Conflict Between Time Dilation and Red/Blue Shift?

In summary: The two twins do not see the same thing - the stay at home sees redshifted light for more than half the trip, while the traveller sees redshifted light for exactly half the trip. Since they don't see the same things they aren't surprised to find that their ages aren't the same....the Earth becomes an ellipsoid because it is foreshortened around a diameter.The circling observer is not in an inertial reference frame and naive intuition from inertial frames does not necessarily carry over.
  • #1
CherylJosie
12
6
TL;DR Summary
Does Red/Blue Shift Indicate Relative Passage of Time?
Experiment 1: Astronaut travels away from Earth at near the speed of light, then travels toward the Earth at near the speed of light.

Einstein tells us she barely aged, but red shift/blue shift don't seem to agree with that.

While traveling away, both Earth and astronaut observe each other red-shifted, moving slowly, and aging slowly.

While traveling toward, both Earth and astronaut observe each other blue-shifted, moving rapidly, and aging rapidly.

Net effect: No change in rate of aging.

Experiment 2: Astronaut circles the Earth at near the speed of light.

Einstein tells us that time slows down for the astronaut and she also becomes foreshortened.

Because the path is circular, there is no relative change in distance, and hence no relative velocity and no red shift or blue shift. Astronaut and Earth age at the same rate. Moreover, from the reference frame of the astronaut, rather than foreshortening toward a lens shape, the Earth becomes an ellipsoid because it is foreshortened around a diameter.

Experiment 3: A photon of a particular frequency is emitted by a light source that is traveling toward an observer at near the speed of light. The velocity of the light source imparts kinetic energy to the photon and blue-shifts its frequency. The photon is blue-shifted so drastically it becomes a gamma ray. It is no longer a photon, and the light source is now an atomic explosion.

I know that these three examples all have different outcomes than what I posited, but I can't reconcile the contradictions and I don't know the math. What am I missing?
 
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  • #2
CherylJosie said:
Net effect: No change in rate of aging.
This is not correct. The two twins do not see the same thing - the stay at home sees redshifted light for more than half the trip, while the traveller sees redshifted light for exactly half the trip. Since they don't see the same things they aren't surprised to find that their ages aren't the same.
CherylJosie said:
Because the path is circular, there is no relative change in distance, and hence no relative velocity and no red shift or blue shift.
In relativity there is a transverse Doppler effect. Essentially, light is redshifted due to the time dilation of the moving source.
CherylJosie said:
Moreover, from the reference frame of the astronaut, rather than foreshortening toward a lens shape, the Earth becomes an ellipsoid because it is foreshortened around a diameter.
The circling observer is not in an inertial reference frame and naive intuition from inertial frames does not necessarily carry over.
CherylJosie said:
The photon is blue-shifted so drastically it becomes a gamma ray. It is no longer a photon, and the light source is now an atomic explosion.
No, a photon is a photon. Just because it's really high energy in your frame doesn't change that. It's always really high energy in someone's frame.
 
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  • #4
CherylJosie said:
Summary:: Does Red/Blue Shift Indicate Relative Passage of Time?

Experiment 2: Astronaut circles the Earth at near the speed of light.

Einstein tells us that time slows down for the astronaut and she also becomes foreshortened.

Because the path is circular, there is no relative change in distance, and hence no relative velocity and no red shift or blue shift. Astronaut and Earth age at the same rate.
I don’t have too much to add to the other excellent answers except to add that this transverse Doppler shift in particular is the subject of many precision tests of relativity. That is partly because it is something that is not just quantitatively but qualitatively different from pre-relativistic physics.
 
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  • #5
CherylJosie said:
I know that these three examples all have different outcomes than what I posited, but I can't reconcile the contradictions and I don't know the math. What am I missing?
You're missing a precise understanding of the theory. All you have is partially understood bits and pieces that you are trying to sew together.
 
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  • #6
CherylJosie said:
Summary:: Does Red/Blue Shift Indicate Relative Passage of Time?
Yes, if you use the formula for the relativistic Doppler effect, not the classical Doppler effect.

