# Galileo's pendulum could show that time is relative

1. Dec 28, 2011

### simplex

Galileo's pendulum could have shown that local time is relative

Making experiments with pendulums Galileo discovered the following properties:
- All pendulums eventually come to rest with the lighter ones coming to rest faster.
- The period is independent of the bob weight.
- The period is independent of the amplitude.
- The square of the period varies directly with the length.
(T=k*sqrt(L))
in a time when the notion of gravity did not exist.

Later experiments done before 1687 showed that k is a function of altitude.

In other words local time passes with different speeds at different altitudes, being slower and slower as the height increases.

If more precise measurements could have been performed then T would have appeared as a vectorial quantity given by the expression:
T=k'*r*sqrt(L)
where
T = pendulum period
k' = a constant
r = the distance from the center of the earth to the pendulum
L = the length of pendulum string

A pendulum rotating around the center of an nonexistent Earth would have indicated a local time:
T1(r) = k1(r)*sqrt(L)

A pendulum rotating around the Earth would have indicated a local time:
T2(r) = k1(r)*sqrt(L) + k'*r*sqrt(L)

If T2(r) is always measured as being zero then
k1(r) = - k'*r

In conclusion, something looking like the theory of relativity would have been possible 300 years ago if somebody really wanted to devise such a theory.

Last edited: Dec 28, 2011
2. Dec 28, 2011

### Vorde

Not really. You could have devised an equation that gave the time it takes for a pendulum to swing with altitude as a variable. But unless you take the time a pendulum takes to swing as the absolute messenger of time (which nobody does), it gives no information on time. Its like pushing a pendulum so it swings faster and saying that the rate of time has increased.

3. Dec 29, 2011

### simplex

That pendulum indicates local time in comparison with absolute time which could have been shown by the positions of Jupiter satellites or by other things.

If one pushes a pendulum it will go faster for half a period but after that it will become an ordinary pendulum. Anyway, a pendulum pushed or perturbed somehow is no longer the pendulum of Galileo but something else.

4. Dec 29, 2011

### Vorde

My point is only that this seems only to be important if one takes the movement of a pendulum (that for instance has a period of 1 second) to be an absolute fundamental representative of time (1 second in this case), which is silly.

5. Dec 29, 2011

### Staff: Mentor

Just for clarity, that variation in k does not match with time dilation. It is a much larger factor. A pendulum clock is nowhere near precise enough to measure time dilation. Furthermore, comparing a pendulum clock to the orbits of Jupiter's moons won't show that time is relative, it just shows that the pendulum clock is wrong.

6. Dec 29, 2011

### simplex

The variation of k is not the time dilatation from the Theory of Relativity (1916) but a different time dilatation.

It is not absolutely necessary to compare the pendulum periods at altitudes r0 and r1 with what the universal clock, formed by Jupiter and its moons, shows.
Using a telescope one can, at least theoretically, compare the period of a pendulum at altitude r1 to the period of an identical pendulum placed at the reference altitude r0.

No matter how relative a quantity is, in any kind of theory of relativity, not necessarily the one from 1916, if you have no means to compare that quantity with anything else there is no way to know the quantity is relative.

7. Dec 29, 2011

### Snip3r

i agree with this but
can you tell me what the above one is? and what nonexistent earth means?

8. Dec 29, 2011

### Staff: Mentor

No! It is not a time dilation, it is just an error, just like heating or cooling a pendulum clock will produce an error.

9. Dec 29, 2011

### simplex

A pendulum placed on a rocket that moves in empty space, far away from any planet, will be characterized by a period T1(r) which depends on how that rocket moves and it is a function of position. This is the meaning of: T1(r) = k1(r)*sqrt(L)

There are two kinds of vectorial local times: one induced by massive bodies, which has a k(r) of the form k'*r and the second induced by the way the pendulum suspension point is forced to move, which this time is just a k(r) and can not be put in the particular form k'*r excepting some cases like circular movement around a planet or about a fixed point, even if the pendulum does not circle any material body, but empty space.

10. Dec 30, 2011

### Snip3r

T=k'*r*sqrt(L)
here r is the distance measured from the center of the earth
but in
T1(r) = k1(r)*sqrt(L)
what is 'r'?

