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Gravitational Time Dilation and a Discussion by Schiff
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Murat Ozer
Apr 25, 2012, 4:17:39 PM4/25/12
to
[[Mod. note -- I have rewrapped the lines. -- jt]]
Hello,
I would like to consult you to resolve an issue in a paper by
L.I.Schiff (On Experimental Tests of the General Theory of Relativity,
Am.J.Phys., 28(1960)340 ). Section II on p.340 reads ? Two identically
constructed clocks are placed at rest, a distance h apart along the
lines of force in a uniform or nearly uniform gravitational field
of acceleration g, as in Fig. 1(a) {In this figure, there is a
round gravitating body and above it are two clocks labeled A
and B separated by a vertical distance h. Clock A is at the top.}
In accordance with the equivalence principle, any comparison of
the periods of these clocks can be made as well in a gravitation-free
region, in which they are accelerated upward with the acceleration
g, as in Fig. 1(b).{In this figure, there are the two clocks A and
B separated by the height h and there is a third clock C placed
next to the clock A. The line of force of g is drawn upward under
the clock B.} We accomplish this by comparing (continued on p.341)
clocks A and B in turn with a third identically constructed clock
C, which is permanently at rest, as they sweep by in near coincidence.
We assume that the fact that A and B are undergoing acceleration
as they pass C does not affect this comparison. Since C is at rest
in a gravitation-free region, it is part of an inertial coordinate
system, and makes a suitable standard for comparing A and B with
each other. Suppose that clock A has upward speed v_A when it
passes C. Then if the period of C is T, the period of A that is
seen by an observer on C is, according to special relativity,
T_A =T(1 ? v_A^2/c^2)^(-1/2) ~ T(1 + v_A^2/2c^2), (1)
where the approximation assumes that the speed of light c is much
greater than v_A. Similarly, when clock B passes C with speed v_B
a second comparison shows that the period of B observed by C is
T_B = T(1 ? v_B^2/c^2) ~ T(1 + v_B^2/2c^2). (2)
On eliminating T between Eqs. (1) and (2), we find to the same
approximation that
T_B ~ T_A[1 + (v_B^2 ? v_A^2)/2c^2] = T_A(1 + gh/c^2), (3)
since v_B^2 = v_A^2 + 2gh. Thus the inertial observer on C can
inform both A and B that the period of clock B exceeds that of clock
A by the fractional amount (gh/c^2).?
My question is this: Isn?t something wrong with the above discussion
because it is known from the gravitational time dilation that between
the two clocks, the period of the one that is at the higher
gravitational potential exceeds the period of the other that is at
the lower potential ? Thus, it must be that the period of clock A
exceeds that of clock B; not the other way around as stated by
Schiff. If you agree with me, I will next point out what is wrong
with Eqs. (1) and (2). Please state your opinions.
Thanks.
Hello,
I would like to consult you to resolve an issue in a paper by
L.I.Schiff (On Experimental Tests of the General Theory of Relativity,
Am.J.Phys., 28(1960)340 ). Section II on p.340 reads ? Two identically
constructed clocks are placed at rest, a distance h apart along the
lines of force in a uniform or nearly uniform gravitational field
of acceleration g, as in Fig. 1(a) {In this figure, there is a
round gravitating body and above it are two clocks labeled A
and B separated by a vertical distance h. Clock A is at the top.}
In accordance with the equivalence principle, any comparison of
the periods of these clocks can be made as well in a gravitation-free
region, in which they are accelerated upward with the acceleration
g, as in Fig. 1(b).{In this figure, there are the two clocks A and
B separated by the height h and there is a third clock C placed
next to the clock A. The line of force of g is drawn upward under
the clock B.} We accomplish this by comparing (continued on p.341)
clocks A and B in turn with a third identically constructed clock
C, which is permanently at rest, as they sweep by in near coincidence.
