I Question about the reciprocity of time dilation

[FactChecker]

I just read this and didn't really grock the last two paragraphs, I'll read it again later, hopefully I can understand.

In your example are you using light clocks where the pulses travel vertically on the graph?

No. The light pulse is horizontal and is just to verify that the Red and Blue clocks are initially synchronized (versus their respective central clocks, R and B) so that the speed of light is c in their respective inertial reference frame.

 

[Chenkel]

No. The light pulse is horizontal and is just to verify that the Red and Blue clocks are initially synchronized (versus their respective central clocks, R and B) so that the speed of light is c in their respective inertial reference frame.

I'm a little confused, the light pulses are horizontal?

Maybe my imagination is lacking a little bit here, but I can assume you most likely know what you are talking about, I'll study it.

 

[FactChecker]

I'm a little confused, the light pulses are horizontal?

Yes. You are thinking about the vertical light used to illustrate the mathematics of the time dilation. I am talking about a (possible too casual) verification that the clocks along the way are initially synchronized so that the speed of light is c in either direction along the relative path. Einstein synchronization is a more formal way of synchronizing clocks that reflects a horizontal pulse so that it goes in a round trip.

Maybe my imagination is lacking a little bit here, but I can assume you most likely know what you are talking about, I'll study it.

Keep in mind that my goal is not to formally derive an equation for time dilation, but only to give intuition into why each inertial reference frame can measure the other's clock (R or B) to be running slow. The quick and dirty explanation is that one reference frame (say Blue) would measure the ticking clock of the other reference frame (R) by comparing it to his own blue clocks as it moves TO THE LEFT of the figure. On the other hand, the Red reference frame would measure the ticking clock B by comparing it to his own red clocks as it moves TO THE RIGHT of the figure. So we are talking about two different sets of clocks in opposite directions from the initial center clock positions.

 

[PeroK]

Looking at the left side from Red's point of view. Red can assume that he is stationary and everything is normal. Suppose Red has measured one light-second to the left of his center. He sees that Blue on the left side is rushing toward the center and will hit a light pulse (from the center) too early. Yet Blue still measures the speed of a left-directed pulse of light as c. How can that be? Blue clocks on that side must be set so that one second of Blue time elapses too early. Blue clocks are set ahead of his Red clocks. The farther to the left you go, the more they are set ahead.

This can't be right. Time dilation depends only on speed, not on direction of motion. In this case, the Blue clocks would show the same time dilation in the Red frame if they are moving away from a light source.

Time dilation is not directly derived from the time delay in the transmission of light signals, which seems to be the implication of your post.

 

[Histspec]

I've heard this is reciprocal and if we situate a clock at either reference frame we will measure the clock in the other reference frame to have a longer period than the clock at "rest."

As explained by others before, one can understand the symmetry of time dilation only in connection with relativity of simultaneity.
So let's say we have two clocks A and B at the same position, both indicating 0 at the start.
In the rest frame S1 of A we have another clock C synchronous with A.
In the rest frame S2 of B we have another clock D synchronous with B.
In order to simplify matters, we chose another frame S3 in which S1 (together with A and C) is moving with +v, while S2 (together with B and D) is moving with -v. This has the advantage that all four clocks have the same time dilation as seen in S3, and we furthermore choose v so that the Lorentz factor becomes 2 in S3.
We also observe that when A and C are synchronous in S1, they are not synchronous in S3 due to relativity of simultaneity, but C is ahead of A by 5 seconds. Likewise, when B and D are synchronous in S2 they are not synchronous in S3 due to relativity of simultaneity, but D is ahead of B by 5 seconds. This is described in the first part of the following image.
In the second part of the image we see the result when S1 and S2 continue their motion for 10 seconds (S3 time) until the clocks coincide. Since the time dilation factor is 2 and all clocks have same speed, all of the clocks will advance by 5 seconds.
We read off from the second image part: B is retarded with respect to C. However, since C is synchronous with A in S1, the observer in S1 concludes that B is retarded with respect to A.
We also read off from the second image part: A is retarded with respect to D. However, since D is synchronous with B in S2, the observer in S2 concludes that A is retarded with respect to B.

