# How to measure time in reference frame with clock?

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• Mike_bb
Mike_bb
TL;DR Summary
Time in reference frame with clock
I considered example of time dilation with light clock. I have a question about measuring time in reference frame with clock.

If we know that clock move from A to B in the reference frame with clock then what time of motion is measured in this reference frame? (In non-moving reference frame time is Δt)

I supposed that time is Δt0 because light reach at end point at this time. But I think it's wrong. I have no more guess.

Thanks.

Mike_bb said:
I supposed that time is Δt0 because light reach at end point at this time. But I think it's wrong.
It's obviously wrong from your diagram, because the distance covered by the light beam in the moving light clock is not ##c \Delta t_0##, it's ##c \Delta t##. Which should make the time it takes in the given frame for the light beam to cover that distance obvious.

Mike_bb
Don't guess, compute!

Where is the light pulse when the clock is at A? Where is the light pulse when the clock is at B? How far has the light travelled, therefore? Hence, how long has it taken?

Mike_bb
Sorry. Picture in the first message is wrong.

Mike_bb said:
Picture in the first message is wrong.
What has changed? The picture you just posted looks the same as the one in the OP.

Mike_bb
PeterDonis said:
What has changed? The picture you just posted looks the same as the one in the OP.
I mean that vΔt is length of path in non-moving reference frame (time in this frame is Δt). And what is time in reference frame with clock?

Mike_bb said:
I mean that vΔt is length of path in non-moving reference frame (time in this frame is Δt).
Yes. So that's the time the light takes to travel from A to B in the non-moving frame, if the clock is moving in that frame.

Mike_bb said:
And what is time in reference frame with clock?
Do you mean the frame in which the clock is at rest? Isn't that what you've drawn in the left part of your diagram?

Mike_bb
PeterDonis said:
Do you mean the frame in which the clock is at rest? Isn't that what you've drawn in the left part of your diagram?
No. I mean moving frame with velocity v and with moving clock.

Mike_bb said:
No. I mean moving frame with velocity v and with moving clock.
Then what does the left part of your diagram represent?

Mike_bb
PeterDonis said:
Then what does the left part of your diagram represent?
It represent rest frame with zero velocity with rest clock. But I mean right diagram when I say about moving frame.

Mike_bb said:
It represent rest frame with zero velocity with rest clock.
And that means it represents any light clock with the same parameters (i.e., the same ##L## between its mirrors) in the frame in which it is at rest.

Mike_bb said:
I mean right diagram when I say about moving frame.
Yes, a frame in which the clock is moving. But now you are asking about a frame in which the clock is at rest. Which is exactly what the left diagram shows. See above.

Mike_bb
PeterDonis said:
But now you are asking about a frame in which the clock is at rest.
Ok. I ask about time in moving frame. I want to know time of motion from A to B in moving reference frame. In the rest frame time is Δt.

Mike_bb said:
I want to know time of motion from A to B in moving reference frame.
What do you mean by "moving reference frame"? You appear to mean a frame that is moving at the same ##v## as the clock. That means the clock is at rest in that frame.

Mike_bb
PeterDonis said:
That means the clock is at rest in that frame.
Sorry, it's my error. Yes. Clock in moving frame is at rest. And I want to know what time is measured by clock in this moving frame.

Mike_bb said:
Clock in moving frame is at rest. And I want to know what time is measured by clock in this moving frame.
Since the clock is at rest in the frame, the left side of your diagram represents what happens. That is the point I have been trying to get across to you. So you can just look at the left side of your diagram to get the answer to your question.

Mike_bb
PeterDonis said:
Since the clock is at rest in the frame, the left side of your diagram represents what happens. That is the point I have been trying to get across to you. So you can just look at the left side of your diagram to get the answer to your question.
I wrote that time is Δt0, is it true? But I'm confused because if we're moving from A to B then our distance is vΔt for our reference frame. And if time is Δt0 then distance is vΔt0. It's wrong result.

Mike_bb said:
I wrote that time is Δt0, is it true?
For the case where the clock is at rest, that is what the left part of your diagram says, yes.

Mike_bb said:
I'm confused because if we're moving from A to B then our distance is vΔt for our reference frame. And if time is Δt0 then distance is vΔt0. It's wrong result.
No, it isn't, you're confusing two different things: the time taken in a frame in which the clock is moving, and the time taken in a frame in which the clock is at rest. Your use of the term "moving frame" to denote a frame in which the clock on the right of your diagram is at rest is probably contributing to your confusion.