We assume that the fact that A and B are undergoing acceleration
as they pass C does not affect this comparison. Since C is at rest
in a gravitation-free region, it is part of an inertial coordinate
system, and makes a suitable standard for comparing A and B with
each other. Suppose that clock A has upward speed v_A when it
passes C. Then if the period of C is T, the period of A that is
seen by an observer on C is, according to special relativity,
T_A =T(1 ? v_A^2/c^2)^(-1/2) ~ T(1 + v_A^2/2c^2), (1)
where the approximation assumes that the speed of light c is much
greater than v_A. Similarly, when clock B passes C with speed v_B
a second comparison shows that the period of B observed by C is
T_B = T(1 ? v_B^2/c^2) ~ T(1 + v_B^2/2c^2). (2)
On eliminating T between Eqs. (1) and (2), we find to the same
approximation that
T_B ~ T_A[1 + (v_B^2 ? v_A^2)/2c^2] = T_A(1 + gh/c^2), (3)
since v_B^2 = v_A^2 + 2gh. Thus the inertial observer on C can
inform both A and B that the period of clock B exceeds that of clock
A by the fractional amount (gh/c^2).?
My question is this: Isn?t something wrong with the above discussion
because it is known from the gravitational time dilation that between
the two clocks, the period of the one that is at the higher
gravitational potential exceeds the period of the other that is at
the lower potential ? Thus, it must be that the period of clock A
exceeds that of clock B; not the other way around as stated by
Schiff. If you agree with me, I will next point out what is wrong
with Eqs. (1) and (2). Please state your opinions.
Thanks.
Anon E. Mouse
Apr 26, 2012, 3:54:37 AM4/26/12
to
On Apr 25, 4:17=A0pm, Murat Ozer <murat.h.o...@gmail.com> wrote:
> [[Mod. note -- I have rewrapped the lines. =A0-- jt]]
> round gravitating body =A0and above it are two clocks labeled A
> and B separated by a vertical distance =A0h. Clock A is at the top.}
>
> In accordance with the equivalence =A0principle, any comparison of
=A0 (2)
>
> since v_B^2 =3D v_A^2 + 2gh. Thus the inertial observer on C can
More elevated clocks appear to run more slowly than lower clocks just
as a planet at higher ellipsis appears to move more slowly than the
same body at closest approach.
The discussion as given is correct. If T is taken to be the small
interval dt. If it is taken to be a long interval such as a journey or
orbital period then integration is necessary and the formulas of the
discussion would not apply so well as higher orbits have longer
periods.
The relations between a malleable space time, velocity and momentum
are complex. For clock A of ABC the dv_A must be less than dV_b for
the equivalence to the gravitational case to be fully valid. Or, if
the distance h is split between A-C and B-C so h is constant and C is
in the middle then the differing potential represented by the greater
elevation of A would show up as a Doppler correction.
I discuss all these different cases because it is not easy to tell
where your understanding may have gone awry. Mr. Schiff's presentation
seems consistent with the principles and formulas of relativity,
including a good approximation of equivalence principle which is all
the formula's claim.
AAG
> [[Mod. note -- I have rewrapped the lines. =A0-- jt]]
>
> Hello,
> I would like to consult you to resolve an issue in a paper by
> L.I.Schiff (On Experimental Tests of the General Theory of Relativity,
> Am.J.Phys., 28(1960)340 ). Section II on p.340 reads ? Two identically
> constructed clocks are placed at rest, a distance h apart along the
> lines of force in a uniform or nearly uniform gravitational field
> of acceleration g, as in Fig. =A01(a) {In this figure, there is a
> Hello,
> I would like to consult you to resolve an issue in a paper by
> L.I.Schiff (On Experimental Tests of the General Theory of Relativity,
> Am.J.Phys., 28(1960)340 ). Section II on p.340 reads ? Two identically
> constructed clocks are placed at rest, a distance h apart along the
> lines of force in a uniform or nearly uniform gravitational field
> round gravitating body =A0and above it are two clocks labeled A
> and B separated by a vertical distance =A0h. Clock A is at the top.}
>
> In accordance with the equivalence =A0principle, any comparison of
> the periods of these clocks can be made as well in a gravitation-free
> region, in which they are accelerated upward =A0with the acceleration
> g, as in Fig. 1(b).{In this figure, there are the two clocks A and
> B separated by the height h and there is a third clock C placed
> next to the clock A. The line of force of g is drawn upward under
> the clock B.} We accomplish this by comparing (continued on p.341)
> clocks A and B in turn with a third identically constructed clock
> C, which is permanently at rest, as they sweep by in near coincidence.