You can derive the formula for the (longitudinal) relativistic Doppler effect in two alternative ways:
  • Multiply the time-dilation factor ##\gamma## with the formula for the classical Doppler effect for the case, that only the receiver is moving relative to the medium (=calculation in the frame of the sender).
  • Alternatively, multiply the time-dilation factor ##\frac{1}{\gamma}## with the formula for the classical Doppler effect for the case, that only the sender is moving relative to the medium (=calculation in the frame of the receiver).
Both approaches lead to the same result. The formula for the (longitudinal) relativistic Doppler effect is:
$$ f_r = f_s \gamma (1- \frac{v}{c}) = f_s \frac{1}{\gamma} \frac{1}{1+ \frac{v}{c}}$$
Source, see Eq. 1.:
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Relativistic_longitudinal_Doppler_effect
CherylJosie said:
Experiment 2:
...
Because the path is circular, there is no relative change in distance, and hence no relative velocity and no red shift or blue shift.
That's wrong. It seems, that you argue here with the classical Doppler effect and so missing a factor for the relativistic time-dilation. For light in vacuum, there exists no classical medium.
 
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  • #7
There's a nicer way; with ##c=1##, if the source has 4-velocity ##u_s^a## and the receiver has 4-velocity ##u_r^a##, then writing ##k^a = \omega_r(u_r^a + n^a)## for some unit vector ##n^a## orthogonal to ##u_r^a## [because ##k^a## is null] and also defining a vector ##V^a## by the orthogonal decomposition ##u_s^a = \gamma(u_r^a + V^a)##, you have ##\omega_s = -k_a u_s^a = -\omega_r\gamma((u_r)_a + n_a)( u_r^a + V^a) = \omega_r \gamma(1 - n_a V^a)##, because ##(u_r)_a u_r^a = -1##, ##(u_r)_a V^a = 0## and ##n_a u_r^a = 0##.

Hence ##\omega_r = \omega_s / (\gamma(1- n_a V^a))##, which is the usual formula. Note that ##n^a## and ##V^a## are 4-vectors orthogonal to ##u_r^a##, so in an orthonormal frame ##(u_r^a, (e_1)^a, (e_2)^a, (e_3)^a)## of the receiver they just take the form ##n^a \equiv (0, \vec{n})## and ##V^a \equiv (0, \vec{V})##
 
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  • #8
Dale said:
t this transverse Doppler shift in particular is the subject of many precision tests of relativity. That is partly because it is something that is not just quantitatively but qualitatively different from pre-relativistic physics.

There's also some history. Herbert Ives of the Ives-Stillwell experiment (which tested the transverse Doppler effect in 1938) believed SR to be incorrect. This was therefore viewed as strong evidence in favor of SR. (Ives had his own interpretation of results, which, to his disappointment, the community did not accept)

And, please, this is a B-level thread!
 
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  • #9
phinds said:
You're missing the point. The OP CLEARLY does not have the experience to make heads nor tails out of your math and all you are likely to do with that approach is to make him/her think that this is the kind of forum that's no good for people without deep math.
Claryfying: This question isn't intended for the brainiacs, though they are welcome to strut their stuff! I'm trying to clarify my hand-waving understanding of relativity by asking basic questions.

I think I've identified the source of my confusion in the concept of inertial reference frame, so I'm going to do some reading up on that. I didn't realize that acceleration would be a factor in time dilation, though I should have sussed that out from the spatial distortion and time dilation of gravity.

I posted at high school level to let everyone know I'm clueless about physics compared to the geniuses that post here. Satisfied and grateful for the excellent responses, so give yourselves a break and a big pat on the back for putting yourselves out there.

The TL;DR for those that care: I am an electrical engineer, honors student turned IC designer now retired, did well in physics undergrad, fine with calculus/advanced math through line integrals, electives in solid state physics, had some difficulty with electromagnetics (partial differentials, boundary conditions, wave guides, A field), communications (transforms), and probability (the whole concept of random variables etc), linear algebra wasn't required for my major when I graduated in 1990 so I faked it, chemistry and history were my weakest grades, and no, nada on relativity. I also rushed through my degree in 3.5 years because I had the opportunity to devote myself to it 100% and I was getting on in years with chronic health issues while hoping to start a career and have a family.

The concept of tunnel diode still mystifies me even though I replaced one in an old Tektronix scope trigger circuit. Probably spent more on the diode than the scope was worth, but it had dual time bases and that was cool to play with while viewing the sync interval of video frames.
 
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  • #10
CherylJosie said:
The TL;DR for those that care: I am an electrical engineer, honors student turned IC designer now retired, did well in physics undergrad, fine with calculus/advanced math through line integrals, electives in solid state physics ...
In that case, there is no excuse for your original post! Your "hand-waving understanding" is full of basic misconceptions.

