s***@yahoo.com

2021-04-21 12:02:50 UTC

Reply

PermalinkThanks,

David Seppala

Bastrop TX

Discussion:

Add Reply

s***@yahoo.com

2021-04-21 12:02:50 UTC

Reply

PermalinkThanks,

David Seppala

Bastrop TX

rotchm

2021-04-21 12:55:48 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0). An identical clock, C1, is traveling in a circular path around that point with angular velocity V as measured in F0. When the traveling clock C1 makes a complete circle, which clock shows the greater elapsed time or do they show the identical elapsed times?

Thanks,

David Seppala

Bastrop TX

If you were to actually do such an exp, how would you go about it, to know "which clock shows the greater elapsed time "?Thanks,

David Seppala

Bastrop TX

Icke Biggers

2021-04-21 13:57:57 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0).

An identical clock, C1, is traveling in a circular path around that

point with angular velocity V as measured in F0. When the traveling

clock C1 makes a complete circle, which clock shows the greater elapsed

time or do they show the identical elapsed times?

Thanks,

David Seppala Bastrop TX

know "which clock shows the greater elapsed time "?

anything, brought back and compared at origo (0,0). And angular velocity

is omega, not V.

s***@yahoo.com

2021-04-21 14:43:33 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0).

An identical clock, C1, is traveling in a circular path around that

point with angular velocity V as measured in F0. When the traveling

clock C1 makes a complete circle, which clock shows the greater elapsed

time or do they show the identical elapsed times?

Thanks,

David Seppala Bastrop TX

know "which clock shows the greater elapsed time "?

anything, brought back and compared at origo (0,0). And angular velocity

is omega, not V.

David Seppala

Bastrop TX

rotchm

2021-04-21 14:46:46 UTC

Reply

PermalinkYou got got, which shows that you don't have the brains to be a 'sensible' person.

Icke Biggers

2021-04-21 15:10:50 UTC

Reply

PermalinkComplaints-To: groups-***@google.com

Injection-Info: google-groups.googlegroups.com; posting-

host=184.160.32.227;

posting-account=BHsbrQoAAAANJj6HqXJ987nOEDAC1EsJ

NNTP-Posting-Host: 184.160.32.22

idiot sepala, you just replied to the troll. Don't do that; DON'T STROKE THE TROLLS.

You got got, which shows that you don't have the brains to be a

'sensible'

person.

around here, as I can read from the other different people. Nobody gives

a damn shit on what you say or what you think you say.

Icke Biggers

2021-04-21 15:04:15 UTC

Reply

PermalinkIf you were to actually do such an exp, how would you go about it, to

know "which clock shows the greater elapsed time "?

disregard anything, brought back and compared at origo (0,0). And

angular velocity is omega, not V.

the radius of the circle very very large. Then during any arc length

the

distance, then compared at origo again. Otherwise your question makes not

much sense.

Rob Acraman

2021-04-22 11:49:50 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0).

An identical clock, C1, is traveling in a circular path around that

point with angular velocity V as measured in F0. When the traveling

clock C1 makes a complete circle, which clock shows the greater elapsed

time or do they show the identical elapsed times?

Thanks,

David Seppala Bastrop TX

know "which clock shows the greater elapsed time "?

anything, brought back and compared at origo (0,0). And angular velocity

is omega, not V.

David Seppala

Bastrop TX

From your OP, let's say Alice is at the origin (0,0), and Bob is travelling in a circle with radius r - so going through (0,r), (r, 0), (0, -r), (-r, 0) and back to (0, r).

So how is that different from Bob instead going in a circle going through (0, 0), (-r, r), (-2r, 0), (-r, -r), and back to (0, 0) ???

The latter treatment matches exactly what Einstein wrote in OTEOMB, where (0,0) is the "point A" :

"If one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the travelled clock on its arrival at A will be 1/2 tv^2/c^2 second slow."

Why would you think that there could be different results from a circle AROUND a "stationary" clock compared to a circle going through that "stationary" clock ?

s***@yahoo.com

2021-04-21 14:32:21 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0). An identical clock, C1, is traveling in a circular path around that point with angular velocity V as measured in F0. When the traveling clock C1 makes a complete circle, which clock shows the greater elapsed time or do they show the identical elapsed times?

Thanks,

David Seppala

Bastrop TX

If you were to actually do such an exp, how would you go about it, to know "which clock shows the greater elapsed time "?Thanks,

David Seppala

Bastrop TX

David Seppala

Bastrop TX

rotchm

2021-04-21 14:44:12 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0). An identical clock, C1, is traveling in a circular path around that point with angular velocity V as measured in F0. When the traveling clock C1 makes a complete circle, which clock shows the greater elapsed time or do they show the identical elapsed times?

