Post by: TyroJack on 30/05/2019 15:50:25
If a clock takes longer to 'tick' - if it takes more time for each 'tick' - then we say it runs slow.
Yet if we measure a clock to be running slow, it is measuring less time than a clock keeping the correct time.
So how can a clock measuring more time be running slow?
If the clock takes longer to tick then it will measure fewer ticks; yet the same clock, stationary relative to observer A and moving relative to observer B is but one clock and both observers are observing and counting the same number of 'ticks'.
So observer B will measure more time for each 'tick' but the same number of 'ticks' and, therefore, a greater quantity of time passing than observer A will measure. So could we say that the clock must have run faster measured by observer B as it measured a greater time? Or slower because each 'tick' took longer?
I am not doubting anything about relativity, only pointing out how difficult it is to apply use the language of classic mechanics when we speak of relativity.
It seems to me that using a term like slowing adds connotations due to the way we use everyday language to describe something counter intuitive.
Post by: Halc on 30/05/2019 17:06:50
This has caused me some puzzlement:Well put. "Ticks slow" is an absolutist term, which means that it takes more absolute time between ticks than does a different clock. In relativistic (non-absolute) interpretation, a clock that ticks slow is inaccurate, and thus not really a clock. There is simply more or less time between a pair of events in one frame as compared to another frame which orders those same events differently. Nothing physical really changes for anything.
If a clock takes longer to 'tick' - if it takes more time for each 'tick' - then we say it runs slow.
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Yet if we measure a clock to be running slow, it is measuring less time than a clock keeping the correct time.Only under an absolutist interpretation is there a concept of 'correct time', and since they cannot produce a standard of it, it seems like a pretty poor interpretation. For instance, how long does it really take for Earth to make one sidereal revolution? Relativity has no 'correct time', so the question is meaningless. An absolute interpretation does have that, so the question is simply unanswered.
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So how can a clock measuring more time be running slow?If a clock ran slow, it would measure less time, not more. It wouldn't really be a clock if it didn't accurately measure what it is supposed to.
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If the clock takes longer to tick then it will measure fewer ticks; yet the same clock, stationary relative to observer A and moving relative to observer B is but one clock and both observers are observing and counting the same number of 'ticks'.I cannot really parse that. A clock that is stationary and local relative to an observer will appear to run at normal full rate, per principle of relativity. Ones that move relative to some frame will measure less time in that frame, but not necessarily appear to run slower to a stationary observer. An approaching clock appears to run faster for instance, but that's mostly due to blue shift.
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So could we say that the clock must have run faster measured by observer B as it measured a greater time? Or slower because each 'tick' took longer?You need to state the situation in terms of frames, not in terms of observers, since a clock not present at the observation has a frame dependent time that it reads but not a frame dependent time that is observed at that observation event.
So under absolute interpretation, the observers have nothing to do with anything. A clock that moves takes longer to do a tick. Under relativity, all clocks measure time accurately: the temporal distance between two events along the worldline of the clock. These are frame independent measurements, and thus nothing 'ticks slow'.
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I am not doubting anything about relativity, only pointing out how difficult it is to apply use the language of classic mechanics when we speak of relativity.Use the relativistic language then. Speak of frames and events, and not objective (lacking a relation) speeds or of appearances.
Post by: Janus on 30/05/2019 23:19:58
This has caused me some puzzlement:I think your confusion arises from trying to treat "time" as being absolute.
If a clock takes longer to 'tick' - if it takes more time for each 'tick' - then we say it runs slow.
Yet if we measure a clock to be running slow, it is measuring less time than a clock keeping the correct time.
So how can a clock measuring more time be running slow?
If the clock takes longer to tick then it will measure fewer ticks; yet the same clock, stationary relative to observer A and moving relative to observer B is but one clock and both observers are observing and counting the same number of 'ticks'.
So observer B will measure more time for each 'tick' but the same number of 'ticks' and, therefore, a greater quantity of time passing than observer A will measure. So could we say that the clock must have run faster measured by observer B as it measured a greater time? Or slower because each 'tick' took longer?
I am not doubting anything about relativity, only pointing out how difficult it is to apply use the language of classic mechanics when we speak of relativity.
It seems to me that using a term like slowing adds connotations due to the way we use everyday language to describe something counter intuitive.
If I measure a clock moving relative to me to be ticking slower than my own clock, It may only tick once for every tow ticks of my own clock, If one tick is one second fro me than the other clock takes two sec between ticks But this only according to me. There is no absolute measure of time against which to judge the other clock. For that clock own perspective, it is ticking once per sec. It is just that the same time period the other clock measures as being one sec, I will measure as being two sec.
In addition, someone at rest with respect to that clock would measure my clock as only ticking once for every two ticks of that clock. His clock measures one sec per tick and mine takes two sec per tick.
As Einstein but it " Time is what clocks measure". In that vein, there is no absolute measure of time only what each clock measures as time.( keeping in mind that we are considering "ideal" clocks; clocks that essentially keep perfect time.)
People many times get the idea that Relativity says that a "moving" clock will run slow, as in the motion physically alters the manner in which the clock works. This isn't what is happening. An observer will measure a clock that has a relative motion with respect to the observer to tick slower than the observer's own clock. It isn't that motion physically effects the clocks, it is that observers in relative motion with respect to each other also measure time differently than each other. And neither observer's measurement of time is any more valid than the other's.
Post by: evan_au on 30/05/2019 23:23:21
Quote from: TyroJack
If a clock takes longer to 'tick' - if it takes more time for each 'tick' - then we say it runs slow.We are used to a clock indicating a time of day; the ticks come in discrete intervals. But that's not the only type of clock.
Another kind of clock is the frequency of a spectral line in a star's radiation or a radio transmitter; the radiation is "continuous".
- We can compare this by saying that the frequency emitted by one source is higher or lower than the frequency emitted by another source, one of which could be a reference in our laboratory (in our frame of reference).