> We assume that the fact that A and B are undergoing acceleration
> as they pass C does not affect this comparison. Since C is at rest
> in a gravitation-free region, it is part of an inertial coordinate
> system, and makes a suitable standard for comparing A and B with
> each other. =A0Suppose that clock A has upward speed v_A when it
> B separated by the height h and there is a third clock C placed
> next to the clock A. The line of force of g is drawn upward under
> the clock B.} We accomplish this by comparing (continued on p.341)
> clocks A and B in turn with a third identically constructed clock
> C, which is permanently at rest, as they sweep by in near coincidence.
> We assume that the fact that A and B are undergoing acceleration
> as they pass C does not affect this comparison. Since C is at rest
> in a gravitation-free region, it is part of an inertial coordinate
> system, and makes a suitable standard for comparing A and B with
> passes C. Then if the period of C is T, the period of A that is
> seen by an observer on C is, according to special relativity,
>
> T_A =3DT(1 ? v_A^2/c^2)^(-1/2) ~ T(1 + v_A^2/2c^2), =A0 =A0 =A0(1)
> seen by an observer on C is, according to special relativity,
>
>
> where the approximation assumes that the speed of light c is much
> greater than v_A. =A0Similarly, when clock B passes C with speed v_B
> where the approximation assumes that the speed of light c is much
> a second comparison shows that the period of B observed by C is
>
> T_B =3D T(1 ? v_B^2/c^2) ~ T(1 + v_B^2/2c^2). =A0 =A0 =A0 =A0 =A0 =A0 =A0=
>
=A0 (2)
>
> On eliminating T between Eqs. (1) and (2), we find to the same
> approximation that
>
> T_B ~ T_A[1 + (v_B^2 ? v_A^2)/2c^2] =3D T_A(1 + gh/c^2), =A0 =A0 =A0(3)
> On eliminating T between Eqs. (1) and (2), we find to the same
> approximation that
>
>
> since v_B^2 =3D v_A^2 + 2gh. Thus the inertial observer on C can
> inform both A and B that the period of clock B exceeds that of clock
> A by the fractional amount (gh/c^2).?
>
> My question is this: Isn?t something wrong with the above discussion
> because it is known from the gravitational time dilation that between
> the two clocks, the period of the one that is at the higher
> gravitational potential exceeds the period of the other that is at
> the lower potential ? =A0Thus, it must be that the period of clock A
> A by the fractional amount (gh/c^2).?
>
> My question is this: Isn?t something wrong with the above discussion
> because it is known from the gravitational time dilation that between
> the two clocks, the period of the one that is at the higher
> gravitational potential exceeds the period of the other that is at
> exceeds that of clock B; not the other way around as stated by
> Schiff. If you agree with me, I will next point out what is wrong
> with Eqs. (1) and (2). Please state your opinions.
>
> Thanks.
As state (correctly) B exceeds A by gh/c*c for the gravitational case.
> Schiff. If you agree with me, I will next point out what is wrong
> with Eqs. (1) and (2). Please state your opinions.
>
> Thanks.
More elevated clocks appear to run more slowly than lower clocks just
as a planet at higher ellipsis appears to move more slowly than the
same body at closest approach.
The discussion as given is correct. If T is taken to be the small
interval dt. If it is taken to be a long interval such as a journey or
orbital period then integration is necessary and the formulas of the
discussion would not apply so well as higher orbits have longer
periods.
The relations between a malleable space time, velocity and momentum
are complex. For clock A of ABC the dv_A must be less than dV_b for
the equivalence to the gravitational case to be fully valid. Or, if
the distance h is split between A-C and B-C so h is constant and C is
in the middle then the differing potential represented by the greater
elevation of A would show up as a Doppler correction.
I discuss all these different cases because it is not easy to tell
where your understanding may have gone awry. Mr. Schiff's presentation
seems consistent with the principles and formulas of relativity,
including a good approximation of equivalence principle which is all
the formula's claim.
AAG
Tom Roberts
Apr 26, 2012, 10:22:53 AM4/26/12
to
On 4/26/12 4/26/12 2:54 AM, Anon E. Mouse wrote:
> On Apr 25, 4:17=A0pm, Murat Ozer<murat.h.o...@gmail.com> wrote:
>> [...]