Thanks,

David Seppala

Bastrop TX

If you were to actually do such an exp, how would you go about it, to know "which clock shows the greater elapsed time "?Thanks,

David Seppala

Bastrop TX

When the moving clock once again crosses the y-axis of F0, record the time of both clocks.

Try again.

Odd Bodkin

2021-04-21 15:40:41 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at

(0,0). An identical clock, C1, is traveling in a circular path around

that point with angular velocity V as measured in F0. When the

traveling clock C1 makes a complete circle, which clock shows the

greater elapsed time or do they show the identical elapsed times?

Thanks,

David Seppala

Bastrop TX

know "which clock shows the greater elapsed time "?

how you’d expect the clock at the center of the circle to know exactly when

the clock at the perimeter of the circle has crossed the y-axis of the

circle, in order to reset the clock.

When the moving clock once again crosses the y-axis of F0, record the

time of both clocks.

Try again.

--

Odd Bodkin -- maker of fine toys, tools, tables

Odd Bodkin -- maker of fine toys, tools, tables

s***@yahoo.com

2021-04-21 21:44:33 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at

(0,0). An identical clock, C1, is traveling in a circular path around

that point with angular velocity V as measured in F0. When the

traveling clock C1 makes a complete circle, which clock shows the

greater elapsed time or do they show the identical elapsed times?

Thanks,

David Seppala

Bastrop TX

know "which clock shows the greater elapsed time "?

how you’d expect the clock at the center of the circle to know exactly when

the clock at the perimeter of the circle has crossed the y-axis of the

circle, in order to reset the clock.

When the moving clock once again crosses the y-axis of F0, record the

time of both clocks.

Try again.

Odd Bodkin -- maker of fine toys, tools, tables

David Seppala

Bastrop TX

Odd Bodkin

2021-04-21 21:53:37 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at

(0,0). An identical clock, C1, is traveling in a circular path around

that point with angular velocity V as measured in F0. When the

traveling clock C1 makes a complete circle, which clock shows the

greater elapsed time or do they show the identical elapsed times?

Thanks,

David Seppala

Bastrop TX

know "which clock shows the greater elapsed time "?

both clocks to zero.

how you’d expect the clock at the center of the circle to know exactly when

the clock at the perimeter of the circle has crossed the y-axis of the

circle, in order to reset the clock.

When the moving clock once again crosses the y-axis of F0, record the

time of both clocks.

Try again.

Odd Bodkin -- maker of fine toys, tools, tables

circle, and put an identical clock at rest where the y axis crosses the

circle. Both of those clocks run at the identical rate. When the moving

clock passes the clock that is sitting at rest on the circumference of

the circle, record the time of that clock at rest and record the time of

the clock traveling around the circumference of the circle. When the

circling clock once again passes the clock that is at rest on the

circumference of the circle record the elapsed time of each of those two clocks.

David Seppala

Bastrop TX

of the circle know “when” the clock at the perimeter has passed any point

out there on the perimeter, in order to record the time showing on the

clock at the center? It’s a good ways away, is it not?

If it helps, think of it this way. Suppose it was your job to watch a clock

and listen for the thunder from a lightning strike from a storm on the

horizon. You hear a thunderclap when the clock near you reads 11:23:07. You

write that down. Have you recorded the time when that lightning strike

happened?

--

Odd Bodkin -- maker of fine toys, tools, tables

Odd Bodkin -- maker of fine toys, tools, tables

rotchm

2021-04-21 23:07:30 UTC

Reply

PermalinkIn inertial reference frame F0, put a clock at rest in the center of the circle, and put an identical clock at rest where the y axis crosses the circle. Both of those clocks run at the identical rate. When the moving clock passes the clock that is sitting at rest on the circumference of the circle, record the time of that clock at rest and record the time of the clock traveling around the circumference of the circle. When the circling clock once again passes the clock that is at rest on the circumference of the circle record the elapsed time of each of those two clocks.

David Seppala

Bastrop TX

That is, C(1,0) and and C1 cross their paths twice, and you want the elapsed times?

Maciej Wozniak

2021-04-21 12:56:42 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0). An identical clock, C1, is traveling in a circular path around that point with angular velocity V as measured in F0. When the traveling clock C1 makes a complete circle, which clock shows the greater elapsed time or do they show the identical elapsed times?

We have GPS now, anyone can check that relativisticbullshit is just some reality enchanting.

Al Coe

2021-04-21 15:00:57 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0). An identical clock, C1, is traveling in a circular path around that point with angular velocity V as measured in F0. When the traveling clock C1 makes a complete circle, which clock shows the greater elapsed time or do they show the identical elapsed times? (When the moving clock crosses the positive y axis of F0, set both clocks to zero. When the moving clock once again crosses the y-axis of F0, record the time of both clocks.)