Note that in either clock, you must disentangle the different rates due to relativity from the (usually much larger) effect of Doppler shift.
One relativistic time variation that is easier to disentangled from Doppler shift is gravitational redshift.
See: https://en.wikipedia.org/wiki/Gravitational_redshift
Post by: TyroJack on 31/05/2019 11:45:58
Event P is the start of a tick and Q is the end of the tick. The tick takes one second.
From Frame B each tick of clock A is measured to be 2 seconds.
Clock A ticks between P and Q
Measured by A this takes 1 second.
Measured by B moving clock A takes 2 seconds to tick, but that is two seconds in frame A. Clock B is an ideal clock in an inertial frame, measured by a resting observer, so will measure the same time (postulate 1) as clock A, 1 second.
So, one might say that from B, A did not tick slower, but measured more time for each tick - so time passed quicker for frame A, measured by frame B, 2 seconds instead of one second, during each tick.
Post by: Halc on 31/05/2019 12:09:16
Clock A's worldline passes through events P and Q.It is one second in frame A, as you stated above. "Measured by A this takes 1 second". It is two seconds in frame B, and frame B is not a proper measurement between events P and Q since the B clock is not present at both events P and Q. The A clock is present at both of them, as you stated on the first line.
Event P is the start of a tick and Q is the end of the tick. The tick takes one second.
From Frame B each tick of clock A is measured to be 2 seconds.
Clock A ticks between P and Q
Measured by A this takes 1 second.
Measured by B moving clock A takes 2 seconds to tick, but that is two seconds in frame A.
You don't specify, so let's say clock B's worldline passes through events P (0), R (0.5) and S(2) in a straight line. In frame A, events Q and R are simultaneous. In frame B, events S and Q are simultaneous. That difference in event ordering is what relativity is about.
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Clock B is an ideal clock in an inertial frame, measured by a resting observer, so will measure the same time (postulate 1) as clock A, 1 second.Both clocks are ideal clocks (how is an ideal clock distinct from a clock anyway?), both in inertial frames, with separate observers next to each if you find that useful, but they're stationary in different frames, so neither clock measures the same length-of-worldline as the other.
Clock A measures the proper time between events P and Q because it is present at both events and its worldline between those events is straight. Clock B is not measuring proper time between those two events because it is not present at both of them, but is instead measuring proper time between events P R and S.
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So, one might say that from B, A did not tick slower, but measured more time for each tick - so time passed quicker for frame A, measured by frame B, 2 seconds instead of one second, during each tick.Or one might stop using absolute language and not talk about tick rates at all.
From frame B, clock A is dilated. From frame A, clock B is dilated. Neither clock ticks slow or measures any different time than one second per second. The time shown is the actual temporal length the worldline measured.
Post by: Janus on 31/05/2019 18:47:24
Clock A's worldline passes through events P and Q.
Event P is the start of a tick and Q is the end of the tick. The tick takes one second.
From Frame B each tick of clock A is measured to be 2 seconds.
Clock A ticks between P and Q
Measured by A this takes 1 second.
Measured by B moving clock A takes 2 seconds to tick, but that is two seconds in frame A. Clock B is an ideal clock in an inertial frame, measured by a resting observer, so will measure the same time (postulate 1) as clock A, 1 second.
So, one might say that from B, A did not tick slower, but measured more time for each tick - so time passed quicker for frame A, measured by frame B, 2 seconds instead of one second, during each tick.
For a little more clarity, I put events P and Q three ticks(sec) apart rather then one. I also added event "R" which represents three sec according to clock B. A and B both start event P. Relative speed is 0.8c This is relative speed. both diagrams below describe exactly the same scenario, just from different frames of reference.
World lines for A and B from the rest frame of A.
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By the in the Time clock A goes from event P to Q( three sec as measured by clock A), clock R has not yet ticked off three sec. Event R has not happened yet according to A and event P happens before event R. Light signals ( yellow lines) leave both A and B after 1 tick has past on each clock. Each signal reaches the other clock when it reads 3 ticks (sec)
World lines for A and B according to B's rest frame.
[ Invalid Attachment ]
In the time it takes for B to go from P to R( three sec as measured by clock B), A has not yet reached ticked off three sec. Event Q has not happened yet according to B, and event R occurs before event Q.
The light signals sent by each clock when it has ticked off one sec each reach the other clock when that clock reads 3 sec.
Post by: jeffreyH on 01/06/2019 10:32:59
Also GPS makes no sense and shouldn't work. Time dilation has physical effects. All the arguments about everything being totally relative dodges the issue.
Post by: Halc on 01/06/2019 13:01:34
I can select a reference frame such that A and B are traveling at identical velocities.Identical speeds at least, yes. Velocity is a vector quantity so in this frame, A and B are travelling at opposite velocities.
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From my point of view both frames have the same clock rate. Who is dilated more 1 A, 2 B, 3 Neither? Someone has to be physically dilated more. Otherwise the twins thought experiment makes no sense.A and B are dilated identically in that frame. You're the one who is dilated not at all. One is never dilated in one's own frame.
The twins scenario cannot be illustrated in that example since A and B will never meet again at that velocity. To meet again, somebody needs to change velocity.
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Also GPS makes no sense and shouldn't work. Time dilation has physical effects. All the arguments about everything being totally relative dodges the issue.The GPS satellites actually run faster than us on Earth, but that's mostly due to GR effects, not SR. The average speed of a GPS satellite is larger than your average speed in any inertial frame, and it being totally relative doesn't change that.
Post by: jeffreyH on 01/06/2019 13:14:16
I can select a reference frame such that A and B are traveling at identical velocities.Identical speeds at least, yes. Velocity is a vector quantity so in this frame, A and B are travelling at opposite velocities.
Yes, speed. My apologies.
