Relativity and Cosmology Blog

Relativity and Cosmology

This is a Blog on relativity and cosmology for engineers and the like. My website "Relativity-4-Engineers" has more in-depth stuff.

You can comment directly on this Blog if the comment/question is relevant. to the topic Comments/questions of a general nature should preferably be posted to the FAQ section of this Blog (http://cr4.globalspec.com/blogentry/316/Relativity-Cosmology-FAQ).

A complete index to the Relativity and Cosmology Blog can be viewed here: http://cr4.globalspec.com/blog/browse/22/Relativity-and-Cosmology"

Regards, Jorrie

Previous in Blog: Twin Paradox - a New Graphical Solution   Next in Blog: Do moving clocks really 'tick slower'? - Resolution
Close
Close
Close
38 comments
Rate Comments: Nested

Do moving clocks really 'tick slower'?

Posted October 05, 2008 11:00 PM by Jorrie

We know the relativistic view: "moving clocks tick slower than stationary clocks". We also know that Einstein's Special Relativity (SR) demands: "there is no absolute movement, only relative movement". This tells us that when two clocks are moving relative to each other, we can view any one of them as the "stationary" reference and the other one as the "moving clock".

An apparent paradox arises out of this opening paragraph: if clock A is considered as the reference and clock B as moving, then B ticks slower than A. However, if B is considered as the reference and clock A as moving, then A ticks slower than B. This is a logical paradox, but it obviously arises out of the wording. It should have been: "if clock A is considered as the reference and clock B as moving, then B ticks slower than A, as observed by A. If B is considered as the reference and clock A as moving, then A ticks slower than B, as observed by B."

What does this tell us about the "real tick rate" of moving clocks? Actually nothing, except perhaps that there cannot be an absolute difference between the tick rate of clocks A and B, provided that they are moving inertially in free space. The differences are relative (or apparent) and caused by a measurement issue. What is this measurement issue?

The isotropy of the propagation of light

It originates in the second postulate of SR – Einstein's assumption that the speed of light must be constant in every direction, irrespective of the inertial movement of source or observer. From this follows the Einstein definition of simultaneity, which boils down to a convention for synchronizing static clocks inside inertial frames.

Einstein made the reasonable assumption that one can send a time-stamped light signal to any observer that is stationary relative to us. The receiver, knowing the coordinate distance from us, can then add the light travel time to that distance and set her clock accordingly. This is known as the "Einstein method" for synchronizing clocks in inertial frames. This, of course, ensures that light will propagate at c in all directions, as measured by clocks thus synchronized.

The rate of clocks

The Einstein synchronization method also ensures that we cannot measure the "real" rate of inertially moving clocks. We need two clocks (say A and C) that are separated in space to compare the rate of the moving clock (say B) to. B will however not agree that clocks A and C are synchronized and is entitled to say: "you guys can tell me nothing about the rate of my clock versus yours, because your two clocks are not correctly synchronized, IMO".

Amazingly, it seems to be impossible to measure the relative rate of two inertially moving clocks during a single pass – you need at least two passes. To achieve that, you must take one (or both) of the clocks out of being inertial – you need a gravitational field or acceleration, both of which destroy SR's applicability.

A different view

As a point for discussion I want to put forward this view: clocks in relative inertial motion all tick at some universal rate, but we have no means of detecting this. What we can detect is the relative, coordinate dependent rates of clocks, which depend on a convention for synchronization of clocks. It is a very good convention, it seems, because SR predicts exactly what we actually measure.

The situation is best described my means of a Loedel diagram[1], as pictured here. Objects static in each frame (A or B) have wordlines progressing at the same rate up the respective time axes. The dotted red and blue lines of simultaneity do however indicate how each view the others clock as "running slow". When the blue observer's clock reads t=4.25, she observes the red clock as reading only t'=4.0 and vice-versa.

Jorrie

[1] The Loedel diagram is due to E. Loedel (Geometric Representation of the Lorentz Transformation. Am. J. Phys. 25: 327, May, 1957). It is just a special case of a Minkowski diagram, with two inertial frames in symmetrical motion in opposite directions relative to some imaginary "neutral" inertial frame. When precisely two inertial frames are required, the Loedel diagram is perhaps the easiest (rigorous) representation of SR. You can read more on Loedel diagrams on my website.

Edit: the Resolution

The view presented above says: "clocks in relative inertial motion all tick at some universal rate, but we have no means of detecting this."

This sounds very compelling (especially if we have no way of checking it!), but there are some clues that it may not be the case. We have no means of directly comparing the tick rates of two clocks in relative inertial motion, as they will have to be brought together at the beginning of the test and then again some time later. This means that at least one of them must be accelerated at least once; this renders that clock to be no longer purely inertial.

However, we can use three purely inertial clocks to perform such a test. In the usual "twin paradox" thought experiment, we substitute (at the turn-around point) the "away clock" with another clock, flying towards home. Now there is no acceleration required and two clocks at a time can be directly compared (and synchronized) as they fly past each other.

The left-hand dates are as recorded by the home clock, the blue line is for the away clock and the red line is for the substitute away-clock. On New Year's Day 2007, the away twin zips past the home twin and sets her clock to read the same date. When her clock reads New Year 2011, another traveler, carrying an identical clock, passes her in the opposite direction, but at the same speed (v=0.6c) relative to home. At the 2011 passing, her "substitute clock" is set to that date as well.

When the substitute away twin passes home (home time 2017), Einstein's special theory of relativity tells us that her clock will read 2015. This means that the combined elapsed time of the away twin and her substitute was less than the elapsed time of the home twin, despite the fact that all three of them were purely inertial during the whole experiment.

Does this tell us that there are some inertial frames that are privileged - inertial frames in which clocks run slower (in an absolute sense) than clocks in some other inertial frames? Not quite. We have no absolute way of knowing what the time reading on the home clock was at the moment that the two away clocks pass each other. The fact that we draw it symmetrically does not guarantee that it must be such a symmetrical case.

