If there is an inertial reference frame, F0, that has a disk centered at (0,0) and there is a clock at rest at (0,0) and another identical clock on the radius of this disk, both clocks run at the same rate. Per Einstein's theory, if the disk starts spinning at some angular velocity V (say V=sqrt(3)/2*c), then the clock on the radius of the disk runs at a slower rate than the clock at the center of the disk as measured by observers in F0.
What happens if there is a power source at the center of the disk along with an ammeter with two wires that extend to the radius of the disk to a resistor and another ammeter. Let's say when the disk is not spinning, both ammeters read one ampere of current continually flowing through the resistor. Now, as in the example above with clocks, if the disk starts spinning with angular velocity V (again around V=c*sqrt(3)/2), what do the two ammeters now read? The ammeters read the coulomb of charge per second. What readings do each of the ammeters now show?
Thanks,
David Seppala
Bastrop TX

sepp623@yahoo.com wrote:

> If there is an inertial reference frame, F0, that has a disk centered at
> (0,0) and there is a clock at rest at (0,0) and another identical clock
> on the radius of this disk, both clocks run at the same rate. Per
> Einstein's theory, if the disk starts spinning at some angular velocity
> V (say V=sqrt(3)/2*c), then the clock on the radius of the disk runs at
> a slower rate than the clock at the center of the disk as measured by
> observers in F0.

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On Saturday, January 8, 2022 at 9:01:51 AM UTC-8, sep...@yahoo.com wrote:
> If there is an inertial reference frame, F0, that has a disk centered at (0,0) and
> there is a clock at rest at (0,0) and another identical clock on the radius of this
> disk, both clocks run at the same rate. Per Einstein's theory, if the disk starts
> spinning at some angular velocity V (say V=sqrt(3)/2*c), then the clock on the
> radius of the disk runs at a slower rate than the clock at the center of the disk
> as measured by observers in F0.

Translation: Consider a solid stationary disk centered at the origin of a system S of inertial coordinates x,y,t, and then suppose the disk is set into rotation, and it undergoes the stresses and strains resulting from this, and reaches a steady-state condition of rotation such that the perimeter has a tangential speed v. The rate in terms of S of an ideal clock attached to the perimeter is dtau/dt = sqrt(1 - v^2/c^2).

> Now suppose there is a power source at the center of the disk along with an
> ammeter with two wires that extend to the radius of the disk to a resistor and
> another ammeter... The ammeters read the coulomb of charge per second. What
> do the two ammeters now read?

This kind of experiment has been performed many times, testing the "centrifugal redshift" between the center and the perimeter of a spinning disk. Charge is conserved, so this is equivalent to just noting the rate of emission/return of electrons at the center with the rate of arrival/departure at the rim. You can do the same thing with photons or wave crests, etc. We always find the expected frequency shift, corresponding to the shift in current. It's like having electrons in a loop, going from center to rim and back. The difference in the rate is just the different in the proper times on the stationary foliation. In terms of S the clock on the rim is slow by the factor sqrt(1 - v^2/c^2), so the number of charges per second is greater by that factor than at the center. That's what the ammeters read.

Special Relativity: 846 ... Barnpole Dave: 0

On Saturday, January 8, 2022 at 12:52:45 PM UTC-6, Al Coe wrote:
> On Saturday, January 8, 2022 at 9:01:51 AM UTC-8, sep...@yahoo.com wrote:
> > If there is an inertial reference frame, F0, that has a disk centered at (0,0) and
> > there is a clock at rest at (0,0) and another identical clock on the radius of this
> > disk, both clocks run at the same rate. Per Einstein's theory, if the disk starts
> > spinning at some angular velocity V (say V=sqrt(3)/2*c), then the clock on the
> > radius of the disk runs at a slower rate than the clock at the center of the disk
> > as measured by observers in F0.
> Translation: Consider a solid stationary disk centered at the origin of a system S of inertial coordinates x,y,t, and then suppose the disk is set into rotation, and it undergoes the stresses and strains resulting from this, and reaches a steady-state condition of rotation such that the perimeter has a tangential speed v. The rate in terms of S of an ideal clock attached to the perimeter is dtau/dt = sqrt(1 - v^2/c^2).
>
> > Now suppose there is a power source at the center of the disk along with an
> > ammeter with two wires that extend to the radius of the disk to a resistor and
> > another ammeter... The ammeters read the coulomb of charge per second. What
> > do the two ammeters now read?
> This kind of experiment has been performed many times, testing the "centrifugal redshift" between the center and the perimeter of a spinning disk. Charge is conserved, so this is equivalent to just noting the rate of emission/return of electrons at the center with the rate of arrival/departure at the rim. You can do the same thing with photons or wave crests, etc. We always find the expected frequency shift, corresponding to the shift in current. It's like having electrons in a loop, going from center to rim and back. The difference in the rate is just the different in the proper times on the stationary foliation. In terms of S the clock on the rim is slow by the factor sqrt(1 - v^2/c^2), so the number of charges per second is greater by that factor than at the center. That's what the ammeters read.
>
> Special Relativity: 846 ... Barnpole Dave: 0
Do both ammeters change their readings, or is it just one or the other? If their readings each were one ampere when the disk was not spinning, what value is now shown on each of the ammeters?
Thanks,
David Seppala
Bastrop TX

On Sunday, January 9, 2022 at 11:08:05 AM UTC-8, sep...@yahoo.com wrote:
> Do both ammeters change their readings, or is it just one or the other?

