Mordred

The answers to the above have been answered if you stop to think about it. The lorentz transformations are based on constant velocity. The minute you accelerate or change direction (acceleration) those formulas must take into account this change. The transforms with rapidity are as follows. [math]\begin{vmatrix}ct'\\x'\\\end{vmatrix}=\begin{vmatrix}cosh\phi&-sinh\phi\\-sinh\phi&cosh\phi\\\end{vmatrix}\begin{vmatrix}ct\\x\\\end{vmatrix}[/math] The formula you posted are Lorentz but only under constant velocity. The above matrix provide the changes due to rapidity. Some articles may refer to it as your hyperbolic rotation. No 4 is not based on negative k or curvature of any form. My advise to you is to step back and learn relativity through the proper stepping stones. 1) equivalence principle 2) relativity of simultaneity 3) The basic Lorentz transforms. 4) the relativistic velocity addition formulas. Master these before trying to tackle the more complex problems such as those on rapidity and hyperbolic rotation. At the moment from reading your responses this is too advanced for you till you master the above. Also get out of this habit of yours in assuming formulas are lacking or incorrect because you think you found some flaw. Relativity is a huge well tested subject. For example symmetry groups/tensors or matrixes. If you can't use the formulas as they are designed to be used. You will get the wrong answers. Not to mention drawing the wrong misconceptions and conclusions. You should focus on what everyone has been trying to teach you on how velocity addition works under relativity instead of assuming they are wrong. For example lets look at the acceleration for simplicity I will use a rocket that turns around and returns to its original start point with maximal turnaround. hyperbolic rotation (rocket acceleration due to direction change first define the four velocity. [latex]u^\mu[/latex] [latex]u^\mu=\frac{dx^\mu}{dt}=(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})[/latex] this gives in the SR limit [latex]\eta u^\mu u^\nu=u^\mu u_\mu=-c^2[/latex] the four velocity has constant length. [latex]d/d\tau(u^\mu u_\mu)=0=2\dot{u}^\mu u_\mu[/latex] the acceleration four vector [latex]a^\mu=\dot{u}^\mu[/latex] [latex]\eta_{\mu\nu}a^\mu u^\nu=a^\mu u_\mu=0[/latex] so the acceleration and velocity four vectors are [latex]c \frac{dt}{d\tau}=u^0[/latex] [latex]\frac{dx^1}{d\tau}=u^1[/latex] [latex]\frac{du^0}{d\tau}=a^0[/latex] [latex]\frac{du^1}{d\tau}=a^1[/latex] both vectors has vanishing 2 and 3 components. using the equations above [latex]-(u^0)^2+(u^1)2=-c^2 ,,-u^0a^0+u^1a^1=0[/latex] in addition [latex]a^\mu a_\mu=-(a^0)^2=(a^1)^2=g^2[/latex] Recognize the pythagoras theory element here? the last equation defines constant acceleration g. with solutions [latex]a^0=\frac{g}{c}u^1,a^1=\frac{g}{c}u^0[/latex] from which [latex]\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0[/latex] hence [latex]\frac{d^2 u^0}{d\tau^2}=\frac{g^2}{c^2}u^0[/latex] similarly [latex]\frac{d^2 u^1}{d\tau^2}=\frac{g^2}{c^2}u^1[/latex] so the solution to the last equation is [latex]u^1=Ae^{(gr/c)}=Be^{(gr/c)}[/latex] hence [latex]\frac{du^1}{d\tau}=\frac{g^2}{c^2}(Ae^{(gr/c)}-Be^{gr/c)})[/latex] with boundary conditions [latex]t=0,\tau=0,u^1=0,\frac{du^1}{d\tau}=a^1=g [/latex] we find A=_B=c/2 and [latex]u^1=c sinh(g\tau/c)[/latex] so [latex]a^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex] hence [latex]u^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex] and finally [latex]x=\frac{c^2}{g}cosh(g\tau /c)[/latex], [latex] ct=\frac{c^2}{g}sinh(g\tau /c)[/latex] the space and time coordinates then fall onto the Hyperbola during rotation [latex]x^2-c^2\tau^2=\frac{c^4}{g^2}[/latex] So regardless of how fast or slow the acceleration is, the rotation itself causes changes to reference frames Now I ask you how much of that do you have a chance in understanding without first understanding the basic Lorentz formulas??? This is just one example of the complexity involved in the more advanced relativity problems Did you see me use a curvature constant k? no its not involved. Also instead of trying to learn this stuff through random google searches try a textbook. Here is a free one. http://www.lightandmatter.com/sr/

