# Adjusting length and period using (inverse) transformations

1. Mar 28, 2014

### ANvH

Trying to see the logic in deriving length contraction and time dilation using the Lorentz transformations and inverse Lorentz transformations. In the following treatise it leads to ambiguities.

Given
$Δ\acute{t}=\gamma(Δt-\beta c^{-1}Δx)$ (1)​
$Δ\acute{x}=\gamma(Δx-\beta c Δt)$ (2)​
and the inverse transformations,
$Δt=\gamma(Δ\acute{t}+\beta c^{-1}Δ\acute{x})$ (3)​
$Δx=\gamma(Δ\acute{x}+\beta c Δ\acute{t})$ (4),​
where the primes refer to the emitter frame (the moving frame of reference) and the non-primes refer to the receiver frame (the stationary frame or rest frame of reference).

Length contraction is defined by measuring the object length $Δ\acute{x}$ in the moving frame while held stationary from the rest frame; moving the line of sight of the receiver accomplishes this and justifies setting $Δt$ to zero. This circumstance can be met by using Equation (2). The resulting equation is $Δ\acute{x}=\gammaΔx$ and it is then rearranged to $Δx=Δ\acute{x}/\gamma$. Since $\gamma > 1$ it suggests a length contraction.

For time dilation the object in the moving frame needs to be stationary, while applying a transform of the starting and ending times in the stationary frame. In this case $Δ\acute{x}$ has to be set to zero, and this can be met by Equation (3) and it leads to $Δt=\gammaΔ\acute{t}$. While this reasoning seams to be straightforward there is a huge inconsistency. To obtain a length contraction, Equation (2) is utilized and the final result is rearranged, contrary to the time dilation derived from Equation (3). Somehow the symmetry of the Lorentz and inverse Lorentz transformations is violated, because Equation (2) belongs to the first set (the Lorentz transformations), and Equation (3) belongs to the second set (the inverse Lorentz transformations).

To exemplify this ambiguity, Equations (1-4) can be reduced to
$Δ\acute{t}=\gammaΔt$ (5)​
$Δ\acute{x}=\gammaΔx$ (6)​
and the inverse transformations,
$Δt=\gammaΔ\acute{t}$ (7)​
$Δx=\gammaΔ\acute{x}$ (8).​

What should one pick: (5) and (6) or (7) and (8)?

2. Mar 28, 2014

### ghwellsjr

This is rather confusing, to me at least. It would make more sense to me if you refer to the stationary (or rest) frame as the one in which an object is at rest and the moving frame as one that is moving at speed β with respect to the first (rest) frame. This makes the object move at speed -β in that frame. Then you only need the first set of equations to get coordinates from the first (rest) frame to coordinates in the second (moving) frame.

This also is confusing to me. The Coordinate Length of an object is the delta between two events at either end at the same Coordinate Time. That means in the moving frame you need to choose the two events at the end points of the object such that they have the same Coordinate Time in the moving frame. In other words, Δt' has to equal zero, not Δt. In the rest frame, it doesn't matter what Δt is because any pair of end point events will be the same distance apart.

When you compare the Coordinate Length as defined this way in the moving frame to the Coordinate Length in the rest frame, you will see the effect called Length Contraction. The Coordinate Length in the rest frame is also called the Proper Length so you can say that the Length Contraction factor, the inverse of gamma, is equal to the Coordinate Length of an object moving in a frame compared to its Proper Length.

Time Dilation refers to the delta Coordinate Time of an object compared to its delta Proper Time for the same pair of events. The delta Proper Time is taken between two events for an object in its rest frame. Then you can transform the Coordinate Times for those same two events into a moving frame and get a new delta Coordinate Time. The ratio of the delta Coordinate Time to the Proper Time is the Time Dilation factor and is equal to gamma.

I think it would be most helpful to look at a pair of spacetime diagrams for a five-foot long object first at rest and then moving at β=0.6. Here's the first spacetime diagram:

Each line represents an end of the five-foot long object. The dots represent 1 nanosecond increments of time. Note that the delta Proper Times are equal to the delta Coordinate Times. I have shown the object over a period of 4 nanoseconds.

Now the spacetime diagram transformed to a speed of -0.6c (β=-0.6). This makes the object move at 0.6c in the diagram:

This diagram was created by taking the coordinates of each event (dot) from the first diagram and transforming them with β=-0.6 (c=1). You can see that the length of the object at any Coordinate Time is 4 feet. For example, at the Coordinate Time of 5 nanoseconds, one end of the object is at the Coordinate Distance of 3 feet and the other end at 7 feet for a delta of 4 feet. Gamma at -0.6c is 1.25 so if we divide the Proper Length of 5 feet by 1.25 we get the Contracted Length of 4 feet.

