How to resolve the contradiction in twin clocks?

[Dale]

The problem is that you do not have a clock to measure the speed u' because the time is defined by the clock.

My previous response was a little rushed and therefore incomplete. Let me expand a bit on why it is an utterly silly objection.

Suppose you had a pendulum clock and nothing else. And suppose that you thought that your pendulum was calibrated for 1s per tick, but your pendulum length was a different length, so it was actually measuring a different amount of time. Then, as long as that is the only thing in the universe then you will never know that it is keeping the wrong time. Just like with the y-axis clock.

Now, suppose that you want to use your pendulum clock to measure decay rates or calculate accelerations and forces or even bake bread. You will find that your measured half life is off, the accelerations and forces are different than what you expected, and the bread is burnt. Just like with the y-axis clock.

From that evidence you will be able to tell that your pendulum clock doesn't keep correct time. It will be physically discernable that the laws of physics are not correctly described with your clock. Just like with the y-axis clock.

So the idea that you could be tricked into thinking that your pendulum clock because there is no other reference to compare it to requires that you not use your clock for ANYTHING, because anything that you do will let you know that it is miscalibrated. You can't even just sit there and watch it tick because of all of the biological processes involved in watching and living that it will get wrong. Just like with the y-axis clock.

So, your objection is absurd because (a) it applies to any clock (b) it relies on an absurd level of ignorance. A clock for which that objection would stand would necessarily be so disconnected from the rest of the universe that you can simply ignore anything about it.

 

[Dale]

Then, Newton can still claim his theory is valid.

No, he cannot:
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

Please read the overwhelming experimental evidence. Once you understand that SR has been experimentally verified and Newtonian physics experimentally contradicted, then you must accept that SR works. You will likely still be confused about how it works, but you at least will not be under the delusion that it is optional for explaining the world.

This is incredibly important. SR was not adopted because people like it, most everybody is bothered by it when they learn it. SR was adopted because the experimental evidence forces us to adopt it.

 

[xinhangshen]

Well, of course I have assumed that special relativity is correct because your original question, all the way back in post #1, was "Can anybody give me an explanation how to resolve the contradiction?" - and there is no contradiction if you assume that special relativity is correct.

The objection that you're raising in the bolded text above suggests that you're starting with a misunderstanding of SR, and that that misunderstanding is leading you to see a contradiction where there is none.

We have started with two identically constructed ball on rod devices. Because they are identically constructed and subject to the same laws of physics, they must operate identically in frames at which they are at rest: B's experience with his ball-rod device is not affected by the fact that A is moving away from him at .6c, just as A's experience with his own ball-rod device is not affected by the fact that B is moving away from him at .6c. This is basically the first postulate of special relativity, and before you reject it out of hand, you might want to consider what would be different (nothing!) for A or B if the other one were suddenly to disappear completely.

Because these are identically constructed devices subject to the same laws of physics, we can be confident that they behave identically in frames in which they are at rest. (We can also verify this by bringing them back together, resetting the position of the ball to zero, and watching them operate side by side with no relative speed).

Therefore, we know that both balls are advancing at a speed of one meter per second as viewed by the observers who are at rest relative to them. In fact, in the theory of relativity, they ARE clocks (Google for Einstein's phrase "time is what a clock measures" and understand what it means), and both A and B can read the current value of their t coordinate from the position of their ball on their rod.

The fact that A sees B's ball moving at .8 the speed of his own, and vice versa, is just another way of saying that B's time is dilated relative to A and vice versa. Both of them agree that their own ball is moving at one meter per second.

(Come to think of it... If I were to bend the rods into circles in the y-z plane, the balls would be describing circles, just as if they were dots on the tip of the hand of a mechanical clock... I could even paint little numbers, 1 through 12 along the rods... and then the clockiness of the bar-rod devices would become even more apparent).