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QuoteFrom my point of view both frames have the same clock rate. Who is dilated more 1 A, 2 B, 3 Neither? Someone has to be physically dilated more. Otherwise the twins thought experiment makes no sense.A and B are dilated identically in that frame. You're the one who is dilated not at all. One is never dilated in one's own frame.
The twins scenario cannot be illustrated in that example since A and B will never meet again at that velocity. To meet again, somebody needs to change velocity.
And there is the problem. Stating that acceleration has to occur for a difference in dilation to be a physical thing. A change in velocity is only required for it to be observable. This guarantees the proximity of observer and observed. If the twin just travels in a straight line for t amount of time and never returns he will still be time dilated. It is just that you will never be able to detect it.
This is the big issue I have with opinions on SR.
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QuoteAlso GPS makes no sense and shouldn't work. Time dilation has physical effects. All the arguments about everything being totally relative dodges the issue.The GPS satellites actually run faster than us on Earth, but that's mostly due to GR effects, not SR. The average speed of a GPS satellite is larger than your average speed in any inertial frame, and it being totally relative doesn't change that.
Post by: jeffreyH on 01/06/2019 13:17:22
Post by: Halc on 01/06/2019 18:08:56
I didn't say that. I said that no objective comparison can be made between clocks not in each other's presence. The acceleration itself does nothing.Quote from: HalcThe twins scenario cannot be illustrated in that example since A and B will never meet again at that velocity. To meet again, somebody needs to change velocity.And there is the problem. Stating that acceleration has to occur for a difference in dilation to be a physical thing.
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BTW This follows from time dilation in SR being a function of velocity and NOT acceleration.Exactly so, yes. I can have two clocks here on Earth, both at identical speeds and depth in gravity well, but one with arbitrarily higher acceleration than the other. SR says they will stay in sync.
Post by: jeffreyH on 02/06/2019 00:29:51
Post by: pzkpfw on 02/06/2019 01:38:15
OK To reiterate: Who is dilated more 1 A, 2 B, 3 Neither? Someone has to be physically dilated more.
You're mixing frames.
If we assume A is the "stay at home" twin and B moves away, then yes there could be a C for whom both A and B are the same speed relative to them. Note that C would also have to be moving, according to both A and B.
According to A, C is moving away at half the speed B is moving away.
According to B, C is moving away at half the speed A is moving away.
According to C, A and B are both moving away at the same speed (in opposite directions).
At this time, according to C, time dilation is the same for A and B.
Then there's the turn-around. B heads back to A. For your addition of C, C will also now need to head back ...
According to A, C is moving closer at half the speed B is moving closer.
According to B, C is moving closer at half the speed A is moving closer.
According to C, A and B are both moving closer at the same speed.
At this time, according to C, time dilation is the same for A and B.
During the two times of plain inertial movement, everyone can consider themselves as at rest, and their clocks will tick at one second per second - according to them. And also, according to them, the other peoples' clocks will tick slower. Your addition of C does not change this.
That B (and C) will be younger than A when they all get back together is not about some "real" physical dilation happening to one but not the other during the travel. That would imply some absolute rest, where A is "really" at rest and B is "really" moving.
It's an effect of one of them staying in a single inertial frame the whole time while the other (or others) do not.
(Although acceleration is needed to have B change direction, the effect is not specifically "caused" by the acceleration. Alternate methods of this experiment involve someone else coming back, and B passing them their clock reading as they cross paths. The clock that returns to A will still read less.)
While B moves away (according to A) the situation is symmetrical, i.e. B can consider A moving way.
While B moves closer (according to A) the situation is symmetrical, i.e. B can consider A moving closer.
(And this is where the "paradox" of "twins' paradox" comes from.)
But ... the overall scenario is not symmetrical, someone (A) just sat in their chair holding their clock and a cup of coffee. Someone else (B) had to change frames; they either spilled their coffee (by accelerating) or had to pass their clock to someone else to return to A.
Post by: jeffreyH on 02/06/2019 12:21:21
OK To reiterate: Who is dilated more 1 A, 2 B, 3 Neither? Someone has to be physically dilated more.Dilation is relative. Nobody is dilated more, absent that relation. The principle of relativity demands this.
So we can pick a frame to make the problem with GPS go away then. If dilation is truly relative this has to be the case. This, however, doesn't solve the problem for the guy on the surface of the earth.
Post by: jeffreyH on 02/06/2019 12:26:38
OK To reiterate: Who is dilated more 1 A, 2 B, 3 Neither? Someone has to be physically dilated more.
You're mixing frames.
If we assume A is the "stay at home" twin and B moves away, then yes there could be a C for whom both A and B are the same speed relative to them. Note that C would also have to be moving, according to both A and B.
According to A, C is moving away at half the speed B is moving away.
According to B, C is moving away at half the speed A is moving away.
According to C, A and B are both moving away at the same speed (in opposite directions).
At this time, according to C, time dilation is the same for A and B.
Then there's the turn-around. B heads back to A. For your addition of C, C will also now need to head back ...
According to A, C is moving closer at half the speed B is moving closer.
According to B, C is moving closer at half the speed A is moving closer.
According to C, A and B are both moving closer at the same speed.
At this time, according to C, time dilation is the same for A and B.
During the two times of plain inertial movement, everyone can consider themselves as at rest, and their clocks will tick at one second per second - according to them. And also, according to them, the other peoples' clocks will tick slower. Your addition of C does not change this.
That B (and C) will be younger than A when they all get back together is not about some "real" physical dilation happening to one but not the other during the travel. That would imply some absolute rest, where A is "really" at rest and B is "really" moving.
It's an effect of one of them staying in a single inertial frame the whole time while the other (or others) do not.
(Although acceleration is needed to have B change direction, the effect is not specifically "caused" by the acceleration. Alternate methods of this experiment involve someone else coming back, and B passing them their clock reading as they cross paths. The clock that returns to A will still read less.)
