How to resolve the contradiction in twin clocks?

[xinhangshen]

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?

 

[Nugatory]

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

This is a classic relativity of simultaneity problem. You're asking the question "When the first observer sees his ball at y=1, where is the second observer's ball at the same time?" and then if the answer to that question is "y' < 1" you're asking "When the second observer see his ball at y=y', where is the first observer's ball at the same time?" Because the two observers do not agree about simultaneity the answer to the second question is not "y=1" but there is no inconsistency.

 

[xinhangshen]

Thanks, Nugatory for your quick response.

If one observer staying with clock 1 to observe the positions of the balls of the two clocks, he will be able to get that the positions of both balls are the same: y1 = y2. You may say that the observer can't see the position of the ball of clock 2 instantly, i.e., there is a time delay. But the time delay is known and can be added back. For example, if we use light as a media to see the position of the ball of clock 2, the time delay will be x2/c which can be added to the the observed position of the ball of clock 2 to make y2 = y1. This is because in the same reference frame, there is no effect from special relativity which only influences values after the reference frame is changed.

When the positions of both clocks in the frame attached to clock 1 are converted to the positions of the balls of the two clocks in the frame attached to clock 2, according to Lorentz Transformation, they will keep the same values, that is, y1' = y1, and y2' = y2. Then, y1 = y2 = y1' = y2'. That is, the time of each clock will remain the same value in both reference frames, which contradicts to the time conversion formula in the Lorentz Transformation.

In this classical debate, it seems that the observer can only measure static objects but no moving objects. That's not true. In special relativity, everything static or moving can be measured in any inertial reference frame just like in a Newtonian reference frame.

 

[Nugatory]

That is, the time of each clock will remain the same value in both reference frames, which contradicts to the time conversion formula in the Lorentz Transformation.

Don't just assert that, try calculating it. The two relevant events are:
Event 1: Observer 1's ball reaches the point x=0,y=1,t=1 using a reference frame in which observer 1 is at rest;
Event 2: Observer 2's ball reaches the point x'=0,y'=1,t'=1 using a reference frame in which observer 2 is at rest.

Lorentz transform those coordinates and see what you get.
(And be sure that you use the Lorentz transformations NOT the time dilation formula)

 

[Dale]

which contradicts the time conversion formula in the Lorentz Transformation.

Of course it contradicts the time dilation formula. The time dilation formula is a simplification of the Lorentz transform for use only when a clock is at rest in one of the frames. The clock you have described is not at rest in either frame so the time dilation formula does not apply and you must use the full Lorentz transform.

You are trying to use a formula where it doesn't apply. Of course you get a mistake.

 

[xinhangshen]

OK, here is what we get according to Lorentz Transformation:

t' = r(t - vx/c2)
x' = r(x - vt)
y' = y
where r = 1/(1 - v2/c2)^1/2

at time 0 in the reference frame attached to clock1:
The position of the ball of clock1: x1 = 0, y1 = 0
The position of the ball of clock2: x2 = 0, y2 = 0
then, at time t in this frame
The position of the ball of clock1: x1 = 0, y1 = t
The position of the ball of clock2: x2 = vt, y2 = t

Now let's convert the events to the moving frame attached to clock2 according to Lorentz Transformation:
(t, x1, y1) => (t1', x1', y1'):
t1' = r(t - vx1/c2) = rt
x1' = r(x1 - vt) = -rvt
y1' = y1 = t
which contradicts the definition of the clock1: y1' = t1'
Similarly,
(t, x2, y2) => (t2', x2', y2'):
t2' = r(t -vx2/c2) = rt(1 - v2/c2) = t/r
x2' = r(x2 - vt) = 0
y2' = y2 = t
which also contradicts the definition of the clock2: y2' = t2'

 

[xinhangshen]

Of course it contradicts the time dilation formula. The time dilation formula is a simplification of the Lorentz transform for use only when a clock is at rest in one of the frames. The clock you have described is not at rest in either frame so the time dilation formula does not apply and you must use the full Lorentz transform.

You are trying to use a formula where it doesn't apply. Of course you get a mistake.

In our case, the clock is not moving but the part of the clock is moving just like a rotating clock on which the needles are moving. As it just tries to record time, the ball of the clock can move at a very slow speed and the time dilation caused by that part can be ignored just like in a rotating clock.