You need to apply the same rigour learning SR as you would learning EE.
 
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  • #11
Vanadium 50 said:
In any event, leaving the OP in the dust is not something we should aspire to. "Just ignore it if you don't like it" could be answered by "if it doesn't belong here, start your own thread". A message that is valuable on its own but does not help the OP better belongs elsewhere.

Actually I was grateful for the elaboration even if I'm not up to the math. I'm in a sticky position where my curiosity exceeds my grasp. Maybe if I keep on posting I'll get a better feel for this and generate less confusion. My first post was a total bomb that got deleted in moderation. I'll take responsibility for the awkwardness as noob. I don't mind. I'm just grateful for a forum like this and all the helpful people here who are so generous with their insights.
 
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  • #12
Here is another alternative, based on a proposal of @PeterDonis. As Einstein calculated in his 1905 paper, the frequency of a photon transforms according to the same formula as it's energy:
Sagittarius A-Star said:
This is interesting and shows, that, according to RT, the energy of a photon transfoms by the same factor as it's frequency, without making use of the QT-formula ##W=h*f##.

In an inertial frame, in which the sender is at rest at least at the point in time, when the photon is sent-out, the energy ##E## of the photon can be transformed in the following way to the receiver:

##E' = \mathbf {P_{photon}} \cdot \mathbf {U_{receiver}} = \begin {pmatrix} E/c \\ \vec {p} \end {pmatrix} \cdot \begin {pmatrix} c \\ \vec {v} \end {pmatrix} \gamma = (E - \vec{p} \cdot \vec{v}) \gamma = (E - p * v * \cos{\varphi}) \gamma##

with ##\varphi## = angle between ##\vec{p}## and ##\vec{v}## in the source-frame. With ##p = E/c ## for photons follows:

##E' = E \gamma (1 - \frac{v}{c} * \cos{\varphi})##.

That is the same factor as in the relativistic doppler shift of the frequency (see Eq. 7):
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Einstein_Doppler_shift_equation
 
  • #13
CherylJosie said:
I didn't realize that acceleration would be a factor in time dilation, though I should have sussed that out from the spatial distortion and time dilation of gravity.

To be clear, acceleration in itself is not a factor in time dilation. In a typical SR "twin paradox" scenario, the role acceleration plays is to make the traveling twin turn around; that affects the geometry of the traveling twin's path through spacetime, which affects the relative aging of the twins, but the acceleration itself does not cause any time dilation. Gravitational time dilation depends on gravitational potential, not acceleration.
 
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  • #14
CherylJosie said:
This question isn't intended for the brainiacs, though they are welcome to strut their stuff!
CherylJosie said:
Actually I was grateful for the elaboration even if I'm not up to the math.

All thread participants, please take note: given these statements by the OP, all conversation about whether more advanced material is or is not appropriate for this thread is hereby declared off topic. The OP says it's OK, so it's OK. Please refrain from any further comments along those lines, and expect some thread cleanup shortly.
 
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  • #15
CherylJosie said:
I didn't realize that acceleration would be a factor in time dilation, though I should have sussed that out from the spatial distortion and time dilation of gravity.
Acceleration isn't a factor in time dilation. It's a factor in making someone non-inertial (it's the definition of being non-inertial, in fact), and that means that you cannot assume that they are at rest in an inertial frame and, therefore, that their perspective isn't the same as that of an inertial frame. Most of the standard results in introductory presentations of special relativity assume everybody is at rest in an inertial frame, and those results don't carry across - which is probably part of where your thinking is going wrong.

I'd recommend a good book. If you've got maths enough to handle an engineering degree, my favourite is Taylor and Wheeler's Spacetime Physics (now free to download from Taylor's website), but other's seem to prefer Morin's Relativity for the Enthusiastic Beginner (the first chapter of which is free to download and can be found with a search).
 
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  • #16
CherylJosie said:
I was grateful for the elaboration even if I'm not up to the math.

To restate the basic point that @etothepi was making in words: the concept of "time dilation" is not really a fundamental concept in relativity. It's just a particular effect that some treatments choose to focus on. The fundamental concept is that a light ray has a path through spacetime, and the things that emit and receive the light ray have paths through spacetime, and at events where a light ray is emitted or received, the light ray and the emitter or receiver both can be described by 4-vectors, and the inner product (the spacetime version of the "dot product" from ordinary vector analysis) of those vectors tells you the emitted or received frequency of the light ray. In other words, it's all just vectors and geometry.
 
