Time dilation: speed relative to what?

[Grimble]

I have been asked about time dilation which I thought I understood, but maybe you can help me provide an answer?
If a spaceship is traveling at 0.866c then time will slow by a factor of 2 due to the speed the spaceship has.

Why?

If it is alone in space, light years away from any other body, how can that speed affect the clock?
What is the speed talked of so glibly actually relative to?

If two spaceships were traveling in opposite directions at 0.433c (lorentz factor = 1.109) then for an observer midway between their clocks would each be running slow at 0.901 x a standard clock, i.e. keeping the same time, yet each spaceship would measure the other's time as being slow by a factor of 2.

So how can one say the clocks are actually running slow, rather than just being measured to run slow by the moving observers?

 

[A.T.]

So how can one say the clocks are actually running slow, rather than just being measured to run slow by the moving observers?

What is the difference between "measured to run slow" and "actually running slow"?

 

[russ_watters]

Whomever told you what the speed is will have to also tell you what it is relative to. Yes, speed only has meaning when you specify the frame of reference from which it is measured.

For the other: have you read-up on the twins paradox? You've proposed a totally symmetrical scenario -- most are not. The symetry or lack thereof matters for who is the one who's clock shows the less elapsed time.

 

[Grimble]

The difference between 'measured to run slow' and 'actually run slow' is that, according to the other spaceship, the clock would be running at half the speed of the one spaceship, while both clocks in the spaceships are running, not at 0.5 but at 0.901 of a standard clock. (Which of course, to comply with Einstein's postulate would run at the same rate in any inertial Frame of Reference). And, as measured by the midway observer both clocks are keeping identical time.

The question specified a lone spaceship, far from any other body, traveling at 0.866c...
I would take it that the implication is that it is relative to space, no other FoR was specified. Am I to tell the questioner that they have to Specify the FoR in which that is measured? Difficult to justify doing that when we are told that time slows when nearing the speed of light, and we have to specify relative to some arbitrary observer...

As for symmetry, there is symmetry. This is not the twin paradox where one has accelerated away from the other, these are two spaceships, each traveling traveling at 0.433c in opposite directions away from a central observer. No one has accelerated, they are inertial Frames of Reference.

Therefore the clocks must be running at the same rate as measured by that central observer - 0.901 of the standard clock held by the central observer who is at rest in space and whose clock must therefore be measuring proper time.

So is 0.901 proper time the true rate of the spaceships clocks? And if so how do they each measure the other's clock to be running at half the rate of their own clock?
(And how many different times can a singe clock measure?)

 

[phinds]

Clocks always run at one second per second unless they are broken. BOTH spaceships see their clocks running at one second per second and the other spaceship's clock running slower.

 

[Grimble]

So, what you are saying is that both clocks are running normally - as the first postulate requires, but when measuring a moving clock, i.e. measuring under different conditions that clock is measured to run slower. Is that not the answer - under those conditions the measurement of the moving clock is affected by the relative speed?
The question is then, how? What is the mechanism by which that measurement is affected...

It cannot be that the clock actually runs slow for their is no connection between the clock and the observer for whom it is in motion.
I.e. if the 'stationary' observer changed his speed there is no connection to the 'moving' clock to promote that change in RELATIVE velocity onto the clock, no way in which the rate of that clock could physically change relative to the observer.
It can only be that the MEASUREMENT BY the remote observer is affected.
The clocks don't change only the measurements? That can't be right surely?

 

[Khashishi]

The gist of relativity is that the view of every observer (in an inertial reference frame) is equally valid. In relativity, there is no "standard" clock.

 

[Bandersnatch]

The clocks don't change only the measurements? That can't be right surely?

It is. Time dilation is purely a coordinate effect. (but differential ageing isn't)

See here:
https://www.physicsforums.com/threads/time-dilation-vs-differential-aging.663977/

 

[phinds]

The clocks don't change only the measurements? That can't be right surely?

In case you don't realize it, you, right now as you read this, are moving at .99999c. You are also moving at .8c You are also moving at .1 c. You are also not moving at all. In each of these frames, your clock appears different, so of course, as others have already pointed out, it IS the measurement that changes.

 

[Ibix]

Both clocks running slow is actually quite closely analogous to the following.

Two cars are doing 30mph side-by-side down a road. They each appear stationary to each other.

One car turns off onto a road that runs at an angle \theta to the first road. That car starts to fall behind according to the one continuing straight on, because its velocity in the straight-on direction is 30\cos\theta. However, the other car can say the exact same thing and for the same reason. So both cars are falling behind according to the other.