While B moves away (according to A) the situation is symmetrical, i.e. B can consider A moving way.
While B moves closer (according to A) the situation is symmetrical, i.e. B can consider A moving closer.
(And this is where the "paradox" of "twins' paradox" comes from.)
But ... the overall scenario is not symmetrical, someone (A) just sat in their chair holding their clock and a cup of coffee. Someone else (B) had to change frames; they either spilled their coffee (by accelerating) or had to pass their clock to someone else to return to A.
This wasn't a twins experiment scenario. In my scenario there is no turn around point. You will never know if anyone's clock is running slower than the others. That is the whole point. I think we should admit when we actually don't know something. Rather than arguing in circles. That would be much more productive.
Knowing what the state of anyone's clock is requires close proximity. If we don't have close proximity then we don't know it's state. We are making an assumption.
Post by: jeffreyH on 02/06/2019 12:42:22
Post by: Halc on 02/06/2019 13:23:26
So we can pick a frame to make the problem with GPS go away then. If dilation is truly relative this has to be the case. This, however, doesn't solve the problem for the guy on the surface of the earth.Dilation due to motion is truly relative. GPS dilation is due mostly to depth of gravity well, which is absolute, so the GPS clocks are less dilated than the Earth clocks in any frame.
This wasn't a twins experiment scenario. In my scenario there is no turn around point. You will never know if anyone's clock is running slower than the others.Sure you can, but you need to pick a frame to do it.
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That is the whole point. I think we should admit when we actually don't know something.The answer is frame dependent. Being frame dependent doesn't mean that you can't know.
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Knowing what the state of anyone's clock is requires close proximity.Proximity is required for an objective comparison since the comparison is the same in any frame. For separated events, the comparison is dependent on how the frame orders those events. That ordering of events is not a physical difference, only an abstract one. Thus neither clock runs physically slower or faster due to relative motion.
Post by: TyroJack on 02/06/2019 13:44:34
This diagram raises another thing that I wonder about.
World lines for A and B according to B's rest frame.Image23.png (7.12 kB . 603x301 - viewed 6320 times)
This is what I believe to be a Minkowski diagram.
Why is the time scale for the moving one out of scale? Is it because the light paths are at 45 degrees?
This bothers me for it seems that is Newtonian mechanics with absolute space and time:
Imagine 2 identical light clocks, A and B, moving apart at speed v.
Seen from clock A the light from clock B moves up at ct while clock B moves away at vt, so the light as seen from the stationary clock can be moving vertically if v is vanishingly small
If B is moving away at approaching light speed, the time measured by clock B, will be almost zero.
If clock B travels at 0.866 c, for 1 second, measured by A, its light will have travelled 1 light second, measured by A, but only 0.5 light seconds, within clock B (this of course is the time clock B is shewing - its proper time).
So clock A, measures 1 time dilated second to have passed for clock B, when clock B reads 0.5 seconds.
Post by: Janus on 02/06/2019 16:17:15
This diagram raises another thing that I wonder about.
World lines for A and B according to B's rest frame.
This is what I believe to be a Minkowski diagram.
Why is the time scale for the moving one out of scale? Is it because the light paths are at 45 degrees?
This bothers me for it seems that is Newtonian mechanics with absolute space and time:QuoteIn an absolute Space and time universe, if you transform from the Frame of A in the fist diagram to the frame of B, you would get this:The light traveling at 45 degree angle is important in that it shows that the speed of light is always c relative to any frame as measured from that frame.
[ You are not allowed to view attachments ]
Imagine 2 identical light clocks, A and B, moving apart at speed v.
Seen from clock A the light from clock B moves up at ct while clock B moves away at vt, so the light as seen from the stationary clock can be moving vertically if v is vanishingly small
If B is moving away at approaching light speed, the time measured by clock B, will be almost zero.
If clock B travels at 0.866 c, for 1 second, measured by A, its light will have travelled 1 light second, measured by A, but only 0.5 light seconds, within clock B (this of course is the time clock B is shewing - its proper time).
So clock A, measures 1 time dilated second to have passed for clock B, when clock B reads 0.5 seconds.
You mean like this.
[ Invalid Attachment ]
However, if you were "riding" along with clock B, the circles representing the expanding sphere of the propagating light, would expand from points at rest with respect to you, and it would be clock A that ticks off only 1 sec for every two of your own.
Post by: yor_on on 02/06/2019 18:05:17
" The graviton sucks"
:)
Fact is, I kind'a agree
Post by: jeffreyH on 02/06/2019 18:24:10
For separated events, the comparison is dependent on how the frame orders those events. That ordering of events is not a physical difference, only an abstract one. Thus neither clock runs physically slower or faster due to relative motion.
So therefore you are arguing that there is never any physical time dilation due to relative motion. Interesting.
Post by: jeffreyH on 02/06/2019 18:26:41
Jeffrey. I won't get involved but you made me laugh
" The graviton sucks"
:)
Fact is, I kind'a agree
I usually get very little time to get involved but I am glad I can raise a chuckle. Thanks yor_on, you made me smile.
Post by: jeffreyH on 02/06/2019 18:39:59
This has been amended to ensure an amount of physical time dilation. Since, apparently, only acceleration causes physical and not abstract time dilation.
Post by: yor_on on 02/06/2019 19:40:39
You lifespan don't change relative your wristwatch
although most everything else becomes as variable
Go for what makes most sense to you
Also called 'logic'
Post by: yor_on on 02/06/2019 19:45:50
Using ones mind is the best there is Jeffrey, learning is the antidote to stagnation.
We all try, as good as we can.
I like it
Post by: TyroJack on 03/06/2019 10:22:31
However, if you were "riding" along with clock B, the circles representing the expanding sphere of the propagating light, would expand from points at rest with respect to you, and it would be clock A that ticks off only 1 sec for every two of your own.
