A Newtonian Argument for Spacetime Curvature

[Murat Ozer]

Hello,
I have found an argument based on Newtonian physics
proving that spacetime must be curved. Consider two identical
simple pendulums (clocks) of length L and bob mass m.
Let one pendulum be on the ground and the other at a height h
above it. For small oscillations (for simplicity only, not necessary
for the argument) the periods of the pendulums are

T(0) = 2PiSqrt(L)/Sqrt(g_0) (on the ground), (1)

T(h) = 2PiSqrt(L)/Sqrt(g_0)*(1 + h/R) = T(0)*(1 + h/R), (2)

where g_0 = GM/R^2, R being the radius of the earth, is
the gravitational acceleration (field) on the surface of the
earth. The expression
g = GM/r^2 = GM/(R+h)^2 =(GM/R^2)/(1+h/R)^2 = g_0/(1 +h/r)
has been used to relate T(h) to T(0).

It is seen that the pendulum higher up in the gravitational field
keeps a different time from the one on the ground. The fractional
time difference is

(T(h) - T(0) )/T(0) = h/R, (3)

which shows that the upper pendulum keeps a longer time
(runs faster) than the lower one.

So far it has been assumed that the length L of the pendulum
is not affected by the gravitational field. This may not be correct.
To allow this possibility we write
T(0) = 2PiSqrt(L(0))/Sqrt(g_0), where L(0) is the length of the
pendulum on the ground. Then

T(h)/T(0) = Sqrt(L(h)/L(0))*r/R = Sqrt(L(h)/L(0))*(1+h/R). (4)

Let us assume that Sqrt(L(h)/L(0)) = (r/R)^p = (1+h/R)^p so that

T(h)/T(0) = (r/R)^(p+1) = (1+h/R)^(p+1) = [1+(p+1)h/R] (5)

The most likely values for p are 1, +/-1/2, and 0. Note that p = -1
is excluded because then T(h)/T(0) = 1 and L(h)/L(0) = (1-2h/R).
In this case time is not curved, only space is curved. But this
would violate the result of the Pound-Rebka-Snider experiments
that time runs faster higher up in a gravitational field. For other
values of p > -1 both space and time are curved. If two
pendulums are run for a month ( T(0) = 30 days) at h = 0 and
h = 100 m, we get

T(h) - T(0) = 30*86400*(p+1)*100/6.38*10^(-6)
= (p+1)* 40.63 s, (6)

where the time durations have been denoted by the same symbols
as the periods. Such a time difference should be measurable easily,
allowing the value of p to be determined. The significance of such an
experiment is self evident. If spacetime were not affected by the
gravitational field, one would get T(h)/T(0) = L(h)/L(0) = 1.
But this is impossible. The experimental result would be
expected to yield a value of p greater than -1, indicating
that both space and time are curved by the gravitational field.
T(h)/T(0) can be expressed in terms of the gravitational
potential Phi(r) = - GM/r as

T(h)/T(0) = (Phi(R)/Phi(r))^(p+1). (7)

The fractional difference in time, instead of Eq. 3, is

(T(h) - T(0))/T(0) = (p+1)h/R = (p+1)[Phi(r) - Phi(R)]/(g_0R)
= (p+1)(g_0h/g_0R) = (p+1)h/R. (8)

That this fractional difference is different from and much
larger than the 'gravitational redshift' result should be no
surprise. In the present case one is dealing with 'macroscopic'
clocks (pendulums) whereas in the gravitational redshift case
the clocks are the photons emitted by the nuclei/atoms.
Comments?

Regards,
Murat Ozer

[cmaj10@yahoo.com]