 

[Dale]

In our case, the clock is not moving

If the clock is not moving then your above description is wrong. Specifically y1≠t and y2≠t. Those equations may only be used if the clock is moving.

Actually, now that I notice it there is another problem. If you are using c=1 then that is the equation of a clock moving at c in the y direction, which is not possible. If you are not using c=1 then the units are inconsistent. Maybe that is the source of your problem.

 

[xinhangshen]

If the clock is not moving then your above description is wrong. Specifically y1≠t and y2≠t. Those equations may only be used if the clock is moving.

Actually, now that I notice it there is another problem. If you are using c=1 then that is the equation of a clock moving at c in the y direction, which is not possible. If you are not using c=1 then the units are inconsistent. Maybe that is the source of your problem.

In the twin clocks system, assuming the ball of a clock moves at a speed of 1 just try to make the formula simple. Of course we can use a variable u to represent the speed of the ball. In that situation, we have:

at time 0 in the reference frame attached to clock1:
The position of the ball of clock1: x1 = 0, y1 = 0
The position of the ball of clock2: x2 = 0, y2 = 0
then, at time t in this frame
The position of the ball of clock1: x1 = 0, y1 = ut
The position of the ball of clock2: x2 = vt, y2 = ut

Now let's convert the events to the moving frame attached to clock2 according to Lorentz Transformation:
(t, x1, y1) => (t1', x1', y1'):
t1' = r(t - vx1/c2) = rt
x1' = r(x1 - vt) = -rvt
y1' = y1 = ut
which contradicts the definition of the clock1: y1' = ut1'
Similarly,
(t, x2, y2) => (t2', x2', y2'):
t2' = r(t -vx2/c2) = rt(1 - v2/c2) = t/r
x2' = r(x2 - vt) = 0
y2' = y2 = ut
which also contradicts the definition of the clock2: y2' = ut2'

The speed u of the ball of a clock is nothing to do with the speed of light. It can be any value.

 

[Janus]

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?

y'= y, but the actual positions of the balls with respect to time will not be the same for each frame.

Look at it this way, each clock has a velocity of v relative to the other, the ball has a velocity of u' in the y direction as measured in the frame of its clock.

What this means is that relative to clock 1 the ball in clock 2 follows a diagonal path as a result of both its motion with respect to clock 2 and Clock 2's motion with respect to clock 1.

To find the speed of ball 2 relative to clock 1, we need to use relativistic velocity addition as it applies to orthogonal velocities.

When U and V are speeds expressed as fractions of c (U'= u'/c and V= v/c), this becomes

S = \sqrt{V^2+U&#039;^2-V^2U&#039;^2}

With a little trig, we can find U, which is the speed of ball 2 in the y direction as measured by clock 1.

U = \sqrt{S^2-V^2}

U= \sqrt{V^2+U&#039;^2-V^2U&#039;^2 -V^2}

U= \sqrt{U&#039;^2-V^2U&#039;^2}

U=U&#039; \sqrt{1-V^2}

\frac{U}{U&#039;}= \sqrt{1-V^2}

Thus the speed of of ball 2 in the y direction is not the same as that of ball 1 as measured by clock 2. (U' is the speed each clock measures its own ball as moving in the y direction and U is the speed that it measures the other clock's ball as moving in the y direction.)

Furthermore, since V=v/c, the ratio in the difference in measured speeds is the same as the time dilation factor.

 

[Dale]

Now let's convert the events to the moving frame attached to clock2 according to Lorentz Transformation:
(t, x1, y1) => (t1', x1', y1'):
t1' = r(t - vx1/c2) = rt
x1' = r(x1 - vt) = -rvt
y1' = y1 = ut

Yes.

which contradicts the definition of the clock1: y1' = ut1'

Therefore the definition of the clock is incompatible with the Lorentz transform. I.e. A clock cannot work that way.

Similarly,
(t, x2, y2) => (t2', x2', y2'):
t2' = r(t -vx2/c2) = rt(1 - v2/c2) = t/r
x2' = r(x2 - vt) = 0
y2' = y2 = ut

Yes.

which also contradicts the definition of the clock2: y2' = ut2'

Therefore the definition of the clock is incompatible with the Lorentz transform. I.e. A clock cannot work that way.