For Time Dilation, we can see that the period from the first event along the blue end of the object to the last event in the moving frame is 5 nanoseconds whereas it was 4 nanoseconds in the rest frame for a ratio of 1.25, the same as the gamma factor.

Now if you want, you can use your second set of equations to transform the coordinates of the events in the second diagram back to the coordinates in the first diagram using the same value of beta, -0.6.

Does this all make perfect sense to you? Any questions?

3. Mar 28, 2014

### ANvH

First, I would like to use the first two equations only; second, what you say sounds confusing to me, so I would like to make sure that I am understanding you. The prime stands for the moving frame and the transformations for $\acute {x}$ and $\acute {t}$ of the moving frame are replaced by relations involving $x$ and $t$ as measured in the rest frame. The equations (1) and (2) were created by $\acute {x}_{2}-\acute {x}_{1}$ and $\acute {t}_{2}-\acute {t}_{1}$. Are we indeed on the same page? I do have questions, but would like to clarify this.

4. Mar 28, 2014

### ghwellsjr

Yes, we are on the same page.

For example, in my first diagram, take the top blue event as the first event and the second up from the bottom on the red line as the second event. Then Δx = 5 and Δt = -3. Remember that β = -0.6, c=1 and γ = 1.25. Plugging these values into your first two equations, we get:

Δt' = γ(Δt - βΔx) = 1.25(-3 + 0.6*5) = 1.25(-3 + 3) = 0
Δx' = γ(Δx - βΔt) = 1.25(5 - 0.6*3) = 1.25(5 - 1.8) = 1.25(3.2) = 4

Notice that Δt' equals zero as it should to get the correct distance between the two ends of the object.

5. Mar 28, 2014

### ftr

But after few days of research it became clear to me that the equations did not just pop into Einsteins head. They are the painstaking work of many people which gave me great confidence in them, and checking them I found them to be consistent.

just some references, enjoy

http://www.mathpages.com/home/kmath571/kmath571.htm

http://www.fourmilab.ch/etexts/einstein/specrel/www/

http://www.dwc.knaw.nl/DL/publications/PU00014148.pdf

http://www.physicsinsights.org/poincare-1900.pdf

you can also dig up Whittaker’s classic book volume 1,2.

“A History of the Theories of Aether and Electricity”

https://archive.org/details/historyoftheorie00whitrich

6. Mar 29, 2014

### ANvH

I appreciate what you show here. So distance is only measured parallel to the Coordinate Distance, irrespective of choosing two events. You can compare the first event of the blue line with the last event of the red line. Always thought to compare points in space-time that have the same event. For your info, the treatise I gave is based on what I have read on the web; it did not came from me (can't find it to provide a link).

It does make a lot more sense after realizing how to read the above diagram and extrapolating more events, such that Δx' between the blue and red lines is zero, yielding Δt' (Trying to use the same logic as done for getting the contracted length). Am I correct when two successive events with respect to the blue line are compared that $δx'$ correctly describes the length contraction? Logic suggests it is. Then,
$\frac{δx'}{δt'}=\gamma^{-2} \frac{δx}{δt}$​
is the contracted speed?

7. Mar 29, 2014

### ghwellsjr

In the rest frame of an object (assuming both are inertial), if you use any pair of events at either end of the object, you will get the same delta distance even though the delta time can be any value. But if you use any of those pairs and transform to a frame where the object is moving, you will get a whole range of delta distances and delta times, which leads to a meaningless idea for delta distance. However, if you pick a pair of events in the rest frame of the object such that when you transform them to the moving frame and the delta time equals zero, then the delta distance shows the length contraction equal to the inverse of gamma.

Another way of understanding this is that if an observer is trying to measure distances to various parts of a moving object, if he doesn't make the measurements "at the same time", he will get all kinds of different answers. Let's say you have a laser range finder and you focus it on one end of an object moving away from you. You will get a succession distances that are changing and getting bigger. If you then focus it on the other end of the object, you will get a bunch more readings. If you just subtract two of the readings for each end of the object, you will get a whole range of lengths, maybe even including zero. What you would have to do is use two laser range finders and focus each on a different end of the object. But even then, you would not necessarily get a meaningful answer unless you could make sure that the time of the measurement is half way between the laser pulse going out and its reflection being received by each range finder. Only when those times match will you get a meaningful result and it will be the Contracted Length of the object.