One error for sure is the velocity after 450 metres. time to fall is 9.579 seconds velocity should be 93.95 m/s. [latex] t=\sqrt{\frac{2h}{g}}[/latex] [latex]v=\sqrt{2gh}[/latex]

What is you velocity of the ball just prior to impact? (I noted I forgot to square seconds above.) should be 9.8 m/sec^2 Your velocity should be higher than 54 metres/sec after falling 450 metres

Hyperbolic rotation with Lorentz calcs requires additional formulas as rotation is acceleration change. The lorentz transforms are designed for constant velocity. So naturally you will get the wrong answer. You must also account for Rapidity

Then you may be doing something wrong in your calculations. Your Ke should return your orginal PE value. (here is a hint) as it falls, its total energy (the sum of the KE and the PE) remains constant and equal to its initial PE. Take the scenario above with the formulas you posted and show your calcs so we can find what your missing (were now dealing specifically with the conservation of energy) involved in the scenario above.

No it isn't. Ok lets take an example a weight sits on a table. Calculate the potential energy of a 1 kg mass sitting 10 feet off the floor with g=9.8 m/s. That weight can obviously perform work after all if you remove the table the weight will fall. Potential energy of course will convert to kinetic energy once the weight starts moving. One thing you will discover both PE and KE are used also in more advanced physics including GR or the formulas for effective action. Its extremely important to properly understand

Your missing the number of photons emitted by all the stars in the course of their lifetime. If expansion didn't occur space would be bright and there would be no nighttime

that works too depends on the book in this go with your textbook. I take it this is homework? If so let us know so we can move this thread to the homework section

Does it? Potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. So in order to calculate PE you have to determine its position from the center of gravity in the case of graviational potential energy.

The scientific definition is "The ability to perform work" Now apply that definition to the problem sets on potential and kinetic energy above.

your welcome

Key added note "any change" so how can you restrict time to simply matter?

Matter applies only to fermionic particles via the Pauli Exclusion principle. Time still passes for bosons as well which isn't considered as matter.

Yeah working from phone didn't copy correct. https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology) All links should be fine now. I should note the last two links explains the metric aspects not so much the fluid portion except via critical density relations. The first two links however provides those details.

To answer this accurately first we must understand an essential detail on galaxy rotation curves in terms of the distribution of the mass. First we will cover the Keplarian curve prior to explaining the mass distribution of due to dark matter. Now take a mass distribution with a center of mass as our focal point. Ie center of our galaxy. With baryonic matter only the mass distribution is greatest towards the center. This distribution of mass will end up with a Keplar rotation curve. (Keplar decline). Dark matter however envelopes and surrounds our galaxies with sufficient mass that it is evenly distributed throughout our galaxy. You no longer have a Keplar decline as your mass is evenly distributed. So yes dark matter influenced rotation curves via their mass/gravity influence but not any other known interaction. I would recommend getting into the habit of remembering energy is the ability to perform work. The term observable energy is rather ambiguous. Anyways lets leave that aside. We don't know the mechanism behind dark energy however we know its influence on expansion. I prefer cosmological constant to dark energy in terminology. To understand DE and its dynamics the best path is to understand its equation of state. Cosmology applies GR as well as the thermodynamic laws. The FLRW metric has two essential components. The metric ie geometry and the Fluid equations, part of those fluid equations being the deceleration/acceleration equation. A brief but rundown on Equations of state can be found here. Its broken down into categories. EoS for matter/radiation Lambda and scalar fields. Each particle species will fall under one of the categories just mentioned. The Eos givez us its density to pressure relations. [latex]w=\frac{p}{\rho}[/latex] For dark energy its EoS is w=-1. https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology) So DE behaves like negative pressure incompressible fluid. (keeping in mind we are treating all cosmological influences under the ideal gas laws.) via the FLRW metric fluid equations. https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.damtp.cam.ac.uk/user/db275/Cosmology/Lectures.pdf&ved=0ahUKEwj128jpubjTAhVHx2MKHXT6BAgQFggcMAE&usg=AFQjCNGxvGWwx9QmCl2FyZLAUmB7lkZ7xQ&sig2=KnbYQXLvMRQDErkRBBnrYw The last link covers everything I have just stated above. Another key detail being universe geometry I wrote a breakdown on this here. http://cosmology101.wikidot.com/universe-geometry page 2 *FLRW metric broken down into dimensions ie 2d/3d and 4d* http://cosmology101.wikidot.com/geometry-flrw-metric/ It will help you understand several chapters of the second link.