Summarizing: B is retarded with respect to A in S1, while A is retarded with respect to B in S2. Time dilation is symmetric.

 

[jartsa]

I've heard this is reciprocal and if we situate a clock at either reference frame we will measure the clock in the other reference frame to have a longer period than the clock at "rest."

I'm a little confused by the reciprocity and I wonder how the period of either clock can be smaller depending on where we set up our rest clock.

Instead of being confused about reciprocity, you should imagine a scenario where observer A is confused about reciprocity.

A says: B moves and B's clock is slow. My clock is normal. Why the heck does B think that my clock is slower than his own clock?

The task is now to explain the thoughts of B in the frame of A.

 

[Sagittarius A-Star]

If you have two light clocks moving relative to each other (A and B) (let's also say they are synchronized when they meet) where each one counts to 4 ticks and claps on each tick and covers itself in paint on the 4th tick lets also say that gamma equals 2.

Please be aware, that only every second light-pulse refection can be a "tick".

Caution, fake light clock!
Occasionally one sees animations with light clocks ticking twice as fast as the ones shown here. For them the counter counts when the pulse reaches the upper mirror and also when it reaches the lower mirror. Such light clocks lead the simple functional principle ad absurdum, because how does the counter know when the pulse arrives at the lower mirror? This information would first have to be laboriously transferred from the lower mirror to the counter. But this transmission cannot be done faster than the speed of light. In particular, the information would not reach the counter before the light pulse itself already arrives again at the upper mirror.

Source:
https://www.einstein-online.info/en/spotlight/light-clocks-time-dilation/

 

[Dale]

@Chenkel I found this good guide on spacetime diagrams

https://cs.westminstercollege.edu/~ccline/courses/phys301/spacetime_diagrams.pdf

 

[Chenkel]

@Chenkel I found this good guide on spacetime diagrams

https://cs.westminstercollege.edu/~ccline/courses/phys301/spacetime_diagrams.pdf

Thanks, I'll check it out.

 

[robphy]

A spacetime diagram is essentially a position-vs-time diagram (as opposed to a spatial diagram of moving boxcars). By convention, the time axis runs upwards instead of to the right, as in PHY 101. The usual stumbling block is how to motivate and determine the ticks along lines that are not parallel to the coordinate axes.

Here's my introduction to spacetime diagrams, motivating a graphical method using light clocks.
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
and more recently
https://www.physicsforums.com/insights/relativity-on-rotated-graph-paper-a-graphical-motivation/
which uses the Doppler effect and what you call "reciprocity" to develop special relativity.

My answer to https://physics.stackexchange.com/questions/383248/how-can-time-dilation-be-symmetric
uses diagrams like this to display the reciprocity

which is analogous to what happens in ordinary Euclidean geometry

(using https://www.desmos.com/calculator/wm9jmrqnw2 with E=-1 ).

Yes.

One way to see this visually is to draw a spacetime diagram. I recommend $\gamma=5/4$ instead of 2. Draw light pulses from each “dump” event, one that is emitted and one that was received at the dumping. Each frame will claim that the other frame’s dump is simultaneous with the event on their own worldline which is halfway between the emitted and received light pulses.

The essence of the first diagram is essentially what @Dale is suggesting, with \gamma=5/4 (i.e. v/c =3/5 and k=2), which is arithmetically nicer than either v/c =1/2 or \gamma=2.

 

[Chenkel]

A spacetime diagram is essentially a position-vs-time diagram (as opposed to a spatial diagram of moving boxcars). By convention, the time axis runs upwards instead of to the right, as in PHY 101. The usual stumbling block is how to motivate and determine the ticks along lines that are not parallel to the coordinate axes.