What we can say is that on the two away clocks the average elapsed time (of the test) is less than the elapsed time on the home clock, because we could measure it absolutely. So what causes this difference? The clocks followed different spacetime paths - somewhere in their respective pasts they had to be accelerated differently in order to move as they did. This change in space versus time orientation apparently causes clocks to run at different rates.

Clear as mud?

Edit (Nov 16, 2009): In order to clear the 'mud' a little, the final statement above:

"... somewhere in their respective pasts they had to be accelerated differently in order to move as they did. This change in space versus time orientation apparently causes clocks to run at different rates"

is not quite mainstream thinking. There is really no way to observe that. It is easier to view the clocks to run at the same rate, but that the one(s) following the longest spatial path record(s) the lesser time differential, in other words, they follow the shorter temporal path. This comes directly from the Lorentz transformations, which have their roots in the invariant (time-like) spacetime interval:

ds2 = dt2 - dx2 - dy2 - dz2

The more distance covered between two events, the less time covered...

-J

Reply

Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#1

Re: Do moving clocks really 'tick slower'?

10/08/2008 4:10 AM

Hi Jorrie. Just a couple of questions that came up in another CR4 discussion 2 weeks ago (a little bit out of this topic though):

1) I claimed that if one travels at a speed near c he will observe a slower "time flow" for the outside universe. Another guy pointed out that as you approach a target (i.e. a planet) you must observe a faster "time flow" on it because of the high blue shift (also a high red shift takes place in the opposite direction). And this must be, also, correct. So, what's going on in this case???

2) Another guy said that if an object travels -aproaching the speed c- it will eventually turn itself into a black hole (relative to us). Initially I had the opinion that it can't be so (thinking that we speak about inertial mass and not gravitational mass), but after a second thought I started to change my mind (and getting confused). As we give more and more energy to this object we increase its kinetic energy (the only difference at speeds near c is that we increase its mass rather than its velocity). But this kinetic energy (which is really huge) must bend the space-time (like the energy of any form will do). And, afterall, SR claims that there is no distinction between the inertial mass and gravitational mass (these are the same thing). So, if we observe that the mass of the object is increased the same must happen to its gravitational field. Does this mean that i.e. a spaceship -travelling at a speed very near c- behaves like a black hole (with respect to us)??? (But it's not so from the point of view of another observer travelling at the same speed e.g. who is in the same inertial frame???) And the driver of the spaceship will conceive us (and everything else) as black holes too (from his point of view)??? Black holes are sth objective (not subjective), aren't they??? So what's really going on???

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#2
In reply to #1

Re: Do moving clocks really 'tick slower'?

10/08/2008 8:53 AM

Hi George. Sorry, I've missed that discussion...

Your 1): Doppler shift in the radial direction is not a relativistic effect per se. It is only the so-called "transverse Doppler effect" which is relativistic, due to the observed time dilation. Doppler shift does indeed cause an apparent speed-up or slow down of clocks, even in Newton dynamics, but for relativistic discussions, we compensate for the Newtonian part and leave it aside...

Your 2): "Another guy said that if an object travels -aproaching the speed c- it will eventually turn itself into a black hole (relative to us). Initially I had the opinion that it can't be so (thinking that we speak about inertial mass and not gravitational mass), but after a second thought I started to change my mind (and getting confused)"

No, objects cannot turn into black holes simply because they have very high kinetic energies, because kinetic energy is a relative thing. In the frame of the moving object, it has zero kinetic energy! It is only when an object is continuously accelerated that an apparent "black hole" can be observed by comoving observers - or rather an event horizon exists in the wake of the observer, from where light cannot reach the observer. This event horizon has no other influence on its surroundings, hence it is not a black hole!

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#3
In reply to #2

Re: Do moving clocks really 'tick slower'?

10/08/2008 11:18 AM

Jorrie, thanks for your reply. I still need some more clarifications:

About (1): Does this mean that (considering the time dilation due to SR and the blue shift due to the Doppler effect) the final result is that we observe a relativistic time dilation??? If it is so, then we observe a slower "time flow" of the outside universe, even on the planet-target (which is in front of us). [I suspect that this time dilation (on the planet) is somewhat different, e.g. it is less slow than the time dilation on the other directions due to the blue shift... and we observe exactly the opposite right behind us due to the red shift...] Am I right???

About (2): I am so happy that a spaceship will not turn into a BH. Hence no increment of the spaceship's gravitational field will take place (due to the increment of its mass). You said:"... It is only when an object is continuously accelerated that an apparent "black hole" can be observed by comoving observers...) What do you mean by saying "comoving observers"??? You said: "... - or rather an event horizon exists in the wake of the observer, from where light cannot reach the observer... " You mean behind the accelerated object??? And why it is so???

Thanks...

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#4
In reply to #3

Re: Do moving clocks really 'tick slower'?

10/09/2008 12:47 AM

George: "Does this mean that (considering the time dilation due to SR and the blue shift due to the Doppler effect) the final result is that we observe a relativistic time dilation???"

No, relativistic time dilation is independent from Doppler shift, as can be tested with pure transverse relative velocities. The equation for radial one-way relativistic Doppler shift ratio is shown on the right,[1] where Tx denotes transmitter, Rx denotes receiver and dot-x is an opening relative velocity between the two. The leftmost expression shows what is measured: a combination of the Newtonian Doppler shift factor (the denominator) and the relativistic time dilation factor (the numerator).

Jorrie:"... It is only when an object is continuously accelerated that an apparent "black hole" can be observed by comoving observers...") George: "What do you mean by saying "comoving observers"???"

It means observers riding onboard the accelerating spaceship, not to be confused with "momentarily comoving inertial observers" that are also sometimes used when analysing accelerated frames of reference.

Behind such a ship there is a cone with a vertex at a certain distance,[2] from where light will take an infinite time to reach the ship. If the acceleration is constant at a m/s2, the distance to the vertex is: c2/a m, so it's quite far behind for normal accelerations. Why this event horizon? The speed of light is not constant for an accelerating observer; it appears to be smaller than c from the rear and greater than c from the front.

Finally, you must be careful when speaking of "mass increase" for moving objects. What increase is their relativistic momentum, while their proper mass just remains constant at the rest mass. One can possible use the momentum of two particles smashing into each other to create a micro black hole (a-la LHC ??), but never a fast-moving particle on its own.

