Mordred

its more commonly used in electrical and electronic textbooks for one thing. Also if the poster is a student his physics class in Highschool will specify flow of charge and not flow of electrons which propogate through a medium at different rates

to add to this your better of following the flow of charge

trolling tactics you are aware I am a Resident Expert on this forum correct ? Any clock at a given reference frame is the coordinate time. Proper time is a clock that follows the Worldline given by the line element of the applicable geometry. usually denoted by the separation distance between two events. For Euclidean geometry ( I did provide links earlier to coordinate time and proper time ) \[ds^2=dx^2+dy^2+dz^2\] The worldline is the null geodesic path which is represented by the line element or if you prefer the affine connections given by the Christoffel connections. The line element is typically easier to understand though lol. Though the geodesic equation includes the Christoffel here is the Minkowskii line element \[ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}\]

your welcome. the criteria to meet is given in the first equation that the distance does not change between any two particles. \[ V(r,t)=V_0(t)+\Omega(t)r\] is the solution to the first equation \[\Omega(t)\] being an antisymmetric tensor that's the criteria. the rest of the article deals with the examination Ok your not familiar with GR terminology fair enough https://en.wikipedia.org/wiki/Coordinate_time please note that the following from the article.

Correct it is precisely about the rigid rod have you looked at the link I posted or did you miss it?

Have you looked at the article I posted yet Lorentz ?

This is mainstream physics my earlier statements stress that under GR there is no rigid rods thus applies to the train no matter how its being examined

Sure I can go with that Here is the Born Rigidity examination https://fnegari.github.io/files/notes/009.pdf From this article it should be clear just how tricky any rigid rod examination can get

No we're still talking about the same topic a train are we to assume the train never turns ? not that it matters in the Lorentz gauge acceleration is a rotation called rapidity.

I still believe your missing the point. Even if every single coordinate was effectively it's own engine and you could contrive some means of synchronization. Once the train needs to turn you would end up with differing accelerations. So once again we're stuck with needing communication. The reason I asked how much GR you understood is that I wanted an idea of your math skills. Are you familiar with the four momentum in mathematical form ?

Nope just examining the situation under GR I do believe this is the relativity forum unless things have changed. I know you ate familiar enough with relativity to recognize the rigid rod conjectures in regards to relativity I was making Lorentz aware that proper time differs from coordinate time in regards to Each engine clock will be in coordinate time. Anyways it will be interesting when the train tries to turn.... so far the examination has been strictly linear.

Ah so now we're changing goalposts. I was showing the problems of Born rigidity. Either way there are no rigid objects under GR.

OK let's examine it. Let's assume the signal is sent by the lead train. Each engine will receive that signal progressively later than the previous engine. Do you consider that simulateneous ? The speed limit of c will always apply it doesn't matter if the signal is through EM frequencies, or transmitted via particle to particle interactions through the train body (which actually transmits less than c) vibration travels at the speed of sound however a hypothetical perfect rigid rod the speed of sound can be treated at the speed of light. So do this assign an event at each box car or engine in your train. Assign any engine or box car as the transmitter. At no point will every event receive the signal simultaneous. That would require instantaneous communication.

How much GR have you studied ? Under GR coordinate time is the time at each event. The proper time follows the worldline. https://en.m.wikipedia.org/wiki/Proper_time What I stated earlier stands

Proper time would follow the wordline between engines so you still have the same problem

The communication between every engine would also be affected so you still wouldn't have simultaneous acceleration not under rigorous treatment with GR being applied.

Unfortunately a simulataneous acceleration of a train would impossible taking into consideration the speed limit of information exchange. This thought experiment would be accurately described via the Rigid Rod under GR. In essence let's make your train one light year hypothetically. If the force for acceleration starts at say the engine. The tail end would not accelerate until one year later. Thats assuming the medium is ideal enough to allow the signal to propagate at c. Which is your speed limit.

Well covered Markus I was going you reply many of the details you have above however you covered everything I was going to say. Along with additional detail +1

Yeah looks like a typing error will have to double check that but yes any divergence would lead to a curvature divergents. Hence it's still viable our universe has a slight curvature. That's still viable for both Plus or minus curvature.

\[\frac{1}{2}\]

There is little to no reason for the density parameter to change as one can accurately treat expansion as a closed adiabatic perfect fluid. lets put some math to that using The FLRW metric. the GR form of the FLRW equation is \[(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\frac{\epsilon(t)}{c^2}-\frac{k c^2}{R^2_0}\frac{1}{a^2(t)}\] k=0 , curvature \[\frac{\epsilon(t)}{c^2}\] is the energy density in the Babera Ryden notation as opposed to mass density \[\rho\] the reason will become clear later on \[\rho_c(t)=\frac{\epsilon(t)}{c^2}=\frac{3H^2(t)}{8\pi G}\] \[H=\frac{\dot{a}}{a}\] critical density value present day value approx 70 km/sec/Mpc \[\rho_c]+\frac{\epsilon_c}{c^2}=\frac{3H_0^2}{8\pi G}=9.2*10^3 g cm^3\]using the 70 km/sec/Mpc \[H^2=\Omega H^2-\frac{kc^2}{R^2_0a^2(t)}\Rightarrow1-\Omega(t)=\frac{kc^2}{H^2(t)a^2(t)R^2_0}\] if \[\Omega=1\] then it equals one at all times since the RHS of the last equation always vanishes for the flat case for the \[\Omega>1,\Omega<1\] the value may change however never change sign ie positive curvature will change but never become negative curvature Now for adiabatic fluid first law of thermodynamics \[dE-PDV+DQ\] the change in internal energy equates to the sum of PDV work and added heat/energy however there is no place for heat/energy to come from or leave the system therefore\[DE+pdV=0\Rightarrow \dot{E}+p\dot{V}=0\] for a commoving sphere \[V=\frac{4\pi}{3}r^3_sa^3{t}\] \[\dot{V}=\frac{4\pi}{3} r^3_s(3a^2\dot{a})=V \frac{\dot{a}}{a}\] \[E=V_\epsilon\] \[\dot{E}=V\dot{\epsilon}+\dot{V_{\epsilon}}\]\[=V\dot{\epsilon}+3\frac{\dot{a}}{a}\epsilon\] with \[\dot{E}+P\dot{V}=0 \]we get \[V\dot{\epsilon}+3\frac{\dot{a}}{a}\epsilon+3\frac{\dot{a}}{a}P=0\] thus \[\dot{\epsilon}+3\frac{\dot{a}}{a}(\epsilon+P)=0\] which is the same as the fluid equation standard notation \[\dot{\rho}+3\frac{\dot{a}}{a}(\rho+P)=0\] there's the first law of thermodynamics as its a closed system according to this examination conservation of energy would apply however this doesn't examine quantum fluctuations or the cosmological constant.

Hrrm interesting thought. Technically the Langrangian paths of the particle interactions will follow the path of least action. So there may very well be some truth in that statement.

Interesting read will have to study it later. Thanks for sharing

Position and momentum are the operators in QM. Klien Gordon equation uses potential and kinetic energy (QFT)

I give up this is the 4th time this got screwed up the last time was when i tried inserting the url for an image