Mordred

How one measures a spacetime region depends on the observer. Different observers will have different spacetime measurements. Remember the geometry changes depending upon observer effects. Observers in the same reference frame will always have identical spacetime conditions and dimensions. (lets include length contraction). You contract length you increase energy density and therefore mass density. Same with giving an object inertia as inertia will also increase the energy density. The (at rest frame) for a planet is accurately described by the Newton approximation under GR. Though once you hit relativistic effects you want the Schwartzchild metric. Though I should note, under GR all frames of reference are inertial.

Griffith will step you into the Dirac notation etc from the list you gave. As you meet the prerequisites with trig and calculus.

Not if you look at observer influence. An outside of reference frame can measure a 1 kg steel ball gain enough inertial mass to equal the gravitational potential of the Earths gravity. Principle of equivalence still applies but you have to watch your observers and which mass applies to which observers. Unfortunately you can't ignore this as its key to understanding how weight is influenced to different observers. Yes I do mean weight as well as mass. There is no inertia mass if your in the same reference frame as the object being measured.

Ok lets simplify this. Your on Earth planet observing a planetoid at high c. That planetoid has two types of mass. Its invariant mass (lets get into the more accepted terminilogy/rest mass. It also has a variant mass we now call inertial mass. (replaces relativistic mass). Now lets take a 1 kg test weight measured on Earth. ( note same reference frame as Observer). inertial mass is zero as it in the same frame as the Earth. Now observer on Earth already noted the planetoid has its (at rest mass) as well as inertial mass. So the planetoid already has greater mass due to relativistic corrections. So when he places that test weight on the Planetoid he can still apply Newtons laws. (after all the planetoid now has a higher effective mass. aka greater gravity. Now if the observer also switches location he removes the inertial mass as he is now back in the same reference frame. ( rule of thumb same reference frame as emitter =no inertial mass.)

David Griffiths "Introductory to Quantum mechanics" https://www.amazon.ca/Introduction-Quantum-Mechanics-David-Griffiths/dp/0131118927 is excellent he assumes a low math skill to start and works you up into the more advanced.

and where is the person being weighted with respect to which observer.?

Define your observers in relation to the planet to properly answer this one

Yes Migl is hitting you with a little trick of observers. It may sound the same as your ball above but in that instance I assumed the observer is remote to the planetoid system. Then applied the equivalence principle. Inertial mass=gravitational mass. However Migl has given a circumstance where you have two effective geometries due to a different observer reference frame. Ok lets try this common question. You see this one on forums often. "If a particle with mass moves close to the speed of light why doesn't it become a blackhole?." The answer is although a distance observer will measure a high inertial mass sufficient to become a BH in regards to its Schwartzchild radius. The invariant (rest mass) does not change. Now to Migl's question. Key note you are in the same frame as the Planet. So you are effectively at rest so you have no inertial mass and neither will the planet as it is treated in its rest frame in this instance.

Ah yes could have saved a bit of time posting that first but thats ok. The info in this thread is still useful to understand.

Yes you can accelerate the ball to have the same mass as the Earth. If we start with a steel ball with the same diameter as the Earth the two spacetime regions will effectively be identical.

Roger stop. Try to at least learn basic physics before making foolish assertions. Pressure is force/unit volume. There is a very simple relation ie equation of state for matter and radiation ie photons. https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology) Please pick up a textbook on physics. As it is my grandaughter who is only 11 has a better understanding that what I've seen posted by you on this thread. Not trying to be offensive but you really need to study. Start by learning the definitions For example why does 1 joule equal 1 Newton?