Murat Ozer wrote:
> Hello,
> I have found an argument based on Newtonian physics
> proving that spacetime must be curved. Consider two identical
> simple pendulums (clocks) of length L and bob mass m.
> Let one pendulum be on the ground and the other at a height h
> above it. For small oscillations (for simplicity only, not necessary
> for the argument) the periods of the pendulums are
>
> T(0) = 2PiSqrt(L)/Sqrt(g_0) (on the ground), (1)
>
> T(h) = 2PiSqrt(L)/Sqrt(g_0)*(1 + h/R) = T(0)*(1 + h/R), (2)
>
> where g_0 = GM/R^2, R being the radius of the earth, is
> the gravitational acceleration (field) on the surface of the
> earth. The expression
> g = GM/r^2 = GM/(R+h)^2 =(GM/R^2)/(1+h/R)^2 = g_0/(1 +h/r)
> has been used to relate T(h) to T(0).
>
> It is seen that the pendulum higher up in the gravitational field
> keeps a different time from the one on the ground. The fractional
> time difference is
>
> (T(h) - T(0) )/T(0) = h/R, (3)
>
> which shows that the upper pendulum keeps a longer time
> (runs faster) than the lower one.
>
> So far it has been assumed that the length L of the pendulum
> is not affected by the gravitational field. This may not be correct.
> To allow this possibility we write
> T(0) = 2PiSqrt(L(0))/Sqrt(g_0), where L(0) is the length of the
> pendulum on the ground. Then
>
> T(h)/T(0) = Sqrt(L(h)/L(0))*r/R = Sqrt(L(h)/L(0))*(1+h/R). (4)
>
> Let us assume that Sqrt(L(h)/L(0)) = (r/R)^p = (1+h/R)^p so that
>
> T(h)/T(0) = (r/R)^(p+1) = (1+h/R)^(p+1) = [1+(p+1)h/R] (5)
>
> The most likely values for p are 1, +/-1/2, and 0. Note that p = -1
> is excluded because then T(h)/T(0) = 1 and L(h)/L(0) = (1-2h/R).
> In this case time is not curved, only space is curved. But this
> would violate the result of the Pound-Rebka-Snider experiments
> that time runs faster higher up in a gravitational field. For other
> values of p > -1 both space and time are curved. If two
> pendulums are run for a month ( T(0) = 30 days) at h = 0 and
> h = 100 m, we get
>
> T(h) - T(0) = 30*86400*(p+1)*100/6.38*10^(-6)
> = (p+1)* 40.63 s, (6)
>
> where the time durations have been denoted by the same symbols
> as the periods. Such a time difference should be measurable easily,
> allowing the value of p to be determined. The significance of such an
> experiment is self evident. If spacetime were not affected by the
> gravitational field, one would get T(h)/T(0) = L(h)/L(0) = 1.
> But this is impossible. The experimental result would be
> expected to yield a value of p greater than -1, indicating
> that both space and time are curved by the gravitational field.
> T(h)/T(0) can be expressed in terms of the gravitational
> potential Phi(r) = - GM/r as
>
> T(h)/T(0) = (Phi(R)/Phi(r))^(p+1). (7)
>
> The fractional difference in time, instead of Eq. 3, is
>
> (T(h) - T(0))/T(0) = (p+1)h/R = (p+1)[Phi(r) - Phi(R)]/(g_0R)
> = (p+1)(g_0h/g_0R) = (p+1)h/R. (8)
>
> That this fractional difference is different from and much
> larger than the 'gravitational redshift' result should be no
> surprise. In the present case one is dealing with 'macroscopic'
> clocks (pendulums) whereas in the gravitational redshift case
> the clocks are the photons emitted by the nuclei/atoms.
> Comments?
>
> Regards,
> Murat Ozer

Einstein himself clearly expressed the fact that it's only the
non-absoluteness of space and time that motivates the geometrized
approach to gravitation. So even according to him, if space is
absolute, it is flat and Euclidean. So either you beat Einstein, or you
didn't truly assume the Newtonian absolute space, or you made a
mistake.

Chris

[Blackbird]

Murat Ozer wrote:
> [...]
> It is seen that the pendulum higher up in the gravitational field
> keeps a different time from the one on the ground. The fractional
> time difference is
>
> (T(h) - T(0) )/T(0) = h/R, (3)
>
> which shows that the upper pendulum keeps a longer time
> (runs faster) than the lower one.

No, it runs slower. In fact, if you move the upper pendulum further and
further away, it will eventually run so slow that it will not run at all,
because the gravitational field is neglible.

[Phillip Helbig---remove CLOTHES to reply]

In article <1138640957.400861.70020@o13g2000cwo.googlegroups.c om>, Murat
Ozer <Murat.H.Ozer@gmail.com> writes:

> Hello,
> I have found an argument based on Newtonian physics
> proving that spacetime must be curved. Consider two identical
> simple pendulums (clocks) of length L and bob mass m.
> Let one pendulum be on the ground and the other at a height h
> above it. For small oscillations (for simplicity only, not necessary
> for the argument) the periods of the pendulums are
>
> T(0) = 2PiSqrt(L)/Sqrt(g_0) (on the ground), (1)
>
> T(h) = 2PiSqrt(L)/Sqrt(g_0)*(1 + h/R) = T(0)*(1 + h/R), (2)
>
> where g_0 = GM/R^2, R being the radius of the earth, is
> the gravitational acceleration (field) on the surface of the
> earth. The expression
> g = GM/r^2 = GM/(R+h)^2 =(GM/R^2)/(1+h/R)^2 = g_0/(1 +h/r)
> has been used to relate T(h) to T(0).
>
> It is seen that the pendulum higher up in the gravitational field
> keeps a different time from the one on the ground. The fractional
> time difference is
>
> (T(h) - T(0) )/T(0) = h/R, (3)
>
> which shows that the upper pendulum keeps a longer time
> (runs faster) than the lower one.

Basically, the period is longer the smaller the acceleration. In fact,
one could use the period to measure the acceleration, if all other
quantities were known. Saying that "the upper pendulum keeps a longer
time than the lower one" seems to me to be just playing with words. This
has absolutely nothing to do with the fact that clocks run slower the
stronger the gravitational field (as seen by an observer at infinity) in
GR, which you seem to be implying. Note also that the sign of the
effect you mention is opposite to that in GR.

> So far it has been assumed that the length L of the pendulum
> is not affected by the gravitational field. This may not be correct.
> To allow this possibility we write
> T(0) = 2PiSqrt(L(0))/Sqrt(g_0), where L(0) is the length of the
> pendulum on the ground. Then
>
> T(h)/T(0) = Sqrt(L(h)/L(0))*r/R = Sqrt(L(h)/L(0))*(1+h/R). (4)
>
> Let us assume that Sqrt(L(h)/L(0)) = (r/R)^p = (1+h/R)^p so that

Where does this assumption come from?

Note that this is not necessary; what you seem to be thinking about is
some sort of material effect of acceleration on the length of the
pendulum. You can idealise the experiment to a perfectly rigid
pendulum.

> T(h)/T(0) = (r/R)^(p+1) = (1+h/R)^(p+1) = [1+(p+1)h/R] (5)
>
> The most likely values for p are 1, +/-1/2, and 0.

Why?

> The fractional difference in time, instead of Eq. 3, is
>
> (T(h) - T(0))/T(0) = (p+1)h/R = (p+1)[Phi(r) - Phi(R)]/(g_0R)
> = (p+1)(g_0h/g_0R) = (p+1)h/R. (8)
>
> That this fractional difference is different from and much
> larger than the 'gravitational redshift' result should be no
> surprise.

It also has nothing to do with it.

> In the present case one is dealing with 'macroscopic'
> clocks (pendulums) whereas in the gravitational redshift case
> the clocks are the photons emitted by the nuclei/atoms.

Completely irrelevant. The slowing down in a strong field in GR applies
to ALL clocks, whatever their size. Your argument would also apply to
an arbitrarily small pendulum.

We all know that a pendulum runs more slowly if the gravitational
acceleration is less. Relating this to some sort of curvature of space,
time or spacetime seems to be just playing with words.

[Ilja Schmelzer]

"Murat Ozer" <Murat.H.Ozer@gmail.com> schrieb
> Hello,
> I have found an argument based on Newtonian physics
> proving that spacetime must be curved. Consider two identical
> simple pendulums (clocks) of length L and bob mass m.
> Let one pendulum be on the ground and the other at a height h
> above it.

As has been mentioned already long time ago, a pendulum
clock, as a physical device in Newtonian theory, is
the clock taken together with the Earth (and, of course,
also taken together with its distance to Earth).

Ilja

[RP]

Murat Ozer wrote:
> Hello,
> I have found an argument based on Newtonian physics
> proving that spacetime must be curved. Consider two identical
> simple pendulums (clocks) of length L and bob mass m.
> Let one pendulum be on the ground and the other at a height h
> above it. For small oscillations (for simplicity only, not necessary
> for the argument) the periods of the pendulums are
>
> T(0) = 2PiSqrt(L)/Sqrt(g_0) (on the ground), (1)
>
> T(h) = 2PiSqrt(L)/Sqrt(g_0)*(1 + h/R) = T(0)*(1 + h/R), (2)
>
> where g_0 = GM/R^2, R being the radius of the earth, is
> the gravitational acceleration (field) on the surface of the
> earth. The expression
> g = GM/r^2 = GM/(R+h)^2 =(GM/R^2)/(1+h/R)^2 = g_0/(1 +h/r)
> has been used to relate T(h) to T(0).
>
> It is seen that the pendulum higher up in the gravitational field
> keeps a different time from the one on the ground.