Here's my introduction to spacetime diagrams, motivating a graphical method using light clocks.
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
and more recently
https://www.physicsforums.com/insights/relativity-on-rotated-graph-paper-a-graphical-motivation/
which uses the Doppler effect and what you call "reciprocity" to develop special relativity.

My answer to https://physics.stackexchange.com/questions/383248/how-can-time-dilation-be-symmetric
uses diagrams like this to display the reciprocity
View attachment 336591
which is analogous to what happens in ordinary Euclidean geometry
View attachment 336592
(using https://www.desmos.com/calculator/wm9jmrqnw2 with E=-1 ).

The essence of the first diagram is essentially what @Dale is suggesting, with \gamma=5/4 (i.e. v/c =3/5 and k=2), which is arithmetically nicer than either v/c =1/2 or \gamma=2.

You gave me a lot to study, I'll look over it. Thanks

 

[Chenkel]

If the Lorentz factor is 2 the two clocks synchronize to an initial reading of zero when they meet then if the rest clock (A) ticks to 10 the other clock (B) will tick to 5.

But at the time the other clock (B) is 5 B will perceive the clock A as ticking to (2.5).

But at the time A is 2.5 A will perceive B as ticking to 1.25

This is one of the issues I run into, the clocks seem to be inconsistent when going back and forth and switching from the perspective of one clock to the perspective of the other.

How can I make the symmetry of time dilation consistent in my mind?

 

[PeterDonis]

How can I make the symmetry of time dilation consistent in my mind?

By including relativity of simultaneity in your analysis. You keep using the phrase "at the time" repeatedly without taking into account that each time you use it, it's a different time--a different pair of simultaneous events.

A good tool for making such situations clear is a spacetime diagram. I would suggest drawing one for your scenario.

 

[Chenkel]

By including relativity of simultaneity in your analysis. You keep using the phrase "at the time" repeatedly without taking into account that each time you use it, it's a different time--a different pair of simultaneous events.

A good tool for making such situations clear is a spacetime diagram. I would suggest drawing one for your scenario.

I'll work on drawing spacetime diagrams. I'll look over the links you guys provided, they look helpful.

 

[Nugatory]

How can I make the symmetry of time dilation consistent in my mind?

Draw a spacetime diagram in which one of the clocks is at rest (time axis vertical, space axis horizontal) and the other is moving (time and space axes slanted). Draw on that diagram a few lines indicating points that happen at the same time according to each clock. Look at how these lines relate to one another.

 

[Chenkel]

Draw a spacetime diagram in which one of the clocks is at rest (time axis vertical, space axis horizontal) and the other is moving (time and space axes slanted). Draw on that diagram a few lines indicating points that happen at the same time according to each clock. Look at how these lines relate to one another.

I'll do that after I grock spacetime diagrams. Thanks for the info.

 

[Sagittarius A-Star]

If the Lorentz factor is 2 the two clocks synchronize to an initial reading of zero when they meet then if the rest clock (A) ticks to 10 the other clock (B) will tick to 5.

But at the time the other clock (B) is 5 B will perceive the clock A as ticking to (2.5).

But at the time A is 2.5 A will perceive B as ticking to 1.25

This is one of the issues I run into, the clocks seem to be inconsistent when going back and forth and switching from the perspective of one clock to the perspective of the other.

How can I make the symmetry of time dilation consistent in my mind?

• With reference to the rest frame of A, clock B moves and clock A is at rest.
• With reference to the rest frame of B, clock A moves and clock B is at rest.

Do you regard this as a contradiction?
Why not?

 

[robphy]

If the Lorentz factor is 2 the two clocks synchronize to an initial reading of zero when they meet then if the rest clock (A) ticks to 10 the other clock (B) will tick to 5.

But at the time the other clock (B) is 5 B will perceive the clock A as ticking to (2.5).

But at the time A is 2.5 A will perceive B as ticking to 1.25

This is one of the issues I run into, the clocks seem to be inconsistent when going back and forth and switching from the perspective of one clock to the perspective of the other.