This happens because the cars no longer agree on what "forward" is. In the case of moving spaceships (and the cars, although their velocity is so low you'll never notice), the two no longer agree what is "forward" through spacetime so they can both say the other is moving more slowly through time.

Health warning: this is only an analogy. It's a good one, but don't go crazy with it.

 

[A.T.]

It can only be that the MEASUREMENT BY the remote observer is affected.

Remoteness has nothing to do with it. Time dilation depends on relative motion, not on distance. And in physics there is no difference between what is measured, and what is "actually". Physics cares only about what is measured.

 

[Rising Eagle]

I'm with the original poster. A photon has no frame of reference needed. It always has no rate of ticking of its internal clock no matter what, even if there are no observers of any kind. This almost implies some kind of universal frame of reference. SR only deals with observers at lower than light speed and how they perceive the rate of clock ticking of other travelers' internal clocks when the observers see those other travelers. But a single lonely traveler at lower than light speed will have some nonzero rate of ticking of the clock. This cannot be measured other than with respect to some universal frame like the CMBR.

 

[PeterDonis]

I'm with the original poster.

You do realize that the OP said nothing at all about photons, right?

This almost implies some kind of universal frame of reference.

No, it doesn't.

SR only deals with observers at lower than light speed

Incorrect. SR can describe light just fine. Light can never be at rest in any inertial frame, but that doesn't mean it can't be described using inertial frames.

This cannot be measured other than with respect to some universal frame like the CMBR.

Incorrect. You can measure clock rates with respect to any inertial frame. And the "frame" provided by the CMBR is not an inertial frame (nor is it a "universal frame"--there is no such thing), because the universe as a whole, including the CMBR and which observers see it as isotropic, is not described by SR, since SR only works when spacetime is flat, and the spacetime of the universe as a whole is curved.

 

[DaveC426913]

Both clocks running slow is actually quite closely analogous to the following.

Two cars are doing 30mph side-by-side down a road. They each appear stationary to each other.

One car turns off onto a road that runs at an angle \theta to the first road. That car starts to fall behind according to the one continuing straight on, because its velocity in the straight-on direction is 30\cos\theta. However, the other car can say the exact same thing and for the same reason. So both cars are falling behind according to the other.

This happens because the cars no longer agree on what "forward" is. In the case of moving spaceships (and the cars, although their velocity is so low you'll never notice), the two no longer agree what is "forward" through spacetime so they can both say the other is moving more slowly through time.

Health warning: this is only an analogy. It's a good one, but don't go crazy with it.

I have never heard it described in such a way. That is very clever and very elegant. Especially the 'why' (bolded).

 

[Grimble]

I see that description by Ibix, referred to by DVEC426913, as an illustration of Minkowski Hyperbolic Rotation - if you know what I mean?

@Khashishi: By a standard clock, I mean one keeping Proper Time, that is one that is measuring time for a particle that it is adjacent to. The same Proper Time that is measured by a stationary clock at the origin of a Frame of Reference.

And in physics there is no difference between what is measured, and what is "actually". Physics cares only about what is measured.

Reference https://www.physicsforums.com/threads/time-dilation-speed-relative-to-what.804008/

So if I measure two people who are 6ft tall but one is further away, and so perspective means that I measure him as only half the height of the other (Let us say they are in floating in space and I view then through the window of my space craft) then those measurements tell me that he IS only half the height of the nearer one?

For that is what you are claiming.

No, the conditions under which a measurement are taken affects that measurement, hence the need for standard conditions, temperatures etc.

As I understand it, taking a measurement from one Inertial Frame of Reference and converting it to make it relative to another Inertial Frame of Reference is done by the Lorentz transformation equations.

So, for example, a proper time in the target iFoR can be converted to a coordinate time in the observer's iFoR.

 

[A.T.]

So if I measure two people who are 6ft tall but one is further away, and so perspective means that I measure him as only half the height of the other

No, you would measure half angular size, which is different from measuring linear height. Relativistic effects have nothing to do with visual effects, like perspective, signal delay etc. They are what still remains after you have already accounted for these effects.

 

[stevendaryl]

I have been asked about time dilation which I thought I understood, but maybe you can help me provide an answer?
If a spaceship is traveling at 0.866c then time will slow by a factor of 2 due to the speed the spaceship has.

Why?

If it is alone in space, light years away from any other body, how can that speed affect the clock?
What is the speed talked of so glibly actually relative to?