As always, each time you use the word "when" to refer to simultaneous separate events, you need to specify the temporal foliation you are referring to. In terms of standard inertial coordinates x,y,t, in which C0 is stationary, and assuming that at t=t0 the readings of both clocks are set to zero and C1 crosses the positive y axis, and the next time C2 crosses the positive y axis is at t=t1, the reading of C0 at t=t1 will be t1, and the reading of C1 at t=t1 will be t1*sqrt(1-V^2/c^2).
Al Coe

2021-04-21 15:15:07 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0). An identical clock, C1, is traveling in a circular path around that point with angular velocity V as measured in F0. When the traveling clock C1 makes a complete circle, which clock shows the greater elapsed time or do they show the identical elapsed times? (When the moving clock crosses the positive y axis of F0, set both clocks to zero. When the moving clock once again crosses the y-axis of F0, record the time of both clocks.)

As always, every time you use the word "when" to refer to simultaneous separate events, you need to specify the temporal foliation you are referring to. In terms of standard inertial coordinates x,y,t, in which C0 is stationary, and assuming that at t=t0 the readings of both clocks are set to zero and C1 crosses the positive y axis, and the next time C2 crosses the positive y axis is at t=t1, the reading of C0 at t=t1 will be t1, and the reading of C1 at t=t1 will be t1*sqrt(1-V^2/c^2).Then during any arc length the traveling clock will show less time as you say then the

clock at the center of the circle.

Again, in terms of inertial coordinate system x,y,t in which C0 is stationary, the rate of elapsed proper time for C0 is dtau/dt = 1 and the rate of elapsed proper time for C1 is dtau/dt = sqrt(1-V^2/c^2).clock at the center of the circle.

During this arc length let another clock [C2] in an inertial reference frame travel in a straight

line with velocity V so that that clock and the clock traveling along the arc both have

almost identical velocities except the clock traveling on the arc travels in a very slightly

curved path.

Yawn. This old canard. Are you, like, 10 years old?line with velocity V so that that clock and the clock traveling along the arc both have

almost identical velocities except the clock traveling on the arc travels in a very slightly

curved path.

Do both of these clocks run at nearly the same rate with the clock traveling

along the arc only very slightly slower?

Both clocks C1 and C2 run at the rate dtau/dt = sqrt(1-V^2/c^2). In general (flat spacetime), any clock moving at speed V in terms of inertial coordinates x,y,t runs at the rate dtau/dt = sqrt(1-V^2/c^2). Of course, in terms of inertial coordinates x',y',t' in which C2 is at rest (and C1 is momentarily at rest), clock C0 is moving at speed V, so it runs at the rate dtau/dt' = sqrt(1-V^2/c^2). A little while later, C1 will be at rest in a slightly different inertial coordinate system, x",y",t", and it will be running at the rate dtau/d" = 1, and clock C0 will be running at the rate dtau/dt" = sqrt(1-V^2/c^2). And so on.along the arc only very slightly slower?

Dono.

2021-04-21 23:07:06 UTC

Reply

PermalinkThanks,

David Seppala

Bastrop TX

Think about this simple exercise this way:

-you add another clock on the circumference of the circle, let's call it C2

-you synchronize C2 with C0, this is very easy because the two clocks are at rest wrt each other

-before you put C1 in motion, you synchronize it with C2. this means that C0,C1 and C2 are all synchronized

-you put C1 in motion

-when C1 completes the circle, you compare it with C2, that should be easy because their positions will coincide

-conform SR, C1 will show less elapsed time than C2 (and than C0 , because C2 and C) show the SAME elapsed time)

DONE

Sylvia Else

2021-04-21 23:29:06 UTC

Reply

PermalinkIn inertial reference frame F0 there is a clock, C0, centered at (0,0). An identical clock, C1, is traveling in a circular path around that point with angular velocity V as measured in F0. When the traveling clock C1 makes a complete circle, which clock shows the greater elapsed time or do they show the identical elapsed times?

Thanks,

David Seppala

Bastrop TX

Let clock C1 travel in a circle. Put clock C0 at some point on that

circle. When C1 and C0 first coincide, and set them to show time = 0.

Compare the shown times when they next coincide.

Sylvia.

Maciej Wozniak

2021-04-22 03:55:33 UTC

Reply

PermalinkThis is needlessly complicated.

Let clock C1 travel in a circle. Put clock C0 at some point on that

circle. When C1 and C0 first coincide, and set them to show time = 0.

Compare the shown times when they next coincide.

You need is to check that the prophesies of Your

relativity are worthless.

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