Yes, when you bend the rod of Clock B into a circle will definitely make the understanding easier. Now we can see that the real measurement of time is the distance (or angle), but not the time itself. This distance or angle is the result of speed multiplied by time. In special relativity, the speed is amplified by a factor γ and the time is scaled down by the factor γ. Then, the result of the multiplication will always be the same before or after transformation. That is, the position of the pointer of Clock B in A's reference frame (ut) will be the same as the position of Clock B in B's reference frame (u't') no matter at what speed Clock B moves away from Clock A because u' = γu and t' = t/γ. Therefore, just reading the clock, the two clocks will always point at the same position (i.e. the time).

Using this time to measure motions in B's reference frame will be exactly the same as in A's reference frame. Therefore, everything follows Newton's laws. You will never notice the effects of special relativity, and special relativity is just an unnecessary manipulation of Newton's mechanics.

 

[Dale]

Therefore, just reading the clock, the two clocks will always point at the same position (i.e. the time).

No, it is only the time in one frame, it is r times the time in the other frame. I.e. it points to the same position, but that position is not the time in both frames.

 

[Nugatory]

Yes, when you bend the rod of Clock B into a circle will definitely make the understanding easier. Now we can see that the real measurement of time is the distance (or angle), but not the time itself.

I could be mistaken, but on reading this it sounds as if you have not yet followed through on my suggestion that you try to understand exactly why Einstein said "Time is what a clock measures".

A and B are both measuring time by the distance (or angle) that their ball moves on the rod. Those are their clocks.

Because A and B are separated by .6 light-seconds (still using my example from a few posts back) the only way that A can know what B's clock reads at the moment that A's clock reads one second is to wait until A's clock reads 1.6 seconds so that the light hitting his eyes was emitted from B when A's clock read one second. The theory of special relativity and a respectable amount of experimental evidence tell us that B's clock will read .8 when the light that reaches A when A's clock reads 1.6 leaves B.
(and likewise if we switch A and B - the situation is completely symmetrical).

 

[xinhangshen]

I could be mistaken, but on reading this it sounds as if you have not yet followed through on my suggestion that you try to understand exactly why Einstein said "Time is what a clock measures".

A and B are both measuring time by the distance (or angle) that their ball moves on the rod. Those are their clocks.

Because A and B are separated by .6 light-seconds (still using my example from a few posts back) the only way that A can know what B's clock reads at the moment that A's clock reads one second is to wait until A's clock reads 1.6 seconds so that the light hitting his eyes was emitted from B when A's clock read one second. The theory of special relativity and a respectable amount of experimental evidence tell us that B's clock will read .8 when the light that reaches A when A's clock reads 1.6 leaves B.
(and likewise if we switch A and B - the situation is completely symmetrical).

You are right when you set the speed of the ball of clock B at 1/$\gamma$ m/sec in A's reference frame.

Now if I set the speed of the ball of both clocks to 1 m/sec in A's reference frame, then observer at A will see the position of the ball of clock B is at 1 meter in A's reference frame after 1.6 seconds. Deducting the time for light to travel, it perfectly matches the position of the ball of clock A. Now let's transform the time-space point of the ball of B from A's reference frame to B's reference frame, we will get

(t, xB, yB) => (tB', xB', yB')
xB' = $\gamma$(xB - vtB)
tB' = $\gamma$(tB - vxB/c2)
yB' = yB

if t = 1 sec and $\gamma$ = 1.25, then xB = 0.6c, yB = ut = 1x1 = 1 m which corresponds to

xB' = 0
tB' = t/$\gamma$ = 0.8 second
yB' = 1 m = u'tB' (i.e., u' = 1.25 m/sec)

In this case, you will see the speed of the ball of B is increased by a factor of $\gamma$ while time is decreased by the same factor $\gamma$. Now let's have a look at a real clock. We can only use the position of the pointer of a clock to tell the time, not the time itself. If we are on a rocket and people ask you what time it is now, you will just use the angle of the pointer of your watch to tell the time. If the hourly arm has an angle of 30 degrees, you will say it's 1 o'clock, 60 degrees, 2 o'clock, etc. You will never ask people, "Wait, please tell me the speed of the rocket first as I have to calculate the new speed of the arm before I can tell you the time."

That is, in the physical world, we always use the position to represent time, while the position is the multiplication of speed and time. If the speed increased by a factor $\gamma$ and time decrease by the same factor $\gamma$, the clock will never notice the change. Then, using such a clock, we will never notice any effects of special relativity.

