B Why and how does the frequency of a moving clock affect its ticking speed?

[Ibix]

Clocks always measure their own proper time. In the case you illustrated, one clock travels along thediagonal paths and one along the straight line AC. Fine.

I'm not clear exactly what your other case, case 2, is. Are you trying to ask about the time between events A and B as determined by the frame you illustrated and by the frame of the moving clock? There's no physical mechanism at work here, they're just measuring different things. Clocks at rest in your diagram's frame measure the vertical distance between A and B. The clock moving from A to B measures the diagonal.

 

[morrobay]

This is about the "time between ticks" in the proper times of both clocks. In case 1. those units in both clocks are equal. Less elapsed time by traveler caused by lesser spacetime path as measured by traveling clock . t (tick) = t' (tick) .The case 2. is a definition : Here time between ticks is not equal, t' = gamma t.

 

[Ibix]

The proper time between ticks of identical clocks is always the same. I suspect you are not clear on the distinction between coordinate time and proper time. Perhaps you could draw a diagram of case 2, since it doesn't appear to derive from part of the spacetime diagram you posted, as I initially thought.

 

[morrobay]

In this diagram t' = gamma t , as viewed from S .

 

[Ibix]

There is no $t'$, $S$, or $S'$ in those diagrams. The only times marked are the $t$ axis and the coordinate times, $t_1$ and $t_2$, of the events $A_1$ and $C_1$. There are no proper times marked. So I'm still confused about what you are asking.

 

[Ibix]

Just to add: if that is meant to be two views of the same scenario from two frames then it's very poorly labelled. The time and space axes are both labelled $x$ and $t$, although they're not the same thing. The positions of the events on the axes are labelled $A$ and $C$ in both diagrams, although they're different. And the events themselves (which are genuinely invariant things) are labelled differently. Also, although this may just be a perspective effect from your camera, the diagonal lines look below 45° to me, which would normally imply spacelike lines.

Can I ask where you found these diagrams?

 

[morrobay]

View attachment 272876 In this diagram t' = gamma t , as viewed from S .

S is left side S' is right side. t' would be t(2) -t(1) * t or t'= gamma t . Edit typo: t'=gamma t

 

[Ibix]

S is left side S' is right side. t' would be t(2) -t(1) * t or t' gamma t

I'm sorry, this makes no sense. Presuming you mean $t_1$ and $t_2$ by t(1) and t(2) then you seem to have written $t'=t_2-t_1t=t'\gamma t$, which is nonsense. You've got terms with units of time and others with units of time squared. Perhaps if you used LaTeX to typeset your maths it would be easier to follow.

 

[morrobay]

I'm sorry, this makes no sense. Presuming you mean $t_1$ and $t_2$ by t(1) and t(2) then you seem to have written $t'=t_2-t_1t=t'\gamma t$, which is nonsense. You've got terms with units of time and others with units of time squared. Perhaps if you used LaTeX to typeset your maths it would be easier to follow.

Given that t' = gamma t .What this means is that from the diagram ,rs, t2 -t1 is numerically equivalent to gamma .And I'll take the task of showing this tomorrow morning.

 

[Ibix]

Defining $\Delta x$ as the distance from A to B and B to C in the left hand diagram:$$\begin{eqnarray*}
t_2&=&\gamma(t+v\Delta x/c^2) \\
t_1&=&\gamma(t-v\Delta x/c^2)\\
t_2-t_1&=&2\gamma\frac{v}{c^2}\Delta x
\end{eqnarray*}$$I do not think this is "numerically equivalent to gamma". And I don't understand how any of this relates to this thread. It's just messing around with coordinates.

 

[PeterDonis]

In case 2. it appears as if the mechanism of actions or proper times is different for the two clocks ?

What makes you say that? In both cases, the proper time is the length along each worldline. The $t$ coordinates in your case 2 are not proper times, so why should you expect them to behave like proper times?

 

[morrobay]

What makes you say that? In both cases, the proper time is the length along each worldline. The $t$ coordinates in your case 2 are not proper times, so why should you expect them to behave like proper times?

Since the graph depicts the relativity of simultaneity . And the coordinates are measuring light flashes from B to A and C then it would seem the proper times can be inferred. Ok good , then can you graph the proper times/worldlines on two identical clocks in S and S' and show that the units (time between ticks) are also identical ? Because this is what the OP is questioning.

 

[robphy]

Possibly useful...