Jorrie

[1] The full explanation of relativistic Doppler shift can be found in Relativity 4 Engineers, section 2.3, page 44. (Psstt, this is the link to the special offer to my newsletter subscribers, now also available to CR4 subscribers.)

[2] Prof. John Mallinckrodt, Cal Poly Pomona, www.csupomona.edu/~ajm/professional/talks/relacc.ppt

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#5
In reply to #4

Re: Do moving clocks really 'tick slower'?

10/09/2008 6:50 AM

Jorrie I should initially ask: "Does this mean that (considering the time dilation due to SR and the blue shift due to the Doppler effect) the final result is that we observe a relativistic time dilation???" I mean that -as we approach the target (planet) at a speed near c- we observe two contrary effects: [1st → the time dilation due to SR] & [2nd → the time contraction due to Doppler effect (blue shift)]. So what is the combination of these two contrary effects???

Does the above equation about the λRx takes into account these two contrary effects that I mentioned above??? What does this equation intuitively represents??? Does this λRx shows us what we finally observe on this planet as we approach it??? (In other words :Do we observe a time dilation on this planet as a final result??? Is this time dilation the same as the time dilation that we observe in any other direction around us... and right behind us...???) I need some more on this.

About the "increment of mass" issue: You said: "What increase is their relativistic momentum, while their proper mass just remains constant at the rest mass." I agree about the relativistic momentum but, anyway, an external observer considers this increment of the momentum (from his point of view) as an increment of the spaceship's mass. This has the meaning that if he tries to accelerate a little bit more the spaceship (by giving external energy to the spaceship) he will find out that it's really hard to do this, as if its mass has increased (it's not an actual increment of the mass... it's just the "estimation" of the observer...) So it's like we are talking about the inertial mass (as it is considered by the observer). The problem arises from the acceptance that the inertial mass and the gravitational mass cannot be distinguished. They are the same thing. But gravitational mass implies a gravitational field. But we don't have any increment of the gravitational field of the spaceship. How is this possible??? Does this mean that, actually, there is a distinction between the inertial mass and the gravitational mass??? Is this result in contradiction with SR???

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#6
In reply to #5

Re: Do moving clocks really 'tick slower'?

10/09/2008 8:18 AM

George: "Does the above equation about the λRx takes into account these two contrary effects that I mentioned above??? What does this equation intuitively represents??? Does this λRx shows us what we finally observe on this planet as we approach it??? "

For a closing radial speed, dot-x = v/c is negative. As an example, let's use v/c = 0.99 and v/c = -0.99. The Newtonian Doppler shifts for a static receiver and moving transmitter work out to 100 and 0.5 respectively. The relativistic component is √(1-v/c2) = 0.142, so the relativistic Doppler shifts work out to 14.1 and 0.071 respectively (the Newton shifts multiplied by the time dilation factor 0.142). You can say that in the case of a positive radial velocity, the two effects work in the same direction and for a negative radial velocity they work in opposite directions.

Intuitively, it simply means that the Newtonian Doppler shift is modified by the relativistic time dilation factor and yes, the total effect is what we observe as we approach or recede from a planet. However, the Newtonian Doppler shift has nothing to with time dilation - it's just a physical stretching or compression of waves!

George: "So it's like we are talking about the inertial mass (as it is considered by the observer). The problem arises from the acceptance that the inertial mass and the gravitational mass cannot be distinguished. They are the same thing. But gravitational mass implies a gravitational field. But we don't have any increment of the gravitational field of the spaceship. How is this possible??? Does this mean that, actually, there is a distinction between the inertial mass and the gravitational mass??? Is this result in contradiction with SR???"

"Relativistic mass" or "moving mass" is not inertial mass (which is the same as rest mass). The "moving mass" is coordinates system (i.e. observer) dependent and hence not an absolute concept. No conflict with SR here...

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#7
In reply to #6

Re: Do moving clocks really 'tick slower'?

10/09/2008 11:28 AM

Thanks, Jorrie.

About (2): No more questions. I'm totally happy... ...

About (1): Nice example. (Sometimes an application clarifies things.) So let's sum up:

a) The spaceship approaches the planet: v/c=-0,99. This gives us: λRx=0,503 λTx and if we take into account the relativity effect we get: λRx=0,071 λTx

b) The spaceship recedes the planet: v/c=+0,99. This gives us: λRx=100 λTx and if we take into account the relativity effect we get: λRx=14,1 λTx

Everything is nice so far. But still sth bothers me: In both cases (a & b) it seems that the relativity effect (which represents the time dilation due to SR) makes the wavelength looks even shorter (0,071 λTx instead of 0,503 λTx & 14,1 λTx instead of 100 λTx). A shorter wavelength (e.g. a greater blue shift) means that the driver of the spaceship must observe several events (which take place on the planet) to happen in shorter time durations (e.g. in a "fast forward mode"). [In other words this means that in this case (e.g. including the relativity effect) he observes a "faster time flow" than he would observe if there was only the Doppler effect (e.g. w/o the relativity effect)] This doesn't make sense because someone should expect that the relativity effect will make things look "slower" (not "faster"). [Or, in other words, it is supposed that because of the time dilation (due to reletivity) the driver should observe a somewhat "slower time flow".] So why we get this weird result???

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#8
In reply to #7

Re: Do moving clocks really 'tick slower'?

10/09/2008 10:47 PM

George: "This doesn't make sense because someone should expect that the relativity effect will make things look "slower" (not "faster"). [Or, in other words, it is supposed that because of the time dilation (due to reletivity) the driver should observe a somewhat "slower time flow".] So why we get this weird result???"

The "weird" result comes from explaining relativistic Doppler shift through the Newton Doppler shift, where there is an absolute frame of reference (aether) in which light propagates at c. The equation I gave is for the planet static in the aether and your spaceship (the receiver) moving, so the receiver's time appears dilated (slower) relative to the planet.