> On Apr 25, 4:17=A0pm, Murat Ozer<murat.h.o...@gmail.com> wrote:
>> My question is this: Isn?t something wrong with the above discussion
>> because it is known from the gravitational time dilation that between
>> the two clocks, the period of the one that is at the higher
>> gravitational potential exceeds the period of the other that is at
>> the lower potential ?
No. Schiff is correct and you are wrong. When comparing two clocks at rest in a
>> because it is known from the gravitational time dilation that between
>> the two clocks, the period of the one that is at the higher
>> gravitational potential exceeds the period of the other that is at
>> the lower potential ?
static gravitational potential, the higher clock is measured to tick faster
(shorter period) than the lower clock.
> As state (correctly) B exceeds A by gh/c*c for the gravitational case.
> More elevated clocks appear to run more slowly than lower clocks just
> as a planet at higher ellipsis appears to move more slowly than the
> same body at closest approach.
tick faster (shorter period) than lower clocks. Your claimed relationship to
velocities of an elliptical orbit is both wrong and irrelevant:
Note that when comparing a clock in circular orbit to a clock at rest on a
planet, one must also take into account the effect of the difference in motion.
In general, for a given planet there is an altitude below which the orbiting
clock is measured to tick slower than the ground clock, and above which the
orbiting clock is measured to tick faster. For the earth, the space shuttle
orbits are below this altitude, and the GPS orbits are above it.
GPS satellite clocks have a nonstandard divider between the
Cs oscillator and the tick counter; more Cs oscillations
are required per tick, in precisely the right ratio so the
satellite clocks can be synchronized with ground clocks.
This provides a universal time coordinate in the ECI frame
common to all GPS clocks, satellite and ground. The standard
divider applies only to clocks at rest on the geoid; all
others need correction.
Note that all standard clocks tick at their standard rate, regardless of their
altitude or motion; this is of course a LOCAL statement. It is only when
COMPARING clocks at different altitudes that this gravitational redshift can be
observed.
Tom Roberts
Murat Ozer
Apr 26, 2012, 10:23:36 AM4/26/12
to
> As state (correctly) B exceeds A by gh/c*c for the gravitational case.
Relativity, p.113), J.B.Hartle considers a similar situation. There are
clocks A (at the top)and B (at the bottom)in a rocket which keep the
time intervals between the two signals emitted by Alice (at the top)
and received by Bob (at the bottom). By employing the Equivalence
Principle, he considers a rocket accelerating vertically upward with
a = g and concludes in Eq. 6.10 that
DT_B = DT_A(1 - gh/c^2),
where I have used D for Delta and T for Tau. He then goes on to say
"The interval at which the pulses are received is smaller by a factor
of (1 - gh/c^2) than the interval at which they are emitted." He also
says at the top of p.119 that "The theorists whose offices are at the
top of the building are older by a few microseconds in 100 yr-"
There are similar treatments of the Gravitational Time Dilation in
various books. They all say that T_top > T_bottom, which is in clear
contradiction with Schiff.
My question is, "Why is Schiff saying just the opposite of what everybody
else is saying?
> More elevated clocks appear to run more slowly than lower clocks just
> as a planet at higher ellipsis appears to move more slowly than the
> same body at closest approach.
>
> The discussion as given is correct.
contradicts that of everybody else's?
Thanks,
MO
Tom Roberts
Apr 27, 2012, 3:35:07 AM4/27/12
to
On 4/26/12 4/26/12 - 9:23 AM, Murat Ozer wrote:
> [In Chapter 6 of his book (Gravity, An introduction to Einstein's General
> Relativity, p.113), J.B.Hartle ] He also
i.e. its period is shorter, i.e. the accumulated time is longer.
> There are similar treatments of the Gravitational Time Dilation in
> various books. They all say that T_top> T_bottom, which is in clear
> contradiction with Schiff.
No, you are confused; they are in agreement with Schiff. You need to be more
careful to distinguish between period and accumulated time. Here T is an
accumulated time, not the period between ticks.