From a little sleuthing, I think @morrobay is using diagrams from

Sanjoy Mahajan (http://web.mit.edu/sanjoy/www/ ) http://www.inference.org.uk/sanjoy/teaching/corpus/IA/2002/relativity-notes.pdf
( http://www.inference.org.uk/sanjoy/teaching/corpus/IA/2002/ )

A.P. French
https://books.google.com/books?id=udA0Gjq8vHIC&q=simultaneity#v=snippet&q=simultaneity&f=false (p 71. Fig 3-2)

...from the caption, we learn that A, B, C are not events but “stations” (parallel inertial worldlines) arranged as described. Looking further on the page

we see that A_1 and C_1 are events, as are A_1{}' and C_1{}'.and Wikipedia
https://books.google.com/books?id=J4-30n591U8C&pg=PA170&lpg=PA170&dq="writing+the+Lorentz+Transformation+and+its+inverse+in+terms+of+coordinate+differences+we+get"&source=bl&ots=Q19MgnBfGz&sig=ACfU3U076-NBpbsruEpqZcltSrcc9MPQ5A&hl=en&sa=X&ved=2ahUKEwiz4ZTEopTtAhUxtDEKHe97CXsQ6AEwBHoECAEQAg#v=onepage&q="writing the Lorentz Transformation and its inverse in terms of coordinate differences we get"&f=false

 

[Nugatory]

Ok good , then can you graph the proper times/worldlines on two identical clocks in S and S' and show that the units (time between ticks) are also identical ? Because this is what the OP is questioning.

It is difficult to do that with a graph, because the surface we're drawig on is Euclidean, not Minkowskian. Thus the length of a line segment in such a diagram is generally not equal to the proper time between the two events at the ends of the segment. The twin paradox diagram from @robphy's post directly above is a good example: If you hold a ruler up to your screen and measure, you will find that the distance AC is less than sum of the distance AB and the distance BC - yet more proper time elapses on the AC path and it is "longer".

 

[Frodo]

The point I was trying to make with my 1 million observers is that it isn't the clock itself which ticks slower. It is the moving observer(s) who measure the clock to tick slower. If 1 million observers each measure the clock to tick at 1 million different rates then it cannot be the clock itself causing the effect - it must be the observers.

My explanation has the clock stationary and the multiple observers moving as it makes the point so clear.

The original poster asked

To make clock ticking slower, frequency of balance wheel must be changed..

That is a fundamental misunderstanding of what is happening and I was trying to illuminate it.

The clock doesn't tick slower because it gets a weaker balance spring, or it gets a larger balance wheel, causing the clock physically to tick slower. The clock always ticks at its proper speed as observed by someone stationary with the clock. There is nothing "funny" about the clock and nothing about the clock changes. The clock has no knowledge that 1 million traveling observers are checking to see how fast it is running. It carries on blissfully unaware, ticking at its normal rate.

It is the observers, who are moving relative to the clock and are traveling across spacetime, who are causing the effect. (Perhaps "experiencing" would be better than "causing" because, of course, it is 'how spacetime works' which is the real cause, and the travellers are experiencing the changes spacetime forces when they travel through it.) As they travel, their view of what is simultaneous at locations separated by distance in space changes. It is these changes which cause them to measure the stationary clock to run slow. And it is why different observers traveling at different speeds all measure different clock rates.

For me, the most interesting item in the Lorentz transformation is the " x " in the time transformation equation. By varying " x ", the distance separating the events, we can make t' as different from t as we want, from microseconds to billions of years. And we can put " x " as positive or negative so t' can go into the future or into the past, relative to t, as we wish.

 

[PeterDonis]

can you graph the proper times/worldlines on two identical clocks in S and S'

No. You can't draw space vs. proper time diagrams that include different clocks in relative motion; they don't work.

 

[robphy]

Here is a more accurate drawing of the spacetime diagram from French's textbook.
(The original diagram doesn't include length contraction...
possibly because that is developed in a later chapter.)

It should be noted that the areas of triangles OA_1C_1 and OA_1{}′C_1{}′ are equal,
since the Lorentz Boost has determinant 1.

Note that Alice measures the time-difference t_{C_1{}′}−t_{A_1{}′},
however, this does not easily tell us the time-dilation factor \gamma.

To see what is going on,
consider a scaled up version of the "light-clock diamond with spacelike-diagonal A_1{}'C_1{}' ",
namely, the causal-diamond with timelike-diagonal OP,
as part of the clock effect diagram based on the diagram by Sanjay Mahajan.

The spacetime diagram below shows the ticking light-clocks belonging to Alice (20 ticks along OZ)
and to non-inertial Bob (8 ticks along OP, followed by 8 ticks along PZ).
Note that all clock diamonds have the same area.

Note that OP has velocity (6)/(10) and
that the area of the causal diamond* with timelike-diagonal OP is (16)(4)=64,
which is equal to the square of the number of clock-diamonds along OP,
which are similar to the causal diamond with diagonal OP. Hence, OP=(8).

(*The causal diamond of OP is the intersection of the causal future of O with the causal past of P.)

By the way, from causal diamond edges (16) and (4),
the square of the doppler factor k^2=(16)/(4)=4...
so, k=2, which is the eigenvalue of this transformation...
so the clock diamond for Bob is stretched in the forward direction by a factor of 2
and shrunk in the opposite direction by a factor of 2.
Then since k=\displaystyle\sqrt{ \frac{1+v}{1−v}},
we have v_{Bob}=\displaystyle\frac{k^2−1}{k^2+1}=\frac{4−1}{4+1}=\frac{3}{5}.