How can I make the symmetry of time dilation consistent in my mind?

The spacetime diagrams in my post offer a geometrical way
to encode what it means that "B will perceive A...".

For B to determine "what pairs of events are simultaneous",
B constructs a line that is spacetime-perpendicular* to B's worldline.
( * perpendicular using the Minkowski metric ).

• In the diagram with the diamonds, follow the spacelike diagonal of one of B's light-clock diamonds.
• Alternatively, along B's worldline, construct the tangent line to the "circle" (a hyperbola in special relativity) whose center is on B's worldline. The Euclidean version is shown in the second diagram.
  Use the desmos visualization with E=+1 to see the special-relativity version.

If you follow the above construction, you'll see that your sequence refers to a sequence of different pairs of events. (Only in the E=0 Galilean case will the sequence of pairs overlap... and do so without inconsistency... but that's not in agreement with special relativity E=+1.)

 

[Chenkel]

• With reference to the rest frame of A, clock B moves and clock A is at rest.
• With reference to the rest frame of B, clock A moves and clock B is at rest.

Do you regard this as a contradiction?
Why not?

I don't see that as a contradiction, relative velocity seems to imply one party is at rest and the other is moving.

 

[Sagittarius A-Star]

I don't see that as a contradiction, relative velocity seems to imply one party is at rest and the other is moving.

Correct.

• With reference to the rest frame of A, clock B is time-dilated by the factor $\sqrt{1-v^2/c^2}$ and clock A is time-dilated by the factor $\sqrt{1-0/c^2}=1$ (=not time-dilated).
• With reference to the rest frame of B, clock A is time-dilated by the factor $\sqrt{1-v^2/c^2}$ and clock B is time-dilated by the factor $\sqrt{1-0/c^2}=1$ (=not time-dilated).

Do you regard this as a contradiction?
Why or why not?

 

[Chenkel]

Correct.

• With reference to the rest frame of A, clock B is time-dilated by the factor $\sqrt{1-v^2/c^2}$ and clock A is time-dilated by the factor $\sqrt{1-0/c^2}=1$ (=not time-dilated).
• With reference to the rest frame of B, clock A is time-dilated by the factor $\sqrt{1-v^2/c^2}$ and clock B is time-dilated by the factor $\sqrt{1-0/c^2}=1$ (=not time-dilated).

Do you regard this as a contradiction?
Why or why not?

That makes sense, I don't see a contradiction.

The rest frame has zero velocity relative to its inertial reference frame, leading to no time dilation for the clock at rest.

 

[Chenkel]

The spacetime diagrams in my post offer a geometrical way
to encode what it means that "B will perceive A...".

For B to determine "what pairs of events are simultaneous",
B constructs a line that is spacetime-perpendicular* to B's worldline.
( * perpendicular using the Minkowski metric ).

• In the diagram with the diamonds, follow the spacelike diagonal of one of B's light-clock diamonds.
• Alternatively, along B's worldline, construct the tangent line to the "circle" (a hyperbola in special relativity) whose center is on B's worldline. The Euclidean version is shown in the second diagram.
  Use the desmos visualization with E=+1 to see the special-relativity version.

If you follow the above construction, you'll see that your sequence refers to a sequence of different pairs of events. (Only in the E=0 Galilean case will the sequence of pairs overlap... and do so without inconsistency... but that's not in agreement with special relativity E=+1.)

Thanks for the info, I think I need to learn more about the geometry of spacetime to understand.

 

[Sagittarius A-Star]

That makes sense, I don't see a contradiction.

The rest frame has zero velocity relative to its inertial reference frame, leading to no time dilation for the clock at rest.

Correct. I would formulate the 2^nd sentence: "With reference to it's inertial rest frame, a clock has zero velocity and therefore no time-dilation."

Time-dilation is a reference frame-dependent effect. The clocks A and B cannot be compared directly with each other when they don't meet.