Actually your not in speculations. However the interparticle interactions you described are involved in drag but there is also drag due to the spin of our sun coupled with gravity. You essentially however accurately described a main part of what is involved in drag. A large factor to consider is how much velocity each particle gains for the amount of force acting upon it. This is all included in the density wave paper I linked though understandably it will take time to fully understand. The drag is described as density waves in the density wave theory. The mathematics are in essence hydrodynamic fluid equations that work for the low densities involved above. Remember a cloud of plasma also has an effective mass. You included the interparticle gravity contributions already which is accurate.

Correct the BB singularity and the singularity of a BH. The energy levels in such a compact finite region and high temperature far exceeds a quanta. A collection of fluctuations within a tight enough volume ie pointlike that exceed a quanta will have an effect. Or for example between two plates. However you must exceed a quanta to gain action.

It does but here is the problem, the energy due to QM fluctuations cause no effective action. Action is motion which GR covers. See the crux? Here is a fairly well detailed formula for 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] We have GR, the Higgs, the electromagnetic and the Yukawa coupling Also particle/antiparticle pairs under the Dirac umbrella. Provided the particles involved have a quanta of energy. Below that threshold we can't even measure the effective action of a fluctuation with the most perfect instrument. The fluctuation individually can cause no action which includes interactive interferance (necessary to detect). As no action is induced how does GR include it ? Recall GR maps freefall motion via geodesics. (Worldlines). (Do not confuse the above with quantum tunnelling ie that used on inflation, different process far different energy levels involved)

It will be interested to see how much you've learned since then. Hopefully you picked up some supportive metrics which is always the best method to model a system. (Pretty much the only way to model any system lol)

A valid question the answer is rather surprisingly detailed under density wsve theory. If you google this you will also find the same theory applies to Saturns rings, Galaxy spiral arms as well as protoplanetary formation. Here is one such article https://www.google.ca/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/1610.05139&ved=0ahUKEwim94iR7rbTAhVHx2MKHXT6BAgQFggfMAA&usg=AFQjCNH46cy9-xC1JU5G0X5D2gNZp-tKmw&sig2=DbzAhUJ0G940ZWr4AoqzRQ Probably the easiest way to explain the gist of the theory is to consider f=ma. It takes more force gravitational or otherwise to move objects of higher mass. This means the elements will also distribute accordingly to their mass. Lighter elements further out as they will have a higher escape velocity. So your hot Jupitors will form in the outer regions with heavier element rich planets forming closer to their star.

Took me less than a minute of google search to realize I was wrong and Argent was correct

What he is referring to is the singularity problem.

http://www.wiese.itp.unibe.ch/lectures/universe.pdf This link works it didn't parse correctly from above. The space between atoms is extremely difficult to describe, all particles are essentially field excitations. This makes it tricky. For example the Proton is made up of two up and 1 down quark but in actuality that is the excess. In point of detail there is a literal sea of quarks/gluons surrounding and enveloping the atom. This is descibed under the S-matrix in QFT. When you get down to it the real illusion is what we think of as "solid". Particles aren't little bullets, but excitations and for the most part have indeterminant volume as a result. This is why you never see volume described as a property of a particle.

Your welcome

Your welcome