Mordred

The bullet particle analogy. Doesn't make any difference when it comes to how a field behaves in terms of kinematics. Are you going to answer the questions?

Ah so your going to ignore the actual questions we posed. I'm certainly positive that Swansort and myself know more about infinitesimal sets than you do. The Levi Civita connection is an infinitesimal field.

Good luck in showing these kinematics of repulsion acting as attraction. I would love to see how you can possibly show this mathematically and follow the conservation rules. Describing infinitesimals does not show the kinematics. Feel free to mathematically show how the Displacement of any object can be in the opposite direction as the acceleration.

Think about it this way. There is no way one can state "my observer measurements are more accurate". Regardless of scenario, all observer measurements are equally accurate.

When you truly think about it the "at rest frame" is meaningless. All frames are equivalent to inertial frames. So why would you believe one is preferred over the other. If you think about it "a preferred frame" is equally meaningless. You don"t base conclusions and definitions from one type of scenario, they must work in every scenario as much as possible.

The elevator portion follows what is predicted by the equivalence principle. If you drew out the path of the two particles in the elevator you can see the deviation from parallel transport. This deviation is a simple consequence of path taken. For the twin paradox portion. The later math shows the frame rotation on the turn around. You recall the video you posted. Recall the region at 5 seconds that did not have lines? that triangle region that was blank? That is the hyperboloid region the math above derives. Yes the math format includes the GR tensors, which is why it looks complex. The key relation is the rotation between v plus to v minus. Velocity as a vector. Outgoing/incoming. The two different directions count as two different frames of reference. Most ppl just getting into SR would look at the twin paradox and count two reference frames A and B. However B has more than one reference frame. Technically if you get precise. B has an infinite number of reference frames due to various acceleration changes. We can reduce that number by reducing precision. These two scenarios help show why we use velocity not acceleration in time dilation. Acceleration induces reference frame changes which can be infinite in number. The best you can do is reduce that infinite number of reference frames by time slices. (instantaneous velocity) to get an approximation of changes. This is true in both scenarios. Always remember velocity is a vector. Put succinctly a change in direction is a change in velocity. A change in velocity is a change in reference frames.

side note, in a free falling particle scenario (Einstein elevator) where is the force? Subject for another thread lol but boils down to "is gravity due to the stress tensor/curvature " vs acceleration argument. Prior answer being the correct one. In the Einstein elevator scenario, just for those interested. Please start a new thread if you have questions on this portion. I worked this up for another thread but its handy to show free fall and how parallel transport relates to geodesics. pS note the equation for "non accelerated motion"

OK The statement "acceleration isn't the cause" is what I am specifically addressing. However I took a moment and looked at the video posted and although 100% accurate, doesn't involve some key math details to show the frame rotation. As this is important to understanding the twin paradox I will include this detail. (leading to the Sagnac effect). The twin paradox is essentially if a see's b time as slow and b see's a time as slow, which twin ages slower? when both observers see the same time dilation between each reference frame. One might think the twins will be the same age on the return trip. However this isn't correct. The video explains the solution as the accelerating observer is the one who will be younger. Which is correct, however it may lead one to assume its the acceleration in GR that's important. (not saying the OP is under this misconception, however we do have other readers.) So lets start with the elevator... This is important as it distinquishes the principle of equivalence. the equivalence principle states the equality between gravitational mass and inertial mass, the problem with this statement occurs when you consider two particles in the elevator. Keeping in mind there is two types of acceleration. Change in direction but not velocity. Change in velocity (magnitude but not direction.) So in the case of the elevator with the two particles. First take a pen and paper, draw a box, with two particles separated on the same vertical elevation. Then draw a line from each particle to the center of the mass. We can see from this drawing that the falling particles undergo two forms of acceleration, change in magnitude and change in direction. The two particles approach one another as they freefall. This is the tidal effect due to the gravitational field. (the deviation from the parallel paths,coincidentally is also used to determine your geodesic equations) This being an example of parallel transport deviation. (curvature). So the equivalence is an expression that holds true only if you treat it at a single point in the gravitational field. Put another way it is locally equivalent. However it isn't equivalent over long distances due to changes in direction. So lets throw some numbers in and describe the tidal effect. [latex]F_a=F_b=\frac{mMG}{r^2}[/latex] let x be the distance between a and b, let [latex]\alpha[/latex] represent the angle between one test body and the center line (center of gravity vertical axis) so from frame a, b experiences a force directed toward a. [latex]F=2F_asin\alpha=2F_a*\frac{x}{2r}=\frac{mMG}{r^3}x[/latex] a then observes b to be accelerating towards him by [latex]F=-md^2x/dt^2[/latex] [latex]\frac{d^2x}{dt^2}=\frac{MG}{r^3}x[/latex] the 1/r^3 is characteristic of tidal forces. That clears up the principle of equivalence a bit.... the key importance is to understand when this relation holds true. the next part will take me a bit to type in... the twin paradox which isn't a paradox at all, essentially can be broken down to the following statement. "we should never have considered the age of a and b to be the same, as the frame of b undergoes accelerations that a does not undergo." so lets look at the acceleration 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 this detail is often overlooked, by the statement :"it doesn't matter how fast observer B accelerates or decelerates" most articles don't mention the rotation as to the reasoning behind that statement.... A lot of attempted solutions try to use the instant accelerated, decelerated argument but ignore frame rotation. (as mentioned by Studiot) this is the problem with placing two rotated frames (outgoing,incoming) onto the same timeline, leads to errors...not to mention ignoring the spatial separation components of doppler vs relativistic doppler. edit::oops almost forgot to reference the above... :Introduction to General Relativity" by Lewis Ryder. Its an excellent book, as it explains details in an easy math level (well easy to me lol) step by step. The author assumes the reader has less previous knowledge, than in many other GR textbooks. (thanks Ajb for bringing it to my attention, its a great addition to my huge collection lol)