The ratio of the readings (at steady state) is what we can infer from the stated conditions. If you are interested in the details of how the effective resistance of a circuit changes when subjected to such a drastic change of state, with the associated dynamical factors, you need to specify the properties of the materials and the dynamics, etc. For example, the disk might break apart, and even if it doesn't, it will certainly be subjected to severe stress and strain, so even radial wires would generally be strained, affecting their resistance, along with the dynamical (centrifugal, Coriolis, and Euler) effects. This is why I carefully mentioned the stress and strain during the spin up in my previous message (which you ignored).

You could just stipulate that you maintain constant current at the hub, sending a fixed number of electrons per second, which would typically require modulating the electromotive force, and then the current at the rim (electrons per second) increases, i.e., is blue shifted. Conversely you could modulate to maintain constant current at at the rim, and then the current at the hub would be reduced. If you take no action to modulate, both currents would likely change, even assuming the circuit remains intact. This all depends on factors that you have (as always) failed to specify... and that are irrelevant to the correctness of special relativity.

> If their readings each were one ampere when the disk was not spinning,
> what value is now shown on each of the ammeters?

Again, we can infer the ratio of the readings in a steady-state condition of spin, assuming a steady-state is achieved, but the absolute values will depend on the detailed stress and strains to which the circuit elements are subjected, which you have failed to specify (as always).

Special Relativity: 847 .... Barnpole Dave: 0

On Sunday, January 9, 2022 at 2:24:26 PM UTC-6, Al Coe wrote:
> On Sunday, January 9, 2022 at 11:08:05 AM UTC-8, sep...@yahoo.com wrote:
> > Do both ammeters change their readings, or is it just one or the other?
> The ratio of the readings (at steady state) is what we can infer from the stated conditions. If you are interested in the details of how the effective resistance of a circuit changes when subjected to such a drastic change of state, with the associated dynamical factors, you need to specify the properties of the materials and the dynamics, etc. For example, the disk might break apart, and even if it doesn't, it will certainly be subjected to severe stress and strain, so even radial wires would generally be strained, affecting their resistance, along with the dynamical (centrifugal, Coriolis, and Euler) effects. This is why I carefully mentioned the stress and strain during the spin up in my previous message (which you ignored).
>
> You could just stipulate that you maintain constant current at the hub, sending a fixed number of electrons per second, which would typically require modulating the electromotive force, and then the current at the rim (electrons per second) increases, i.e., is blue shifted. Conversely you could modulate to maintain constant current at at the rim, and then the current at the hub would be reduced. If you take no action to modulate, both currents would likely change, even assuming the circuit remains intact. This all depends on factors that you have (as always) failed to specify... and that are irrelevant to the correctness of special relativity.
> > If their readings each were one ampere when the disk was not spinning,
> > what value is now shown on each of the ammeters?
> Again, we can infer the ratio of the readings in a steady-state condition of spin, assuming a steady-state is achieved, but the absolute values will depend on the detailed stress and strains to which the circuit elements are subjected, which you have failed to specify (as always).
>
> Special Relativity: 847 .... Barnpole Dave: 0

If the radius is very large or the angular velocity is very small aren't all those factors you ask about insignificant? If there was a clock on the edge of the disk, are suggesting that the time shown on the clock very much depends on the clock's construction (mechanical or electrical) more so than Einstein's concept of relative velocity?
David Seppala
Bastrop TX

On Monday, January 10, 2022 at 7:47:38 AM UTC-8, sep...@yahoo.com wrote:
> If the radius is very large or the angular velocity is very small aren't all those
> factors you ask about insignificant?

No, not at all. For example, the circumference of the disk (at a given radius) would have to be reduced by a factor of two if the rim speed is sqrt(3)/2, regardless of the radius and angular velocity. [Please note that "angular velocity" means the angular rate of rotation, not the velocity of the rim, which you have specified.] The disk will need to drastically deform (strain) radially and/or circumferentially to arrive at a rotating stable configuration. Again, this applies for the specified rim speed, regardless of radius and angular speed (noting that the product is consistent with the specified rim speed). You shouldn't have to have simple facts like this explained to you.

> If there was a clock on the edge of the disk, are [you] suggesting that the time shown
> on the clock very much depends on the clock's construction (mechanical or electrical)
> more so than Einstein's concept of relative velocity?

Your brain is completely malfunctioning. I specifically said that the clock rates are the things that are fully specified, and that what is unspecified in your scenario is how the effective resistance and overall configuration of the disk and wires is drastically altered by the deformation resulting from the spin-up. If you wanted to describe a scenario that avoids all these factors, then just imagine a clock moving around a circle at speed v, and imagine sending electrons (or photons or wave crests) from the hub to the revolving clock, and ask if special relativity predicts the frequency shift that special relativity predicts. The answer (duh) is yes. Understand?

Special Relativity: 848 .... Barnpole Dave: 0