But they can be compared to the coordinate-time of a reference-frame, which could be thought of as represented by a grid of Einstein-synchronized clocks, that are all at rest in this frame.

According to your scenario in posting#39:

• With reference to the rest frame of A, clock B needs 8 ticks of the coordinate time to covers itself in paint and clock A needs only 4 ticks of the coordinate time to covers itself in paint.
• With reference to the rest frame of B, clock A needs 8 ticks of the coordinate time to covers itself in paint and clock B needs 4 only ticks of the coordinate time to covers itself in paint.

 

[Chenkel]

I've been reading the spacetime diagram article that Dale linked, its a little bit confusing but I have a question about the symmetry of time dilation, you guys may have already pointed me in the right direction but I want to be sure.

If a spaceship is launched from earth and the spaceship aims a telescope at earth it will observe people aging slower than people in the spaceship, if a person on earth aims a telescope at the spaceship the person on earth will observe people on the spaceship aging slower than people on earth, why is this not contradiction?

Sorry if this question is a little bit basic, I'm just trying to see what I need to understand/learn to resolve this seeming paradox in my head.

Maybe you guys already told me what I have to study to understand it (spacetime diagrams) but I'm wondering if there's a more simple explanation.

Thanks for all the help so far.

 

[PeterDonis]

If a spaceship is launched from earth and the spaceship aims a telescope at earth it will observe people aging slower than people in the spaceship

Only if the spaceship is moving away from earth. If the spaceship turns around and moves back towards earth, it will observe, through the telescope, people on earth aging faster than people in the spaceship. And even if the spaceship is moving away from earth, it does not observe, through the telescope, the slower aging rate that the time dilation calculation tells you. What is observed through the telescope is determined by the relativistic Doppler effect. The time dilation calculation is what you get if you take what is actually observed, according to the relativistic Doppler effect, and factor out the effects of light travel time.

if a person on earth aims a telescope at the spaceship the person on earth will observe people on the spaceship aging slower than people on earth

Same comment as above; what is actually observed depends on the direction of travel and is determined by the relativistic Doppler effect.

why is this not contradiction?

Why do you think it is a contradiction? Just a vague feeling is not enough. Think carefully and see if you can find a logical reason why it should be a contradiction. And if you can't find one (which you won't be able to), why is the question quoted above even a question?

If it helps, separate out the Doppler effect--what is directly observed--and the time dilation effect, which involves adjusting for light travel time.

 

[Chenkel]

Only if the spaceship is moving away from earth. If the spaceship turns around and moves back towards earth, it will observe, through the telescope, people on earth aging faster than people in the spaceship. And even if the spaceship is moving away from earth, it does not observe, through the telescope, the slower aging rate that the time dilation calculation tells you. What is observed through the telescope is determined by the relativistic Doppler effect. The time dilation calculation is what you get if you take what is actually observed, according to the relativistic Doppler effect, and factor out the effects of light travel time.Same comment as above; what is actually observed depends on the direction of travel and is determined by the relativistic Doppler effect.Why do you think it is a contradiction? Just a vague feeling is not enough. Think carefully and see if you can find a logical reason why it should be a contradiction. And if you can't find one (which you won't be able to), why is the question quoted above even a question?

If it helps, separate out the Doppler effect--what is directly observed--and the time dilation effect, which involves adjusting for light travel time.

You gave me some things I can study, thanks.

 

[Dale]

the spaceship aims a telescope at earth it will observe

You have to be very careful here. This is one of the things that is often done didactically that is very potentially confusing.

Sometimes the word "observe" is used to mean the raw observations, what someone actually sees visually. Other times the word "observe" is used to mean what someone calculates actually happened after accounting for the finite speed of light in their reference frame.

I think in this case you are using it in the "raw observations" meaning since you mention the telescope.

If a spaceship is launched from earth and the spaceship aims a telescope at earth it will observe people aging slower than people in the spaceship, if a person on earth aims a telescope at the spaceship the person on earth will observe people on the spaceship aging slower than people on earth, why is this not contradiction?