We could also have viewed the spaceship as static in the aether and the planet moving. Then the planet's time would have been dilated (slower) as judged by the (static) ship. Note that the Newtonian factors differ for the two two cases, but bring in the relativistic time dilation and they give the same final result, i.e., the influence of the aether disappears! Then enters the apparent paradox that each observer (planet and spaceship), will observe the others time to be dilated ("slowed down").

It is obviously never required to go through the Newtonian step, but it is good to build the right understanding of the underlying dynamics.

Jorrie

PS: In chapter 2 of Relativity 4 Engineers a lot of interesting numerical examples are calculated as "applications".

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#9
In reply to #8

Re: Do moving clocks really 'tick slower'?

10/10/2008 5:44 AM

Nice Jorrie. I liked that.

But still there is a problem. You insist on saying: "... Then enters the apparent paradox that each observer (planet and spaceship), will observe the others time to be dilated ("slowed down")." But, as I pointed out in my previous post, the results show exactly the opposite. For example, in the case that the spaceship approaches the planet: Considering only the Newtonian factor you get λRx=0,503 λTx . When you take into account the Relativity factor you get λRx=0,071 λTx . As you see, we have λRxRx hence the light coming from the planet (as it is observed by the driver) have an even shorter wavelength (even greater blue shift) when the relativity factor is taken into account. This, inevitably, leads to the conclusion that the driver must observe an even "faster time flow" on the planet (events on the planet happens in an even "faster mode" from his point of view). This comes in contradiction with your claim that "... each observer will observe the others time to be dilated ("slowed down")." [btw I, also, expected the relativity factor (when it comes into the game) to make things "going slower" not "faster"]

Do you get my point??? The results show exactly the opposite of what we were expected. That's the whole confusion. Any comments on this???

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#10
In reply to #9

Re: Do moving clocks really 'tick slower'?

10/10/2008 9:00 AM

Hi George

I think the confusion comes in when you insist on including the Newtonian Doppler effect into a concept of "the apparent rate of clocks". In Galileo and Newton's days, they knew about the Doppler effect and none of them connected that with the rate of clocks. Obviously, because clock rates could not depend on whether the relative radial velocity was positive or negative!

The observed relativistic time dilation has nothing to do with the standard Doppler effect. What is true is that the relativistic time dilation modifies the Newtonian Doppler effect by a time dilation term. It is an "over and above" effect. It also does not matter if the relative radial velocity is positive or negative.

George: "When you take into account the Relativity factor you get λRx=0,071 λTx . As you see, we have λRxRx hence the light coming from the planet (as it is observed by the driver) have an even shorter wavelength (even greater blue shift) when the relativity factor is taken into account."

Correct, but then do not read anything about clock rates or time flow into it - as explained above! As a matter of fact, I argued in the opening post that there is no velocity induced difference between the clock rates of the planet[1] and the spaceship, except that they measure each others clock rates to have slowed down (reciprocally, after the Newtonian Doppler shift has been taken out). It's a measurement problem, caused by the different definitions of simultaneity.

BTW, have you checked out that chapter that I referenced?

Jorrie

[1] Except as may be caused by the gravity of the planet and its parent star, which is a much tougher story...

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#11
In reply to #10

Re: Do moving clocks really 'tick slower'?

10/10/2008 11:29 AM

Thanks Jorrie. Now it's perfectly clear. "Time flow" has nothing to do with the Doppler effect. Any change of the "time flow" has to do only with the relativity. (Of course, as you perfectly show me, the relativity influences the λRx but this is another issue.) Hence, as we travel (at a speed near c) we observe the same time dilation (caused by the relativity) in any direction of the outside universe (even on a planet right in front of us or on another planet right behind us). Am I right??? (Please confirm.)

(Sorry but the other guy who proposed this issue deceived me, making me believe that the (i.e.) blue shift (that we observe as we approach the planet) will make us, also, observe "events going faster" (e.g. faster "time flow") on this planet. But these two things are not related to each other.)

[Thanks again for your patience.]

__________________
George
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#12
In reply to #11

Re: Do moving clocks really 'tick slower'?

10/13/2008 2:57 AM

Jorrie, please, confirm the above results on my post #11.

Thanks again.

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#13
In reply to #11

Re: Do moving clocks really 'tick slower'?

10/13/2008 5:12 AM

George: "Hence, as we travel (at a speed near c) we observe the same time dilation (caused by the relativity) in any direction of the outside universe (even on a planet right in front of us or on another planet right behind us). Am I right??? (Please confirm.)"

Correct. However, your correspondent is right that there is an apparent (non-relativistic) "speed-up" of observed events in the case of a closing velocity and vice-versa for an opening velocity. If however, you fly for some time away from a planet and then for an equal time back towards the planet, the non-relativistic Doppler effects will not cause any difference in elapsed times on the ship and on the planet. The relativistic effect will cause an absolute elapsed time difference (hence the twin "paradox").

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#14
In reply to #13

Re: Do moving clocks really 'tick slower'?

10/13/2008 8:11 AM

But, Jorrie, you just said (in your previous post) that the approach of the planet (and the related blue shift) is not related to any apparent change of "time flow" on the planet, e.g. a "speed up" of the observed events on it. So, o.k., after this I was convinced that you are right (and I was wrong) on this... And now you take it back again... And we are going back to the begining... So what's really happening???... ...

[Afterall, this was exactly what I said from the begining of our conversation, e.g. two opposite effects: 1) a non-relativistic "speed up" of the events (and a blue shift) that we observe on the planet (because we are moving towards the planet) and 2) the relativistic time dilation (due to SR). And my initial question was: what is the final combination of these two opposite effects concerning our "final perception of the time flow" on the planet???]

I feel that I'm going "towards and backwards" all the time (in a vicious circle) without being able to have a clear picture... ...

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#15
In reply to #14

Re: Do moving clocks really 'tick slower'?

10/13/2008 10:03 AM

George: "But, Jorrie, you just said (in your previous post) that the approach of the planet (and the related blue shift) is not related to any apparent change of "time flow" on the planet, e.g. a "speed up" of the observed events on it."