I would like you to rethink of it deeply with a completely open mind!

 

[Nugatory]

Now if I set the speed of the ball of both clocks to 1 m/sec in A's reference frame then...

That is physically impossible if the two clocks are identically constructed. If they are identically constructed, then the speed of B's ball in a frame in which B and his clock are at rest must be the same as the speed of A's ball in a frame in which A is at rest.

 

[pervect]

I really don't understand what you think your point is.

That is, in the physical world, we always use the position to represent time. While the position is the multiplication of speed and time. If the speed increased by a factor γ and time decrease by the same factor γ, the clock will never notice the change. Then, using such a clock, we will never notice any effects of special relativity.

Your statement isn't literally true - we could use a number of things besides position to represent time if we wanted to. In fact it's routine to see clocks with LED readouts that represent time by electronic states. You probalby own several of them.

I'm not sure why you are limiting yourself to saying that only position can possibly represent time. It's not true - it's not a good sign that your argument starts with an untrue point :-(.

Focusing on measuring time only via position makes the problem only slightly more complex but apparently it makes it just complex enough that you can confuse yourself :-(

If you can imagine a radio receiver and an electronic clock, that can encode the current time and broadcast it via a radio signal (such a clock is encoding time by a means other than position, obviously), then we can make the problem so simple that it would be more difficult to get the wrong answer.

We have two observers, A and B, moving by each other. At time T=0, both are at the same place, and we reset both their clocks.

At time T A emits a timestamped radio signal that encodes the message:"Clock A, time=T". At some time k*T, k being the doppler shift factor, B receives the signal and broadcasts a reply, which says. "Received signal from clock A=T. Time of reception B=k*T"

Because of relativity, the doppler factor k for sending a signal from A to B is the same as the doppler factor k' for sending a signal from B to A

Thus, at a time k^2, clock A receives the above signal from B

Using the fact that the speed of light is a constant, "c", knowing that the signal was sent at T and that the echo/retransmission arrived at k^2*T, A concludes that the time in A"s frame at which the rebroadcast occurred is (1+k^2)*T/2, exactly halfway between the time of transmission and the time of reception.

If this isn't immediately obvious, drawing a space-time diagram can help.

This is obviously different than the reading of B's clock, which was k*T. Hence we know that A's clock and B's clock cannot keep time at the same rate. This is independent of exactly HOW we encode time, whether we do it electronically, with an analogue readout, or via any other means.

This is a short outline of the derivation of the Lorentz transforation using Bondi's K-calcululs approach. One source of this is Bondi's book "Relativity and common sense". It's one of the simplest approaches to SR, requiring only high school algebra.

 

[Mentz114]

We can only use the position of the pointer of a clock to tell the time, not the time itself.

What does this mean ?

You fail to understand that relativity is not about clocks, but time itself. We define a clock to be that which measures the proper interval along a worldline, like an odometer measures the spatial interval.

If your ball 'clock' does not measure the proper interval it is not a clock, however you care to present it.

I would like you to rethink of it deeply with a completely open mind!

Your argument is based on the conviction that SR is wrong or irrelevant, together with ignorance of the meaning of SR. I suggest you do some learning before trying to do something that many great minds have failed to do.

 

[ghwellsjr]

I am pretty confused in the following situation:

Two identical clocks moving at a constant speed v from each other in x-direction. If each clock is made up of a ball moving at a constant speed of 1 on a ruler in y-direction, then the position of the ball of a clock is the time of the clock. According to special relativity, y' = y no matter at what speed the two inertial reference frames move away from each other. Thus, the two clocks will always have the same time in both reference frames if they start from the same time at the same position, which contradicts the time conversion formula in the Lorentz Transformation.

Can anybody give me an explanation how to resolve the contradiction?

Since you are not being persuaded by any of the many excellent answers that you have been given, I would like to try a different tact which is to recast your scenario into a much simpler, but maybe equivalent scenario in hopes that it might get down to the core of your issue.

You have constructed a clock defined from the point of view of a stationary ruler and against which a ball moves at a constant speed. The time is read off by the markings on the ruler adjacent to the ball.