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  • #17
Where I'm having immediate difficulty with the math is in interpreting the editor's mathematical notation. Bear with me, for a former engineer I'm sort of a Luddite by virtue of economic necessity. The code is rendering as code rather than notation in my FOSS Ubuntu 20.04 with Firefox and Chromium.

I see a button below the editor window called 'LaTeX Guide':

Learn how to type mathematical equations using LaTeX
LaTeX is the standard language used by mathematicians, physicists, etc. for typesetting mathematical expressions. PF uses https://docs.mathjax.org/en/latest/mathjax.html, a Javascript-based engine, to process LaTeX code in posts and render it in the way you would see it in a textbook.​
Note: the PF apps for iOS (iPhone, iPad) and Android can only display raw LaTeX code as plain text. You must use PF via a web browser in order to see properly-rendered equations.​
If you're on a computer, you can see the LaTeX code for any equation displayed in the forum by right-clicking (under Windows) or control-clicking (under Mac OS) on it. This brings up a contextual popup menu. Choose Show Math As, then TeX Commands.​

I also have an iPhone, and an iPad. Apparently that's no help.

In Ubuntu I appear to have a choice of installing MathJax and/or TexLive libraries.

A cursory web search has turned up little guidance on MathJax, but the MathJax notes in Synaptic Package manager says it does all the work for me once it is installed:
MathJax was designed with the goal of consolidating the recent advances in web​
technologies into a single, definitive, math-on-the-web platform supporting​
the major browsers and operating systems.​
It requires no setup on the part of the user (no plugins to download or software to install), so the page author can write web documents that include mathematics and be confident that users will be able to view it naturally and easily. Simply include MathJax and some mathematics in a web page, and MathJax will do the rest.​

This doesn't make clear if the package renders the code as notation in a browser, or if it enables the code to be served without being mangled in a web page editor.

Selecting the libjs-mathjax package automatically adds the fonts-mathjax package but leaves the fonts-extras and other packages that turned up in the mathjax search unselected.

Alternatively...

I haven't done much web search investigation into TexLive yet. I only discovered it while looking up MathJax.

Selecting the texlive-latex-base package automatically selects a long list of other packages, but leaves even more of them in the search unselected:

These packages are either mandated by the core LaTeX team, or very widely​
used and strongly recommended in practice.​
This package includes the following CTAN packages:​
ae -- Virtual fonts for T1 encoded CMR-fonts​
amscls -- AMS document classes for LaTeX​
amsmath -- AMS mathematical facilities for LaTeX​
atbegshi -- Execute stuff at \shipout time​
atveryend -- Hooks at the very end of a document​
auxhook -- Hooks for auxiliary files​
babel -- Multilingual support for Plain TeX or LaTeX​
babel-english -- Babel support for English​
babelbib -- Multilingual bibliographies​
bigintcalc -- Integer calculations on very large numbers​
bookmark -- A new bookmark (outline) organization for hyperref​
carlisle -- David Carlisle's small packages​
colortbl -- Add colour to LaTeX tables​
epstopdf-pkg -- Call epstopdf "on the fly"​
etexcmds -- Avoid name clashes with e-TeX commands​
fancyhdr -- Extensive control of page headers and footers in LaTeX2e​
fix2col -- Fix miscellaneous two column mode features​
geometry -- Flexible and complete interface to document dimensions​
gettitlestring -- Clean up title references​
graphics -- The LaTeX standard graphics bundle​
graphics-cfg -- Sample configuration files for LaTeX color and graphics​
grfext -- Manipulate the graphics package's list of extensions​
hycolor -- Implements colour for packages hyperref and bookmark​
hyperref -- Extensive support for hypertext in LaTeX​
intcalc -- Expandable arithmetic operations with integers​
kvdefinekeys -- Define keys for use in the kvsetkeys package​
kvoptions -- Key value format for package options​
kvsetkeys -- Key value parser with default handler support​
l3backend -- LaTeX3 backend drivers​
l3kernel -- LaTeX3 programming conventions​
latex -- A TeX macro package that defines LaTeX​
latex-bin -- LaTeX executables and man pages​
latex-fonts -- A collection of fonts used in LaTeX distributions​
latexconfig -- configuration files for LaTeX-related formats​
ltxcmds -- Some LaTeX kernel commands for general use​
ltxmisc -- Miscellaneous LaTeX packages, etc​
mfnfss -- Packages to typeset oldgerman and pandora fonts in LaTeX​
mptopdf -- mpost to PDF, native MetaPost graphics inclusion​
natbib -- Flexible bibliography support​
oberdiek -- A bundle of packages submitted by Heiko Oberdiek​
pslatex -- Use PostScript fonts by default​
psnfss -- Font support for common PostScript fonts​
pspicture -- PostScript picture support​
refcount -- Counter operations with label references​
rerunfilecheck -- Checksum based rerun checks on auxiliary files​
tools -- The LaTeX standard tools bundle​
uniquecounter -- Provides unlimited unique counter​
url -- Verbatim with URL-sensitive line breaks​
I'm leaning toward MathJax since it's only two libraries. I'd prefer not to break my installation with an ill-advised library and end up in DLL hell, not that it seems likely for such a crucial function as mathematical symbol rendering. It just seems safer to experiment with the simpler option first, provided it doesn't cause conflicts with TexLive if I ultimately need to go that route.