I do not consider Doppler shift as an apparent change of "time flow" (and I do not think any scientist does), that's why I made the original statement. In the last post I simply implied that your correspondent could perhaps argue that it is an apparent change in time flow. The moment one adds "apparent" into the sentence, it is guaranteed that anything goes...

The crux of the matter is that when one of the observers is accelerated (to fly away and return) and the other one remains inertial, the relativistic effect causes real (absolute) elapsed time difference between them. The Newtonian Doppler shifts do nothing of that sort - they remain purely apparent time rate effects.

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#16
In reply to #15

Re: Do moving clocks really 'tick slower'?

10/14/2008 4:17 AM

O.k. Jorrie, so let's "disconnect" the Newtonian Doppler shift from any change of the "time flow".

Now I'll try to describe the whole issue of the time "speed up" on the planet as we approach it: let's assume that the driver (A) have a friend (B) who is located on the planet and (B) sends to (A) light pulses every 1 sec. Then we have:

Case a) the spaceship is steady (in the inertial frame of the planet): (A) receives the light pulses every Δt1=1sec and we have υ=c=S1/Δt1 where S1 is the distance of two sequential pulses (from (A) point of view) which is 3x108m.

Case b) the spaceship is moving in a very high speed towards the planet: (A) receives the light pulses every Δt2<1sec. That is because we have υ=c=S2/Δt2 where S2 is the distance of two sequential pulses (from (A) point of view) which is S2<S1 [as (A) meets sequential pulses while he has moved a little bit in the opposite direction], hence (A) must receive, correspondingly, the sequential pulses in a time duration Δt2<Δt1. So (A) must observe that the event of the pulses' transmission by (B) happens in a somewhat "faster mode" (or in other words we could say that (A) considers that "(B)'s 1sec" is somewhat "shorter" than "(A)'s 1sec"). In the same way (A) should observe any other event on the planet to happen in a somewhat "faster mode" (because -like the light pulses- the view of any event happening on the planet, also, travels at speed c and observed by (A) )

[Of course, exactly the opposite happens if (A) travels away from the planet e.g. Δt2>1sec or a "slow down" of the "time flow" on the planet as it is observed by (A).]

So I think that this apparent "speed up" of the "time flow" that you mentioned in your previous post is what I described above. Am I right???

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#17
In reply to #16

Re: Do moving clocks really 'tick slower'?

10/14/2008 5:32 AM

George: "So I think that this apparent "speed up" of the "time flow" that you mentioned in your previous post is what I described above. Am I right???"

Yep, that's right. But take note of what I wrote in my previous post about the "real difference" between Newtonian Doppler shifts and relativistic time dilation in the case of a two-way flight, where clocks can be compared directly, a-la the "twin paradox".

BTW, what do you think about the proposal in the opening post, i.e., that the time dilation in purely inertial frames are also just apparent and not "real" - that is until you bring in acceleration or gravity so that two clocks can be compared directly more than once?

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#18
In reply to #17

Re: Do moving clocks really 'tick slower'?

10/14/2008 11:50 AM

Jorrie: "BTW, what do you think about the proposal in the opening post, i.e., that the time dilation in purely inertial frames are also just apparent and not "real" - that is until you bring in acceleration or gravity so that two clocks can be compared directly more than once?"... As far as I know, SR claims that there is no such thing as "absolute time". Being in an inertial frame, your time is very real and the time that you observe in all other things (other inertial frames) around you is very real too. On the other hand I'm thinking about what you said in your previous posts e.g. that the mass increment that we observe on a fast moving object (relative to us) is, also, apparent and not real. Maybe a similar phenomenon takes place concerning the time dilation that we observe in other fast moving inertial frames. And, maybe, afterall there is a "universal, absolute time". But, as you said, in order to make an absolute estimation of time you need two passages and this destroys the relativity (not inertial frames any more). So there is no way for such an estimation and the absolute time remains a rather philosophical meaning. (In another conversation you criticized me that I was philosophizing... remember???... ...) Afterall everything in the universe is moving in relation with everything else (and the universe itself is expanded). Everyone observes everything else around him just a little bit "distorted" (even in a tiny, negligible way) due to SR. I think that the only way of having an "absolute image" of the world is to bring "ether" back to the game and being in its inertial frame. And then we should ask: "is there any real need to bring ether back (beside the fact that it may be convenient)"???... ...

Anyway I think that your "approach" (or proposal) is a good food for thought for the physicists (not me)... ...

(BTW I like that you always use the "she" whenever you talk about the scientists or observers in your imaginary experiments. Yeap, an imaginary lab full of women... This must be the lab of my dreams... ...)

You still avoid (???) to give me an answer in my initial question (a question presented, persistently, in some of my previous posts): What is the combination of these contrary effects that coexist e.g. the non reletivistic, apparent "speed up" of time (on the planet as we approach it) and the relativistic time dilation (due to SR)??? What is the the final result??? Which of these two effects is "stronger" (as we travel very near c)??? I mean, what kind of "time flow" on the planet we will (finally) observe??? (I will be grateful if I have a final answer from you... thanks...)

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#19
In reply to #18

Re: Do moving clocks really 'tick slower'?

10/15/2008 12:40 AM

George:"Which of these two effects is "stronger" (as we travel very near c)??? I mean, what kind of "time flow" on the planet we will (finally) observe??? "

The fully relativistic Doppler shift is given by the far right side of the equation I gave above. As an example, for v/c = -0.6, the full relativistic effect (λrx/λtx) is 0.5, while the time dilation effect alone is 0.8. At v/c = -0.99c, the corresponding figures are 0.071 and 0.141. So, at high relative velocities, bringing in the Doppler effect seems to either half or double the time dilation effect. Which effect wins? Depends on how you define the "competition", i.e. whether you divide or subtract, I suppose...

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#20
In reply to #19

Re: Do moving clocks really 'tick slower'?

10/15/2008 12:04 PM

Sorry, Jorrie, but I have so much work... I work in relativistic speeds... And I observe all those weird phenomena: The working day seems to be sooo long (because of the time dilation I suppose) and things cannot "go faster" (due to the increment of the mass I suppose)... ...

I'll write again tomorrow (it's time to go home)... I think the whole issue is not completely closed yet...

