The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0 m).
To simplify the calculation let's say the building is at the equator, mean sea-level.

The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0 m) ?
To simplify the calculation let's say the building is at the equator, mean sea-level.

"qbit" wrote in message
...
The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0
m) ?
To simplify the calculation let's say the building is at the equator, mean
sea-level.


what is the speed difference? how thick is the earth? what is the rotational
speed?
will the time dilation keep people younger at the top floors?
will people on the top floors get more radiation from space?

"qbit" wrote in message
...
: The Empire State Building rises to 381 m at the 102nd floor
: (including broadcast antenna it reaches 443 m).
: How much is the net time dilation effect per day in the top floor
: (let's say at height 380 m) compared to the first floor (ie. at height 0
m) ?
: To simplify the calculation let's say the building is at the equator, mean
sea-level.


"If we assume that the result proved for a polygonal line is also valid for
a continuously curved line, we arrive at this result: If one of two
synchronous clocks at A is moved in a closed curve with constant velocity
until it returns to A, the journey lasting t seconds, then by the clock
which has remained at rest the travelled clock on its arrival at A will be
1/2tv^2/c^2 second slow. Thence we conclude that a balance-clock at the
equator must go more slowly, by a very small amount, than a precisely
similar clock situated at one of the poles under otherwise identical
conditions."

Key word: "assume".

We will assume Einstein was an idiot, therefore Einstein was an idiot.

We will assume Einstein's disciples are all idiots, therefore Einstein's

disciples are all idiots.



"Tom Roberts" wrote in message
om...

This is PHYSICS, not math or logic, and "proof" is completely irrelevant.
--Tom Roberts

We don't assume Tom Roberts is an idiot, he proves it all by himself.

"Dear Leader" wrote
"qbit" wrote in message

The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0 m) ?
To simplify the calculation let's say the building is at the equator, mean
sea-level.

what is the speed difference?

Do you mean the between the 102nd floor and the first floor?
Hmm. isn't it 0 ? :-)

how thick is the earth?

Mass of Earth : 5.973600E+24 kg
Mean Radius : 6372797 m

what is the rotational speed?

Just calculate it:
v = 2*pi*(R+h) / 86164.091
(86164.091 seconds per sideral day; one can also take 24*60*60=86400 sec)

will the time dilation keep people younger at the top floors?
will people on the top floors get more radiation from space?

Yeah, that's all interessting questions, but the topic here is just time dilation.
Can you calculate it?

On Aug 23, 12:23 pm, "qbit" wrote:
The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0 m).
To simplify the calculation let's say the building is at the equator, mean sea-level.

The equation you are looking for is he

http://en.wikipedia.org/wiki/Gravita..._time_dilation

"Igor" wrote
On Aug 23, 12:23 pm, "qbit" wrote:

The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0 m).
To simplify the calculation let's say the building is at the equator, mean sea-level.

The equation you are looking for is he
http://en.wikipedia.org/wiki/Gravita..._time_dilation

I used the equation under the heading "Circular orbits" the

M = 5.973600E+24
R = 6372797
t0 = 86400 ; time elapsed on earth
h = 380 ; top floor (102nd) of the Empire State Building
r0 = 2 * G * M / c^2 ; Schwarzschild radius
tf = t0 / sqrt(1 - 3 / 2 * r0 / (R + h))
GravTD = t0 - tf
= -0.0000901858739
= -90.186 microsec

Hmm. time in the 102nd floor (380m) of the Empire State Building
seems to go 90 microsec/day SLOWER than on ground! (???)
Can this be true? I would expect time go FASTER in that height.

On Aug 23, 4:24 pm, "qbit" wrote:
"Igor" wrote

On Aug 23, 12:23 pm, "qbit" wrote:

The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0 m).
To simplify the calculation let's say the building is at the equator, mean sea-level.

The equation you are looking for is he
http://en.wikipedia.org/wiki/Gravita..._time_dilation

I used the equation under the heading "Circular orbits" the

I don't think that's appropriate, since neither a person at
the top nor the bottom of the building is in a free-falling
orbit.

M = 5.973600E+24
R = 6372797
t0 = 86400 ; time elapsed on earth
h = 380 ; top floor (102nd) of the Empire State Building
r0 = 2 * G * M / c^2 ; Schwarzschild radius
tf = t0 / sqrt(1 - 3 / 2 * r0 / (R + h))

According to the Wiki page, tf is time for an observer
infinitely far from the gravitating body. You want to compare
two times that are both within the gravitational field. So
I think that you want to work out the different t0's for
(a) an object at the foot of the building, radius = R, and
(b) an object at the top of the building, radius = R+h.

GravTD = t0 - tf
= -0.0000901858739
= -90.186 microsec

Hmm. time in the 102nd floor (380m) of the Empire State Building
seems to go 90 microsec/day SLOWER than on ground! (???)
Can this be true? I would expect time go FASTER in that height.

No, you've compared the empire state building to time outside
the solar system, not to the ground.