umm you have to be careful in thinking acceleration. I will dig up a lesson in one of my textbooks. This particular textbook is one Ajb mentioned to me. It has the best solution. Just as soon as I dig it out and have my coffee

I never win

The time dilation portion is [latex]\Delta\acute {T}=\frac{\Delta T_o}{\sqrt {1-\frac{v^2}{c^2}}}[/latex] or alternately [latex]\acute{T}=\gamma T_0[/latex] so if you set the velocity as constant the time dilation portion will stay constant. The additional delay is due to seperation distance. Not time dilation. The spacetime interval which is defined as the seperation distance between events includes both distance and time deviation. The time dilation length contraction is a path deviation. Without time dilation this portion is simply the distance. With time dilation the distance follows a curved path. The intervals in your link has two influences upon distance. Spatial seperation and curvature due to dilation. It is the curvature influence were intetested in. look at each specific formula for the distance of seperation. https://en.m.wikipedia.org/wiki/Time_dilation honestly this has been mentioned several times. ((by myself and others)) SO I have no idea how one can miss the seperation distance itself. Unfortunately the link that includes those graphs did not mention the pertinant detail of seperation distance. (at least it didn't go into the mathematical detail of how to distinquish between the Doppler portion (toward/away from) redshift to the time dilation influence on the Doppler). Those formulas are on the link above. Nor does that link resolve the twin paradox itself. (or even describe the actual paradox) which is resolved by Pams rotating frame/acceleration/deceleration. lets try this angle.... [latex]\acute{x}=\gamma(x-vt),\acute{y}=y,\acute{z}=z,\acute{t}=\gamma(\frac{t-vx}{c^2})[/latex] [latex]\gamma=(1-\frac{v^2}{c^2})^{-\frac{1}{2}}[/latex] [latex]\acute{x}=\gamma(x-vt)[/latex] if [latex]\gamma=0[/latex] then the seperation along the x axis is simply the spatial components, as there is no length contraction or time dilation. [latex]\acute{x}=x,\acute{y}=y,\acute{z}=z,\acute{t}=t[/latex] so separation between events follows Pythagoras theory, in the former case as the x axis and time axis changes due to gamma, Pythagoras theory no longer holds true. (this deviation is your length contraction and time dilation)

Those time stamps contain two details. Not one. The time dilation portion of each time stamp is the gamma factor. (Lorentz boost). If the ship is moving at constant velocity the Lorentz boost itself is constant. The other detail in those time stamps is seperation distance. This is the additional delay. This seperation distance portion isn't time dilated. It simply increases the time of a signal to reach a to b. Not due to dilation but due to change in distance.

nice back of the envelope. +1

the effect isn't ignored. You need to account for doppler shift to get the time dilation portion. You seem to have difficulty seperating the individual cause and effects. change in clock rates due to time dilation change in percieved clock rates due to distance change.