Pick ANY inertial coordinate system. Then a moving clock will experience time dilation relative to that coordinate system. How can it be simultaneously time-dilated according to EVERY single inertial coordinate system? Well, it is.

For simplification, let's just consider 1-dimensional motion in a single direction. Then the time dilation formula is, in infinitesimal form:

\delta \tau = \sqrt{\delta t^2 - \frac{1}{c^2}\delta x^2}

meaning that if a clock moves from the point x at time t to the point x+\delta x at time t+\delta t, then the time on the clock will advance by an amount \delta \tau as given above. The faster the clock is traveling, the bigger \delta x will be, the smaller \delta \tau will be.

The essential thing to realize about relativity is that even though \delta t is different in different reference frames, and \delta x is different in different reference frames, the combination \sqrt{\delta t^2 - \frac{1}{c^2}\delta x^2} is the same in EVERY reference frame. So every single inertial reference frame agrees that the above time dilation formula works.

So "relative to what" has the answer: "relative to ANY inertial reference frame".

 

[robphy]

Along the lines of what Ibix (above) said,
this is a variant of what I made for my relativity class this semester.
(You can actually pursue the analogy quite far... a paper has been in the works.)

Here is an interactive version (also try the E-slider to see the analogy)
https://www.desmos.com/calculator/ti58l2sair

Here is a physical description of the above
based on an older post (https://www.physicsforums.com/threads/time-dilation.667761/#post-4248601)

Take a hyperbola (events equal intervals in time from a common meeting event).
Imagine a field of runners with a wristwatch... running from a given starting event with all possible velocities in the x-direction. Ask each runner to mark with firecracker when his watch reads 1 sec. On a position-vs-time graph, you get a hyperbola [in special relativity]... (but a vertical line in Galilean relativity). Let say you do this again when your watches read 70 years, to obtain a similar larger hyperbola.

Draw two radii (These represent each of your worldlines corresponding to equal elapsed times according the respective wristwatches.) to the hyperbola.
At the tip of each radius, construct the tangent line to this hyperbola (the analogue of the circle in spacetime).
Note that each tangent meets the other radial direction before the hyperbola.
Each tangent represents events simultaneous with the tip of its radii according to that observer.
In other words, when you reach your 70th birthday on your worldline, the event on your friend's worldline that you say is simultaneous with your 70th birthday is her 50th birthday.
...and she will say the same thing about your 50th birthday on your worldline being simultaneous with her 70th birthday on her worldline.

As I mentioned earlier, if we did this experiment in Galilean physics, we would get a vertical line [as we might expect] of "my watch reads 1 sec" events. More accurately, we have not-so-precise clocks and we travel at not so large speeds that we extrapolate our limited experiment to a vertical line. If we proceed to draw the tangents to this "circle" [the vertical line], we would see that our sets of events simultaneous with our 70th birthdays coincide. This is absolute simultaneity in ordinary Galilean physics.

in the interactive graph above, use v1=tanh(arccosh(70/50))=0.69985=(approx 0.70) and v2=0 and note that 50/70=0.714.
(Center the intersection point and zoom in. By clicking near the intersection, the graph gives the coordinates.)

 

[Grimble]

I beg to differ. The statement that was made, viz. '... 'in physics there is no difference between what is measured, and what is "actually". ...' was a general statement of principle, not a statement about relativity.

Reference https://www.physicsforums.com/threads/time-dilation-speed-relative-to-what.804008/

 

[Ibix]

You aren't measuring height, though, you are measuring angular size. I think that was AT's point.

 

[Grimble]

The point isn't about what you are measuring, but the fact that measuring is subject to a host of different factors and the conditions under which it is done WILL affect those measurements. So to say what is measured is what is real HAS to be qualified by HOW it is measured.

 

[Grimble]

Let us make this very simple, for those of us who visualize in simple terms:

Let us imagine a space station far away from any other body in space with an atomic grandfather clock, where the local time is displayed on a large clock face.
This clock, at the origin of the space station's Frame of Reference is keeping Proper Time.

Now imagine an observer traveling at 0.866c and viewing that clock (allowing for the time it would take for light to travel to that observer)
What will that observer see when he looks at the clock?
He will see the time that is displayed on the clock face!
BUT due to the Lorentz transformation equations he knows that he has to convert that time to be relative to his (the observer's) Frame of Reference. And that will tell him that when the clock reads that 6 minutes have elapsed, twelve minutes will have elapsed in the Grandfather Clock's Frame of Reference as measured by the traveling observer.

