Two observers must agree what each clock reads....

In summary, the clocks X and Y appear to advance slower than the clock on the spaceship according to the spaceship observer. When the spaceship passes over clock Y, it reads a later time than the spaceship's clock. This may seem contradictory since clocks X and Y are synchronized in the Earth frame, but this is due to the difference in frames of reference and the concept of length contraction. Clock Y is actually ahead of the spaceship's clock when it passes over it, as observed by the spaceship observer.
  • #1
Phynos
31
4

Homework Statement


You are gliding over Earth's surface at a high speed, carrying your high-precision clock. At points X and Y on the ground are similar clocks, synchronized in the ground frame of reference. As you pass over clock X, it and your clock both read 0.

(a) According to you, do clocks X and Y advance slower or faster than yours?

(b) When you pass over clock Y, does it read the same time, an earlier time, or a later time than yours? (Make sure your answer agrees with what ground observers should see)

(c) Reconcile any seeming contradictions between your answers to part (a) and (b).

Homework Equations



Lo = L/γ

t' = γ(t - vx/c2)
or Δt' = Δt/γ

γ = (1-u2/c2)-1/2

The Attempt at a Solution



(a) Clocks moving in your (inertial) frame of reference appear to be ticking slower than clocks stationary in your frame. So the clocks X and Y appear to advance slower.

(b) This is where I get confused. I see clocks X and Y running slowly, although an observer in that frame observes my clocks running more slowly.

Say I'm traveling at (31/2/2)c so γ is 2, the observers on the ground measure my journey from above X to above Y to be t0. So when I pass over clock Y, they observe clock Y to be t0. (Since we should agree, I should also see t0, yet from the reasoning below I don't)

They observe my clocks running half as fast, thus my clock only reads t0/2 when I am above clock Y.

For me, the length of my journey is shorter due to length contraction. So in their frame I travel (t0) / (31/2/2c) or (2t0c) / (31/2), but in mine I travel half that. So my clock does register t0/2 due to the path being half as long. That part makes sense (I think?).

Yet they appear slowed to me, by a factor of 2, from this I can conclude their clock should read t0/4! That makes no sense. My reasoning is twisted somewhere.
 
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  • #2
You are given that clocks X and Y are synchronized in the Earth frame. Are X and Y also synchronized in your frame?
 
  • #3
TSny said:
You are given that clocks X and Y are synchronized in the Earth frame. Are X and Y also synchronized in your frame?

Synchronized clocks that are stationary in one frame should not be synchronized in a frame in which they are moving. You probably asked that so I'd have some insight that would fix my issue, but I can't think of it...

How can I apply that understanding to make sense of my fault? I'm failing to problem solve effectively within the framework of special relativity.

As far as I'm concerned in the spaceship, clock Y should be ahead of clock X?

...Because if we had an observer on the ground and a light pulse was emitted from each clock at t=0, he would say they arrive at the same time. Since the arrival of both is a single event, I must agree. However what I observe is the observer on the ground (and the clocks) moving along the vector that connects clock Y to X. Since the observer is gaining on the signal from X while receding from the signal from Y, I must observe Y emit the signal sooner so they reach him at the same time, thus it's time is further ahead.
 
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  • #4
Phynos said:
Synchronized clocks that are stationary in one frame should not be synchronized in a frame in which they are moving. You probably asked that so I'd have some insight that would fix my issue, but I can't think of it...

How can I apply that understanding to make sense of my fault? I'm failing to problem solve effectively within the framework of special relativity.

As far as I'm concerned in the spaceship, clock Y should be ahead of clock X?

...Because if we had an observer on the ground and a light pulse was emitted from each clock at t=0, he would say they arrive at the same time. Since the arrival of both is a single event, I must agree. However what I observe is the observer on the ground (and the clocks) moving along the vector that connects clock Y to X. Since the observer is gaining on the signal from X while receding from the signal from Y, I must observe Y emit the signal sooner so they reach him at the same time, thus it's time is further ahead.
Yes. So, at the instant you pass X, what does clock Y display simultaneously (according to you)?
 
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  • #5
TSny said:
Yes. So, at the instant you pass X, what does clock Y display simultaneously (according to you)?

Ohhh so Y is ahead by t0 minus the trip time (In my FOR). That's the key! Wow I really should have picked up on that.

(b) It reads a later time.

(c) In my frame of reference t != 0 at Y, t > 0, which explains why even though I see their clocks tick at a slower rate, the clock at Y is ahead of my own when I am above it.

Thanks for the help.
 
  • #6
OK. Good work!
 

FAQ: Two observers must agree what each clock reads....

What is the significance of two observers agreeing on the time shown on a clock?

This concept is known as time dilation and is a fundamental principle in special relativity. It states that time is not absolute and can vary depending on the relative motion of the observer and the object being observed. Therefore, two observers may perceive different times for the same event due to their relative motion.

How does time dilation affect the measurements of time?

Time dilation can cause time to appear to pass slower for objects in motion compared to objects at rest. This effect becomes more significant as the relative velocity between the observer and the object increases. It also plays a crucial role in the accuracy of GPS systems, as the satellites that provide the time signal are moving at high velocities relative to the Earth's surface.

What factors influence the amount of time dilation?

The amount of time dilation depends on the relative velocity between the observer and the object, as well as the gravitational potential of the object. The closer an object is to a massive body, the stronger its gravitational pull, and the more time dilation will occur.

Can time dilation be observed in everyday life?

Yes, time dilation is a well-established phenomenon that has been observed and confirmed through various experiments, such as the famous Hafele-Keating experiment. However, the effects are only noticeable at extremely high speeds or in the presence of strong gravitational fields.

What are the implications of time dilation in space travel?

Time dilation has significant implications for space travel, as it means that time passes slower for astronauts on a high-speed spacecraft compared to those on Earth. This could result in astronauts aging at a slower rate than people on Earth, leading to the possibility of time travel in the future. It also poses challenges for precise navigation and communication in space.

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