Now I would like you to consider a similar clock from the point of a stationary ball and against which a ruler moves at a constant speed. The time is again read off by the markings on the ruler adjacent to the ball.

We are not concerned with the issue of whether these two clocks tick at the same rate, only that they are physically equivalent clocks, working on the same physical mechanism. Agreed?

Now I'd like to consider that we have two of our ruler/ball clocks oriented identically and traveling towards each other at some arbitrary but constant speed. As they pass each other, an observer located at the conjunction of the two balls notes that they display the same time on their respective rulers.

Now the question is: will the observer continue to actually see the times on the two ruler/ball clocks remaining equal to each other? What would Newton say? What do you say? And why?

 

[xinhangshen]

Since you are not being persuaded by any of the many excellent answers that you have been given, I would like to try a different tact which is to recast your scenario into a much simpler, but maybe equivalent scenario in hopes that it might get down to the core of your issue.

You have constructed a clock defined from the point of view of a stationary ruler and against which a ball moves at a constant speed. The time is read off by the markings on the ruler adjacent to the ball.

Now I would like you to consider a similar clock from the point of a stationary ball and against which a ruler moves at a constant speed. The time is again read off by the markings on the ruler adjacent to the ball.

We are not concerned with the issue of whether these two clocks tick at the same rate, only that they are physically equivalent clocks, working on the same physical mechanism. Agreed?

Now I'd like to consider that we have two of our ruler/ball clocks oriented identically and traveling towards each other at some arbitrary but constant speed. As they pass each other, an observer located at the conjunction of the two balls notes that they display the same time on their respective rulers.

Now the question is: will the observer continue to actually see the times on the two ruler/ball clocks remaining equal to each other? What would Newton say? What do you say? And why?

In this situation, the observer standing at the middle of the two clocks will always see the two clocks have exact the same time no matter whether you use special relativity or Newtonian mechanics because of the symmetry.

Here, your clocks are completely equivalent to my clocks in telling time. Actually, Nugatory's circle rod clock is also equivalent to our clocks. Nugatory's clock is actually the traditional mechanical clock. Since special relativity says that the speed of the hourly arm will increase by the factor of $\gamma$ if the clock moves, then the mark pointed by this arm will represents different time when the speed at which the clock moves is different. That means, the time the clock shows is incorrect once the clock moves. Therefore, special relativity leads to the conclusion that all mechanical clocks can't work correctly if they are moving.

 

[Nugatory]

Since special relativitysays that the speed of the hourly arm will increase by the factor of $\gamma$ if it moves, then the mark pointed by this arm will represents different time when its speed is different. That means, they are incorrect once it moves. Therefore, according to special relativity, all mechanical clocks can't work correctly if they are moving.

It's not just mechanical clocks, it's all time-dependent physical processes. Instead of the moving ball or the hands of a clock, we could use the melting of a block of ice or the evaporation of water in a bowl, the progressive decay of a sample of radioactive material...

But I think we may have found the source of your underlying confusion. Special relativity says that the physics must remain consistent whether we say that A is at rest while B is moving at a speed v relative to A; or B is at rest while A is moving in the other direction. Thus, SR does not allow us to say that a clock is right "until it starts moving" - every clock is always moving relative to some observers and at rest relative to others, always.

 

[Dale]

we always use the position to represent time, ...

I would like you to rethink of it deeply with a completely open mind!

I urge you to have an open mind also, particularly keep an open mind as you read through the experimental evidence which supports SR and contradicts Newtonian physics. Your own derivation shows that there are relativistic effects for the "y-axis" clock.

However, I take issue with your statement that we always use position to represent time. It is not correct, and even when it is correct it is always some sort of cyclical position. There is no example that I am aware of where a clock measures time in the way you have described. You are not making a general analysis of clocks here.