Ok even without accounting for seperation distance mentioned above. Lets look at those two numbers. 4 years and 2 years. This represents the total time elapsed. This isn't rate of time. Rate of time is the time for each second to tick. Total time is the accumulation of those ticks. It is the ticks of the clock that synchronizes when the clock is in the same reference frame. Not the total time elapsed. As mentioned above there is also a difference between time it takes to recieve each tick, when you involve direction. As mentioned you need to account for this. This is where understanding the formulas used to generate those graphs becomes important. its also important to keep track of who the observer is... I suspect this is common, a person with familiarity with the subject can more readily deduce the correct relations. Speaking of relations. Just a side note any time I see a linear graph I instantly think [latex]y=mx+c [/latex]. I tend to do this also with common waveforms (gotta love Laplace transformation). There is a lesson I learned years ago. "The most complex problems can be simplified if you reduce it to its individual parts" when I look at a graph I start reducing it.

no, the reference frame of rest is the vertical column. If Jim is the observer on the diagram on the left. His clock is along the y axis. Not the diagonal. That diagonal line is Pams clock. Which if you draw a vertical line at 2012 to 2011 you can see Pams clock running slower than Jims clock. For Pams clock the Pam is the sender image on right. Same thing Pam sees Jims clock running slower. AT no point does an observer see another clock running faster than his own... DO NOT CONFUSE time elapsed with time rate. They are two very different units of measure. The first is accumulated time. The second is time per a unit. The sender of the greeting Jim on the left sees his clock as the fastest. The sender on the right Pam sees her clock running the fastest. Jim sender on the left diagram sees his own clock as fastest. Pam sender on right diagram sees her clock as fastest. The quoted section is accurate.

yes different rates... however the sender of the greetings still sees his own clock as the fastest.

you will see different rates but tbe observer at rest or in his own reference frame will always see his clock as the fastest clock.

Positions being properly termed events. Each event observer and emitter has positions [latex](\acute{ct},\acute{x},\acute{y},\acute{z})[/latex] and [latex](ct,x,y,z)[/latex]

The two terms "dilated" and "contracted" are not symmetrical lol... Dilated can mean contracted or expanded. however the Lorentz boost for both outgoing and ingoing will be contracted in both directions. Ie the observer at rest watching the moving observer will always see the other clock as slower. Regardless of direction

look at the Lorentz boost formula for time dilation. Lets start with the same reference frame (equivalent to at rest). In this case between observers if both observers are at rest there is no time dilation. Both clocks tick at the same rate. When one clock goes to a different frame time dilation occurs. The tick rate is different from the clock at rest. Over a period of time you count the number of ticks for each clock. Now bring the second clock back to the same reference frame. You no longer have any lorentz boost. No longer have any time dilation between clocks. The clocks both tick at the same rate. However the recorded number of ticks (time elapsed) will be different. As I stated the accumulated time elapsed will be different. However the rate of ticks will restore. (reading your posts this aspect is where your getting confused.)

there is a difference between rate of time and accumulated number of ticks. The duration between clock ticks restore to being the same. The number of ticks will be different. You keep mixing these two up. It is the tick rate that is restored. Not the amount of time each clock records.

Correct however every country should have a range of amateur radio frequencies. Check your local regulations.

excellent +1particularly (locally chaotic/globally homogeneous and isotropic aka uniform)

Your missing a key detail on mass, which as you already know is "resistance to inertia change". however although the photon has no "rest mass" now called invariant mass. The photon has inertia mass equivalence. What used to be called relativistic mass. the formula [latex]e=mc^2[/latex] isn't complete, when you add the momentum term the total energy of the particle becomes. [latex]e^2=p^2c^2+m^2c^4[/latex] so lets use an everyday real world test... a particle accelerator fires two protons, as the protons gain inertia they gain total energy. This allows us to create particles will far higher mass than the rest mass of the two protons. Ever wonder how two protons could make a Higg's Boson? This applies to relativistic radiation. We apply the total energy of all particles, also the sequence that particles drop out of thermal equilibrium also depends on total energy of the particle. So the photon has one of the highest "total mass" values. Due to its extreme kinetic energy... https://en.wikipedia.org/wiki/Mass#Inertial_vs._gravitational_mass