__________________
George
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#21
In reply to #19

Re: Do moving clocks really 'tick slower'?

10/16/2008 11:43 AM

Hi again, Jorrie. In your previous posts you insisted on saying to "disconnect" (in my mind) the concepts of "Doppler effect" and "apparent change of time flow". You said that "doppler effect" has nothing to do with any apparent "speed up" or "slow down" of the events observed on the planet (by the driver). And now, in order to give me an answer about this "apparent change of time flow", you go back to these Doppler effect equations. I don't get it. You say that (at v/c=-0,99) we have λRxTx=0,071 (and it's right). But what this tells us about the "time flow"??? (from what you have already said, I thought that this tells us nothing.) This "0,5" is just the relation between λRx and λTx. I think (or I thought) that these equations show us only the influence of relativity on the "λRx" observed by the driver (meaning that the "Doppler effect influenced by the relativity" gives us λRxTx=0,071 instead of 0,5 that we get concerning only the "pure Doppler effect".) BTW, does this "0,141" (i.e. the denominator of the equation or, in other words, the "relativistic factor") represent the "pure time dilation due to relativity"???

So, what this "λRxTx=0,071" tells us about the "apparent change of time flow" on the planet (as it is perceived by the driver)???

__________________
George
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#22
In reply to #21

Re: Do moving clocks really 'tick slower'?

10/17/2008 10:46 AM

Jorrie, are you still there??? I would be so grateful if you could give me some more help. (See my previous post.)

Thanks.

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#23
In reply to #22

Re: Do moving clocks really 'tick slower'?

10/18/2008 1:11 AM

Hi George. We are on a two week tour of the "Wild Coast" of South Africa, where there is only very slow and intermittent GPRS connections available. Be patient, I'll get back to you when I find a stable internet connection...

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Anonymous Poster
#24
In reply to #23

Re: Do moving clocks really 'tick slower'?

10/19/2008 2:50 PM

I hope Sir the world will find a stable internet connection.but it will be irrealistic in nature. your phil

Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#25
In reply to #21

Re: Do moving clocks really 'tick slower'?

10/20/2008 11:43 PM

Hi George. In purely inertial frames, it is so that the Newtonian Doppler effect and the relativistic time dilation are equally "apparent" in terms of clock rates, although they work in opposing directions for negative radial velocities. We know that the Doppler shift does not mean an absolute change in clock rates. Likewise, relative velocities (alone) cannot cause an absolute change in clock rates.

The moment either acceleration of gravity (or both) enters for one of the frames, the situation changes, because one frame is still permanently inertial, while the other one is not. This means that the spacetime histories of the two frames are different, causing different flows of time. The interesting part is that only the time dilation part plays a role here, not the Doppler shift part (see the sketch below, where the red arrows indicate the Newtonian Doppler shift and the black arrows the relativistic Doppler shift scenario). This is the reason why I say that the rates of clocks are not influenced by Newtonian Doppler shift.

Recall that I also said that when we talk "apparent changes", anything goes. After all, perspective makes objects look smaller at a distance – a mere apparent effect on size, as we know. Doppler shift is a similar apparent effect on time, without any bearing on "real time".

The most interesting (and controversial) view is that in the case of purely inertial frames of reference, the relativistic time dilation is also just an apparent effect. It is only when two frames have different spacetime "travel histories" that real differences in elapsed time can be observed. This is the real topic of this thread…

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#26
In reply to #25

Re: Do moving clocks really 'tick slower'?

10/23/2008 5:16 AM

Jorrie, thanks for your reply. Your point of view is perfectly clear. Let me summarize:

1) The Doppler effect has an apparent influence on "time flow" of a target (e.g. a planet) (i.e. an apparent "speed up" as you -the driver- approach the planet and an apparent "slow down" as you recede from the planet). This was exactly what I asserted from the beggining of our discussion. I never claimed that it's sth real.

[In my post #16, where I described this apparent "speed up" with the concept of the "light pulses transmission", it was obvious that this apparent "speed up" is connected (in a way) with the Doppler effect (i.e. the blue shift) because you can "replace" the light pulses with the crests and troughs of the electrmagnetic wave (i.e. light coming from the planet). That's why I couldn't understand your persistence to point out that "there is no "connection" between a change of "time flow" and the Doppler effect". Now I understand that we were initially agreed on this, because (by saying this) you actually meant that: "there is no "connection" between a real change of "time flow" and the Doppler effect"... (because this change in only apparent.)... Afterall, it is obvious that this change of time flow (observed on the planet by you) is only apparent because -if you leave the planet, travel away and rerurn back- the "slow down" of the time flow on the planet (as you go away) overwhelm the "speed up" of the time flow on the planet (as you come back).]

2) As you said: "...in the case of purely inertial frames of reference, the relativistic time dilation is also just an apparent effect."... Well, that was sth brand new for me. As you see from my post#18, my perception (always) was that this relativistic time dilation (that we observe on another inertial frame which is moving very fast relative to our inertial frame) is (in a way) very real. And I presented some of my thoughts on this in my post#18. BTW, it is perfectly clear that when sth "go away and come back" (relative to us) we observe -finally- a very real time dilation. I totally agree on this. But I'm still very sceptical about your claim that, concerning pure inertial systems, this time dilation is only apparent and not real.

(I'll come back on this later... I still work in relativistic speeds... ...)

__________________
George
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#27
In reply to #26

Re: Do moving clocks really 'tick slower'?

10/30/2008 11:17 AM

Jorrie, I must ask again my original question which has not been answered yet. So, in my post #18 I asked: "What is the combination of these contrary effects that coexist i.e. the non reletivistic, apparent "speed up" of time on the planet as we approach it (related, in a way, to the Doppler effect) and the relativistic time dilation (due to SR)???"

Then you refered again to this fine equation which gives the "λRxTx".

At your post #19 you said: "At v/c = -0.99c, the corresponding figures are 0.071 and 0.141." and, afterwards, you compare these two results. I don't understand why... I think that you should compare the "0,071" (which is the overall effect, i.e. the Doppler effect influnced by the relativity) and the pure time dilation itself (due to SR).

