# A moving clock lags behind a stationary clock

• myoho.renge.kyo
In summary: B lags behind the stationary clock at A by the amount... (1/2)*(t*v^2/c^2)...every time the moving clock at B is synchronized with the stationary clock at A would be wrong. In summary, Einstein explains in The Principle of Relativity that according to the equations x = vt and T = (t - v*x/c^2)/sqrt(1 - v^2/c^2), the time marked by a clock in a stationary system will appear slower when observed by a moving clock. This leads to the peculiar consequence that two initially synchronous clocks in the stationary system will no longer be synchronized when one is moved with a velocity v. The moving clock will lag behind the stationary clock by (1/2
myoho.renge.kyo
A. Einstein states the following on page 49 in The Principle of Relativity:

"Between the quantities x, t, and T, which refer to the position of the clock, we have evidently, x = vt and

T = (t - v*x/c^2)/sqrt(1 - v^2/c^2).

Therefore,

T = t*sqrt(1 - v^2/c^2) = t - (1 - sqrt(1 - v^2/c^2))*t

whence it follows that the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2).
From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B."

it is important to note here that "the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2)." for example, if the observer moving with the clock at A measures the value of t using the clock at A (the observer at A and the clock at A are at rest relatively to each other), and the stationary observer at B measures the value of t using the stationary clock at B, the value of t obtained by the obsever at A equals the value of t obtained by the observer at B.

but if the stationary observer at B measures the value of t using the stationary clock at B, and then the same stationary observer at B measures the value of t using the moving clock at A, then the value of t obtained by the stationary observer at B using the stationary clock at B is not equal to the value of t obtained by the same oberver using the moving clock at A. the moving clock lags behind the stationary clock "by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order)." but i don't know what magnitudes of fourth and higher orde is.

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myoho.renge.kyo said:
A. Einstein states the following on page 49 in The Principle of Relativity:

"Between the quantities x, t, and T, which refer to the position of the clock, we have evidently, x = vt and

T = (t - v*x/c^2)/sqrt(1 - v^2/c^2).

Therefore,

T = t*sqrt(1 - v^2/c^2) = t - (1 - sqrt(1 - v^2/c^2))*t

whence it follows that the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2).
From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B."

it is important to note here that "the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2)." for example, if the observer moving with the clock at A measures the value of t using the clock at A (the observer at A and the clock at A are at rest relatively to each other), and the stationary observer at B measures the value of t using the stationary clock at B, the value of t obtained by the obsever at A equals the value of t obtained by the observer at B.
It seems like all you're saying with this last part is that the two clocks at rest in the stationary system are synchronized, so that at any single moment in the stationary system's rest frame, according to that frame's definition of simultaneity, the clocks at A and B will both show the same time (which is almost a tautology, because these clocks are used to define the coordinate system that is 'the stationary system's rest frame').
myoho.renge.kyo said:
but if the stationary observer at B measures the value of t using the stationary clock at B, and then the same stationary observer at B measures the value of t using the moving clock at A,
If t is taken to mean the time-coordinate in the stationary system's rest frame, then you can't "measure the value of t using the moving clock"--the time t of an event in the stationary system's rest frame is defined in terms of the reading of one of the clocks which is at rest and synchronized in that frame, whichever clock was next to the event as it happened. The moving clock's readings are represented using the symbol T, not t, according to Einstein's quote above.

JesseM said:
It seems like all you're saying with this last part is that the two clocks at rest in the stationary system are synchronized, so that at any single moment in the stationary system's rest frame, according to that frame's definition of simultaneity, the clocks at A and B will both show the same time (which is almost a tautology, because these clocks are used to define the coordinate system that is 'the stationary system's rest frame').

yes. it is almost a tautology, but it isn't. the point is that to say that for all of us the moving clock at A is slower than the stationary clock at B is not correct because some of us are moving along with the clock. if the color of the clock at A is red and the color of the clock at B is blue, those moving along with the red clock (let's call them the red team) would say that the blue clock is slower than the red clock. the comon A and B time would be the red (and blue) clock at rest relatively to the red and blue team respectively.

JesseM said:
If t is taken to mean the time-coordinate in the stationary system's rest frame, then you can't "measure the value of t using the moving clock"--the time t of an event in the stationary system's rest frame is defined in terms of the reading of one of the clocks which is at rest and synchronized in that frame, whichever clock was next to the event as it happened. The moving clock's readings are represented using the symbol T, not t, according to Einstein's quote above.

yes. you are right. i should have used T for the time value of the event using the moving clock. thanks.

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## What is the concept of a moving clock lagging behind a stationary clock?

The concept of a moving clock lagging behind a stationary clock is a phenomenon in which a clock that is in motion, relative to an observer, appears to run slower than a clock that is stationary relative to the observer. This is due to the effects of time dilation, a principle in Einstein's theory of relativity.

## What causes a clock to lag behind when it is in motion?

The cause of a clock lagging behind when it is in motion is time dilation, which is a result of the speed of light being constant in all reference frames. As an object approaches the speed of light, time slows down for that object relative to a stationary observer. Therefore, a clock in motion appears to run slower than a stationary clock.

## How does the speed of the clock affect its lag behind a stationary clock?

The speed of the clock directly affects its lag behind a stationary clock. The faster the clock is moving, the greater the effect of time dilation, and the more it will lag behind a stationary clock. This is why the difference in time between a moving and stationary clock is only noticeable at extremely high speeds, such as those close to the speed of light.

## Does the direction of the clock's motion affect its lag behind a stationary clock?

No, the direction of the clock's motion does not affect its lag behind a stationary clock. The effects of time dilation are the same regardless of the direction of motion. However, the difference in time between a moving and stationary clock may be more noticeable if the clock is moving directly towards or away from the observer, as opposed to moving perpendicular to the observer's line of sight.

## Is the concept of a moving clock lagging behind a stationary clock just a theory?

No, the concept of a moving clock lagging behind a stationary clock is not just a theory. It has been proven through numerous experiments and is a fundamental principle in Einstein's theory of relativity. This phenomenon has also been observed in various real-world scenarios, such as in GPS satellites, where time dilation must be taken into account for accurate positioning calculations.

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