# How to resolve the contradiction in twin clocks?

1. Jul 8, 2013

### 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?

2. Jul 8, 2013

### Staff: Mentor

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.

3. Jul 8, 2013

### 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.

Last edited: Jul 8, 2013
4. Jul 8, 2013

### Staff: Mentor

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)

Last edited: Jul 8, 2013
5. Jul 8, 2013

### Staff: Mentor

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.

6. Jul 8, 2013

### 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'

7. Jul 8, 2013

### xinhangshen

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.

Last edited: Jul 8, 2013
8. Jul 8, 2013

### Staff: Mentor

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.

9. Jul 8, 2013

### xinhangshen

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.

Last edited: Jul 8, 2013
10. Jul 8, 2013

### Janus

Staff Emeritus
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

$$S = \sqrt{V^2+U'^2-V^2U'^2}$$

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

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

$$U= \sqrt{V^2+U'^2-V^2U'^2 -V^2}$$

$$U= \sqrt{U'^2-V^2U'^2}$$

$$U=U' \sqrt{1-V^2}$$

$$\frac{U}{U'}= \sqrt{1-V^2}$$

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.

11. Jul 8, 2013

### Staff: Mentor

Yes.

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

Yes.

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

12. Jul 8, 2013

### Samshorn

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.

13. Jul 9, 2013

### harrylin

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/ ).

14. Jul 9, 2013

### xinhangshen

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?

15. Jul 9, 2013

### Staff: Mentor

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

16. Jul 9, 2013

### xinhangshen

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

17. Jul 9, 2013

### Staff: Mentor

t2' = r(t -vx2/c2) = rt(1 - v2/c2) = t/r
And
y2' = y2 = ut
So
y2' = u r t2'

18. Jul 9, 2013

### Staff: Mentor

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.

19. Jul 9, 2013

### xinhangshen

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.

20. Jul 9, 2013

### Samshorn

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.

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.