example. "Take a field and assign a coordinate at every point in that region." If you don't place a limit of measurement scale you end up with an infinite number of coordinates. Much like you can devide a metre in half then devide the two halfs again. No matter how many times you devide each section by 1/2 you will never have any section reach zero. There are thousands of examples where infinity can occur,

A big confusion most people have is thinking mass is from just one source. Mass is essentially resistance to inertia... What this entails is that any interactions that has a binding or attractive force can provide that resistance. This leads to numerous possible sources of mass. For example the Higgs field only interacts with a small list of the SM particles. W+, W-,Z bosons. These bosons are the the mediator gauge bosons. Quarks and leptons gain a small portion of mass from the Higgs field due to their interaction with the weak field. Of which the mediator is above. So although they acquire mass from the Higgs field it not due to directly interacting with the Higgs field but via the mediator bosons. take for example quarks confined in a proton. Only a miniscule amount of the protons mass comes from the quarks interactions with the Higgs field (albeit indirect). Protons mass. 938.272046 MeV/c^2 made up of 2 up and one down quark. Let's look at mass of each quark. up quark. 2.3 MeV/c^2*2=4.6 MeV/c^2 down quark. 4.8 MeV/c^2 Total 5.4 MeV/c^2. Works out to less than 1% the mass of the proton is from the Higgs field indirect interaction. The majority of the rest mass of the proton comes from the strong force, which has no interactions with the Higgs sector. The rest mass of a proton is a measure of its combined resistance to inertia due to its various field interactions. In essence it has more than one source of mass. When you interact with spacetime (gravity) this provides another source of gaining mass. At one time we called this relativistic mass but that term caused too much confusion. Now it's commonly called inertial mass. The atom itself has electromagnetic mass. hope this helps Now the Higgs sector itself The Higgs field has 4 components. One can treat these as four seperate scalar fields. [latex]\phi_1,\phi_2,\phi_3,\phi_4[/latex]. Prior to electroweak symmetry break these fields all have a non zero field value. when the weak symmetry break occurs. [latex]\phi_2,\phi_3,\phi_4[/latex] Couple to the w+,W-,and z gauge bosons. Essentially they become gobbled up by the weak field. The non zero vacuum today portion is the [latex]\phi_1[/latex] field

You have to look at the type of quarks for composite particles. For example a Neutron though electric charge neutral isn't its own antiparticle as it is composed of antiquarks. Which gives it an opposite baryon number. Quarks having electric charge

No idea about that but the photon is also it's own antiparticle.

No your missing the point. Pop ScI articles states far too many wrong statements to even go into listing them all. I rarely EVER see a pop ScI article properly describe any cosmology related topic. They all make oversimplifications that mislead the reader into the wrong impressions. If you wish to place your faith in them you will always get the wrong understanding. I showed you a plausible and mathematically accurate possibility where Inflation and dark energy can fit with the standard model particles and thermodynamic processes. Yet you refused to even consider the possibility based on your Blind faith in your uninformed personal philosophy. Then when we pints out the math you accused us of having blind faith in LCDM. When we were pointing out basic physics principles.

I've always felt if you keep exercising your brain it will have a better chance staying healthy. I'm on my late 40's and I'm still studying numerous textbooks. I've probably read and studied well over a hundred diligently.

If you don't like LCDM fine pick up a textbook on LQC. However at least learn the rudimentary physics first. Which from your previous posts you definitely lack on. Those rudimentary physics are not specific to LCDM. A good example is the physics behind redshift. Redshift is not specific to LCDM or any other model. Your biggest mistake is assuming that every formula LCDM uses is only used by LCDM. Its not that case at all. For example the FLRW metric is used by other models such as MOND. Or Poplowskii universe in the event horizon of a BH. The impression I've gotten is your more interested in ranting how you don't agree with something rather than the study. Or learning ( there isn't a single formula used by LCDM that anyone can point to and state this is an LCDM only formula)

Yeah through diligence and effort one can gain a very comprehensive understanding. However there is no easy route. Judging from your background you probably have a decent math background so that definetely helps. I found the biggest help outside a university was to buy the textbooks. When I ran into a problem with a particular section of a textbook I could ask for forum help. Next is digging up specific PEER reviewed articles. They help supplement the textbooks and vise versa. The textbooks give you a handle on what is considered main stream. If you are interested in a particular field first let us know, I can certainly offer resources and recommendations on good quality textbooks. Some understanding of your math skills will help on the recommended material.