 

[Samshorn]

Of course we can use a variable u to represent the speed of the ball. In that situation, we have:
at time 0 in the reference frame attached to clock1:
The position of the ball of clock1: x1 = 0, y1 = 0
The position of the ball of clock2: x2 = 0, y2 = 0
then, at time t in this frame
The position of the ball of clock1: x1 = 0, y1 = ut
The position of the ball of clock2: x2 = vt, y2 = ut

That's not right. You said the clocks are identical, so each ball moves (in the y or y' direction) at the speed u relative to the rest frame of the respective clock. This implies that the ball in clock2 is moving with velocity u*sqrt(1-v^2) in terms of the reference frame attached to clock1.

By the same token, the ball in clock1 is moving with speed u*sqrt(1-v^2) in terms of the reference frame attached to clock2.

 

[harrylin]

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?

y' = y in that description refers to the transformation of fixed system position coordinates of those systems - for example the coordinates of the rulers that are in rest in each inertial system. Therefore, it does not refer to the positions of balls that are moving along y and y'.

A similar case can be found in Einstein's 1905 derivation: y'=y does not refer to the positions of light rays along y and y' at time t (see "An analogous consideration" in §3 of http://www.fourmilab.ch/etexts/einstein/specrel/www/ ).

 

[xinhangshen]

y'= y, but the actual positions of the balls with respect to time will not be the same for each frame.

Look at it this way, each clock has a velocity of v relative to the other, the ball has a velocity of u' in the y direction as measured in the frame of its clock.

What this means is that relative to clock 1 the ball in clock 2 follows a diagonal path as a result of both its motion with respect to clock 2 and Clock 2's motion with respect to clock 1.

To find the speed of ball 2 relative to clock 1, we need to use relativistic velocity addition as it applies to orthogonal velocities.

When U and V are speeds expressed as fractions of c (U'= u'/c and V= v/c), this becomes

S = \sqrt{V^2+U&#039;^2-V^2U&#039;^2}

With a little trig, we can find U, which is the speed of ball 2 in the y direction as measured by clock 1.

U = \sqrt{S^2-V^2}

U= \sqrt{V^2+U&#039;^2-V^2U&#039;^2 -V^2}

U= \sqrt{U&#039;^2-V^2U&#039;^2}

U=U&#039; \sqrt{1-V^2}

\frac{U}{U&#039;}= \sqrt{1-V^2}

Thus the speed of of ball 2 in the y direction is not the same as that of ball 1 as measured by clock 2. (U' is the speed each clock measures its own ball as moving in the y direction and U is the speed that it measures the other clock's ball as moving in the y direction.)

Furthermore, since V=v/c, the ratio in the difference in measured speeds is the same as the time dilation factor.

Hi Janus, your explanation is perfect according to special relativity with graceful mathematics which is also the reason so many people like and believe special relativity.

However, in the real physical world, we don't have an abstract clock to tell the exact time in an inertial reference frame. The twin clocks I mentioned here are the clocks that the observers use to check the time. That is, the position of the ball on the ruler is already calibrated as the time of the clock. If the position of the ball of clock1 is exactly the same as the position of the ball of clock2, then the observers will agree that they are at the same time. Special relativity just tries to reduce the speed by a factor and increase the time by the same factor to produce the same position of the ball of a clock. Thus, the final result is the same as that of a Newtonian mechanics.

Does this example indicate that the effects of special relativity can't be noticed in the real physical world or there is no need to have special relativity at all?

 

[Dale]

Does this example indicate that the effects of special relativity can't be noticed in the real physical world or there is no need to have special relativity at all?

No, this example demonstrates that the real world doesn't work the way you think it should. In the real physical world relativistic effects are definitely noticeable, the physical evidence is overwhelming.

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[xinhangshen]

No, this example demonstrates that the real world doesn't work the way you think it should. In the real physical world relativistic effects are definitely noticeable, the physical evidence is overwhelming.

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

So, can you explain how the two observers notice the time dilation in the twin clocks here?

 

[Dale]

So, can you explain how the two observers notice the time dilation in the twin clocks here?