The only difference is how the two add available energy to our spacetime system. *assuming were specifically dealing with a multi particle system. The first two equations apply in this case. In either case your correlating the first two equations in my first post to the effective mass of the system. It doesn't matter if your involving inertial or rest mass both are forms of mass so your spacetime system will be affected. How to mathematically define spacetime will differ simply to derive your metric. [latex]G_{\mu\nu}.[/latex] a system that strictly comprised of static particles will not require the stress/momentum tensor components of the stress tensor. (recall its a matrix) A system of particles in motion will require us to use the stress tensor to derive your metric tensor. However once you have defined your metric this then determines an individual particles motion. Hrrm probably easiest to see on the stress tensor itself. [latex]T_{\mu\nu}=\begin{pmatrix}\rho&0&0&0\\0&p&0&0\\0&0&p&0\\0&0&0&p\end{pmatrix}[/latex] So lets look at [latex] T_{00}[/latex] this is your rest mass energy density. [latex] T_{01} [/latex] is your pressure term. Particles with momentum generate pressure via an equation of state. [latex] w=\rho/p [/latex]

Roger I really wish you would pay attention to the basic physics definitions. You will find physics much simpler to understand if you did. First off define force. Ie define a Newton of force. Then define energy and mass as well as inertia. If you do the above you will realize your posts makes no sense whatsoever. I mentioned before that lasers can generate force. Lasers can move objects if strong enough. Yet light is not matter. A gravitational wave is another example. Temperature or pressure can also generate force. None of these involve matter. Matter is fermionic particles it does not include bosons. (I mentioned this before) yet bosons can and do generate force. This is 100% wrong period. The very definition of energy should have told you that. Just look at the unit for energy the Joule. "the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves one meter in the direction of action of the force, equivalent to one 3600th of a watt-hour."

friction which will involve heat. If you have an infrared camera you can see the heat of each footstep.