Why would this be a contradiction? What is being contradicted?

Try writing down some mathematical symbols for what you think is contradictory. A contradiction would be something like $A>B$ and $B>A$, but remember that what the spaceship sees looking at earth in the telescope and what the earth sees looking at earth without the telescope are not both $A$.

 

[Sagittarius A-Star]

If a spaceship is launched from earth and the spaceship aims a telescope at earth it will observe people aging slower than people in the spaceship, if a person on earth aims a telescope at the spaceship the person on earth will observe people on the spaceship aging slower than people on earth, why is this not contradiction?

As others mentioned, this scenario describes the relativistic Doppler effect.

An optical clock sends out light with a certain frequency. Assume, that on earth and at the spaceship are equal optical clocks. Call the frequency of such an optical clock in it's rest-frame $f_0$.

The relativistic longitudinal Doppler effect, as calculated in the receiver's frame, is:$$f_R =f_0 {1 \over \gamma (1+{v / c})}$$The factor 1/$\gamma$ is the influence of time-dilation of the moving sender, the rest of the term behind it is the influence of increasing distance. In case of decreasing distance, the sign of $v$ would be negative.

A calculation in the sender's rest-frame leads to a formula, that is equivalent to the above one:$$f_R =f_0 \gamma (1-{v / c})$$ I propose, that you do a calculation to check the equivalence of both formulas.

Source:
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_in_an_arbitrary_direction

Animations of the Doppler effect:
https://www.einstein-online.info/en/spotlight/doppler/

The relativistic longitudinal Doppler effect formula can be derived in the rest frame of the receiver or the rest frame of the sender by combining the non-relativistic (classical) Doppler effect formula with the $\gamma$-factor, to account for time-dilation.

 

[Chenkel]

Only if the spaceship is moving away from earth. If the spaceship turns around and moves back towards earth, it will observe, through the telescope, people on earth aging faster than people in the spaceship.

That kind of goes against my intuition, I thought the time dilation formula $T_m = {\gamma}T_r$ tells you that the moving object has a larger period than the clock in the rest objects rest frame.

I thought time dilation always occurs when there is any relative velocity at all, so the telescope on the spaceship should observe earth to tick slower than the clock on the spaceship, but now there is this Doppler effect you're invoking which I need to look at for your logic to make sense.

I trust your logic does make sense, it's just a lot of this stuff is over my head, physics is hard.

 

[PeterDonis]

That kind of goes against my intuition

That's why I specifically distinguished the relativistic Doppler effect, which is what you directly observe, with time dilation, which is what you calculate by correcting what you observe for light travel time. You need to take the time to understand that distinction and retrain your intuitions accordingly.

I thought the time dilation formula $T_m = {\gamma}T_r$ tells you that the moving object has a larger period than the clock in the rest objects rest frame.

It tells you what the calculated period of the moving clock is in the rest clock's rest frame, after correcting for light travel time. The rest clock cannot directly observe the moving clock's clock rate because the moving clock is not co-located with the rest clock: the rest clock can only observe the light signals from the moving clock, and the fact that the moving clock is moving relative to the rest clock means that the light travel time in the rest clock's frame is not constant. If the moving clock is moving away from the rest clock, the light travel time increases with each successive light signal. If the moving clock is moving towards the rest clock, the light travel time decreases with each successive light signal. You have to take that into account to relate what the rest clock directly observes with the calculated time dilation.

 

[Herman Trivilino]

But at the time the other clock (B) is 5 B will perceive the clock A as ticking to (2.5).

"At the time" according to who? In which frame do they occur at the same time?

But at the time A is 2.5 A will perceive B as ticking to 1.25

Again, "at the time" according to who? In which frame do they occur at the same time?

Events that are simultaneous in one frame aren't simultaneous in the other.

