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

[Phynos]

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

L_o = 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 (3^1/2/2)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) / (3^1/2/2c) or (2t₀c) / (3^1/2), 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.

 

[TSny]

You are given that clocks X and Y are synchronized in the Earth frame. Are X and Y also synchronized in your frame?

 

[Phynos]

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.

 

[TSny]

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)?

 

[Phynos]

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.

 

[TSny]

OK. Good work!