Post by: TyroJack on 03/06/2019 10:53:56
However, think about it happening physically.
Clock A will tick once every second - that is the physical clock ticking - the light in clock A hitting the mirror and returning, as the clock is in an inertial frame.
Clock B will also tick once every second - that is the physical clock ticking - the light in clock B hitting the mirror and returning - clock B is also in an inertial frame, like clock A:
identical clocks in inertial frames will follow the same scientific laws and will behave identically - postulate 1.
The measurement of the light taking the diagonal path, is the measurement from clock A of the time the light will take to travel the extra distance - two light seconds - the dilated time; yet clock A has only ticked 1 second while it measures clock B to be time dilated, taking two seconds for each tick.
It is the rate that the inertial clock measures the moving clock to run at - its relative time.
Both physical clocks have ticked one second while each will measure the other, moving clock, to have measured two seconds. That is not the moving clocks slowing, it is each clock measuring time to pass fast in the other moving clock.
And is that not why it is called Time Dilation?
Post by: TyroJack on 05/06/2019 17:00:48
The time displayed by a moving clock is slow.
Yet, it being the proper time for that clock, it is the correct time for that clock as an inertial clock; so, compared with what exactly, is that clock slow? The clock time of the observer relative to whom it is moving, or is it the time measured by that observer in the moving frame?
Surely, this is not some airy fairy property a moving clock has in relativistic Spacetime, but very basic Spacetime mechanics.
It seems to me that even in -
Relativity: The Special and General Theory 1920,
XII. The Behaviour of Measuring-Rods and Clocks in Motion,
Einstein used the Lorentz transformation equations to shew that - "As a consequence of its motion the clock goes more slowly than when at rest."
Why is it not shewn using simple Spacetime mechanics?
Post by: Colin2B on 05/06/2019 18:24:56
Why is it not shewn using simple Spacetime mechanics?What do you mean by simple spacetime mechanics? It can’t get much simpler than the way Einstein described it, Lorentz transform is just the maths that allows you to calculate the effect.
Post by: chiralSPO on 05/06/2019 18:32:49
Post by: jeffreyH on 05/06/2019 19:19:53
Post by: Halc on 05/06/2019 19:38:51
It seems that learning about relativity is not what you're after in making these posts. If I am incorrect about that, then reread all those responses, and perhaps respond to the unclear parts instead of just posting stuff like that.
Almost everything you say is wrong. For instance:
The time of a moving clock is dilated - there is more of it.Dilated means there is less of it.
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The time displayed by a moving clock is slow.It is exactly correct.
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it is the correct time for that clock as an inertial clock;They're both inertial in your example
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so, compared with what exactly, is that clock slow? The clock time of the observer relative to whom it is moving, or is it the time measured by that observer in the moving frame?All rates are relative to a frame of reference and have nothing to do with observers, present or not.
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Surely, this is not some airy fairy property a moving clock has in relativistic Spacetime,You actually got this one correct.
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Einstein used the Lorentz transformation equations to shew that - "As a consequence of its motion the clock goes more slowly than when at rest."Wrong. The equations might be used to calculate dilations and such, but they do not show it. It was deduced from (shown by) the premise of the constant speed of light in any reference frame, and required no Lorentz transformation for the demonstration.
Post by: Colin2B on 06/06/2019 07:36:29
Why have you started 2 topics on the same subject?
Unless you have a very good reason they will be merged.
Post by: Fussball on 06/06/2019 12:26:54
Post by: TyroJack on 06/06/2019 16:36:56
In other threads, you seem to totally ignore almost all replies, including some very good ones.I have been studying SR for several years now and I have not responded to replies, some very good as you say, that state what I already understand.
It seems that learning about relativity is not what you're after in making these posts. If I am incorrect about that, then reread all those responses, and perhaps respond to the unclear parts instead of just posting stuff like that.
Almost everything you say is wrong.
What I still find awkward is trying to use the correct terms as my natural language is not scientific and in such areas, as in these forums, I often received replies that are, not criticizing exactly, but often correcting my use of terms that have precise scientific meanings rather than addressing the point I am making.
Let me see if I can be any more precise:
Yes I can see that is how it is taken, yet it seems to me that if frame A is moving relative to frame B, then if clock A ticks 5 times between event 1 and event 2, then being in frame A the difference will be time only: Proper time, the time displayed on the clock, 5 unitsThe time of a moving clock is dilated - there is more of it.Dilated means there is less of it.
While observed from frame B, each tick will be longer by the Lorentz factor, so after 5 ticks the time B will measure = γ5 units.
So more time has passed between events 1 & 2 for a clock in Frame B, when it is measured from Frame A relative to which frame B is moving,
Yes, but why then do we talk about clocks slowing – including Einstein who wrote:The time displayed by a moving clock is slow.It is exactly correct.
Quote from: Einstein: Relativity: The Special and General Theory. XII.
As a consequence of its motion the clock goes more slowly than when at rest.If the moving clock is keeping exact time, as is the resting clock, how is it that "As a consequence of its motion the clock goes more slowly than when at rest"?
Einstein was correct and you are correct, I know that; but I would like you to say how both statements can be correct about the same moving clock. With respect to what is Einstein's clock going slow?
Indeed they are but why was that worth commenting on? My point was that as an inertial clock the first postulate was relevant and the clock therefore had to be displaying proper time. My comment made no reference to the other clock, whether it was inertial or not.it is the correct time for that clock as an inertial clock;They're both inertial in your example
Hmmm. OK. So rates are only relevant to frames or reference, not observers within those frames?so, compared with what exactly, is that clock slow? The clock time of the observer relative to whom it is moving, or is it the time measured by that observer in the moving frame?All rates are relative to a frame of reference and have nothing to do with observers, present or not.
Both the resting frame and the moving frame are inertial frames - (I know you agree with that :) )
So both have the same time rate.