I'm not currently participating in any other forums that utilize LaTex or other text-based rendering code, so any increased functionality that TexLive may have is of zero value to me at this time. But if it's preferable over MathJax, ...?

Suggestions?
 
  • #18
You should just be able to type maths, with two # characters before and after for inline maths (e.g. ##F=ma##) or two $ characters for paragraph maths - e.g. $$F=ma$$You don't need to install anything unless you want to write LaTeX in documents outside the forum.

Click the reply button in this message to see the source code for the maths above.
 
  • #19
CherylJosie said:
The code is rendering as code rather than notation in my FOSS Ubuntu 20.04 with Firefox and Chromium.

Do you have Javascript disabled? The rendering of LaTeX into math on PF doesn't require any libraries to be installed on your end (all those Ubuntu packages you are looking at are for doing rendering of LaTeX documents you write on your computer using programs on your computer, not in your browser), but it does require Javascript.
 
  • #20
Ibix said:
I'd recommend a good book. If you've got maths enough to handle an engineering degree, my favourite is Taylor and Wheeler's Spacetime Physics (now free to download from Taylor's website), but other's seem to prefer Morin's Relativity for the Enthusiastic Beginner (the first chapter of which is free to download and can be found with a search).

Thanks, downloaded. I don't have the maths any more, but my ambition is to recoup that skill eventually. I still have good grasp of the concepts, at least those I did well on originally.

I'm also interested in string theory, so any similar text on quantum mechanics would be equally appreciated.
 
  • #21
PeterDonis said:
Do you have Javascript disabled?
DOH!

Umatrix had disabled javascript. Thanks, fixed. Much nicer. Time to re-read the thread.

I'm using umatrix with Firefox to speed up page loading and prevent tracking/hacking since Linux desktop doesn't have much available in antivirus/antispyware FOSS (that I am aware of anyway). I only enable the bare minimum required to load the page, and when it's still not working, I use Chromium without umatrix instead because sometimes just having it installed but disabled still breaks some web pages.
 
  • #22
So next time I make a similar post, how should I categorize it? I've got another burning question about black holes.
 
  • #23
Ibix said:
You should just be able to type maths, with two # characters before and after for inline maths (e.g. ##F=ma##) or two $ characters for paragraph maths - e.g. $$F=ma$$You don't need to install anything unless you want to write LaTeX in documents outside the forum.

Click the reply button in this message to see the source code for the maths above.
Cool, thanks. Much better!
 
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  • #24
CherylJosie said:
So next time I make a similar post, how should I categorize it? I've got another burning question about black holes.

Please start a new thread in this section.
 
  • #25
CherylJosie said:
So next time I make a similar post, how should I categorize it? I've got another burning question about black holes.
Vanadium 50 said:
Please start a new thread in this section.
...probably labelled I level, if that's what you're asking. Definitely not A level unless you want differential geometry thrown at you.
 
  • #26
CherylJosie said:
I'm also interested in string theory, so any similar text on quantum mechanics would be equally appreciated.
A good book seems to be "A First Course in String Theory" (Second Edition) by Barton Zwiebach. It covers classical relativistic strings in D-dimensional Minkowski space and later covers quantized relativistic strings. But the math is not easier than that of standard SR. :-) Additional information:
Sagittarius A-Star said:
I found with Google:

Source:
https://www.liverpool.ac.uk/~mohaupt/Strings07.pdf

I found with https://scholar.google.comSource:
https://iopscience.iop.org/article/10.1088/0264-9381/5/3/001/meta

I think with "classical" they mean "not quantized".

