Galileo's pendulum could show that time is relative

[simplex]

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

Making experiments with pendulums Galileo discovered the following properties:
- Pendulums nearly return to their release heights.
- 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.

 

[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.

 

[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.

 

[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.

 

[russ_watters]

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.

 

[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.

 

[Snip3r]

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

i agree with this but

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

can you tell me what the above one is? and what nonexistent Earth means?

 

[russ_watters]

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

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

 

[simplex]

can you tell me what the above one is? and what nonexistent Earth means?

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.

 

[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'?

 

[simplex]

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'?

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).

 

[Snip3r]

do you really think pendulum will swing in outer space?

 

[simplex]

do you really think pendulum will swing in outer space?

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.

 

[Vorde]

\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.

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

 

[simplex]

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

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.

 

[Vorde]

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.

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.

 

[simplex]

Of course a pendulum swings faster in a centrifuge ... This is not time dilation, but ... an error.

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.

 

[Dadface]

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.

 

[Snip3r]

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.

let me keep it simple. if you are stationary in outer space and there is no massive body around a pendulum won't swing what's your comment on that?time doesn't exist there?

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

NO,as altitude increases time moves faster

 

[simplex]

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.

I have already answered all this objections.

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 length L', the vectorial time will be different.

 

[Vorde]

Ok, but you do realize the theory has no physical application except for predicting the period of a pendulum at differing heights?

 

[russ_watters]

Put another way: do you realize that a pendulum clock and an atomic clock will not show the same variation?

 

[simplex]

Ok, but you do realize the theory has no physical application except for predicting the period of a pendulum at differing heights?

It is more than that.

Exemple:
Let's suppose we have two massive bodies 1 and 2 and a pendulum, somewhere in their vecinity.

The total vectorial time the pendulum sees is:

T(r,t) = k(r,t)*sqrt(L) + k1*r*sqrt(L) + k2*(r-r12)*sqrt(L) [eq.1]

where:
r = the position vector of the pendulum suspension point
r12 = the position vector of the center of the second body relative to the center of the first body.
k1*r*sqrt(L) = vectorial time induced by the massive body no. 1
k2*(r-r12)*sqrt(L) = vectorial time induced by the massive body no. 2
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.

Now we put two conditions:
a) T(r,t) = 0 (the pendulum does not swing)
b) k(r,t)*sqrt(L) = 0 (the suspension point of the pendulum either do not move or move in such a way that the pendulum does not swing)

with (a) and (b), [eq.1] turns into:
k1*r*sqrt(L) + k2*(r-r12)*sqrt(L) = 0 [eq.2]

To simplify, we will consider k1 = k2 (the two massive bodies are identical) which plugged into [eq.2] will lead to the solution:

r = 0.5*r12

In other words, the pendulum should be in between the two massive bodies (Lagrange point 1).

 

[Vorde]

Well I'll start off by saying I don't understand what you mean when you say 'total vectorial time', but assuming I'm following what you're saying (which I'm not totally, but that's my fault not yours), this seems at best to be simply a way to figure out which body out of two is pulling on the pendulum faster, and to add to that, only if this is some mathematically idealized pendulum, as a physical representation of this would not work at all.

 

[D H]

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.

You are talking nonsense.

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