The pendulum clock will tick slower at higher altitudes. Also of note is
that the operation of the pendulum clock depends in a direct way upon
the Earth's gravitational field, and thus this type of clock actually
consists of the the oscillating mass and Earth taken together as a
system. IOW, the Earth is itself an internal component of the clock. An
ideal clock OTOH will oscillate independently of the presence of
external gravitational and/or electromagnetic fields, or IOW, via
completely internal mechanisms.

Richard Perry

The fractional
> time difference is
>
> (T(h) - T(0) )/T(0) = h/R, (3)
>
> which shows that the upper pendulum keeps a longer time
> (runs faster) than the lower one.
>
> So far it has been assumed that the length L of the pendulum
> is not affected by the gravitational field. This may not be correct.
> To allow this possibility we write
> T(0) = 2PiSqrt(L(0))/Sqrt(g_0), where L(0) is the length of the
> pendulum on the ground. Then
>
> T(h)/T(0) = Sqrt(L(h)/L(0))*r/R = Sqrt(L(h)/L(0))*(1+h/R). (4)
>
> Let us assume that Sqrt(L(h)/L(0)) = (r/R)^p = (1+h/R)^p so that
>
> T(h)/T(0) = (r/R)^(p+1) = (1+h/R)^(p+1) = [1+(p+1)h/R] (5)
>
> The most likely values for p are 1, +/-1/2, and 0. Note that p = -1
> is excluded because then T(h)/T(0) = 1 and L(h)/L(0) = (1-2h/R).
> In this case time is not curved, only space is curved. But this
> would violate the result of the Pound-Rebka-Snider experiments
> that time runs faster higher up in a gravitational field. For other
> values of p > -1 both space and time are curved. If two
> pendulums are run for a month ( T(0) = 30 days) at h = 0 and
> h = 100 m, we get
>
> T(h) - T(0) = 30*86400*(p+1)*100/6.38*10^(-6)
> = (p+1)* 40.63 s, (6)
>
> where the time durations have been denoted by the same symbols
> as the periods. Such a time difference should be measurable easily,
> allowing the value of p to be determined. The significance of such an
> experiment is self evident. If spacetime were not affected by the
> gravitational field, one would get T(h)/T(0) = L(h)/L(0) = 1.
> But this is impossible. The experimental result would be
> expected to yield a value of p greater than -1, indicating
> that both space and time are curved by the gravitational field.
> T(h)/T(0) can be expressed in terms of the gravitational
> potential Phi(r) = - GM/r as
>
> T(h)/T(0) = (Phi(R)/Phi(r))^(p+1). (7)
>
> The fractional difference in time, instead of Eq. 3, is
>
> (T(h) - T(0))/T(0) = (p+1)h/R = (p+1)[Phi(r) - Phi(R)]/(g_0R)
> = (p+1)(g_0h/g_0R) = (p+1)h/R. (8)
>
> That this fractional difference is different from and much
> larger than the 'gravitational redshift' result should be no
> surprise. In the present case one is dealing with 'macroscopic'
> clocks (pendulums) whereas in the gravitational redshift case
> the clocks are the photons emitted by the nuclei/atoms.
> Comments?
>
> Regards,
> Murat Ozer
>

[Joe Fischer]

(Phillip Helbig wrote:

>Ozer writes:
>[snip]
> g = GM/r^2 = GM/(R+h)^2 =(GM/R^2)/(1+h/R)^2 = g_0/(1 +h/r)
>> has been used to relate T(h) to T(0).
>>
>> It is seen that the pendulum higher up in the gravitational field
>> keeps a different time from the one on the ground. The fractional
>> time difference is
>>
>> (T(h) - T(0) )/T(0) = h/R, (3)
>>
>> which shows that the upper pendulum keeps a longer time
>> (runs faster) than the lower one.
>
>Basically, the period is longer the smaller the acceleration. In fact,
>one could use the period to measure the acceleration, if all other
>quantities were known.

Of course, perhaps he needs a minus sign before the F.

Pendulums are accelerometers first, and always,
they were the first gravimeters, and will not swing without
acceleration.

http://www.ngs.noaa.gov/PUBS_LIB/Geodesy4Layman/TR80003C.HTM

I apologize for saying surface gravity is like acceleration.

Joe Fischer