For the ratio between values of t0 at different r in the
same gravitational field, if I understand the equations correctly,
I get this (doing a bunch of small-number approximations):

t1/t2 = sqrt( 1 - r0/r1) / sqrt(1 - r0/r2)

~ ( 1 - 0.5*r0/r1 )*( 1 + 0.5*r0/r2 )

~ 1 + 0.5*r0*(1/r2 - 1/r1)

with 1/r2 - 1/r1 = 1/(R+h) - 1/R
~ [R - (R+h)]/R^2

~ -h/R^2

So I get t1/t2 ~ 1 - 0.5*h*r0/R^2

Using r0 = 0.0089 m (is that what you got?)
this gives me t1/t2 ~ 1 - 4.2E-14.

The fraction 4.2E-14 corresponds to 3.6 nanoseconds.

- Randy

"Randy Poe" wrote in message
ups.com...
: On Aug 23, 4:24 pm, "qbit" wrote:
: "Igor" wrote
:
: On Aug 23, 12:23 pm, "qbit" wrote:
:
: The Empire State Building rises to 381 m at the 102nd floor
: (including broadcast antenna it reaches 443 m).
: How much is the net time dilation effect per day in the top floor
: (let's say at height 380 m) compared to the first floor (ie. at
height 0 m).
: To simplify the calculation let's say the building is at the
equator, mean sea-level.
:
: The equation you are looking for is he
: http://en.wikipedia.org/wiki/Gravita..._time_dilation
:
: I used the equation under the heading "Circular orbits" the
:
: I don't think that's appropriate, since neither a person at
: the top nor the bottom of the building is in a free-falling
: orbit.
:
: M = 5.973600E+24
: R = 6372797
: t0 = 86400 ; time elapsed on earth
: h = 380 ; top floor (102nd) of the Empire State
Building
: r0 = 2 * G * M / c^2 ; Schwarzschild radius
: tf = t0 / sqrt(1 - 3 / 2 * r0 / (R + h))
:
: According to the Wiki page, tf is time for an observer
: infinitely far from the gravitating body.


Oh well, we'll just have to send all bigots and trolls
infinitely far from the Solar system to do their "observing"
then.

"Randy Poe" wrote
On Aug 23, 4:24 pm, "qbit" wrote:
"Igor" wrote

On Aug 23, 12:23 pm, "qbit" wrote:

The Empire State Building rises to 381 m at the 102nd floor
(including broadcast antenna it reaches 443 m).
How much is the net time dilation effect per day in the top floor
(let's say at height 380 m) compared to the first floor (ie. at height 0 m).
To simplify the calculation let's say the building is at the equator, mean sea-level.

The equation you are looking for is he
http://en.wikipedia.org/wiki/Gravita..._time_dilation

I used the equation under the heading "Circular orbits" the

I don't think that's appropriate, since neither a person at
the top nor the bottom of the building is in a free-falling
orbit.

M = 5.973600E+24
R = 6372797
t0 = 86400 ; time elapsed on earth
h = 380 ; top floor (102nd) of the Empire State Building
r0 = 2 * G * M / c^2 ; Schwarzschild radius
tf = t0 / sqrt(1 - 3 / 2 * r0 / (R + h))

According to the Wiki page, tf is time for an observer
infinitely far from the gravitating body. You want to compare
two times that are both within the gravitational field. So
I think that you want to work out the different t0's for
(a) an object at the foot of the building, radius = R, and
(b) an object at the top of the building, radius = R+h.

GravTD = t0 - tf
= -0.0000901858739
= -90.186 microsec

Hmm. time in the 102nd floor (380m) of the Empire State Building
seems to go 90 microsec/day SLOWER than on ground! (???)
Can this be true? I would expect time go FASTER in that height.

No, you've compared the empire state building to time outside
the solar system, not to the ground.

For the ratio between values of t0 at different r in the
same gravitational field, if I understand the equations correctly,
I get this (doing a bunch of small-number approximations):

t1/t2 = sqrt( 1 - r0/r1) / sqrt(1 - r0/r2)
~ ( 1 - 0.5*r0/r1 )*( 1 + 0.5*r0/r2 )
~ 1 + 0.5*r0*(1/r2 - 1/r1)

with 1/r2 - 1/r1 = 1/(R+h) - 1/R
~ [R - (R+h)]/R^2
~ -h/R^2

So I get t1/t2 ~ 1 - 0.5*h*r0/R^2

Using r0 = 0.0089 m (is that what you got?)
this gives me t1/t2 ~ 1 - 4.2E-14.

The fraction 4.2E-14 corresponds to 3.6 nanoseconds.

Yes, I got the same r0.
But which of the t1 and t2 is the time for the top of the building?
And does r1 mean R, and does r2 mean R+h ?
And does your "h" mean the distance from (R+h) to R ?
And: where do you use 24*60*60 = 86400 ? This should the floor time, right?
Then the time at top should be more than 86400, right?