You already know the radius of the event horizon. Which gives you the point of no return hyperslice. That is the equation I posted earlier. [latex]r_s=\frac{2GM}{c^2}[/latex] that is a specific coordinate hyperslice. It does not represent the hyperslice further away from a BH Of the same mass. Yes this is complex. Welcome to relativity. There is no easy way to understand all the relations without intensive mathematical study. When you ask for a curvature relations you must specify by which coordinates are you referring to. If you set the time coordinate at a constant value say at rest.you get a different curvature than an observer at a different time reference frame. One that is moving.

The event horizon in 3D for a Schwartzchild blackhole is a sphere with all the same mathematics that can be applied to a sphere. The radius of that sphere is given by the Schwartzchild radius formula posted previously. The curvature of that sphere is the last formula I posted. That curvature is at any point along the circumference of that radius. The 2d hyperslice curvature as you move away from the Schwartzchild radius would follow the Flamms parabloid https://en.m.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid. describing curvature as you approach a BH in 3d is rather complex. So you will have to often use slices. Nearly impossible in 4d. No particle will follow the Flamms parabloid as the time coordinate is set as constant. Well here is a handy list of representations corresponding to some of the different geometry changes. http://www.google.ca/url?q=http://www.rpi.edu/dept/phys/Courses/Astronomy/CurvedSpacetimeAJP.pdf&sa=U&ved=0ahUKEwiGorW_1fPLAhVkn4MKHX_jBK8QFgglMAY&usg=AFQjCNEQurpyze2nnu2lCkZ7nnuVmeW5Fg has a good math break of the different relations involved depending on which coordinates are changing. (You can see from it some of the shapes can get extremely complex.)

The Higgs boson is its own antiparticle.

[latex]k=\frac{1}{r}[/latex] So correct the curvature varies with the radius.

Does a ball of one metre in radius have the same curvature as a ball 10 metres in radius ? No spacetime coordinates do not fall into a BH.

Sure we may not be able to measure every single aspect or the exact value of pi. Just like we can't predict the precise motion of every particle. However we can make reasonable approximations. For example averaging.

Measuring is of primary importance in science.

The formula calculates the radiated power of a gravity wave. No it's not so simple for gravitational mass. That's what Newtonian theory uses. However this theory ran into a problem with Mercury. Gravitational mass works fine until you add significant relativity effects. Say you use that formula on two objects moving slowly, with a decent separation distance. Extremely accurate in this case. However if both objects are moving at 0.5 c you would get the wrong radiated power. ( This is hinting at the radiated mass value in the Ligo paper disceptancy you asked about) Now let's really blow your mind away. Let's compare the Schwartzchild (uncharged, non rotating) metric to the Kerr metric (uncharted, rotating. Schwartzchild case. Event horizon is [latex]r_s=\frac{2GM}{c^2}[/latex] Kerr metric. (You have not one but two event horizons,one inner and one outer). Outer. [latex]r_o=m+\sqrt{m^2-a^2}[/latex] Inner [latex]r_i=m-\sqrt{m^2-a^2}[/latex] In the Kerr metric form the last equations are given differently. https://en.m.wikipedia.org/wiki/Kerr_metric. they use the Schwartzchild radius to calculate the inner/outer event horizons. The formulas on the wikilink are more accurate as it will show that the event horizons are slightly flattened. Wider on the equator. ( The first two equations are a simplified form ignoring frame dragging and working with the singularity mass.)

Roflmao