Your thinking in terms of strictly the Lorentz transforms. GR doesn't quite work the same. In the Lorentz transforms the metric is Euclidean flat. [latex]G_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}[/latex] in this instance the stress/momentum term is not required as spacetime itself is Euclidean flat. The h tensor is you individual particle deviation from the Minkoskii tensor [latex] \eta [/latex]. one sec going to copy paste to save time. Start with the metric [latex](dx^0)^2-(dx^1)^2-(dx^2)^2-(dx^3)^2[/latex] (please note this equation is a direct translation of Pythagorous theory applied to our coordinates) apply it to the equation of motion of a particle in a gravitational field with geodesic equation [latex]\frac{d^2x^\mu}{ds^2}=\Gamma^\mu_{\lambda\nu}\frac{dx^\lambda}{ds}\frac{dx^\nu}{ds}=0[/latex] [latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex] [latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex] so how does this work with the boosts and rotations. lets first show a boost in the x direction. [latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex][latex]=\begin{pmatrix}\gamma&-\beta\gamma&0&0\\-\beta\gamma&\gamma&o&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex] now a boost in the y direction [latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&-\beta\gamma&0\\0&0&1&0\\-\beta\gamma&0&\gamma&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex] in the z direction [latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}\gamma&0&0&-\beta\gamma\\0&1&0&0\\0&0&1&0\\-\beta\gamma&0&0&\gamma\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex] now stop and think about the coordinate ct which gives your length with the appropriate length contraction. remember the inertial frames will not have identical coordinate frames and we need to handle the frame transforms. We use pythagorous theory relations to reflect this in GR. the above three boosts can be compactly written as [latex]\acute{x}=\Lambda(v) x[/latex] the problem gets further complicated when you get acceleration which generates rapidity. [latex]\epsilon^\phi=\gamma(1+\beta=1+v/c[/latex] which gives Lorentz transforms as [latex]ct-x+\epsilon^{-\phi}(\acute{ct}-\acute{x})[/latex] [latex]ct=x+\epsilon^{-\phi}(\acute{ct}+\acute{x})[/latex] [latex]y=\acute{y}[/latex] [latex]z=\acute{z}[/latex] which is a hyperbolic rotation of your IF frames. [latex]\gamma=cosh\phi=\frac{\epsilon^\phi+\epsilon^{-\phi}}{2}[/latex] [latex]\beta\gamma=sinh\phi=\frac{\epsilon^\phi-\epsilon^{-phi}}{2}[/latex] and therefore [latex]\beta=tanh\phi\frac{\epsilon^\phi-\epsilon^{-\phi}}{\epsilon^\phi-\epsilon^{-\phi}}[/latex] which in matrix form is [latex]\begin{pmatrix}\acute{ct}\\ \acute{x}\\\acute{y}\\\acute{z}\end{pmatrix}[/latex]][latex]=\begin{pmatrix}cosh\phi&-sinh\phi&0&0\\-sinh\phi&cosh\phi&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex][latex]\begin{pmatrix}ct\\x\\y\\z\end{pmatrix}[/latex] Now the above is your key Lorentz transformations. As our global metric is Euclidean flat we can use the Minkowskii metric. When you have a stress tensor the Minkowskii tensor is no longer used and we now use the metric tensor [latex] G_{\mu\nu} [/latex] Space tells matter how to move Which is what is happening above. Recall that a freefall particle the rate that particle falls is spacetime itself. We don't use force. If you also recall the mass term cancels out in freefall motion. So for the individual particle the mass makes no difference. As far as the equivalence principle itself study the Einstein elevator. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.einstein-online.info/spotlights/equivalence_principle&ved=0ahUKEwjCiP-88c7TAhWBKGMKHTHyALEQFghdMA0&usg=AFQjCNHrNM7avF6jzeB_-E7qPRENhHvuRA&sig2=S-AG7TC0e-9pImciIrpyeQ The key detail is [latex] f=m_{inertial}a=m_{gravitational}[/latex] g. g itself is also acceleration therefore m_i=m_g and the above are equivalent. The Einstein elevator explains why [latex] f=ma [/latex] [latex] f=mg [/latex] [latex] m_{i}=m_g [/latex] For now ignore the tidal force though that becomes important in your intrinsic curvature. You will have tidal force when you have curvature not when your in Euclidean flat. the key point is that the stress/momentum tensor is the average of all particles in a region which tells spacetime how to curve. Spacetime itself tells an individual particle how to move. The stress tensor sets your global metric. That global metric tells individual particles how to move not the stress tensor. Hope that helps. Actually lets add a thought statement. Take two particles and let them freefall. If your global metric is Euclidean flat they will fall in a Parallel path to each other. (excusively via principle of equivalence) If your global metric is curved they will have a deviation to their freefall determined by the centre of gravity.(tidal force) Ie both paths will get closer and closer together. (The rate of change is the Kronecker delta function just an FYI). A general rule to follow, when transforming between inertial frames the (at rest frame is your local geometry) the particle motion is the deviations from that reference frame. This includes switching observers. ie one observer at a different gravitational potential. [latex]G_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}[/latex] when you have curvature you would use the stress tensor but it is used to determine your geometry. It is the geometry that determines the freefall path.

You don't need the stress tensor to see the difference.. [latex] e=m_oc^2 [/latex] rest mass no monentum term. [latex] e^2=pc^2+(m_oc^2)^2 [/latex] total energy rest mass+momentum. The difference is whether or not your looking at the momentum. Remember gravitational mass is the same as inertial mass as far as the equivalence is involved. Now lets look closer at the field equations. In particular your metric tensor vs the stress tensor. The stress tensor tells space how to curve but is spacetime that tells how individual particles move. [latex] G_{\mu\nu}=\eta_{\mu\nu}+T_{\mu\nu}[/latex] it is the metric tensor that tells an individual how to move via its geodesic equation. However the stress tensor for the localized area tells space how to curve from the Euclidean metric [latex] \eta_{\mu\nu}[/latex]

No it must simply contain energy. The definition of energy is the ability to perform work. (energy is a property not a thing unto itself). The ability to perform work can generate force. Matter is fermionic particles not bosons such as light. Yet light can generate force. all these can generate force sufficient to induce kinematic action. [latex]\stackrel{Action}{\overbrace{\mathcal{L}}} \sim \stackrel{relativity}{\overbrace{\mathbb{R}}}- \stackrel{Maxwell}{\overbrace{1/4F_{\mu\nu}F^{\mu\nu}}}+\stackrel{Dirac}{\overbrace{i \overline{\psi}\gamma_\mu\psi}}+\stackrel{Higg's}{\overbrace{\mid D_\mu h\mid-V\mid h\mid}} +\stackrel{Yugawa-coupling}{\overbrace{h\overline{\psi}\psi}}[/latex] Notice we include the electromagnetic,strong, Higgs, relativity and particle/antiparticle pairs via Dirac.