The observer in Frame A measures proper time along its world line where only time is changing.
The observer in Frame B measures dilated time along A's world line because, seen from B's frame, both the time and location of clock A are changing.
Now it seems to me that the rate observer B, in frame B, measures for Clock A, in frame A, is different from both the rate in frame B measured from frame B and the rate in frame A measured from frame A.
The language gremlin bites again!Einstein used the Lorentz transformation equations to shew that - "As a consequence of its motion the clock goes more slowly than when at rest."Wrong. The equations might be used to calculate dilations and such, but they do not show it. It was deduced from (shown by) the premise of the constant speed of light in any reference frame, and required no Lorentz transformation for the demonstration.
Yes I know that you are right for Special Relativity was deduced from Einstein's two Postulates.
If it is wrong to say that Einstein used the Lorentz Translation Equations to shew that a clock goes more slowly, it is certainly true that he used them to argue that point in chapter XII. He made no reference as far as I can see to the constant speed of light in that chapter.
But then perhaps I am being stupid and misreading it again...
Post by: TyroJack on 06/06/2019 17:49:20
@TyroJackMaybe that would be best as they have both tuned in the same direction... ;)
Why have you started 2 topics on the same subject?
Unless you have a very good reason they will be merged.
Post by: Halc on 06/06/2019 19:05:00
What I still find awkward is trying to use the correct terms as my natural language is not scientific and in such areas, as in these forums, I often received replies that are, not criticizing exactly, but often correcting my use of terms that have precise scientific meanings rather than addressing the point I am making.Fair enough. Keep in mind that science is a precise language and meanings change when wordings change.
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By this I presume that clock A is stationary in frame A, and ditto for B, and both are inertial. If clock A is present at events E1 and E2, then clock B is present at only one of them at best, and thus hasn't measured the inertial temporal separation between those two events. So OK, let's say the two events are separated by 5 seconds. (I am assuming units/ticks to be one second each)Quote from: HalcYes I can see that is how it is taken, yet it seems to me that if frame A is moving relative to frame B, then if clock A ticks 5 times between event 1 and event 2, then being in frame A the difference will be time onlyQuote from: TyroJackThe time of a moving clock is dilated - there is more of it.Dilated means there is less of it.
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Proper time, the time displayed on the clock, 5 unitsAll clocks read proper time, so yes. Clock A displays 5 seconds at event E2 assuming zero at E1.
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While observed from frame B, each tick will be longer by the Lorentz factor, so after 5 ticks the time B will measure = γ5 units.From frame B, seconds on clock B are the same length, 1 second each. But it is present only at event E1 I presume, so it reaches event E3 and E4 when clock B measures 2.5 and 10 seconds respectively (assuming γ is 2). In frame B, events E4 and E2 are simultaneous, and in frame A, events E3 and E2 are simultaneous. The difference is due to the fact that after E1, the two clocks are no longer in each other's presence, and hence the ordering of all these events is frame dependent.
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So more time has passed between events 1 & 2 for a clock in Frame B, when it is measured from Frame A relative to which frame B is moving,OK, I see what you're saying. It reads 10 seconds at the event simultaneous with E2 in frame B, yes.
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Yes, but why then do we talk about clocks slowing – including Einstein who wrote:Quote granted. Einstein uses that wording it seems when speaking of dilation. The comment sans context lacks frame references, which are there (frame K) if context is restored. It 'goes more slowly' in a frame in which it has more motion than does a clock which, in that frame, has less or none.Quote from: Einstein: Relativity: The Special and General Theory. XII.As a consequence of its motion the clock goes more slowly than when at rest.
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If the moving clock is keeping exact time, as is the resting clock, how is it that "As a consequence of its motion the clock goes more slowly than when at rest"?It keeps exact proper time, not exact time, since there is no such thing. I meant to convey that. Under GR there is sort of an exact time, but no clock rate seems to ever be compared to it. Nobody can build a clock that compensates and thus reads actual time, if there is such a thing.
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Making sure you know the difference between the term inertial and stationary. Only the latter is a relation. You used the term in a context which seemed to imply it being the stationary one.Quote from: HalcIndeed they are but why was that worth commenting on?Quote from: TyroJackit is the correct time for that clock as an inertial clock;They're both inertial in your example
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My point was that as an inertial clock the first postulate was relevant and the clock therefore had to be displaying proper time.OK, but even a non-inertial clock will display proper time. A non-inertial clock might be present at the same two events as an inertial clock but read a different duration between those two events due to a different temporal length of the two respective worldlines.
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OK. So rates are only relevant to frames or reference, not observers within those frames?The rates are relevant to observers, but those rates don't reflect what those observers will observe. A moving clock might run slow in my frame, but if it is coming towards me, I will observe it running faster than my local clock. If a ship a lightyear away departs for Earth at 0.9c, the trip will take 1.11 years in my frame, but 40 days to an observer because the light from the departure event takes a year to get here. The clock on the ship will log 159 days to make the trip, so if that clock is observed from Earth, it will appear to run 159 days in 40 days, or about 4x faster than an Earth clock. That's what I mean by the difference between how a frame orders events and how events are observed
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Both the resting frame and the moving frame are inertial frames - (I know you agree with that :) )Yes, but I might have issue with any frame being called 'the resting frame'.
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So both have the same time rate.See, that's why I don't like observers. The observer in one frame observes a faster rate for the incoming moving clock, which is quite different than saying that the moving clock is dilated (slower) in that frame. He can of course compute how far away each event is and compensate, but then it becomes a computation, not an observation. I just find it easier to skip observation and say that time is dilated for objects moving relative to a given frame.
The observer in Frame A measures proper time along its world line where only time is changing.
The observer in Frame B measures dilated time along A's world line because, seen from B's frame, both the time and location of clock A are changing.