On the observer's clock, that traveling with him, and also keeping Proper Time, only 6 minutes will have passed..

Yet how can that be? For keeping proper time, both clocks will have be reading the same!

The answer of course is that the observer's measurement of the Grandfather clock's passage of time is COORDINATE time, which at 0.866c is twice as fast as Proper Time.

Surely Einstein's first postulate tells us that time will pass the same in all inertial Frames of Reference, it is only when the added factor of the relative speed of an observer is added, i.e. the time is measured by a moving observer that a different, coordinate time is measured?

The Grandfather clock cannot display different times depending on who is reading it - for that is physically and logically impossible.

 

[phinds]

The point isn't about what you are measuring, but the fact that measuring is subject to a host of different factors and the conditions under which it is done WILL affect those measurements. So to say what is measured is what is real HAS to be qualified by HOW it is measured.

But, just to be clear since you have not indicated that you get the difference, it does matter if you are measuring apples using units of oranges and think that your measurement is meaningful. Height and angles are not the same thing.

 

[phyzguy]

The Grandfather clock cannot display different times depending on who is reading it - for that is physically and logically impossible.

Grimble: You're still not getting it. Try putting your disbelief aside and listen to what the people here are telling you. The clock can display different times depending on who is reading it, because they don't agree on when they are reading it. Try re-reading post #10. Just like the two cars don't agree on which way is forward, two observers in relative motion don't agree on when something is happening. Everyone's first reaction on learning about time dilation is, "How can each observer think the other one's clock is moving slower, that's impossible." But it is not impossible, it is the way things are. Space-time diagrams help a lot for visualizing these things. Try drawing two space-time diagrams, one for each of two observers in relative motion, and you will see how they do not agree on when things are happening.

 

[phinds]

Grimble: You're still not getting it. ... Try re-reading post #10.

Try rereading post #9 as well, since it says the same thing. We've ALL been telling you, Grimble, that you are not getting it and you keep not listening and insisting that you are right and we are all wrong. Is that really likely?

 

[nitsuj]

The Grandfather clock cannot display different times depending on who is reading it - for that is physically and logically impossible.

Of course the grandfather clock cannot display different times depending on who's looking at it; how would it even know who IS looking at it.

The variables being measured are displaced.

My preferred perspective is space-time keeps the sequence of "happenings", the causal structure intact via a maximum speed. Due to that logical and physical requirement, there must be disagreements on where/when the happenings took place. That is the variables being measured can be displaced by comparative motion, but the happenings cannot be; they just as you say with the face the grandfather clock must be (when measured) whatever they happen to have been.

Relativity of simultaneity has no physical significance, so it's logically fine if everyone disagrees on the clocks reading due to comparative motion, at any arbitrary "now" moment.

 

[phyzguy]

nitsuj: I think why you said is true - the causal structure of space-time is not changed by a relative velocity. Let me ask you these two questions:

(1) Do you agree that what time the clock displays depends on when you read it?
(2) Do you accept that two observers in relative motion can disagree on the meaning of "now".

If you answer yes to these two questions, then you will see how the clock can display different times "right now" depending on who is reading it.

 

[1977ub]

People don't always realize that you can't measure that slower speed of a moving clock unless you have confidence in remote simultaneity within your frame, as determined by Einstein synchronization for instance. The "dots" you connect to decide that the moving clock runs slow are particular pings at particular distances and times, and one applies a bit of mathematics to come up with the "slower clock".

 

[nitsuj]

nitsuj: I think why you said is true - the causal structure of space-time is not changed by a relative velocity. Let me ask you these two questions:

(1) Do you agree that what time the clock displays depends on when you read it?
(2) Do you accept that two observers in relative motion can disagree on the meaning of "now".

If you answer yes to these two questions, then you will see how the clock can display different times "right now" depending on who is reading it.

Yes definately, I had hoped my reply illustrated that the clock simply does what it does and people can observe what it is doing. And to touch on the idea of causal struture and a maximum speed being physically significant, while comparative measures of where/when are not.

I was being funny with the choice of grammar the OP used
"The Grandfather clock cannot display different times depending on who is reading it - for that is physically and logically impossible." That sentence suggests the clock will change its display depending on who is looking at it.

Also the last sentence in the post you referred to makes it clear that observers can read different times on the grandfather clock; as it is physically not significant if they differ.