I hope you meant to say "Always thought to compare events in space-time that have the same time" because the way you said it doesn't make any sense. You should be clear that in relativity, an event is a point in space at an instant of time.

Did you make a copy of it? If not, are you sure you are remembering it exactly?

I'm not sure why you said this. If you're talking about contracted length, then you should say "Δx' between the blue and red lines yields the contracted length when Δt' is zero".

If you're talking about Time Dilation, you don't want to go between the blue and the red lines, each line has its own Time Dilation and it's not necessarily when Δx' is zero. The Proper Time is when the two events are at the same Coordinate Distance in the rest frame of the object but in the moving frame, both Δx' and Δt' are non-zero.

You can see in the rest frame, the blue line depicts a coordinate interval of 4 nanoseconds and the red line also depicts a coordinate interval of 4 nanoseconds. You can think of each line having its own clock that ticks through 4 nanoseconds of Proper Time equal to 4 nanoseconds of Coordinate Time.

In the moving frame, the blue line events go from the Coordinate Times of 0 to 5 nanoseconds for a Coordinate Time interval of 5 nanoseconds (while the Coordinate Distance goes through 3 feet). The red line events go from the Coordinate Times of 3.75 to 8.75 nanoseconds for a Coordinate Time interval of 5 nanoseconds (while the Coordinate Distance also goes through 3 feet). As I said before, we determine the Time Dilation as the Coordinate Time interval divided by the Proper Time interval. In these cases (for the blue and red lines), the Coordinate Time intervals are 5 nanoseconds while the Proper Time intervals are 4 nanoseconds making the Time Dilation factors for both the blue and red lines be 1.25.

The only sense in which the logic for Time Dilation is the same as for Length Contraction is in taking the ratio of a Coordinate interval (length or time) to a Proper interval (length or time). But the way they are applied is completely different as I explained previously.

No. Two successive events on the blue line have to do with Time Dilation, not Length Contraction. Length contraction always involves two different world lines, not just one. The world lines are marking a succession of events and events are instants in time and points in space having no length associated with them.

I don't know how you got that equation but in any case, it has nothing to do with contracted speed. Speeds don't contract. There's no such thing, at least I never heard of it. If you have two inertial objects or observers moving with respect to each other, they will each calculate or measure the speed of the other one to be the same, using laser range finders, for example (just in opposite directions).

But this has nothing to do with frames. You can determine the speed of an object in different reference frames by doing the Lorentz Transformation process on the coordinates of various events. For example, in the second diagram in my previous post, you can determine the speed of the blue world line by taking the Δx'/Δt' which is 3 feet divided by 5 nanoseconds or 0.6 feet per nanosecond or 0.6c. You can do the same for the red world line.

8. Mar 29, 2014

### ANvH

http://resonanceswavesandfields.blogspot.com/2011/07/derivation-of-length-contraction-and.html

As it should, my mistake. It is crystal clear, I appreciate your time and effort. There is only one thing that is bothering me, or maybe not. In the primed frame both length contraction and time dilation occur, or is it only time dilation or length contraction?

I am asking this because of the Muon experiment. It is explained as a dilated half life of the muon in the muon's moving frame and the muon sees a length contraction when the muon's frame is considered a rest frame with respect to a moving earth frame.

There is no mention of the apparent length contraction in the muon's moving frame, or the time dilation of the moving earth's frame.

Again, your lengthy explanations are highly appreciated.

-Alfred

9. Mar 29, 2014

### ghwellsjr

That reference is correct as long as you read the whole thing to realize that they have the object moving in the stationary frame and stationary in the moving frame. I had assumed that it was the other way around in my discussions.

In any frame (primed or unprimed), objects that have length will be contracted if they are moving and time will be dilated for them.

Since we don't consider the size (or length) of the muon, we aren't concerned about it's contraction in the earth's rest frame (although it is there), we only focus on it's Time Dilation. In the muon's rest frame, we aren't concerned about the time on the earth (although it is dilated), we only focus on the contracted distance.

Here is a thread where I have drawn some spacetime diagrams regarding the muon:

10. Mar 29, 2014

### ANvH

"Moving in the stationary frame" and "stationary in a moving frame" makes it very confusing.