Look at your own results:
t2' = r(t -vx2/c2) = rt(1 - v2/c2) = t/r
And
y2' = y2 = ut
So
y2' = u r t2'

 

[Nugatory]

So, can you explain how the two observers notice the time dilation in the twin clocks here?

Let's try some numbers.

Assume the two observers (call them A and B) are moving apart at .6c so $\gamma$ is 1.25. The balls are set up to move in the y direction at 1 m/sec, both observers have placed their balls at y=0 and started them moving as they passed each other at the origin.

After 1.6 seconds of his time, A sees (light reaches his eyes) the position of B's ball after one seconds of A's time has passed and B is .6 light-seconds away. At this point A reasons as follows:
- The light took .6 seconds to traverse the .6 light-seconds that separated us.
- Therefore I am seeing the state of B's ball and rod as it was .6 seconds ago.
- At that moment, .6 seconds ago, my ball was at the position y=1m (calculated in either of two ways: "after 1 sec my ball will have moved 1 meter", or "my ball is now at 1.6 m so .6 seconds ago it was at 1meter).

However, A will observe that B's ball is at .8 meters - that is, B's ball was at .8 when the light that A received at time 1.6 left B's ball. (This is the Lorentz transformation calculation that we've been begging you to make since the beginning of this thread).

Because distance in the y direction is unaffected by the transformation, A concludes that while one second of his time passed, only .8 seconds of B's time passed. This is time dilation it's symmetrical - B can reason in the exact same way about A.

 

[xinhangshen]

Look at your own results:
t2' = r(t -vx2/c2) = rt(1 - v2/c2) = t/r
And
y2' = y2 = ut
So
y2' = u r t2'

That is, y2' = u' t2' where u' = u r, which has exactly the same format as in y. Therefore, the observer will think the clock is still the same as its twin clock.

Since the observer uses the position of the ball of the clock as the corresponding time (not the abstract time t2' which can never be measured without a physical clock), using the time of this clock (i.e. the position of its ball) to measure speeds of other objects will produce results not different from those measured on the other reference frame. Actually, this is the common way we are doing in all experiments. Thus, it means that in physical world we can't notice the time dilation.

 

[Samshorn]

In the real physical world, we don't have an abstract clock to tell the exact time in an inertial reference frame. The twin clocks I mentioned here are the clocks that the observers use to check the time. That is, the position of the ball on the ruler is already calibrated as the time of the clock.

Yes, for each twin the position of the ball on his co-moving ruler is a measure of the time coordinate of a system of inertial coordinates in terms of which he is at rest. For consistency, the two twins would use something like identical guns (at rest in their respective frames) to shoot the balls, so they have the same speeds in terms of their respective co-moving inertial coordinate systems.

If the position of the ball of clock1 is exactly the same as the position of the ball of clock2, then the observers will agree that they are at the same time.

That doesn't follow logically, and, as a matter of empirical fact, it isn't true. To understand this, you need to be more clear and precise about what it means to say that two spatially separate events "are at the same time". You claim that two specific events, namely (a) ball1 reaches y* (a specific value of y), and (b) ball2 reaches y*, occur "at the same time". But that's an ambiguous statement, and it's actually false in the sense that you have in mind. To be clear and accurate, you should say:

If (x,y,t) are inertial coordinates (defined just as Newton would have defined them) in which twin1 is at rest, and if (X,Y,T) are inertial coordinates in which twin2 is at rest, and if t1* is the time when ball1 reaches y*, and T2* is the time when ball2 reaches y*, then t1* = T2*. That is a true statement (assuming they used identical guns, for example, to fire the balls).

But you are claiming something very different. Let t2* denote the time when ball2 reaches y*, and let T1* denote the time when ball1 reaches y*. You are asserting that t1* = t2*, and that T1* = T2*. That would be true in Newtonian physics, but it happens to be empirically false, essentially because all forms of energy have inertia (something which Newton didn't realize) and consequently relatively moving inertial coordinate systems are not related to each other in the way Newton thought. This is clearly explained in any good book on relativity.

 

[Dale]

That is, y2' = u' t2' where u' = u r, which has exactly the same format as in y. Therefore, the observer will think the clock is still the same as its twin clock.