 

[xinhangshen]

Now Let us concentrate on resolving the above contradiction. As I mentioned, the clocks used in my thought experiment are just general physical clocks that can be most accurate atomic clocks but just have a special way to display the time (actually you can use the circular traditional display for the clocks as well but need two coordinates: y and z positions). Since we always have y = y' and z = z' in Lorentz Transformation, then we have the contradiction:

if you use the display (i.e. y position for my clocks or y and z positions for circular clocks) as the time in Special Relativity, then the displayed time is an invariant in Lorentz Transformation which contradicts the time conversion formula in Lorentz Transformation;

if you say that the display is not the time in Special Relativity because the ball moving speed or the arm rotating speed has been changed after Lorentz Transformation which makes the displayed time on general physical clocks incorrect in Special Relativity, then the time in Special Relativity becomes mysterious and Special Relativity is no longer a theory of physics.

DaleSpam, could you please give an explanation to resolve the contradiction?

 

[Nugatory]

if you use the display (i.e. y position for my clocks or y and z positions for circular clocks) as the time in Special Relativity, then the displayed time is an invariant in Lorentz Transformation which contradicts the time conversion formula in Lorentz Transformation;

You are (still) confusing two things:
- Proper time, which is invariant and what the position of the clock's hands (or the progress of any physical process: fraction of a radioactive sample that has decayed between two observations, number of oscillations of a cesium atom between two observations, number of my hairs which have turned gray between two observations) measures.
- Coordinate time, which is different for different observers using different coordinate systems (also known as "frames of reference"). The Lorentz transformations describe how to convert one observer's coordinates, including coordinate time, to another observer's coordinates in a way that preserves the laws of physics and especially ensures that the relationship between the position of the hands of the clock and each observers' coordinate time is consistent with the physical process moving the hands of the clock.

Edit: an exercise that you might find helpful would be to state what it means to say that two events at two different locations are simultaneous without using the words "at the same time". Once you do that, you can apply it to the two events "First clock's hand is pointing straight down" and "Second clock's hand is pointing straight down".

 

[harrylin]

Now Let us concentrate on resolving the above contradiction. As I mentioned, the clocks used in my thought experiment are just general physical clocks that can be most accurate atomic clocks but just have a special way to display the time (actually you can use the circular traditional display for the clocks as well but need two coordinates: y and z positions). Since we always have y = y' and z = z' in Lorentz Transformation, then we have the contradiction:

if you use the display (i.e. y position for my clocks or y and z positions for circular clocks) as the time in Special Relativity, then the displayed time is an invariant in Lorentz Transformation which contradicts the time conversion formula in Lorentz Transformation; [..]

That's erroneous, as I explained in post #13 already:
https://www.physicsforums.com/showthread.php?p=4439896

In a nutshell, the moving y positions do not correspond to the fixed y coordinates of the inertial frames of the Lorentz Transformation - it's that simple!

 

[Dale]

As I mentioned, the clocks used in my thought experiment are just general physical clocks

No, they are not. Clocks measure proper time along their worldline. The "clocks" used in your thought experiment do not. They also are not equivalent to any clock I am aware of, certainly they cannot be called "general physical clocks".

if you use the display (i.e. y position for my clocks or y and z positions for circular clocks) as the time in Special Relativity, then the displayed time is an invariant in Lorentz Transformation which contradicts the time conversion formula in Lorentz Transformation;

The measured value is an invariant on any measurement apparatus. For clocks, that means that the displayed time is invariant I.e. proper time is invariant. This is well-known and not at all in contradiction with relativity.

The relativistic effect is that the frame invariant proper time is only equal to the frame variant coordinate time for an inertial frame where the clock is at rest.

DaleSpam, could you please give an explanation to resolve the contradiction?

There is no contradiction.
1) your "clocks" are not clocks
2) proper time is invariant
3) coordinate time is not invariant
4) your "clocks" don't violate any relativistic effects, the velocity of your "clock" is different in different frames, transforming according to the Lorentz transform exactly as it should, as shown above.

 

[nitsuj]

if you say that the display is not the time in Special Relativity because the ball moving speed or the arm rotating speed has been changed after Lorentz Transformation which makes the displayed time on general physical clocks incorrect in Special Relativity,
then the time in Special Relativity becomes mysterious and Special Relativity is no longer a theory of physics.

What's the physical significance of this understanding you have anyways? An observation has no consequence.

Misunderstood or not. The "contradiction" is conceptual, not physical. So your musing is no longer about physics.