Tom Roberts
> [In Chapter 6 of his book (Gravity, An introduction to Einstein's General
> Relativity, p.113), J.B.Hartle ] He also
> says at the top of p.119 that "The theorists whose offices are at the
> top of the building are older by a few microseconds in 100 yr-"
Yes. This means the clock at the top is ticking faster than the lower clock --
> top of the building are older by a few microseconds in 100 yr-"
i.e. its period is shorter, i.e. the accumulated time is longer.
> There are similar treatments of the Gravitational Time Dilation in
> various books. They all say that T_top> T_bottom, which is in clear
> contradiction with Schiff.
careful to distinguish between period and accumulated time. Here T is an
accumulated time, not the period between ticks.
Tom Roberts
Ralph Hartley
Apr 27, 2012, 3:35:10 AM4/27/12
to
On 04/26/2012 10:23 AM, Murat Ozer wrote:
> On Thursday, April 26, 2012 10:54:37 AM UTC+3, Anon E. Mouse wrote:
>> On Apr 25, 4:17=A0pm, Murat Ozer<murat.h.o...@gmail.com> wrote:
..
> On Thursday, April 26, 2012 10:54:37 AM UTC+3, Anon E. Mouse wrote:
>> On Apr 25, 4:17=A0pm, Murat Ozer<murat.h.o...@gmail.com> wrote:
>>> [Quoting Schiff]:
>>>> T_B ~ T_A(1 + gh/c^2) [cleaned up]
Here T_A and T_B are the *periods* of clock A and B as seen by a
particular observer. B is slower.
>>> it is known from the gravitational time dilation that between
>>> the two clocks, the period of the one that is at the higher
>>> gravitational potential exceeds the period of the other that is at
>>> the lower potential?
potential is negative?
..
> In Chapter 6 of his book (Gravity, An introduction to Einstein's General
> Relativity, p.113), J.B.Hartle considers a similar situation.There are
> clocks A (at the top)and B (at the bottom)in a rocket which keep the
> time intervals between the two signals emitted by Alice (at the top)
> and received by Bob (at the bottom). By employing the Equivalence
> Principle, he considers a rocket accelerating vertically upward with
> a = g and concludes in Eq. 6.10 that
>
> DT_B = DT_A(1 - gh/c^2),
..
> time intervals between the two signals emitted by Alice (at the top)
> and received by Bob (at the bottom). By employing the Equivalence
> Principle, he considers a rocket accelerating vertically upward with
> a = g and concludes in Eq. 6.10 that
>
> DT_B = DT_A(1 - gh/c^2),
> "The interval at which the pulses are received is smaller by a factor
> of (1 - gh/c^2) than the interval at which they are emitted."
The emission interval is measured by Alice's clock, and the reception
> of (1 - gh/c^2) than the interval at which they are emitted."
interval by Bob's clock. B measures a shorter interval, so B is slower.
Here DT_A and DT_B are the number of *ticks* of the two clocks.
> He also
> says at the top of p.119 that "The theorists whose offices are at the
> top of the building are older by a few microseconds in 100 yr-"
ticks more.
Ralph Hartley
holog
Apr 28, 2012, 10:38:03 AM4/28/12
to
On Apr 27, 3:35 am, Ralph Hartley <hart...@aic.nrl.navy.mil> wrote:
[[Mod. note -- 34 excessively-quoted lines snipped here. -- jt]]
holog
[[Mod. note -- 34 excessively-quoted lines snipped here. -- jt]]
> The emission interval is measured by Alice's clock, and the reception
> interval by Bob's clock. B measures a shorter interval, so B is slower.
>
> Here DT_A and DT_B are the number of *ticks* of the two clocks.
>
> > He also
> > says at the top of p.119 that "The theorists whose offices are at the
> > top of the building are older by a few microseconds in 100 yr-"
>
> They are older because their clocks ticked more times. A faster clock
> ticks more.
>
> Ralph Hartley
yet both clocks are the same age
> interval by Bob's clock. B measures a shorter interval, so B is slower.
>
> Here DT_A and DT_B are the number of *ticks* of the two clocks.
>
> > He also
> > says at the top of p.119 that "The theorists whose offices are at the
> > top of the building are older by a few microseconds in 100 yr-"
>
> They are older because their clocks ticked more times. A faster clock
> ticks more.