 

[nitsuj]

So if I measure two people who are 6ft tall but one is further away, and so perspective means that I measure him as only half the height of the other (Let us say they are in floating in space and I view then through the window of my space craft) then those measurements tell me that he IS only half the height of the nearer one?

That's the thing with comparatives; you can compare anything you wish. Even apples to oranges. Or this coordinate chart where the clock reads 12:00 to that coordinate chart where the clock reads 18:00 at the same arbitrary now moment. And a comparative is not a measurement.

 

[Ibix]

Let us imagine a space station far away from any other body in space with an atomic grandfather clock, where the local time is displayed on a large clock face.
This clock, at the origin of the space station's Frame of Reference is keeping Proper Time.
<snip snip snip>
On the observer's clock, that traveling with him, and also keeping Proper Time, only 6 minutes will have passed..

Yet how can that be? For keeping proper time, both clocks will have be reading the same!

The answer of course is that the observer's measurement of the Grandfather clock's passage of time is COORDINATE time, which at 0.866c is twice as fast as Proper Time.

Grimble, you look to me to be confused about what proper time is. It is not some absolute time that clocks track. "Proper" in this context means "its own", from the Latin proprius. It's related to "appropriate". It does not mean "correct" or "true" or anything like that. A clock always shows proper time ("its own time") - that's more or less the definition of a clock. It does not mean that two clocks have to agree.

You may see texts telling you that proper time is an invariant quantity. This simply means that everyone can agree on what a given clock will show at some event. It does not mean that all clocks tell The Proper ("Right") Time - there's no such thing. For example, if the clock breaks down at the point that it reads 12 noon, clocks passing by at that very instant might show 1pm or 11am. However, everyone will agree that they expect the broken clock to read 12 noon at the time it broke down, based on its motion. As you noted - if that weren't the case something would be badly wrong.

 

[Grimble]

A very basic question, but how does one explain it to someone with this train of thought?

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of their own (lorentz factor 2)
If a neutral observer, C, is permanently mid-way between them then each will me moving at 0.433c (lorentz factor 1.1) and their clocks will each, therefore, be running at 0.9 the rate of the neutral observers clock.
So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.
And if we add two more observers, D & E, permanently positioned between A and C, and B and C, then each of A, B and C will be moving at 0.2165c relative to D and E... and so on.
Extending this line of logic we can prove that all clocks must be running at the same rate. The 'slowing' of fast moving clocks is all relative to the moving observer. It is an effect caused by their relative motion, where the relative speed/velocity is no more than a factor in that measurement.
All clocks run the same but are subject to the conditions (relative velocity) under which they are measured.
Which is why we talk of taking measurements from one Frame of Reference and transforming them by means of the Lorentz transformation equations to determine the measurement relative to a moving observer.

So how is it that we talk of clocks physically slowing? Of two clocks physically slowing each with respect to the other at the same time?

Note that this is a symmetrical situation. There is no acceleration. Any Length contraction would be symmetrical too.
Relativity of simultaneity would also seem to be irrelevant here as everything is reciprocal depending on where it is viewed from...

 

[Orodruin]

with standard clocks are traveling at 0.866c relative to one another,

C, is permanently mid-way between them then each will me moving at 0.433c

No. You are trying to use classical addition of velocities here. C will not see each of them moving at 0.433c.

So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.

Yes, what you need to look up is relativity of simultaneity. Different observers will not agree on which events are simultaneous. In C's frame, the events A's clock shows 1 PM and B's clock shows 1 PM are simultaneous. In the frames of A and B they are not.

Relativity of simultaneity would also seem to be irrelevant here

No, relativity of simultaneity is of very high importance here!

 

[nitsuj]

So how is it that we talk of clocks physically slowing; of two clocks physically slowing each with respect to the other at the same time?

Note that this is a symmetrical situation. There is no acceleration. Any Length contraction would be symmetrical too.
Relativity of simultaneity would also seem to be irrelevant here as everything is reciprocal depending on where it is viewed from...

You use the terms "reciprocal" & "symmetrical", and those terms require a unique concept in SR. "at the same time" or simultaneous is also a unique concept in SR for spatially separated events.

TL:DR - The observation that "two clocks physically slowing each with respect to the other at the same time" is physically meaningless. (assumption that they're spatially separated, otherwise one's broken )

Orodruin beat me to it

Reread the OP and there is also a misunderstanding of "reciprocal", as it lead to an incorrect assumption about RoS. Same could be said for "symmetry".

 

[harrylin]

A very basic question, but how does one explain it to someone with this train of thought?