 

[Herman Trivilino]

Sometimes the word "observe" is used to mean the raw observations, what someone actually sees visually. Other times the word "observe" is used to mean what someone calculates actually happened after accounting for the finite speed of light in their reference frame.

I was under the impression that the word "observe" always referred to the latter in textbooks and in the literature, to carefully distinguish it from what we see.

 

[robphy]

That kind of goes against my intuition…

….
I trust your logic does make sense, it's just a lot of this stuff is over my head, physics is hard.

You’ve tried words, logic, formulas, intuition…
in my opinion, it might be time to invest more effort in learning to draw spacetime diagrams.

A lot of relativity problems are analogues of geometry problems… and they will help build your relativistic intuition and explain what the various formulas mean physically.

Would you solve a geometry problem without a diagram?

My $0.02.

 

[Herman Trivilino]

If a spaceship is launched from earth and the spaceship aims a telescope at earth it will observe people aging slower than people in the spaceship, if a person on earth aims a telescope at the spaceship the person on earth will observe people on the spaceship aging slower than people on earth, why is this not contradiction?

As long as you keep comparing only one clock at rest in one frame to only one clock at rest in another frame, you will continue to be confused. You have to think in terms of events. An event occurs in the spaceship, and then later another event occurs in the spaceship. The time that elapses between those events is always less than the time that elapses between those events as measured on Earth. These observations get distilled into one statement, that the clocks on the ship are running slow, but it is that distillation that is the cause of your confusion.

Now consider an event that happens on Earth, and then later a second event occurs on Earth. The time that elapses between those events is always less than the time that elapses between those events as measured in the spaceship. These observations get distilled into the statement that clocks on Earth are running slow. Again, that distillation is the cause of your confusion.

Note that the two events occurring on the spaceship, call them A and B, are not the same events that occur on Earth, call them C and D. $\Delta t_{AB}$ is smaller than $\gamma \Delta t_{AB}$, and $\Delta t_{CD}$ is smaller than $\gamma \Delta t_{CD}$. That is not a contradiction!

 

[Dale]

I was under the impression that the word "observe" always referred to the latter in textbooks and in the literature, to carefully distinguish it from what we see.

Some authors are careful, but others are sloppy. And different authors, even careful ones, may use different words. But here I am more concerned about what @Chenkel intended. I am not sure whether they were making such a distinction or not.

 

[Chenkel]

I see if I am to consider one clock at rest and the other as moving in order to determine what the rest clock sees in terms of frequency of ticks I must also take into account the doppler effect.

I'm also going to try to get good at spacetime diagrams, thanks for the help everyone.

 

[Sagittarius A-Star]

I see if I am to consider one clock at rest and the other as moving in order to determine what the rest clock sees in terms of frequency of ticks I must also take into account the doppler effect.

I'm also going to try to get good at spacetime diagrams, thanks for the help everyone.

You can find in chapter 11 of Morin's book about classical mechanics:

• page XI-19: chapter "11.4 The Lorentz transformations"
• page XI-30: chapter "11.7 Minkowski diagrams"
• page XI-32: chapter "11.8 The Doppler effect"

Source:
https://scholar.harvard.edu/files/david-morin/files/cmchap11.pdf
via:
https://scholar.harvard.edu/david-morin/classical-mechanics

 

[Chenkel]

You can find in chapter 11 of Morin's book about classical mechanics:

• page XI-19: chapter "11.4 The Lorentz transformations"
• page XI-30: chapter "11.7 Minkowski diagrams"
• page XI-32: chapter "11.8 The Doppler effect"

Source:
https://scholar.harvard.edu/files/david-morin/files/cmchap11.pdf
via:
https://scholar.harvard.edu/david-morin/classical-mechanics

Thank you for the references.

 

[Chenkel]

@Chenkel I found this good guide on spacetime diagrams

https://cs.westminstercollege.edu/~ccline/courses/phys301/spacetime_diagrams.pdf

I just found this post again, I'll try to study that article again.