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Now it seems to me that the rate observer B, in frame B, measures for Clock A, in frame A, is different from both the rate in frame B measured from frame B and the rate in frame A measured from frame A.Of course. One's own proper rate is just '1', a comparison between two identical things. The rate of A in B's frame (at least under SR) is the same as the rate of B is A's frame. Not so under GR where two clocks can remain at a constant separation and yet have their time diverge, with one clock clearly racking up more time than the other no matter the frame.
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The language gremlin bites again!I think it is since the transform is the conclusion of the argument (the prediction), not one of the premises from which the behavior was derived.
Yes I know that you are right for Special Relativity was deduced from Einstein's two Postulates.
If it is wrong to say that Einstein used the Lorentz Translation Equations to shew that a clock goes more slowly,
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it is certainly true that he used them to argue that point in chapter XII. He made no reference as far as I can see to the constant speed of light in that chapter.Agree, the effect was already established at that point and the paper is turning to quantification. Here is how to compute exactly what clocks will do given this or that motion. From that, experiments like Hafele–Keating could be performed with a range of expectations that would verify or falsify this new theory.
The Lorentz transformation was obviously already around at the time, else it would have been named after Einstein. The theory was a cumulation and generalization of the work of several contributors.
Post by: Colin2B on 06/06/2019 22:18:42
STR and GR are mathematical models of reality that do not get into the mechanics of phenomena. It would take a new theory to explain mechanics of gravity or time dilation. But vibrations in a clock are subject to external forces like gravity.This is incorrect. For example, SR very clearly describes the mechanism of time dilation.
Post by: PmbPhy on 07/06/2019 01:57:23
Post by: Fussball on 07/06/2019 08:46:07
For example, SR very clearly describes the mechanism of time dilation.Where does SR get into the mechanics of a clock or its vibrations? Unless you're talking about a light clock. I agree that a horizontal light clock, moving horizontally, has its tick rate slowed due to motion. I am skeptical of the vertical light clock, where the theory of aberration applies.
Post by: Fussball on 07/06/2019 09:12:44
Post by: pzkpfw on 07/06/2019 09:45:37
A moving clock slows down or does it tick normally because in its frame it is always stationary. Therefore, a moving clock will tick normally (without slowing). Does my question make sense?
Say you and a friend are drifting in space. You are both wearing watches.
The distance between you is changing, so there is relative motion.
There is no absolute concept of "moving" or "not moving", so both of you can consider yourself as at rest, and the other is the one who is moving.
For both of you, your own watch (no matter what kind of clock it is) is ticking at 1 second per second.
For both of you, the watch on the other is slow.
Post by: Fussball on 07/06/2019 10:07:31
For both of you, your own watch (no matter what kind of clock it is) is ticking at 1 second per second.That's not possible. It either ticks slow for all or it ticks normally for all. I say that it ticks normally for all, because a clock ticks according to its own frame. And in its own frame it is always at rest and therefore it ticks normally for all. Where is the possibility of recording the slowing down of a clock when in its own frame it is always at rest?
For both of you, the watch on the other is slow.
Post by: Colin2B on 07/06/2019 10:08:58
Where does SR get into the mechanics of a clock or its vibrations?If you think SR has to do with the mechanics of clocks then you have completely misunderstood relativity and ought not to be answering questions in this section
A moving clock slows down or does it tick normally because in its frame it is always stationary. Therefore, a moving clock will tick normally (without slowing). Does my question make sense?Within its own frame a clock is ticking normally, but other clocks moving relative to it will be measured as slow
Post by: Fussball on 07/06/2019 10:17:31
Within its own frame a clock is ticking normally, but other clocks moving relative to it will be measured as slowIf all clocks are at rest in their own frame, where does the slowing down take place?
Post by: Colin2B on 07/06/2019 10:24:43
That's not possible. It either ticks slow for all or it ticks normally for all. I say that it ticks normally for all, because a clock ticks according to its own frame.Then you are again mistaken.
I suggest you stop posting in topics you know nothing about.
If all clocks are at rest in their own frame, where does the slowing down take place?It takes place in the measurements between frames.
I suggest you start by reading up about Galilean relativity and then try to understand the difference with SR.
Until you have done that, please don’t participate in this thread as your comments are only confusing the OP’s questions.
Post by: TyroJack on 07/06/2019 18:17:31
I created a webpage to provide the derivation. See: http://www.newenglandphysics.org/physics_world/sr/time_dilation.htmThank you, that is an excellent derivation - it is exactly how I think and wanted to write it down! Great!
There is one further point though, that I believe is worth making using your diagram:
From S', the 1 second measured to have passed in S is measured to have taken 7.09 seconds. The clock is measured to take longer to tick in frame S.
Yet an identical light clock in frame S' would also have ticked in 1 second, measured in S', while S' measured the greater time to pass in S.
What we see is that in each frame two times are measured - τ for the time measured on that frames clock and the longer time t measured for the tick of the clock in the other moving frame.
This is no surprise if we examine the invariant Spacetime interval between the two events.
Yes, the spacetime interval which is invariant whichever frame it is measured from.
Measured from the clocks frame the spacetime interval s² = c²τ² =1
Measured from a different frame from which the clock is moving, the spacetime interval s² = c²t² - v²t²
so c²τ² = c²t² - v²t²
τ² = t(c² - v²)
τ = t(1 - v²/c²)^-½
or τ = t/γ
So the time for a tick of the moving clock = γ seconds, measured from the frame relative to which it is moving.
Post by: Colin2B on 07/06/2019 23:23:00
I created a webpage to provide the derivation. See: http://www.newenglandphysics.org/physics_world/sr/time_dilation.htmThanks Pete, I was trying to find that for @TyroJack but couldn’t locate it.
There is also this http://www.newenglandphysics.org/physics_world/gr/grav_red_shift.htm
Which although it refers to gravitational redshift does give a useful diagram of the clock differences.