 
  • #27
etotheipi said:
There's a nicer way
Maybe, it's easier to do a Lorentz-transformation for the time-component of the four-frequency (##\vec{n}## is a unit vector in the travel direction of the light-wave):
##\mathbf N =
\begin{pmatrix}
N_t \\
N_x \\
N_y \\
N_z \end{pmatrix}
= \nu
\begin{pmatrix}
1 \\
\vec{n} \end{pmatrix}
##.

Lorentz-transformation:
##N'_t = \gamma (N_t - \beta N_x)##
$$ \nu' = \gamma ( \nu - \beta \nu \cos {\varphi}) = \nu \gamma (1 - \frac{v}{c} * \cos {\varphi})$$
 
  • #28
If you only want the Doppler shift of the frequency (of light) it's sufficient to stay in one arbitrary reference frame. All you need to know is that the Doppler effect (in the vacuum) only depends on the relative velocity of the light source and the observer and that ##(k^{\mu})=(k,\vec{k})## is a four-vector and ##c k=\omega##. Then you need the (normalized) four-velocity of the source ##u^{\mu}=\gamma(1,\vec{\beta})##. In the rest frame of the source ##u^{* \mu}=(1,0,0,0)## and thus ##u^{*\mu} k_{\mu}^{*}=u^{\mu} k_{\mu}=\omega^*/c##, i.e., the frequency of the em. wave as measured in the rest frame of the light source. Thus we have
$$\omega^*=c u^{\mu} k_{\mu} = c \gamma (k-\vec{k} \cdot \vec{\beta})=\omega \gamma(1-\beta \cos \alpha),$$
where ##\alpha## is the angle between the direction of wave propation and velocity of the source. So the frequency for the observer in the frame, where the light sources moves with ##\vec{v}##, you get
$$\omega=\omega^* \frac{\sqrt{1-\beta^2}}{1-\beta \cos \alpha}.$$
The numerator is due to time dilation, while the denominator is due to the Doppler shift if analyzed within Newtonian physics (which of course doesn't apply for light to begin with).

The case ##\theta=\pi/2##, i.e., if you observe the frequency shift when observing the light coming from a direction perpendicular to the velocity of the light-source. This transverse Doppler effect does not occur in the Newtonian treatment and is indeed due to the time dilation between the clocks used to measure the frequency of the light in the rest frame of the source vs. the frequency measured with a clock in the rest frame of the observer, where the source moves with ##\vec{v}##.
 
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  • #29
vanhees71 said:
$$\omega=\omega^* \frac{\sqrt{1-\beta^2}}{1-\beta \cos \alpha}.$$
Sagittarius A-Star said:
$$ \nu' = \gamma ( \nu - \beta \nu \cos {\varphi}) = \nu \gamma (1 - \frac{v}{c} * \cos {\varphi})$$
Your ##\alpha## is the angle in the (un-star-ed) receiver-frame, while my ##\varphi## is the angle in the (unprimed) source-frame. If I set ##\varphi = 90°##, then I get ##\nu' = \gamma \nu##, that means translated to you formula: ##\omega = \gamma \omega^*##. If I put this into your formula (and changed the sign of ##\beta## to the sender frame), then I get the https://en.wikisource.org/wiki/Translation:On_the_Electrodynamics_of_Moving_Bodies#%C2%A7_7._Theory_of_Dopplers_principle_and_aberration.:
##\gamma \omega^*=\omega^* \frac{\sqrt{1-\beta^2}}{1+\beta \cos \alpha}##
##1+\beta \cos \alpha = 1-\beta^2##
$$\cos \alpha = -\beta$$
 
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  • #30
No. The relations between the angles are of course different. You find the formulae for both Doppler effect and aberration, using the standard Lorentz boost for the plane em. wave field, in Sect. 3.2.1 here:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Note that the IRF ##\Sigma'## is the rest frame of the light source and thus ##\vec{\beta} c## the velocity of the light source in the IRF ##\Sigma##.
 