SR has a postulate that "builds in" all mechanical physics as it applies to motion.

A ruler in comparative motion is not a "proper" ruler, same goes for the clock.

those two statements are all that need to be said for the above.

 

[Dale]

Here is how to completely work this problem and show that there is no contradiction. Consider a frame where the x velocity of the "xinhangshen clock" is 0 and the y velocity is k (in units where c=1). In this frame the worldline of the "xinhangshen clock" is given by $(t,x,y,z)=(t,0,kt,0)$. The display on the "xinhangshen clock" would read $t_{xinhangshen}=t=y/k$. A physical clock traveling with the "xinhangshen clock" would read $t_{physical}=\sqrt{t^2-(kt)^2}=t \sqrt{1-k^2} = t_{xinhangshen} \sqrt{1-k^2}$. So we see immediately that the proposed "xinhangshen clock" does NOT represent a general physical clock and does NOT keep proper time as a standard physical clock.

Now, if we transform to a primed reference frame moving at velocity v in the x direction wrt the unprimed frame. Then we find that the worldline of the clock is $(t',x',y',z')=\left(\frac{t}{\sqrt{1-v^2}},\frac{vt}{\sqrt{1-v^2}}, k t, 0\right) = (t',t'v, t' k \sqrt{1-v^2},0)$. Thus, at a time t' the clock reads a time $t_{xinhangshen}=y/k=y'/k=t'\sqrt{1-v^2}$. So even though the "xinhangshen clock" does not keep proper time, it still time dilates as expected. Furthermore, the physical clock would read $t_{physical}=\sqrt{t'^2-(t'v)^2-(t'k \sqrt{1-v^2})^2}=t' \sqrt{1-k^2}\sqrt{1-v^2} = t_{xinhangshen} \sqrt{1-k^2}$, so the error between the "xinhangshen clock" and the physical clock is the same in both frames.

 

[JM]

Clocks and Light

I am pretty confused in the following situation:

Two identical clocks moving at a constant speed v from each other in x-direction. If each clock is made up of a ball moving at a constant speed of 1 on a ruler in y-direction, then the position of the ball of a clock is the time of the clock. According to special relativity, y' = y no matter at what speed the two inertial reference frames move away from each other. Thus, the two clocks will always have the same time in both reference frames if they start from the same time at the same position, which contradicts the time conversion formula in the Lorentz Transformation.

Can anybody give me an explanation how to resolve the contradiction?

You assert two identical clocks in relative motion. Einsteins Special Relativity also is based on identical clocks in relative motion. His transforms, and the formula, result from his Light Postulate, which states that the speed of light is independent of the motion of the source. Its the light that causes the time differences given by the time dilation formula. A good example of how this works is given by Feynman in ' Six not-so-easy Pieces' pages 59-63.

 

[ghwellsjr]

What's the physical significance of this understanding you have anyways? An observation has no consequence.

I thought all of physics was based on observations. What do you mean when you say they have no consequence?

Misunderstood or not. The "contradiction" is conceptual, not physical. So your musing is no longer about physics.

SR has a postulate that "builds in" all mechanical physics as it applies to motion.

A ruler in comparative motion is not a "proper" ruler, same goes for the clock.

I thought every ruler measures proper length and every clock measures proper time. What do you mean by these statements?

those two statements are all that need to be said for the above.

 

[Samshorn]

I thought ... every clock measures proper time.

What do you mean by that? The literal readings on an actual clock need not correspond to the proper time along the worldline of the clock. (Actual clocks run fast or slow.) Do you just mean that an ideal clock, i.e., a device constructed in such a way as to read the proper time along its worldline, will read the proper time along its worldline? True enough, but circular. Or do you mean that the temporal state of a clock will progress in proportion to proper time, even though this may not correspond to the literal readings on the clock? If so, then it doesn't need to be a "clock", you could just as well refer to any physical system, and then you would need to say what is meant by "temporal state", which you can't define with reference to elapsed proper time or it is circular.

Telling people that "every clock measures proper time" is not good, because it could only be literally true if we simply defined 'proper time' to be whatever any clock reads, which of course would be utterly incoherent. That's what beginners tend to think you must mean, which totally sends them down the wrong track.