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of their own (lorentz factor 2)
If a neutral observer, C, is permanently mid-way between them then each will me moving at 0.433c (lorentz factor 1.1) and their clocks will each, therefore, be running at 0.9 the rate of the neutral observers clock.
So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.
And if we add two more observers, D & E, permanently positioned between A and C, and B and C, then each of A, B and C will be moving at 0.2165c relative to D and E... and so on.
Extending this line of logic we can prove that all clocks must be running at the same rate. The 'slowing' of fast moving clocks is all relative to the moving observer. It is an effect caused by their relative motion, where the relative speed/velocity is no more than a factor in that measurement.
All clocks run the same but are subject to the conditions (relative velocity) under which they are measured.
Which is why we talk of taking measurements from one Frame of Reference and transforming them by means of the Lorentz transformation equations to determine the measurement relative to a moving observer.

So how is it that we talk of clocks physically slowing? Of two clocks physically slowing each with respect to the other at the same time?

Note that this is a symmetrical situation. There is no acceleration. Any Length contraction would be symmetrical too.
Relativity of simultaneity would also seem to be irrelevant here as everything is reciprocal depending on where it is viewed from...

I can't follow your reasoning, but perhaps that is due to a lack of definition of what you mean with "running at the same rate". According to all Galilean reference frames except the one in which the clock is not moving, the clock is "running at a slower rate".
Relativity of simultaneity is essential. In fact, the rate at which you "observe" a distant clock to be ticking (as well as the length contraction) is a function of your assumption of the one-way speed of light; simultaneity is defined such that light speed appears to be isotropic relative to your frame of choice.

The only sure and unambiguous way to compare clock rates is to compare clocks side by side. Other people who reasoned that all clocks must be running at the same rate, concluded that Einstein's calculation about an accelerating clock must be wrong (his calculation assumes that acceleration has no direct effect on clock rate):
"by the clock which has remained at rest the traveled clock on its arrival at A will be ½tv²/c² second slow".
- §4 of http://fourmilab.ch/etexts/einstein/specrel/www/

What does your line of reasoning give?

 

[Grimble]

OK, let us say that the relative speed of C to A and to B is v.

I can't follow your reasoning, but perhaps that is due to a lack of definition of what you mean with "running at the same rate". According to all Galilean reference frames except the one in which the clock is not moving, the clock is "running at a slower rate".

Reference https://www.physicsforums.com/threads/do-clocks-really-run-slow.810605/

So clocks, A & B, are running at the same rate as measured by C,

Reference https://www.physicsforums.com/threads/do-clocks-really-run-slow.810605/

What is your problem here? If C is permanently at the mid point of the line AB, then AC = BC and the speed of each relative to C will be the same - or the distances AC and BC would not continue to be equal.
If the speed of each relative to c is identical then their Lorentz factors will be the same and the slowing of those clocks, relative to C's clock will be the same, therefore clock A must read the same as clock B as measured by observer C. How is that not symmetry?

So although A and B both show 1pm observed by C, they show different times observed from each other. That is due to relativity of Simultaneity. Can you explain how that works differently for one than for the other? It seems like a reciprocal arrangement to me...

 

[nitsuj]

It seems like a reciprocal arrangement to me...

That's seems like a "privileged" perspective right? Analysis of all the results at the same time is different from when it is playing out in a continuum. This is the same perspective where you mentioned something happening "at the same time".

 

[Dale]

All ideal clocks run at a rate of one second per second of proper time (trivially $d\tau/d\tau = 1$ or barely less than trivially $c\,d\tau/ds=1$). Ideal clocks run at a rate of less than one second of proper time per second of coordinate time in any inertial frame in which they are moving ($d\tau/dt \le 1$).

 

[Nugatory]

From the initial post in this thread:

Extending this line of logic we can prove that all clocks must be running at the same rate.

I must confess that I'm not seeing the proof here. If you could spell it out in a bit more detail how to extend that line of logic to the conclusion that "all clocks must be running at the same rate?

I suspect that when you do, you will find that you are making an assumption:

If an observer finds that clocks A and B are both in motion relative to him and running at the same rate; and a second observer finds that clocks B and C are both in motion relative to him and running at the same rate then there exists some observer for whom all three clocks are running at the same rate.​

This is not correct, except in the uninteresting special case in which at least two of the three clocks are at rest relative to one another. However, every time that you add another clock "in between" (BTW, have you noticed that the physical positions of the clocks is irrelevant? They don't need to go "in between", they can go anywhere as long as they have the right relative velocities) that clock is in motion relative to the clocks on either side of it as well as all the other clocks, so this special case doesn't apply.

