How can the Universe know when you change directions?

[Dale]

Clocks are always some form of oscillator. Math is using time as a sequence marker to describe the oscillation or periodic effect. ... How can I see time as more than numbers on a page?

Is the oscillator a number on a page? The math describes the oscillator, but the oscillator is not numbers on a page.

 

[Dale]

The point of Einstein's "time is what clocks measure" is to suggest that you stop philosophising and follow the evidence. Clocks are apparently measuring something because we get repeatable, predictable and consistent measurement from them.

Personally, I think that it is even more than that. The thing that clocks measure is a useful quantity. It deserves a name. That name is "time". If we didn't call it "time" then we would have to call it something else, and as a practical matter that would be the word that we use for talking about when things happen. Time.

 

[cianfa72]

The math describes the oscillator, but the oscillator is not numbers on a page.

You mean that the math provides "a map" that maps what clocks read into a real number.

 

[Nugatory]

Ok, therefore the flocks of clocks entirely filling the space you were talking about, are actually "at rest w.r.t. each other" as measured for instance by constant round-trip travel time of bouncing light/radar pulses exchanged between them. Otherwise which could be a different physical criterion to define "at rest w.r.t. each other" ?

This is drifting into a thread hijack, but be aware that @Ibix is just providing an abridged description of Taylor and Wheeler's ("Spacetime Physics") model of an inertial frame with associated simultaneity convention.

To avoid further drift, you might want to review that section of Taylor and Wheeler and then if you have followup questions, start a new thread.

 

[jnhrtmn]

Back to Doppler, it seems overcomplicated. Using 0.6c added to c gives 1.6c, (contracted length) / (speed multiplier), 0.8/1.6=0.5. The other is 0.8/0.4=2 This works throughout. Is it because1.6c can't be real, but even sqrt 1.6 is still 1.265. Using 1 instead of contracted length is the classical version. I don't understand the use of source and observer in the math if the relative velocity between the frames is the only factor.

 

[Sagittarius A-Star]

Back to Doppler, it seems overcomplicated. Using 0.6c added to c gives 1.6c, (contracted length) / (speed multiplier), 0.8/1.6=0.5. The other is 0.8/0.4=2 This works throughout.

Yes.
$\sqrt{1+v/c \over 1-v/c} = \sqrt{(1+v/c)(1-v/c) \over (1-v/c)(1-v/c)} = {1 \over \gamma (1-v/c)} [= \gamma (1+v/c)]$

Usually, the $\gamma$ factor in the relativistic Doppler formula is explained via time-dilation, but because of $c=\lambda f$ also an argumentation via length contraction is possible.

Is it because1.6c can't be real

The closing speed between 2 objects can be greater than $c$. Only the speeds of each object relativ to the choosen inertial reference frame cannot be greater than $c$.

I don't understand the use of source and observer in the math if the relative velocity between the frames is the only factor.

See
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

 

[Dale]

it seems overcomplicated.

Sometime things are complicated. Nature has no obligation to be convenient or simple.

The theoretically predicted and experimentally validated formula is derived here: https://www.mathpages.com/home/kmath587/kmath587.htm

 

[Ibix]

I don't understand the use of source and observer in the math if the relative velocity between the frames is the only factor.

You need to be careful here. If you and I stand stationary near each other, we share a rest frame. I throw a ball to you. The ball has a different rest frame with a single velocity with respect to our frame. But to you, light from the ball is blue shifted while it is red shifted to me. So the $v$ in the Doppler shift formula cannot be a frame velocity since we couldn't get different Doppler shifts if it were. And we could get opposite Doppler shifts by reversing our $x$ axes so the velocity switched from $+v$ to $-v$.

The $v$ in the Doppler shift formula is the closing rate (or separation rate, depending on your sign convention) of the source with respect to a specific observer, not a frame velocity. It will typically be equal to plus-or-minus the velocity of the source's rest frame in the observer's frame, but it is a distinct concept.