Ok, yet there is in my view a discrepancy when comparing spatial contraction and temporal dilation. If a waveform is subject to the Lorentz transformations then the angular frequency and wavenumber domains are increased by γ(1+β). This means the period and the wavelength are shortened, i.e. both domains of the waveform are blue shifted. I am always making the association that the wavelength is contracted, yet a time dilation means the period of the wave should be increased, suggesting I am wrongly associating the period (temporal domain) of the wave with time dilation.

With respect to the muon, I have tried to treat the muon as a wave: the half life is the temporal aspect of the muon wave and the spatial decay, deduced from the vertical depth of the atmosphere and the flux of energy loss amounts to a decay of one muon per 10 km. While such a wave treatment may not be valid, yet the thought experiment put into would allow an analysis. The outcome may not be what you want. I agree I am not interested in the size of the muon, I am interested in the path it is following.

I have studied that one, but it does not answer why time dilation holds for an object and a Doppler blueshift holds for the frequency domain, perhaps the OP of that thread has the same issue.

Because this would be a difference in light-like and space-like processes?

11. Mar 30, 2014

### ghwellsjr

No, this has nothing to do with the difference between light-like and space-like processes. You are talking about the difference between Doppler shifts and Time Dilation. They are two completely different things. Doppler shifts are observable and symmetrical between two inertial observers and are independent of any reference frame. Time Dilation is not observable and changes according to each reference frame. It is only symmetrical (equal) for two observers if they happen to be traveling in opposite directions at the same speed in a particular reference frame.

Here, let me illustrate with a couple spacetime diagrams showing two inertial observers approaching, then passing each other at a relative speed of 0.6c. They each have a clock that ticks once a second and they can each see the other ones clock as well as their own. Here's the spacetime diagram for the rest frame of the red observer as he is watching the ticks on the blue observer's clock:

At the bottom half of the diagram, the blue observer is approaching the red observer and the red observer sees the blue observer's clock ticking twice as fast as his own. We can calculate this Doppler factor using your formula, γ(1+β). Since at β=0.6, γ=1.25, and the formula evaluates to 1.25(1+0.6) = 1.25(1.6) = 2.

But note that the time dilation of blue's clock is 1.25 (gamma) which you can see by observing that in 5 seconds of Coordinate Time, blue's clock has ticked out 4 seconds and 5/4 = 1.25.

After blue passes red, red observes blue's clock ticking at one-half the rate of his own clock which is the inverse of the previous Doppler factor and yet the Time Dilation of the blue clock remains at 1.25.

Note that there is no Time Dilation of red's clock, it ticks at the same rate as the Coordinate time because it is at rest in this frame.

Now let's look at how the blue observer looks at the red observer's clock:

Note that during their approach, blue sees red's clock ticking at twice the rate of his own, even though red's clock is not Time Dilated in this frame. And after they pass each other, blue sees red's clock ticking at half the rate of his own.

Each observer sees the same Doppler shifts in the other observer's clock but neither observer is aware of the Time Dilation.

12. Mar 30, 2014

### ANvH

I think this statement is not correct, because from blue's perspective the red clock should be time dilated.

This statement is not helpful as if time dilation does not exist. You know, I have seen this in many posts here. First you get the explanation that time dilation exists, is real, therefore the rate constant of the muon decay is physically slower. And then later it is said that observers are not aware of each other's time dilation. Sorry, that I am picky here.

Actually, no. The reason is that the Doppler shift is obtained by the same Lorentz transformations, while these transformations are also used to draw diagrams to explain time dilation and length contraction. It then becomes completely confusing because you say Doppler is real, is symmetric, as time dilation should be too. (Both red and blue observers, know that time dilation occurs in the opposing frame).

But let's focus on the Lorentz transformations. When applied to a wave the full forms
$t'=\gamma(t-\beta c^{-1}x)$​
$x'=\gamma(x-\beta ct)$​
are used to get the transformed waveform, which will give a Doppler shift that equals γ + γβ. To get at the length contraction and time dilation these forms are used
$Δt'=\gamma Δt$​
$Δx'=\gamma Δx$,​
and they look symmetrical, because Δt and Δx look both contracted. I know you explained by the diagrams that this is not true, and I grasped everything, yet it remains a mystery for me why the discrepancy exist. The only two things I can think of is that $\beta c^{-1}x$ and $\beta ct$ were set to zero for length contraction and time dilation, while the waveform transform procedure does not need this "simplification" because the relation $ω=kc$ takes care of grouping temporal and spatial aspects. In addition, another thought is that the wave is traveling at $c$, used to signaling the clock rates, yet the observers travel at 0.6c. Funny that the clocks do not show the time dilation (and length contraction) by their signaling waves.