 

[Vanadium 50]

Grimble, I expect this is going to go in the exact same direction as all your other threads on this. Making increasingly complicated scenarios, guessing at the solution, and then debating it until the thread gets locked hasn't helped in the past, and is unlikely to help if we try it one more time.

A different approach is needed - taking advantage of physics being a quantitative science. You need to explicitly state what calculation would convince you that clocks run slow, and then we can do it together. For example, X observes Y's clock to read t3 when X's reads t1, and observes Y's clock to read t4 when X's clock reads t2. X will say Y's clock runs slow if (t4-t3)/(t2 - t1) < 1.

Issues about symmetries need to be applied in the solution of the problem, not the posing of the problem. As this (and the other threads) show, starting on the solution while still posing the question only adds to the confusion. Give us a clearly posed question.

 

[pervect]

A very basic question, but how does one explain it to someone with this train of thought?

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of
So how is it that we talk of clocks physically slowing? Of two clocks physically slowing each with respect to the other at the same time?

I would recommend the thread "Symmetrical time dilation implies the relativity of simultaneity", https://www.physicsforums.com/threa...on-implies-relativity-of-simultaneity.805210/

Symmetrical time dilation is what I call the situation where A says B's clock is slow, and B says A's clock is slow. This notion is not consistent with the notion of absolute time.

I rather strongly suspect here that the issue is that Grimble is implicitly assuming the existence of absolute time. This thread is my best attempt to show why absolute time is incompatible with symmetrical time dilation, and how what's known as "the relativity of simultaneity" fixes the issue.

 

[harrylin]

[..] if the 'stationary' observer changed his speed there is no connection to the 'moving' clock [... omit part that I cannot follow...] It can only be that the MEASUREMENT BY the remote observer is affected.
The clocks don't change only the measurements? That can't be right surely?

I agree with that part of your reasoning: the measurements are done with clocks and rulers.

[..]
So although A and B both show 1pm observed by C, they show different times observed from each other. That is due to relativity of Simultaneity. Can you explain how that works differently for one than for the other? It seems like a reciprocal arrangement to me...

Explaining such a complex arrangement with mere words is complicated and will be difficult to do without ambiguity. The best explication is with numbers - and it avoids wasting time on long discussions. Please prepare an example calculation. Then if you do not already find the answer from it yourself, you can present it here.

 

[Ibix]

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of their own (lorentz factor 2)
If a neutral observer, C, is permanently mid-way between them then each will me moving at 0.433c (lorentz factor 1.1) and their clocks will each, therefore, be running at 0.9 the rate of the neutral observers clock.
So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.

As has been noted before, the observer C would not be traveling at $\sqrt{3}c/4$ because velocities do not add linearly in a relativistic universe. It turns out that, if A is stationary in some frame in which B is approaching at $\sqrt{3}c/2$, then the observer C must be approaching A at $c/\sqrt{3}$ so that in C's rest frame, both A and B are approaching at the same speed in opposite directions.

So, here is a Minkowski diagram of that setup, as seen by C:

Apologies for the lack of labelling. A Minkowski diagram is basically a displacement-time graph, except that time runs vertically upwards and x position is shown horizontally. The vertical gridlines are one light second apart; the horizontal ones are one second apart.

The world-line of C is the purple line right up the middle. C is at rest in this frame, so the x position never changes. It starts at the origin at time zero (marked by a solid purple square) and stays right there.

The world-line of A is represented by the red line. It starts at x=-8ls at time zero (marked by a solid red square), and moves with velocity $c/\sqrt{3}$ - so as time goes on (up the page) it gets closer to C, and eventually meets at the top of the diagram.

The world-line of B is in blue, and is a mirror image of the world-line of A. It starts at x=8ls and moves with velocity $-c/\sqrt{3}$ and reaches C at the same time as A does.

As you noted, the situation is symmetrical as viewed from his frame. I've also added markers to each line showing when, according to C, clocks moving with A, B and C would tick. The purple ticks are every second; the red and blue ticks occur at the same time as each other, but you can see that they are spaced out more - their clocks tick slower in this frame.