Post by: Halc on 07/06/2019 23:25:14
This is no surprise if we examine the invariant Spacetime interval between the two events.Indeed, there is an invariant spacetime interval between any two events, but there are move than two events depicted in that example, even if none of them are explicitly labeled. I count 6 events in all, but two of them (emit, detect) are called out, and yes, there is a frame independent interval between them.
Yes, the spacetime interval which is invariant whichever frame it is measured from.
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Measured from the clocks frame the spacetime interval s² = c²τ² =1That can't be.
τ is say a microsecond and x (distance) is 300 meters, so c²τ² - x² = 300m² - 300m² = 0, so s = √0 = 0, not 1. The interval between any pair of events separated in a light-like manner is always zero.
Edit: I was computing the interval between the bottom and top, not a detector at the bottom waiting for the reflected signal to return. Yes, there is a nonzero interval between those two events, which in my example is 600m if the mirrors are 300m apart in the light clock.
Post by: TyroJack on 08/06/2019 14:07:17
Yes, the spacetime interval which is invariant whichever frame it is measured from.c = 1
Measured from the clocks frame the spacetime interval s² = c²τ² =1
Measured from a different frame from which the clock is moving, the spacetime interval s² = c²t² - v²t²
so c²τ² = c²t² - v²t²
τ² = t(c² - v²)
τ = t(1 - v²/c²)^-½
or τ = t/γ
So the time for a tick of the moving clock = γ seconds, measured from the frame relative to which it is moving.
τ = 1 sec - given in the description.
t = 7.09 - given in the description.
v = 0.99c - given in the description.
γ = 7.09
1/γ = 0.141
Substituting for τ = t/γ we get: 1 = 7.09 x 0.141
1 = 0.99969
The clock in frame S ticks once a second.
The clock in frame S' ticks once every second.
The relative speed of the frames is 0.99c
Event 1. The two clocks start their ticks when they are adjacent.
Event 2. The clock in frame S completes its tick.
Event 3. The clock in frame S' completes its tick.
The invariant spacetime interval between event 1 and event 2 is one second.
Relative to frame S event 1 and Event 2 are at the same location.
Relative to frame S' event 1 and event 2 are separated by 1 light second.
Relative to frame S' event 1 and event 3 are at the same location.
Relative to frame S event 1 and event 3 are are separated by 1 light second.
Relative the distance between event 3 and event 4 (that is the distance between the clocks after 1 second, measured on either clock) is 0.99 light second.
The invariant spacetime interval between event 1 and event 2 is 1 light second.
The invariant spacetime interval between event 1 and event 3 is 1 light second.
Ah! I can see what is leading you astray, Halc. Event 1, is the beginning of the tick and event two is the end of the tick, not emitting the light and receiving it - that is the mechanism of the clock. We could just as easily be talking about a pendulum clock! For out purposes it is just a clock. After 1 second only the time coordinate has changed.
It is confusing because we use the light clock to shew how the moving clock has longer ticks. The light travel is also the time measured. We are concerned with the movement of the clocks, not the lights within those clocks.
Post by: TyroJack on 11/06/2019 14:26:16
No, 1=1. There are only 3 digits of precision being used.Yes indeed 1=1 silly me!
c times one second actually, so a light second, not a second. Seems to be a typo since you get it right everywhere else.yes, thank you
There was an event 4? None was defined.My mistake there is no need for an event 4; 0.99 is the distance between event 2 and event 3 - the distance frame S' has travelled from frame S.
Yes, in either frame, the distance between clocks 1 second after event 1 is .99 light seconds, but they're not the same two events being compared.
The invariant spacetime interval between event 1 and event 2 is 1 light second.
The invariant spacetime interval between event 1 and event 3 is 1 light second.
But I am not comparing the invariant spacetime interval between event 1 and event 2 with the invariant spacetime interval between event 1 and event 3.
Event 1 is the start of the ticks of each clock.
Events 2 and 3 are the ends of those ticks.
Each invariant spacetime interval can be measured two ways:
from S the interval between 1 and 2 is time only s²=(ct)² but measured from S' that interval s²=(ct')² - (vt')² . Similarly
from S' the interval between 1 and 3 is time only s²=(ct)² but measured from S that interval s²=(ct')² - (vt')² (note: t is the time in the frame that is not moving and t' is the time in the moving frame in each scenario)
All straight forward and reciprocal as it should be...
Ah! I can see what is leading you astray, Halc. Event 1, is the beginning of the tick and event two is the end of the tick, not emitting the light and receiving it - that is the mechanism of the clock.I have no idea what you mean by this statement. How are those different things? Does not light get emitted at event 1? At a different time or location?[\quote]
I was responding to this:
Edit: I was computing the interval between the bottom and top, not a detector at the bottom waiting for the reflected signal to return. Yes, there is a nonzero interval between those two events, which in my example is 600m if the mirrors are 300m apart in the light clock.If we were to look at the light in the clock it does indeed travel 600m as you say, but if we are examining the clocks then the clock in S does not move and the spacetime interval is the one second elapsed for that clock. s²=(ct)²
Post by: Halc on 11/06/2019 16:19:27
Frames don't have locations, so they don't travel. I know what you mean by that however, and yes, an object moving at .99c in some frame is by definition .99 light seconds away after one second in that frame, but events 1 and 3 are more than 1 second apart in that frame. The moving clock has been moving for 7.09 seconds in that frame, not 1.There was an event 4? None was defined.My mistake there is no need for an event 4; 0.99 is the distance between event 2 and event 3 - the distance frame S' has travelled from frame S.
Events 1 and 2 occur at the same location in space and event 3 happens 7.09 seconds after event 1, so events 2 and 3 are about 7 light seconds apart in space in either frame, but not necessarily in other frames.
I didn't respond to the rest because I don't disagree with any of it.