xip14 wrote:
> Start out at x = 0 with a pocket watch reading t = 0 and walk 10 meters in
> 10 seconds,

According to whom?

> ie, walking speed 1 meter per second.

Relative to what?

> When you reach x = 10 your watch reads t = 10.

But only for you and everything/everyone that is at rest relative to you.

> Now your buddy, Walker-B, walks along side, double file.

Then they are *at rest* relative to you, so (at rest) in *the same* inertial
reference frame.

> His pocket watch runs slowly.

*Why*, for crying out loud?

> Every one of your seconds is only half of his, or, equivalently, the
> duration of his 1 second is 2 of your seconds. Time dilation.

No, *no*, and NO.

For that to happen (γ ≈ 2), Walker-B would have to move at the speed
v ≈ 0.8667 c *relative to you*. Evidently they are not doing that,
they are moving *alongside* you.

You have developed a misconceptions about (special) relativity from the
unfortunate (and wrong) pop-sci statement “moving clocks run slow”.

That is NOT what special relativity says. Instead, it says:

*Less* *proper* time *elapses* in inertial reference frames that are in
motion *relative to* the inertial reference frame which defines proper time
(by measuring it with clocks *at rest* in *that* frame).

This is a simple consequence of the Lorentz transformation between inertial
reference frames (which we know – as has been proved by Lorentz/Poincaré and
Einstein independently – has to be used, as a consequence of the constancy
of c):

t = γ (t' − v/c² x'), γ = 1/√(1 − v²/c²)
⇒ Δt = γ (Δt' + v/c² Δx').

The "moving" clocks are at rest in their inertial frame:

Δx' = 0
⇒ Δt = γ Δt'
⇔ Δt' = Δt/γ < Δt.

Or using the Minkowski metric, γ and "time dilation" jump out more naturally:

−c² dτ² = η_ab dx^a dx^b; [η]_ab = diag(−1, 1, 1, 1), x⁰ := ct, (x^i) = x⃗.

⇔ dτ² = −1/c² η_ab dx^a dx^b

⇔ dτ = √(−1/c² η_ab dx^a dx^b) = 1/c √(−η_ab dx^a dx^b)

⇔ dτ/dt = 1/c √(η_aa dx^a/dt dx^a/dt) = 1/c √(c² − v⃗²) = √(1 − v⃗²/c²)
= 1/γ

⇒ Δτ = ∫_W dt/γ = Δt/γ.

It then *appears* (superficially) to be so that clocks in *relative* motion
run _slower_ than at *relative* rest. But that is NOT so; they simply
measure *less* *elapsed* *proper* time because actually less proper time
elapses where they are.

> Alternatively, tell him when he starts out to go 1 meter every second on
> his clock.

This is not feasible, as your meter (rest length) and his meter (moving
length) are not the same. He will measure your meters to be length
contracted in the direction of his motion as for him – who rightfully
considers himself at rest – everything else is moving (relative to him).

--
PointedEars

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