Question about the reciprocity of time dilation

[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.

 

[Mister T]

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.

 

[Mister T]

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.

 

[Mister T]

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.