>
> Ralph Hartley
holog
Anon E. Mouse
Apr 28, 2012, 10:38:21 AM4/28/12
to
>
> Yes. This means the clock at the top is ticking faster than the lower clock --
> i.e. its period is shorter, i.e. the accumulated time is longer.
>
>
Yes, the crux of the difficulty.
Watches used to be regulated by pendulum's or springs and gears. If
you had a short tick, the clock ran fast, this was common knowledge
once.
AAG
Anon E. Mouse
May 3, 2012, 7:10:39 AM5/3/12
to
> My question is, "Why is Schiff saying just the opposite of what everybody
> else is saying?
>
> > More elevated clocks appear to run more slowly than lower clocks just
> > as a planet at higher ellipsis appears to move more slowly than the
> > same body at closest approach.
>
> > The discussion as given is correct.
>
> If Schiff's discussion is correct, then why is it that his finding
> contradicts that of everybody else's?
>
> Thanks,
> MO
>
> If T is taken to be the small
Your question is very good. I have sat in on, or participated in
> else is saying?
>
> > More elevated clocks appear to run more slowly than lower clocks just
> > as a planet at higher ellipsis appears to move more slowly than the
> > same body at closest approach.
>
> > The discussion as given is correct.
>
> If Schiff's discussion is correct, then why is it that his finding
> contradicts that of everybody else's?
>
> Thanks,
> MO
>
> If T is taken to be the small
several discussion of this type and find I must work hard to make sure
my own statements conform to my own understanding. If I consider the
point of view of others, even more so.
Having recently thought about these issues and referencing original
papers, I think that these observations may help you in your own
efforts to comprehend.
First there is the tick vs period consideration. I use the following
rule, short ticks cause fast clocks, fast clocks measure standard
runners as slow runners. This rule covers Lorentz time dilation of
ticks - dt, The definite integral T of dt (the method I use), or co-
variant transform of GRT, and the effect of these have on real
observables, i.e. motions. I recommend you develop your own rules of
thumb to use as a means of preventing falling into the many possible
conceptual errors.
The more subtle point I discovered due to the manner in which Schiff
develops his reasoning and the responses of others to that, the time
like gradient tangent to the radius of a massive body - particle
system and the time like gradient parallel to the body - particle
system are not the same gradients. Nor, do they seem to be co-variant.
They appear to me to be contra-variant Meaning dt_r * dt_c obey a
product rule under GRT for a given local patch.
If mass M streches space-time radially, it mutually and reciprocally
compresses space-time along the circle. Thus, Schiff who presents a
sort of thought experiment involving clocks magically suspended in a
gravitational field comes to same place as everyone else who typically
follow the orbital path.
Then there is Tom Roberts and his GPS system. If Newtonian mechanics
were rigorously correct then the orbital energy of the space craft and
the gravitational field of the Earth would exactly balance and there
would no effect on clocks.
Instead the Equivalence Principle applies and these qualities of space-
time, first and second type balance or not depending on orbital
altitude. Apparently at GPS orbital altitude the balance is in favor
of Cs atomic vibrations seeming fast so more ticks at that altitude
and velocity are needed to synch these short ticked, faster clocks
with ground clocks.
Which, according to my rule of thumb for fast clocks would seem to say
that these satellites would seem a little slow to Newton. They are
able to maintain their orbits in spite of seeming to have longer
orbital periods as measured by their onboard Cs clocks than they
should according to mechanics alone.
This analysis neglects any effect the G field or the crafts motion may
have (does have) on the rate of Cs atomic vibration which is the
probable source of the lack of co-variance or equivalence between the
orbiting clocks and the ground clocks.
Thus the proper analysis of the GPS system under GRT seems to me
extremely complicated and as a result, I'm glad it is someone else's
responsibility.
Sincerely,
AAG
holog
May 20, 2012, 3:31:18 AM5/20/12
to
On May 3, 7:10=A0am, "Anon E. Mouse" <agall...@gmail.com> wrote:
> > My question is, "Why is Schiff saying just the opposite of what everybo=
its like asking who is younger, the mother who walks a block to work
or the father who drives 30 miles back and forth to work, but they
both have dinner at the same time.
holog
> > My question is, "Why is Schiff saying just the opposite of what everybo=
or the father who drives 30 miles back and forth to work, but they
both have dinner at the same time.
holog
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