We have the numerator ( √1-(v/c)2 ) which is related to the pure Doppler effect which is 0,141. This term is "shortening" the received wavelength and this shows us an apparent "speed up" of the events (very logical). We have, also, the other term ( 1/(1-v/c) ) which is the relativity factor (which influences the Doppler effect) which is 0,5025. This relativity factor is also "shortening" the received wavelength, so it shows us, also, an apparent "speed up" of the events. So it cannot be related to the time dilation due to SR (the time dilation should "lengthen" the received wavelength). So, what this relativity factor actually means (or represents)??? [Afterall (and as a final result) we get the "0,071" (from the whole equation). This shows us that this relativity factor is further "shortening" the received wavelength (i.e. it gives us "0,071" instead of "0,141"). (This is sth that I had already mentioned in my previous posts.) This shows us a further overall "speed up" of the events. So, how this term (0,5025) can be related to the time dilation??? (In other words, this time dilation should influence the Doppler effect by somewhat "lengthening" the received wavelength... not by further "shortening" it...)]

(Now, can you understand my comfusion???)

(Note: And we talk about "apparent" and not "real" influence on the "time flow")

[Let's express again my initial question in other words: The time dilation due to SR gives us a red shift... we have, also, a blue shift (apparent speed up of the events) due to our approach to the planet... which is the final result of these two contrary effects???... a blue shift or a red shift???...]

I know that you are exhausted by me but I would be very obliged to you if I could get a final and clear answer.

Thanks.

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#28
In reply to #27

Re: Do moving clocks really 'tick slower'?

11/02/2008 1:39 AM

Hi George, I'm back home now and have more utility as far as internet is concerned. I hope we can put this one to bed quickly now...

You wrote: "We have the numerator ( √1-(v/c)2 ) which is related to the pure Doppler effect which is 0,141."

This is the "pure relativistic effect", not the "pure Doppler effect". The Newtonian Doppler effect is either 1/(1-v/c) or (1-v/c), depending on which party is "static relative to the aether". This very "aether" postulate of Newton's time is probably throwing you off the track, I think.

In SR, the situation is very simple: it does not matter who is considered "moving", the relativistic Doppler ratio for wavelength is always √(1+v/c)/√(1-v/c), where v is a positive radial ("opening") velocity.

So the issue is: is it even valid to ask the question "how does the "pure Doppler" and the "pure relativistic" part compare? The middle parts of the equations that I gave in post #8 above must be read with the (invalid) aether in mind, making it risky to drill down into the various factors!

If you insist on doing that, you have to be careful in choosing the reference frame to be the same as the aether frame - otherwise the middle parts of the equations are not valid! I fail to see what it is that bothers you (possibly because of the aether confusion?) I suggest you reread post #8 and preferably also chapter 2 of Relativity 4 Engineers.[1] Then please rephrase your question, if it is still unclear.

Regards, Jorrie

[1] If you haven't obtained the eBook, the chapter is available for download here.

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#29
In reply to #28

Re: Do moving clocks really 'tick slower'?

11/02/2008 1:59 AM

Correction of typo: my "The Newtonian Doppler effect is either 1/(1-v/c) or (1-v/c), depending on which party is "static relative to the aether"" should read: The Newtonian Doppler effect is either 1/(1-v/c) or (1+v/c), depending on which party is "static relative to the aether". This is for wavelength ratio and v is a positive radial speed (opening velocity). So for a spacehip approaching a target, v is negative.

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#30
In reply to #29

Re: Do moving clocks really 'tick slower'?

11/04/2008 11:13 AM

Jorrie, thanks for your reply. I had considered the opposite for these two terms of the equation. Thanks for the correction. But the issue that bothers me is still here. So, let me rewrite my previous post correctly. (Please, read carefully and you'll understand what bothers me.)

(And let's forget about the whole "ether issue"…)

We have the term ( 1/(1-v/c) ) which is the pure Doppler effect which is 0,5025. This term is "shortening" the received wavelength, so it shows us, also, an apparent "speed up" of the events on the planet (very logical).

We have, also, the other term ( √1-(v/c)2 ) which is related to the relativity factor (which influences the Doppler effect) which is 0,141. This term is, also, "shortening" the received wavelength and this shows us, also, an apparent "speed up" of the events on the planet.

It seems that this "0,141" cannot be related to the time dilation due to SR (because the time dilation should "lengthen" the received wavelength). So, what this relativity factor actually means (or represents)???

[Let's put it this way: Afterall (and as a final result) we get the "0,071" (from the whole equation). This shows us that this relativity factor is further "shortening" the received wavelength (i.e. it gives us "0,071" instead of "0,5025"). (This is sth that I had already mentioned in my previous posts.) This shows us a further overall "speed up" of the events. So, how this "0,141" (i.e. the relativity factor) can be related to the time dilation (due to SR)??? (In other words: if this "0,141" is representing the influence of the time dilation (due to SR), it should influence the Doppler effect by somewhat "lengthening" the received wavelength... not by further "shortening" it...)]

I hope that my confusion is now very clear. And I hope that you'll help me on this.

Thanks.

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#31
In reply to #30

Re: Do moving clocks really 'tick slower'?

11/05/2008 12:43 AM

Hi George, you wrote:

"We have the term ( 1/(1-v/c) ) which is the pure Doppler effect which is 0,5025. This term is "shortening" the received wavelength, so it shows us, also, an apparent "speed up" of the events on the planet (very logical)."

Correct, but remember that you are implying by the (Newtonian) "term ( 1/(1-v/c) " that the planet (the transmitter) is static in the reference frame and that the spaceship (the receiver) is moving. You are now forced to analyze the situation from the planet's point of view.

"We have, also, the other term ( √1-(v/c)2 ) which is related to the relativity factor (which influences the Doppler effect) which is 0,141. This term is, also, "shortening" the received wavelength and this shows us, also, an apparent "speed up" of the events on the planet.

No, since the planet is static in the reference frame, it shows a slowdown of the events on the spaceship, as observed by the planet, which translates to a speedup of the events on the planet as viewed by the spaceship (not the relativistic view, but rather a quasi-Newtonian view, if we insist on going through Newton).