There is a non-circular way of correctly saying what you are probably trying to say, but it's quite a bit more subtle and complicated than just saying "every clock reads proper time".

 

[ghwellsjr]

Telling people that "every clock measures proper time" is not good, because it could only be literally true if we simply defined 'proper time' to be whatever any clock reads, which of course would be utterly incoherent. That's what beginners tend to think you must mean, which totally sends them down the wrong track.

You don't understand--I am a beginner and most of what I've learned is from experts like these:

As Ibix pointed out, every clock measures its own proper time.

In SR and GR there is a well-defined mathematical procedure to calculate proper time for moving objects along trajectoreis through spacetime (as measured by a co-moving clock).

I prefer to define "proper time" mathematically, as a property of a timelike curve in spacetime, and then take one of the axioms that define SR to be "A clock measures the proper time of the curve in spacetime that represents its motion".

Proper time is measured by a single clock and can be used only for events that occur locally, right next to the clock.

The proper time, in both SR and GR, is the time actually measured by a single clock.

So I think you will have to agree, I'm not just sending beginners down the wrong track, I've been sent down there with them.

There is a non-circular way of correctly saying what you are probably trying to say, but it's quite a bit more subtle and complicated than just saying "every clock reads proper time".

So then why don't you say it in a way that all of us beginners can understand?

 

[Dale]

True enough, but circular. ...

There is a non-circular way of correctly saying what you are probably trying to say, but it's quite a bit more subtle and complicated than just saying "every clock reads proper time".

I would be interested to hear it because personally I think the statement you are objecting to is fine.

 

[WannabeNewton]

I agree. This is unequivocally a pointless exercise in semantics. See the following passage from Wald: http://postimg.org/image/aq9amdtkh/

Also see here (from MTW): http://postimg.org/image/s16qrlpnj/

 

[atyy]

I agree. This is unequivocally a pointless exercise in semantics. See the following passage from Wald: http://postimg.org/image/aq9amdtkh/

I think Samshorn has a point that a clock is defined as a device that reads proper time. The theory says such a device can be made since proper time is a coordinate-independent quantity along a worldline.

An example of a "clock" that does not read proper time is a pendulum.

In our modern age, we have the luxury of defining atomic vibrations as clocks, then it is indeed derived, not defined, that those read proper time.

 

[Enigman]

Greetings,
Okay, crash course of SR
All laws of physics hold true in inertial frames- no observer can tell if (s)he is moving.
There is NO correct time, time flows differently for different inertial frames; ie. All observers disagree on the matter of time and length and as far their frame is concerned they are ALL correct.
Now proper time is time measured by a non accelerated clock which pases through both events-this is the closest thing SR has to "correct" time - but pretty irrelavent to this post.
Proper length (not "correct" length -as all observers believe they are correct...) is length measured in the frame where object being measured has zero relative velocity.
Okay coming back to the post :
let the clocks start with 2#u# relative velocity in opposite directions from origin along x axis

For the observer at origin both clocks would show the same time as both have same magnitude of velocity wrt to O viz. #u#. But then if there were to be a similarly synced clock at O it would show a different time as the other clocks (note: magnitude of oscillation would be same ie. y=y' time taken to reach max displ. would differ ie. Time period)

Let's take another observer who is at rest wrt to one of the clock let's say A
To that observer the origin is moving away with velocity #u#
- and hence the time period of clock at O would greater than that clock A. This time period will increase by the same factor that O thinks A -clock has increased by. As for the other clock at let's say B, will have still greater time period as it moves at2#u#

B will have the same oppinion about A.
The calculations have aldready been done by Janus

Appologies; if there are any mistakes I've started SR only in the last weekend (that too by a book which calls Gallileo the father Of modern physics and talks about Einstein in present tense...)

Regards

 

[ghwellsjr]

I think Samshorn has a point that a clock is defined as a device that reads proper time. The theory says such a device can be made since proper time is a coordinate-independent quantity along a worldline.

An example of a "clock" that does not read proper time is a pendulum.