Similarly, PZ=(8) and OZ=(20).

Note the time-dilation factor is (by counting) \gamma=(10)/(8).

The left corner of the causal diamond has t_{M{}′}=2 and the right has t_{N{}′}=8,
so t_{N{}′}−t_{M{}′}=6, which doesn't easily lead to \gamma=(10)/(8).

(For more information on these diagrams, look at my Insights.)UPDATE:

Why is the area of the causal diamond
equal to the square-interval of its timelike diagonal?
Look at this diagram...

• take the Minkowski-right-triangle with hypotenuse OP
  and slide the diamonds along the legs in the lightlike direction of the black arrow to determine
  the width of the causal diamond of OP as the sum \Delta t +\Delta x in clock diamond edges
• then slide along the other lightlike direction to determine
  the edge-height of the causal diamond of OP as the difference \Delta t -\Delta x in clock diamond edges
• the area of the causal diamond of OP is the product,
  which equals the square-interval (\Delta t)^2 - (\Delta x)^2
  in clock diamond areas.

 

[vanhees71]

It is difficult to that with a graph, because the surface we're drawig on is Euclidean, not Minkowskian. Thus the length of a line segment in such a diagram is generally not equal to the proper time between the two events at the ends of the segment. The twin paradox diagram from @robphy's post directly above is a good example: If you hold a ruler up to your screen and measure, you will find that the distance AC is less than sum of the distance AB and the distance BC - yet more proper time elapses on the AC path and it is "longer".

I think in Minkowski diagrams one should always give the unit-lengths tix on the axes on the two systems, exactly for the reason that you must forget about Euclidean geometry of the plane and think in terms of the Minkowskian/Lorentzian geometry. The unit tix have to be constructed with help of hyperbolae (Lorentzian) rather than circles (Euclidean).

It's also important to emphasize the indefiniteness of the fundamental form, establishing the causality structure of spacetime.

 

[robphy]

I think in Minkowski diagrams one should always give the unit-lengths tix on the axes on the two systems, exactly for the reason that you must forget about Euclidean geometry of the plane and think in terms of the Minkowskian/Lorentzian geometry. The unit tix have to be constructed with help of hyperbolae (Lorentzian) rather than circles (Euclidean).

It's also important to emphasize the indefiniteness of the fundamental form, establishing the causality structure of spacetime.

And this is exactly what the "rotated graph paper" does for a Minkowski spacetime diagram.

• In fact, it can easily show unit lengths in multiple frames of reference systems
  (the above clock-effect diagram posted earlier shows three systems;
  see the elastic collision diagram in my PF Insight https://www.physicsforums.com/insights/relativity-rotated-graph-paper/ which shows unit lengths for five systems).
• The underlying hyperbolas at each event are encoded by
  "the family of equal-area causal diamonds with a common corner event",
  where the area of a causal diamond is equal to the square-interval.
  (See the hyperbola in the above Insight, which shows light-clock diamonds for nine systems.)
• The causal diamond encodes the causal structure,
  since its edges are lightlike and determined by the light cones,
  and its diagonals (one timelike and the other spacelike) are Minkowski-orthogonal
  and have the same absolute magnitude.
  (The timelike diagonal is essentially the radius vector,
  and the spacelike diagonal is along the tangent line to a hyperbola.
  Thus, the "tangent line is orthogonal to the radius".)
• The light-clock diamonds (the basic unit of the rotated graph paper) are determined
  by the light-clocks of inertial observers.
  The reshaping of the clock diamond
  (stretching by k in one direction and shrinking by k in the other direction)
  is a Lorentz boost transformation,
  where k (the Doppler factor, which is exp(rapidity) )
  is the larger eigenvalue of the particular boost (the smaller eigenvalue is its reciprocal 1/k)...
  and this k encodes the velocity and the time-dilation factor.
• So, from counting clock diamonds, one can do quantitative calculations
  [especially when k is rational, leading to simple arithmetic involving Pythagorean triples].
  (You are secretly working in light-cone coordinates, in the spirit of the Bondi k-calculus.)
• and it's easy to get rotated graph paper
  [no need for (just-)two-observer diagrams,
  no need for hyperbolic-graph-paper (which distinguishes an event)]
• and one can (and should) emphasize the physical interpretation to the measurements read off from the diagram, and demote the algebraic Lorentz-transformation formulas [until needed].
• ...and you don't have to forget everything about Euclidean geometry...
  You can keep the "affine structure" (parallel lines, addition of vectors, invariance of planar areas).
  You just have to get the invariant square-magnitudes
  by counting up the number of light-clock diamonds in a causal diamond.

 

[vanhees71]

Sure, that's using light-cone coordinates implicitly. Here the hyperbolae have the equations x^+ x^-=\text{const}. and this leads to the equal-area prescription of the diamonds. It's a pretty elegant method.