"It seems that this "0,141" cannot be related to the time dilation due to SR (because the time dilation should "lengthen" the received wavelength). So, what this relativity factor actually means (or represents)???"

The simplest relativistic procedure is to choose the frame of interest (the spaceship's) as the reference. The planet is now moving at v = –0.99c in this frame. The Newtonian Doppler factor is 1+v/c = 0.01 for wavelength received (a speedup, as expected). The relativistic wavelength factor is 1/0.141=7.09 (a slowdown, as expected). The net effect is a speedup of 0.071, as before. No issues here!

I hope this helps!

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#32
In reply to #31

Re: Do moving clocks really 'tick slower'?

11/06/2008 4:22 AM

O.k. Jorrie, now the things become clear.

The whole problem is that I was trying to understand how things work by looking at the equation λRxTx=√[1-(v/c)2]/[1-(v/c)]. This assumes that the planet is static (in the ether) and the spaceship is moving. Since the whole issue is very "reference dependent", I should rather try to understand how things work by looking at the equation λRxTx=[1+(v/c)]/√[1-(v/c)2]. This assumes that the spaceship is static (in the ether) and the planet is moving. So, it's more appropriate to use this equation, as we are trying to understand "how things seem to be" from the driver's point of view (i.e. from the spaceship's reference frame). By doing this we get all that you said in your last paragraph: The Newtonian Doppler factor is 0.01 showing as a "speedup". The relativistic wavelength factor is 7.09 showing us a "slowdown". The combination of these two (i.e. the pure Doppler effect and the time dilation due to SR) is a final "speedup" of 0.071.

[BTW, from what you said in your 2nd paragraph, it seems that it's wrong to use the first equation (i.e. where the planet is static in the ether) trying to understand how things (on the planet) are observed by the driver's point of view... because (by using this 1st equation) we conclude that -concerning the relativity factor- the driver observes a "speed up" of the events on the planet... and this conclusion is wrong (as the relativity factor should "slowdown" the events)... Am I right???...]

Please, confirm all the above.

Thanks.

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#33
In reply to #32

Re: Do moving clocks really 'tick slower'?

11/06/2008 11:49 AM

Hi GK, you are right now, except here:

"BTW, from what you said in your 2nd paragraph, it seems that it's wrong to use the first equation (i.e. where the planet is static in the ether) trying to understand how things (on the planet) are observed by the driver's point of view..."

My para 4 of #31: "No, since the planet is static in the reference frame, it shows a slowdown of the events on the spaceship, as observed by the planet, which translates to a speedup of the events on the planet as viewed by the spaceship (not the relativistic view, but rather a quasi-Newtonian view, if we insist on going through Newton)."

It says that you can analyze it from the planet's view and you will correctly get an apparent relativistic slow down of the spaceships events. The only thing that is relativistically wrong is to then say: "but this looks like a speed up of events from the spaceship point of view", because the equation does not give the view from the moving spaceship.

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#34
In reply to #33

Re: Do moving clocks really 'tick slower'?

11/07/2008 3:29 AM

Jorrie, you said: "It says that you can analyze it from the planet's view and you will correctly get an apparent relativistic slow down of the spaceships events. The only thing that is relativistically wrong is to then say: "but this looks like a speed up of events from the spaceship point of view", because the equation does not give the view from the moving spaceship."

This is exactly what I said in my previous post. I didn't say that it's wrong to use this equation in general. I said that it's right to use it in order to understand "how things seem to be" on the spaceship from the planet's point of view (i.e. the "slowdown" of the events on the spaceship, which is correct). But it's "wrong" to use it if we are trying to understand "how things seem to be" on the planet from the spaceship's point of view. And that's because this leads us indirectly to a wrong conclusion concerning the driver's point of view (i.e. we thing that he observes a "speed up" of the events on the planet, which is obviously wrong).

So we totally agree. Don't you think???

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#35
In reply to #34

Re: Do moving clocks really 'tick slower'?

11/07/2008 5:42 AM

Hi George: "So we totally agree. Don't you think???"

Yep, I think so too. The whole thing shows the dangers of going through the Newtonian step en-route to the relativistic value. I've used it in my eBook because it gives engineers something familiar to "hook onto". I now realize I should also have also told them to immediately forget about it, because the Newtonian view is misleading with its aether dependence! (I do touch upon that in the chapter, though).

Jorrie

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#36
In reply to #35

Re: Do moving clocks really 'tick slower'?

11/07/2008 6:55 AM

Thanks, again, for your time... ...

__________________
George
Reply
Guru
Engineering Fields - Aerospace Engineering - Retired South Africa - Member - The Rainbow-nation Engineering Fields - Engineering Physics - Relativity & Cosmology Popular Science - Cosmology - The Big Picture!

Join Date: May 2006
Location: Pretoria, South Africa
Posts: 3804
Good Answers: 69
#37
In reply to #36

Re: Do moving clocks really 'tick slower'?

11/16/2009 3:50 AM

Hi George,

After a long discussion on a different forum, I've decided to edit my original opening post slightly by adding something to the bottom...

It does not change what we have discussed in the rest of the exchange.

-J

__________________
"Perplexity is the beginning of knowledge." -- Kahlil Gibran
Reply
Guru
Hobbies - Musician - New Member Greece - Member - New Member

Join Date: Mar 2007
Location: Greece / Athens
Posts: 722
Good Answers: 28
#38
In reply to #37

Re: Do moving clocks really 'tick slower'?

11/19/2009 9:33 AM

Thank you, Jorrie... ...

(I wish I had the time to participate on other forums too... ...)

__________________
George
Reply
Reply to Blog Entry 38 comments
Copy to Clipboard

Users who posted comments:

Anonymous Poster (1); G.K. (20); Jorrie (17)

Previous in Blog: Twin Paradox - a New Graphical Solution   Next in Blog: Do moving clocks really 'tick slower'? - Resolution
You might be interested in: IC Clocks, Inertial Sensors, Inertial and Gyro Systems

Advertisement