But what does this look like according to A? We can use the Lorentz transforms to calculate the coordinates of each of the points in the Minkowski diagram as seen from a frame in which A is at rest and re-draw the diagram. That turns out to look like:

You can see that this time A (in red) is stationary, so the position is always the same, giving a vertical line. You can see that the red clock ticks are spaced exactly the same as the grid - they are 1s apart. You can see that there are the same number of ticks on each line as there were in the first diagram, and you can see that this time the purple clock is running slow and the blue clock is running really slow. You can also see why this isn't a problem - in this frame the three observers did not start their clocks at the same time.

So, yes, the clocks are running slow - it's not an illusion (although it is only an effect of co-ordinate choice). The different observers disagree about the time the clocks were started so that there is no contradiction when they meet up and can unambiguously compare clocks.

 

[Grimble]

To be sure that I can follow your explanations, that with the number of replies can become a little confusing with so many different replies to consider, let me check a few basic premises to be sure that I am correctly appreciating the concepts.
1. The view of space from any Inertial Frame of Reference is at rest; i.e. there are a set of axes that give a fixed set of coordinates for any point in space.
2. A full set of four coordinates specifying a fixed point in space, at a specific point in time, constitutes an event.
2. Every event can be mapped, that is given a unique set of coordinates, in any Inertial Frame of Reference.
3. Events, being fixed in time cannot move.
4. There must be a fixed point with a set of unique spatial coordinates, midway between any two events, in every Inertial Frame of Reference.
5. If light from two events, arrive at the fixed point midway between them at the same time, (which constitutes a single event), then those events were simultaneous. (Einsteins definition of simultaneity).

 

[harrylin]

Probably you meant it like that, but just to make sure I added some words:

5. If light from two events, arrive at the fixed point midway between them at the same time, (which constitutes a single event), then those events were simultaneous according to that inertial reference system. (Einsteins definition of simultaneity).

PS. this is just the example of Einstein here: http://www.bartleby.com/173/9.html

Personally I find it easier to consider the inverse: a light bulb in the middle of a train gives off a flash of light in both directions. The times that the light rays are thought to reach the ends of the train depend on the used reference system.

 

[Grimble]

Thank you, yes that is correct. I just wanted to be sure that I was understanding this.

So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?
And there would be a set of spatial coordinates that would define a fixed point in each and every inertial frame of reference that was spatially mid-way between points A and B?
And that the lights emitted at event A and event B would meet at that said mid-point?
If the lights from A and B arrive together, as a single event and be so measured by an observer at rest at the mid points between event A and event B, would that not imply that they could be measured to be simultaneous, in any, indeed in all, inertial frames of reference?
However it must be at a different mid-point in each frame, as the stationary mid-point in one frame would be moving away in any other frame.
So the observer in any frame would declare that he was the only one who could determine simultaneity...

Now, this is my problem: what is it I am getting wrong here? That all seems so very clear and logical and no matter how many times I have gone over it, I cannot see it!
Please help me and explain it?

 

[Mentz114]

So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?

(my emphasis)
Events are events. They exist, therefore they exist in all frames. If I meet someone at the station at 8pm that is an unchangeable physical fact. My worldline intersected with the other partys worldline. Everyone will agree that we met, furthermore everyone will agree that our clocks read 8pm when we met.

 

[Nugatory]

those two events A and B would occur in every inertial frame of reference?

Some of your confusion may be that you're thinking that things happen "in" frames. They don't. They happen, and then if it is convenient we assign them time and space coordinates, and a frame is just an arbitrarily chosen convention for assigning these coordinates. When you hear someone saying something like "this event happened in frame F at time t and position x", that's a convenient but sloppy shorthand for the more precise "There's this event. We can choose any frame we want to assign time and space coordinates to that event, and if we choose frame F, we'll end up assigning time coordinate t and space coordinate x to it, but of course if we had chosen a different frame we would have assigned different x and t values".

An event is like a pencil mark on a piece of paper - it's there whether we draw a set of coordinate axes on the paper or not. If it's convenient to describe the location of the point using x and y coordinates, we can draw an x-axis and a y-axis on the sheet of paper as well to create a "frame", and we can choose to draw the axes anywhere we please.

 

[Dale]

And that the lights emitted at event A and event B would meet at that said mid-point?

This can only be true in one frame. In all other frames the light emitted at A and B would not meet at the midpoint.

 

[Ibix]

Bear in mind that in at least one frame, the light sources are moving. What does "half way between" the sources mean when they don't emit simultaneously and have moved between one emission event and the other? Do you mean that (assuming that the sources are at opposite ends of a rod) the midpoint of that rod? In that case the pulses will cross there but, because the rod is moving, that's not the same as the point half way between the spatial locations of the emission events.