Perhaps it will help you to think of two separate clocks. In the diagram you have drawn, the clock on the left is at rest and the clock on the right is moving. So in that frame, the time it takes the light beam in the left clock is ##t_0##, and the time it takes the light beam in the right clock is ##t##.

But you could also draw another diagram, in the frame in which the right clock is at rest. In this diagram, on the right there would be a clock at rest (the one that's on the right and moving in your original diagram), and on the left there would be a clock moving to the left with speed ##v## (the one that's on the left in your original diagram). And in that diagram, drawn in a different frame, the time it takes the light beam in the right clock would be ##t_0##, and the time it takes the light beam in the left clock would be ##t##.

Mike_bb
PeterDonis said:
the time taken in a frame in which the clock is moving, and the time taken in a frame in which the clock is at rest.
Why do you say about light beam? I mean another case. Say what time do you measure in moving frame when you travel from A to B?

Mike_bb said:
Why do you say about light beam?
Um, because your diagram is a diagram of a light clock?

Mike_bb said:
I mean another case. Say what time do you measure in moving frame when you travel from A to B?
The light clock is a clock; it measures time--the time it takes the light beam to go back and forth between its mirrors. That time is the time you have been asking about.

Mike_bb
This conversation seems enormously confusing.

The diagram in #1 has two clocks, the one on the left which is stationary in the frame in which the diagram is drawn, and the one on the right which is moving to the right with velocity ##v##. In this frame, the period of the left clock is ##\Delta t_0## (well, twice that actually because the light pulse needs to travel down and up, but it doesn't matter here). The period of the clock on the right is ##\Delta t##, which turns out to be ##\left(1-v^2/c^2\right)^{-1/2}\Delta t_0##.

As far as I can work out, OP is asking what are the periods of the two clocks in a frame where the right clock is stationary and the left clock is moving to the left with speed ##v## (i.e. velocity ##-v##). If that is correct, then the answer is simply that the periods swap: now the right clock has period ##\Delta t_0## and the left clock has period ##\Delta t##.

Mike_bb
Ok. I'll try to rephrase.

Let's write equation for non-moving reference frame:

##(cΔt)^2 = (cΔt0)^2 + (vΔt)^2##

How to write equation for moving reference frame?

Mike_bb said:
How to write equation for moving reference frame?
Do you mean the frame where the right hand clock is at rest? And how are you arriving at that equation? Are you just applying Pythagoras' Theorem to the triangle I've marked in red below?

Mike_bb
Thanks to all! My mistake was that I tried to find time of motion in moving frame but distance in this frame is 0 because clock in moving frame is at rest.

Motore

## What is a reference frame in the context of measuring time?

A reference frame is a perspective from which an observer measures and records events, including the passage of time. It can be stationary or moving, and the measurements of time can differ depending on the relative motion between different reference frames.

## How does a clock measure time in its own reference frame?

A clock measures time in its own reference frame by counting the regular intervals of a periodic process, such as the oscillations of a pendulum or the vibrations of a quartz crystal. In its own reference frame, a clock runs at its normal rate, unaffected by any relative motion to other reference frames.

## What is time dilation and how does it affect time measurement in different reference frames?

Time dilation is a phenomenon predicted by Einstein's theory of relativity, where time runs slower for an object in motion relative to a stationary observer. This means that a clock moving at high speed relative to an observer will measure time more slowly compared to a clock at rest in the observer's reference frame.

## How do you synchronize clocks in different reference frames?

Synchronizing clocks in different reference frames can be challenging due to the effects of relative motion and time dilation. One common method involves using light signals. Observers can exchange light signals and account for the travel time of the signals to adjust their clocks accordingly. However, perfect synchronization is often impossible due to the relativity of simultaneity.

## What role does the speed of light play in measuring time across reference frames?

The speed of light is a constant in all reference frames and plays a crucial role in the theory of relativity. It acts as a limiting speed for information and matter. When measuring time across reference frames, the constancy of the speed of light ensures that the effects of time dilation and length contraction are consistent and predictable, allowing for accurate calculations of time differences between moving and stationary observers.

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