In our modern age, we have the luxury of defining atomic vibrations as clocks, then it is indeed derived, not defined, that those read proper time.

I don't understand your post. First you agree with the idea "that a clock is defined as a device that reads proper time" and then you say that it is "not defined" that clocks "read proper time".

Also, Special Relativity cannot account for gravity and therefore it cannot account for a pendulum clock.

 

[ghwellsjr]

Greetings,
Okay, crash course of SR

You've done quite well for only having started on SR since last weekend. However, there are a few "mistakes".

All laws of physics hold true in inertial frames- no observer can tell if (s)he is moving.
There is NO correct time, time flows differently for different inertial frames; ie. All observers disagree on the matter of time and length and as far their frame is concerned they are ALL correct.
Now proper time is time measured by a non accelerated clock which pases through both events-this is the closest thing SR has to "correct" time - but pretty irrelavent to this post.

This description is related to the spacetime interval of a timelike pair of events and is one example of Proper Time but it completely misses the point of Proper Time which is that an accelerated clock which passes through the same two events will accumulate a different Proper Time. This is the whole point of the so-called Twin Paradox--two clocks start at the same event with the same Proper Time and then take different paths through spacetime (at least one accelerates) and finally end up at the second event with different Proper Times on them.

Proper length (not "correct" length -as all observers believe they are correct...) is length measured in the frame where object being measured has zero relative velocity.

This also is too restrictive. In this situation, the Proper Length is equal to the Coordinate Length but even when the object is moving, its Proper Length can be measured by a ruler that is comoving with it even though both of them are not equal to the Coordinate Length.

Okay coming back to the post :
let the clocks start with 2#u# relative velocity in opposite directions from origin along x axis

For the observer at origin both clocks would show the same time as both have same magnitude of velocity wrt to O viz. #u#. But then if there were to be a similarly synced clock at O it would show a different time as the other clocks (note: magnitude of oscillation would be same ie. y=y' time taken to reach max displ. would differ ie. Time period)

Let's take another observer who is at rest wrt to one of the clock let's say A
To that observer the origin is moving away with velocity #u#
- and hence the time period of clock at O would greater than that clock A. This time period will increase by the same factor that O thinks A -clock has increased by. As for the other clock at let's say B, will have still greater time period as it moves at2#u#

A will not see or measure the speed of B to be 2#u# but something less as determined by the relativistic velocity addition formula (or by applying the Lorentz Transformation process to the different scenarios).

B will have the same oppinion about A.
The calculations have aldready been done by Janus

If you're going to mention something like this, it would be nice if you would provide a link or reference.

Appologies; if there are any mistakes I've started SR only in the last weekend (that too by a book which calls Gallileo the father Of modern physics and talks about Einstein in present tense...)

Regards

 

[Nugatory]

I think Samshorn has a point that a clock is defined as a device that reads proper time. The theory says such a device can be made since proper time is a coordinate-independent quantity along a worldline.

An example of a "clock" that does not read proper time is a pendulum.

In our modern age, we have the luxury of defining atomic vibrations as clocks, then it is indeed derived, not defined, that those read proper time.

I've seen this discussion before, and to me it always comes down to people (myself included) not being completely clear about the distinction between coordinate time along the worldline of an observer at rest at the spatial origin of a coordinate system and proper time along the same worldline. The distinction between the two usually isn't very useful; we generally try to choose coordinate systems in which the value of the time coordinate for an inertial observer following a given worldline is the same as proper time; or equivalently $g_{tt}$ expressed in that coordinate system is equal to 1 along that worldline.

The readings of an ideal clock give us both coordinate time in that coordinate system and proper time; as WbN points out above they're equal so discussing which the clock is measuring is sterile.

On the other hand, a non-ideal clock still provides a perfectly good time coordinate; it labels each point on that worldline with a unique value and with appropriate choice of simultaneity convention will supply a time coordinate for points off that worldline as well. All that's going on is that the imperfections of the clock are encoded in the value of $g_{tt}$ along its inertial worldline - when the metric tensor is expressed in coordinates in which the clock is providing the t coordinate. This can still be a perfectly flat spacetime; the non-unity metric components are compensating for the less than ideally simple choice of coordinates..