You can certainly notice that $u\ne u'$. The fact that the form is the same doesn't negate the fact that the value is different, they will not think it is the same.

 

[xinhangshen]

You can certainly notice that $u\ne u'$. The fact that the form is the same doesn't negate the fact that the value is different, they will not think it is the same.

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

Generally we should be aware that there is no abstract time in real physical world. All we have just motions of different objects. We use the motion of one object as the standard motion to define time with which to measure the motion of the observed object to see how fast the observed object is moving. According to special relativity, the abstract time dilation will influence all the motions in the same way in an inertial reference system. Then, the measured result will never been changed no matter what time dilation the reference frame has. For example, if we define a time unit as the time during which the standard object moves one distance unit, the speed of the observed object will be the distance the observed object has moved in such a time unit. for example, 5 distance units/time unit, which will never been changed by so called time dilation. Therefore, the time dilation is completely redundant.

 

[Samshorn]

You do not have a clock to measure the speed u' because the time is defined by the clock.

The speed u' is measured by clocks at rest with respect to twin1, located all along the path of twin2, and synchronized in the frame of twin1 in accord with the Newtonian definition of inertial coordinates.

 

[Dale]

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

This is a patently silly objection. You can take any type of clock and miscalibrate it and, if you don't compare it to anything else, then you won't know. This is trivial.

 

[xinhangshen]

Let's try some numbers.

Assume the two observers (call them A and B) are moving apart at .6c so $\gamma$ is 1.25. The balls are set up to move in the y direction at 1 m/sec, both observers have placed their balls at y=0 and started them moving as they passed each other at the origin.

After 1.6 seconds of his time, A sees (light reaches his eyes) the position of B's ball after one seconds of A's time has passed and B is .6 light-seconds away. At this point A reasons as follows:
- The light took .6 seconds to traverse the .6 light-seconds that separated us.
- Therefore I am seeing the state of B's ball and rod as it was .6 seconds ago.
- At that moment, .6 seconds ago, my ball was at the position y=1m (calculated in either of two ways: "after 1 sec my ball will have moved 1 meter", or "my ball is now at 1.6 m so .6 seconds ago it was at 1meter).

However, A will observe that B's ball is at .8 meters - that is, B's ball was at .8 when the light that A received at time 1.6 left B's ball. (This is the Lorentz transformation calculation that we've been begging you to make since the beginning of this thread).

Because distance in the y direction is unaffected by the transformation, A concludes that while one second of his time passed, only .8 seconds of B's time passed. This is time dilation it's symmetrical - B can reason in the exact same way about A.

The problem in your reasoning is that you have already assumed special relativity is correct and set the y-direction speed of the ball of B in A's static reference frame to be smaller than the y-direction speed of the ball of A (i.e. uB = uA/$\gamma$). With this speed difference, of course, you will get the time difference (or more directly the y position difference).

 

[russ_watters]

The problem in your reasoning is that you have already assumed special relativity is correct...

Um...that's how science works. You assume a theory is correct and make predictions with it, then see if those predictions match reality.

 

[xinhangshen]

Um...that's how science works. You assume a theory is correct and make predictions with it, then see if those predictions match reality.

If the time dilation is created like this, Newtonian mechanics can also say there is no time dilation at all. If you set the speed of the ball of B in A's reference frame to 1 m/sec (equal to A's speed), then no difference of the positions of the balls in y-direction will be produced. Then, Newton can still claim his theory is valid.

 

[WannabeNewton]

...

 

[Nugatory]

The problem in your reasoning is that you have already assumed special relativity is correct and set the y-direction speed of the ball of B in A's static reference frame to be smaller than the y-direction speed of the ball of A (i.e. uB = uA/$\gamma$). With this speed difference, of course, you will get the time difference (or more directly the y position difference).

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).

 

[russ_watters]

If the time dilation is created like this, Newtonian mechanics can also say there is no time dilation at all. If you set the speed of the ball of B in A's reference frame to 1 m/sec (equal to A's speed), then no difference of the positions of the balls in y-direction will be produced. Then, Newton can still claim his theory is valid.

Except that when you actually test it, you find that Newton's predictions are wrong and Relativity's predictions are right.

 

[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.