13. Mar 30, 2014

### ghwellsjr

If by "from blue's perspective" you mean "in blue's rest frame", then it is true that the red clock is time dilated in that frame as shown here:

Each frame establishes a different Time Dilation to each observer's clock but they all indicate the same Doppler for each observer.

Here is another frame transformed from the first one to a speed of 0.333c:

In this frame, both clocks are Time Dilated to 1.06 but the Doppler remains the same.

Time Dilation is a coordinate effect. Coordinates are not real. They are man-made constructs which enable us to describe scenarios. Would you say that the origin of a coordinate system is real? Or that the directions that the axes point in are real? Or the units that we use to mark off the dimensions, as if nature was aware of these constructs?

Just like the speed of an object is different in different coordinate systems, so is Time Dilation. This doesn't render speed or Time Dilation to the realm of non-existence.

When I say that observers are not aware of each others Time Dilation, I mean that they cannot measure, see, or observe it directly like they can Doppler shifts. In order to become aware of the Time Dilation of a moving clock, they have to make radar measurements and apply Einstein's convention that the time of the outgoing radar signal is equal to the time of the returning echo signal. They have to keep track of a lot of measurements and then after the fact they can compile all the information and calculate the Time Dilation (and Length Contraction) of the other clocks and objects. So they can become aware of the others Time Dilation but only after doing a lot of work. Then when they get all done, they can transform from their own rest frame (in which they assumed the propagation of light was c in all directions) to any other frame and get a different set of Time Dilations, even for their own clock.

Your opinion on Time Dilation is tantamount to saying that each observer's own rest frame is preferred. But no frame is preferred in relativity. They are all equally valid.

You should abandon the short-cut formulas for Time Dilation and Length Contraction since, as you noted, they cannot be right in all situations. The Lorentz Transforms that you presented always work. That's what I use.

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14. Mar 30, 2014

### ANvH

You mean the Doppler shift of 2, as established by γ=1.25, β=0.6? But I think that that is not correct. You also need to transform the wave using β=0.333c and γ=1.06 →shift =1.41. So the Doppler shift changes too.

I agree, that is why the Doppler shift should change too if you create a new frame that is associated with a time dilation of 1.06.

I am not sure. You have shown me how to get to the right form of time dilation and length contraction using the Lorentz transformations and I am pleased with this. You clearly showed that one get to this by using one set of Lorentz transformations. The shortcut allows a quick calculation of the value of time dilation and length contraction with respect to the relative speed between two frames of references. You do the same.

I guess I have to live with the apparent discrepancy.

15. Mar 30, 2014

### ghwellsjr

No, you have merely shown that another shortcut formula doesn't work in all cases but the diagrams show that the Doppler is the same in all frames as determined by the Lorentz Transformation. If they didn't, the Lorentz Transformation wouldn't comport with reality and would have to be abandoned.

Actually, I prefer a different shortcut formula for Doppler because it depends only on the relative speed between the two clocks:

SQRT((1+BETA)/(1-BETA))

16. Mar 30, 2014

### ANvH

I am sorry to bother again, but I think you misread what I wrote. I think you are ambiguous when you say that the Doppler is the same in *all frames* and showing 3 diagrams and avoiding an answer. With respect to two frames having a relative speed of 0.6c moving toward each other, the Doppler shift is
$\gamma(1+\beta)=\sqrt{\frac{1+\beta}{1-\beta}}$.​
Once the two frames pass each other the Doppler becomes red-shifted
$\gamma(1-\beta)=\sqrt{\frac{1-\beta}{1+\beta}}$​
Using other frames will require again transforms. You gave the impression with your third diagram moving at 0.333c relative to the the first diagram that the Doppler remains 2 or 0.5, which cannot be true.
I don't think we disagree on this, nor do I think we disagree on time dilation and length contraction. All I am trying to make clear is that I fully understand the Doppler shift, and that I fully understand the diagrams providing length contraction and time dilation.

I am also trying to make clear that Doppler involves a blueshift (or a redshift) of both the frequency and wavenumber domains. The blueshift of the frequency domain translates to a shortening of the period, which seems paradoxical to a time dilation; the redshift of the wavenumber domain translates to a wavelength increase, which seems paradoxical to a length contraction.