Correct

Correct on all the above and yes virtual particles can have less mass than the real. Also very good descriptive on the VP. +1 Now lets look closer at As there is insufficient momentum of individual VP to cause action. No instrument can measure individual VP. We can detect them under confined regions of a collective quantity to generate measurable action. ie greater than a quanta. For example on confined regions the proton itself. The Prof site link posted above also mentions the 2 up and 1 down quark is only the excess color charge. It is a literal sea of fluctuations/virtual quarks and gluons. Further detail can be found by googling S-Matrix QFT I always add pdf at end you get better results.

No plants do not interfere with the double slit anymore than any other specific material. The interference is due to slit size vs light path. Lets clarify quantum observable as defined under QFT. In order to cause kinetic motion on even the smallest known particles theoretical or otherwise. You require a minimal amount of energy of 1 quanta of energy. This kinetic motion can be treated under "action". Now there are always two types of interfence. (constructive, destructive). Waveforms that combine are constructive. Waveforms where the sum energy is less than the combination of the two is destructive. Under QFT this forms your creation/annihiation operators. Virtual particles have less than a quanta of energy so cannot cause action.

Don't get your heart set in stone on a set distance. The key is when you have sufficient difference in potential to cause action.

In essence yes though GR is more applicable as you will have numerous variations within that region. Its simply more accurate to map multiple variations via GR than in SR in this instance.

Depends on the size of the anistropy region. SR as applied to gravitational potential variation. SR does not apply via gravity relations in a homogeneous and isotropic global metric. Lets explain that. You need a change in gravitational potential in a given volume to apply SR. Event A will be at a different gravitational potential than event B. Example stars, Galaxies, etc. For cutoff distance on the range of local from a galaxy the scientific community afiak still accept a cutoff where the effective force or amount of action due to gravity is 100 times that of the citical density. Critical density in joules is roughly [latex] 7.2*10^{-10} [/latex] joules per cubic metre. (I ran the calcs previous with you on how to calculate that number). The cutoff itself is due to effective action in a given region. (action equates to kinematic motion.) That will be your region where gravitational SR relations takes effect and is considered local. The global being homogeneous and isotropic will have no effective differences in gravitational influence between any two event coordinates. (that's your global metric) ie FLRW metric. You require a Difference in gravitational potential to apply SR relations due to gravity

Your question on what causes expansion has been answered in numerous other threads that you started. Including a rather lengthy one on what causes expansion. The answer is thermodynamic relations that can be equated to kinetic vs potential energy relations via the equations of state for each particle contributor of their own self gravity vs their inherent kinetic energy. Geordiof posted a section where I showed the scalar field modelling of Higgs inflation. In the universe today tbe dominant contributor is the cosmological constant. I recall supplying all the gritty details before with you. I'm not into repeating previous efforts to teach the same individual the same lessons from his own previous threads on more than one occassion. Look over those previous threads you were involved in.

Lets do simple math. Speed of light in vacuum in km/sec is 300,000 km/sec. 70 km/sec/Mpc multiplied be 4400 units. is 308,000 km/sec Oh were already faster than c recessive velocity. If you devide c by 70 you will get 4286 Mpc. This is the point where recessive velocity equals c. How much simpler can you get. A little understanding and grade 4 math. There is your Hubble horizon using 70 km/sec/Mpc precisely 4286 Mpc away. Simple mathematics. [latex]v_{recessive}=H_o d [/latex] Where [latex]H_o =70 km/sec/Mpc [/latex] good ole Hubbles law. [latex]70 km/sec/Mpc*4286[/latex] Mpc is 300,020 km/sec. No conflict here just simple math