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  • #31
vanhees71 said:
No. The relations between the angles are of course different. You find the formulae for both Doppler effect and aberration, using the standard Lorentz boost for the plane em. wave field, in Sect. 3.2.1 here:

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
The aberration formula I wrote is identical to that in Einstein 1905, §7, with ##\beta## in the source frame and the case that the angle in the source frame ist 90 degree. He also wrote ##\cos {\alpha} = -v/c##.

It is the same as your formula (3.2.62) for ##\alpha' = 90°##, except the sign of ##\beta##.
 
  • #32
The different sign comes simply from the different meaning of ##\vec{v}##. In description by Einstein, the observer is moving with ##\vec{v}_E## in the rest-frame of the light source, while in my description ##\vec{v}## is the velocity of the light source in the rest frame of the observer, i.e., ##\vec{v}=-\vec{v}_E##. Thus mine and his equations are equivalent.
 
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  • #33
One simple way to understand both of these cases qualitatively is to remember the Einstein equivalence principle that acceleration and gravity are locally equivalent, and that there is accelerational time dilation. First derived by Einstein in 1907 (using only SR), it says e.g. that a clock near the front of an accelerating rocket must run slightly faster than a clock near its tail.

CherylJosie said:
Summary:: Does Red/Blue Shift Indicate Relative Passage of Time?

Experiment 1: Astronaut travels away from Earth at near the speed of light, then travels toward the Earth at near the speed of light.

What you are ignoring are the accelerations at the launch time and the turnaround time.

At launch time, while accelerating (let's say at 1 G), the astronaut imputes a time dilation field to the whole universe, with clocks running faster in the "up" (ahead of her) direction and slower in the "down" (behind her), and the size of the effect dependent on distance. Because the Earth is close, the effect is pretty small, and I'll neglect it here.

At turnaround time, the astronaut must accelerate toward Earth to slow down and stop, and an equal amount to speed up in the return direction. During this entire acceleration, Earth is "above" the astronaut, by roughly the full distance of travel, and so she perceives clocks on Earth to be running much faster than her own clocks. This is a big chunk of the final clock difference when she returns. Changing the G-force doesn't affect this much; if you doubled the acceleration, you'd have twice the effect for half the time, and get about the same total time shift.

CherylJosie said:
Experiment 2: Astronaut circles the Earth at near the speed of light.

Again in this case, the astronaut is accelerating, and the Earth is in the "up" direction, so its clocks are running faster in her (non-inertial) frame.
 
  • #34
H_A_Landman said:
there is accelerational time dilation

This is a somewhat misleading way to put it. The time dilation is caused by a difference in height in a "gravitational field", not by acceleration. Acceleration can "create" a "gravitational field", but it is the difference in height in the field, not the acceleration itself, that causes the time dilation. (Note that in the standard "twin paradox", the stay-at-home twin, whose clock is supposed to be running very, very fast during the turnaround, is in free fall.)

Also, as my use of the quotes indicates, the "gravitational field" that is "created" by acceleration turns out to be a frame-dependent thing and not a good basis on which to construct a general theory. While it works qualitatively in these particular cases, it doesn't generalize well.
 
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  • #35
PeterDonis said:
The time dilation is caused by a difference in height in a "gravitational field", not by acceleration. Acceleration can "create" a "gravitational field", but it is the difference in height in the field, not the acceleration itself, that causes the time dilation.

The difference in time rates is (to first order) proportional to both the height ##h## and the acceleration ##g##, and thus to their product ##gh##. Yes? The weak field approximation is $$T_d = 1 + gh/c^2$$ in which they both matter. So I don't understand your quibble here; it seems incorrect, or perhaps poorly stated. Are you saying that if you hold the acceleration constant then it's only a function of height? That's true but incomplete. If you hold the height constant, then it's only a function of the acceleration.

The goal here was to provide a simple qualitative Level I answer to show the questioner "where they went wrong". Ignoring the effects of the terminal accelerations is one way to go very wrong. (There are other ways.) Whether I succeeded or not should primarily be judged by how useful (or useless) my answer was to her.

Einstein's 1907 paper derives accelerational time dilation first (in section 18), and then using the EEP concludes that there must also be gravitational time dilation. The initial derivation doesn't depend on any theory of gravity at all; it only uses SR. Today we would tend to start with GR and work downwards, but that's not required. It doesn't depend on GR, only SR. The existence and (first order) magnitude of gravitational time dilation doesn't depend on GR either; it's already forced by SR + EEP.
 

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