Just to make clear that I am utilizing the Lorentz transforms to a wave:

$\omega t'-kx'=\gamma[\omega(t-\beta c^{-1}x)-k(x-\beta ct)]$​

Gathering the temporal components and the spatial components, and using $ω=kc$ we get

$\omega t'-kx'=\gamma(1+\beta)(\omega t-kx)$​

17. Mar 31, 2014

### ghwellsjr

Frames do not pass each other, the two observers with their clocks pass each other. I already twice described the inverse Doppler relationship when they pass in post #11.

Of course it is true and of course we disagree. Look at any of the diagrams. You will see that prior to the passing, between each pair of dots representing an observer's clock there is another thin line which shows that the Doppler factor is 2. After the passing, you will see that there is an additional dot between each pair of thin lines, which shows a Doppler factor of 0.5. These ratios are identical in all three frames. That is absolutely unambiguous.

I previously suspected that you were treating the rest frame of an observer as preferred but now it appears that you are equating the rest frame of an observer with the observer, otherwise, I cannot understand how you would say "the two frames pass each other". Am I correct? Is that why you think my third diagram is invalid, because there is no at-rest observer in it?

Last edited: Mar 31, 2014
18. Mar 31, 2014

### ANvH

I do not understand why the Doppler shift is "absolute" when a Lorentz transform is applied with β=0.333. The Doppler shift was first established by a Lorentz transform with β=0.6, correct? Then we established the time dilation related to β=0.6 using the first diagram. Then we evaluated that both red and blue observers see the Doppler shift of 2 before the clocks pass each other and then the shift of 0.5 after the clocks pass. I have no problem with that, and the time dilation of the moving frame remains 1.25 before and after.

Each clock has its own frame of reference, so when a clock moves relative to the other clock, then the frame associated with the moving clock moves too. When the clock passes, its frame passes (This is wrong?).

A clock emits a light wave at every tick, and if a clock's ticking is subject to a Lorentz transform (the frame without a rest observer, β=0.333, has a time dilation of 1.06), so should be the light wave that was emitted in another frame of reference. The way I understand you suggests that I am wrong and I don't get it, sorry.

Maybe the misunderstanding is just semantics, I don't know.

19. Mar 31, 2014

### ghwellsjr

It's the coordinates of events that are subject to the Lorentz Transformation when we go from one frame to another, not a clock's ticking, or Doppler shifts, or time dilation. Nothing changes except the values of the coordinates.

If you set up a scenario according to the coordinates of one frame, you are describing where (spacial coordinates) each observer, object, clock, etc is at each moment in time (temporal coordinates). If the object (observer, clock) is moving in that frame, then the object is subject to time dilation, not the frame. If the object is not moving, then its time dilation factor is 1 (or you could say it has no time dilation, they mean the same thing).

Light propagates at c along 45 degree diagonals the way we draw our diagrams. I put tick marks (dots) to show the progress of time on the thick world line for an object (observer or clock). These tick marks are subject to Time Dilation. I typically draw thin lines from some of these tick marks to show the image of the time on the clock as it propagates to the world lines of other objects (observers or clocks). The Doppler is shown by the ratio of thin lines to dots along a world line.

When you transform to a different frame at some speed with respect to the defining frame, you get a new set of coordinates for all the events (dots, intersections of lines, etc) but it doesn't change any information about what is happening to the various objects. This includes Doppler.

So you can transform the coordinates to another frame moving at any speed with respect to the defining frame, it doesn't have to be a speed where any particular object is at rest. You don't even have to have any object at rest in the defining frame, it's totally arbitrary, you do whatever you want.

I hope this clears up any semantic issues.

20. Mar 31, 2014

### ANvH

Thanks,

I also hope that this will clear up a number of semantic issues. The coordinates of the events are transformed and mapped to the frame. One can go from one frame to another. If a clock is moving in a particular frame then it is subject to time dilation. I get this and I assume you will agree with my repeat of your words.

How I see it when it comes to Doppler:
The clock is emitting a signal, which consists of a light wave with a wavelength of say 600 nm (as measured from a defining frame). If the clock is moving in a frame, it is subject to time dilation and appears to tick slower. At each tick a light wave is emitted and the wavelength of 600 nm is (should be) Doppler shifted. If we transform to a frame that will further increase the speed of the clock, the time dilation factor is increased further, the clock ticks even slower, and at each tick the wavelength of the signaling wave is (should be) more Doppler shifted.

What I think how you responded:
My guess is that you do not agree with the Doppler shift assessment. I have tried to remove semantic issues. If I understand you correctly, the defining frame is the frame from where the clock is emitting the signal and any frame transformation should not, does not, and cannot affect the Doppler shift.