11. Dec 30, 2011

### simplex

Both r are the distance measured from the center of the earth. r is the position vector of the pendulum.
In general, you can choose the origin of r anywhere if the pendulum moves in empty space far away from any celestial bodies. If the pendulum moves around the earth, it is convenient to select the origin of r as the center of the earth, otherwise the vectorial time induced by a massive body (earth) will no longer be given by T=k'*r*sqrt(L) but by a slightly more complicated expression:
T=k'*(rp-re)*sqrt(L)
where:
rp = position vector of pendulum relative to an arbitrary point in space
re = position vector of the center of the earth relative to the same arbitrary point

The position vector of the pendulum relative to the center of the earth will be
r = (rp-re).

12. Dec 30, 2011

### Snip3r

do you really think pendulum will swing in outer space?

13. Dec 30, 2011

### simplex

Yes, as long as the laws of mechanics give a precise expression for the period of a pendulum swinging anywhere. The pendulum can swing very fast even if the nearest massive body is 100 million km away. It depends on how the suspension point of the pendulum moves.

Last edited: Dec 30, 2011
14. Dec 30, 2011

### Vorde

I don't think so, simply on the basis of common sense.

15. Dec 30, 2011

### simplex

Inside the cockpit of a centrifuge, the type used to train jet fighter pilots, a pendulum swings faster. This is experimentally demonstrated. Not only it moves quicker but it swings in a plan nearly parallel to the earth surface, if the rotation speed of the centrifuge is not excessively low, which means that movement induces a supplemental vectorial time component different from that induced by a massive body.

16. Dec 30, 2011

### Vorde

The idea of a centrifuge is to simulate a higher gravitation field. In fact (and someone correct me if I am extending my knowledge too far here), I believe one of the cornerstones of General Relativity is that an acceleration and gravity are two ways of looking at the same thing.

Of course a pendulum swings faster in a centrifuge, it is getting pulled 'downwards' by a greater force. This is not time dilation, but as russ_watters said, an error.

17. Dec 30, 2011

### simplex

You name it error.
I name it vectorial local time.
Most of the people know that:
T = T (a,g) = 2*pi*sqrt(L/|a+g|).
where g = local gravitational acceleration and a = acceleration due to the movement.

But, I repeat, you could not talk about g in the time of Galileo because the concept did not exist.

An equivalent theory to the one Newton developed could have been devised earlier, at least in theory, using the concept of vectorial local time and no forces.

Do not think in terms of Newtonian mechanics because this theory of vectorial local time, measured with pendulums, it is not based on Newtonian concepts regarding gravity.

18. Dec 30, 2011

Simplex,
Imagine using two different types of watch to measure the time period of a pendulum at the place where you are now and then repeating your experiment at a place where the gravitational field strength is very much weaker.You will not detect any time dilation effects,all you will show is that things accelerate less and pendulums swing more slowly in weaker gravitational fields.

19. Dec 30, 2011

### Snip3r

let me keep it simple. if you are stationary in outer space and there is no massive body around a pendulum wont swing whats your comment on that?time doesn't exist there?
NO,as altitude increases time moves faster

20. Dec 30, 2011

### simplex

You are talking about the concept of gravitational fields.
You are talking about the concept of time dilatation from the Theory of Relativity (1916).
I do not use the two concepts. I am talking about a theory that could have been built before 1687 without them and that just looks like a theory of relativity.

The concept of vectorial local time induced by massive bodies, I have mentioned, is defined by the mathematical relation:T(r) = k'*r*sqrt(L) and that is all.

I have already mentioned that in order to measure T at a certain altitude you need a universal time indicator that can be Jupiter and its satellites.
So, any point in space is characterized by 2 kind of times: (1) the universal time, t, that is a scalar and the same for every point and (2) the vectorial local time which varies in space and also is dependent, for the general case, of the time, t.
For a pendulum moving somehow in the vecinity of an earth like body the total vectorial time will be:

T(r,t) = k(r,t)*sqrt(L) + k'*r*sqrt(L)

where:
k'*r*sqrt(L) = vectorial time induced by the massive body
k(r,t)*sqrt(L) = vectorial time induced by the movement of the pendulum suspension point
T(r,t) = total vectorial time in which the pendulum lives.

This local vectorial time is quite relative because it also depends on L. If you change the pendulum and use one with a string of lenght L', the vectorial time will be different.