Spacetime is sometimes depicted as a 2 dimensional surface with pits
representing massive bodies which curve the surface. I assume that
non-accelerating clocks on the flat surface of such a representation
experience no time dilation, and clocks sitting on curved regions do
experience some slowing.

(Please feel free to correct my understanding if I am wrong!)

My real question is this: if a clock was placed at the bottom of one of
the pits produced by a massive body where the surface is essentially
flat, would it run slow or would it run at the same speed as a clock on
the flat surface?

An associated question would be: is there an association between time
dilation and tidal forces such that a clock will only run slow if there
are tidal forces (assuming that the clock is not accelerating)?

Hopefully my questions are not completely nonsensical!

TIA,

E. Jones.

there are, mostly becus you dont mention why the answers should be
significant

its unpolite too, shame on you

Very good questions. I wish more folks would consider them.


Gravity is maximum at the earths surface. Surprise ?
All the earth's mass in one side of your body.

Pendulum clocks run slower at altitude or depth.
Vibrating mass clocks (accelerometer type ) run faster at altidude or
depth.
Masslesses clocks are expected to be gravity and motion insensitive.

http://tf.nist.gov/timefreq/cesium/parcs.htm
http://bigben.stanford.edu/sumo/status.htm
http://bigben.stanford.edu/sumo/

As for time...
It waits for no man, only women. ;-)

Sue...

Enkidu Jones wrote:
Spacetime is sometimes depicted as a 2 dimensional surface with pits
representing massive bodies which curve the surface.

That picture, common as it is, is entirely wrong, so don't trust any
conclusion you base on it. :-)

I assume that
non-accelerating clocks on the flat surface of such a representation
experience no time dilation, and clocks sitting on curved regions do
experience some slowing.

This is difficult to answer because most solutions to general relativity
can't be drawn as a 2D surface to begin with. The only exception of note is
the Schwarzschild solution for a spherically symmetric gravitating body. In
that particular case, it's true that a clock on a curvy part will
effectively run slower than a clock on a flat part. In general, you can't
tell from looking at the surface. In fact, I could concoct a solution which
looked exactly like the Schwarzschild solution when depicted as a surface,
but which had opposite time dilation properties (clocks on the curved part
run faster).

The "surface with pits" picture that you usually see doesn't look anything
like the Schwarzschild geometry, by the way. It's not a solution to GR at
all, but a plot of the Newtonian gravitational potential which at some point
(decades ago) was misinterpreted by someone (Eddington?) as being a
spacetime diagram. The mistake has been perpetuated by generations of
popular books written by people who learned physics from other popular books.

My real question is this: if a clock was placed at the bottom of one of
the pits produced by a massive body where the surface is essentially
flat, would it run slow or would it run at the same speed as a clock on
the flat surface?

The best way to think about this is actually in terms of the Newtonian
potential. Is there a net energy cost involved in transporting a massive
object from the center of the earth out to infinity? Clearly so: there's a
positive cost to go from the center to the surface, and a further positive
cost to go from the surface to infinity. For roughly the same reason, a
photon emitted at the center of the earth (imagining the earth to be
transparent) will lose energy as it rises, which means that it will redshift
(E=hf), and as the photon's frequency slows down, so slows everything else.
So a clock at the center of the earth will appear to tick slowly when seen
from the surface, and even more slowly from outer space. There's a
reciprocal blueshift of clocks higher up when seen from lower down, so it
makes sense to think of this as a "real" slowing of the clock.

Since the surface-with-pits drawings actually show the Newtonian potential,
a good rule of thumb is that clocks lower down in the pits tick slower --
it's height, rather than slope, that matters.

-- Ben

"Ben Rudiak-Gould" wrote in message ...
Enkidu Jones wrote:
Spacetime is sometimes depicted as a 2 dimensional surface with pits
representing massive bodies which curve the surface.

That picture, common as it is, is entirely wrong, so don't trust any
conclusion you base on it. :-)

I assume that
non-accelerating clocks on the flat surface of such a representation
experience no time dilation, and clocks sitting on curved regions do
experience some slowing.

This is difficult to answer because most solutions to general relativity
can't be drawn as a 2D surface to begin with. The only exception of note is
the Schwarzschild solution for a spherically symmetric gravitating body. In
that particular case, it's true that a clock on a curvy part will
effectively run slower than a clock on a flat part. In general, you can't
tell from looking at the surface. In fact, I could concoct a solution which
looked exactly like the Schwarzschild solution when depicted as a surface,
but which had opposite time dilation properties (clocks on the curved part
run faster).

The "surface with pits" picture that you usually see doesn't look anything
like the Schwarzschild geometry, by the way. It's not a solution to GR at
all, but a plot of the Newtonian gravitational potential which at some point
(decades ago) was misinterpreted by someone (Eddington?) as being a
spacetime diagram. The mistake has been perpetuated by generations of
popular books written by people who learned physics from other popular books.

My real question is this: if a clock was placed at the bottom of one of
the pits produced by a massive body where the surface is essentially
flat, would it run slow or would it run at the same speed as a clock on
the flat surface?

The best way to think about this is actually in terms of the Newtonian
potential. Is there a net energy cost involved in transporting a massive
object from the center of the earth out to infinity? Clearly so: there's a
positive cost to go from the center to the surface, and a further positive
cost to go from the surface to infinity. For roughly the same reason, a
photon emitted at the center of the earth (imagining the earth to be
transparent) will lose energy as it rises, which means that it will redshift

Is that the current interpretation of GPS launch presets ?
http://scitation.aip.org/getabs/serv...cvips&gifs=yes

Sue...


(E=hf), and as the photon's frequency slows down, so slows everything else.
So a clock at the center of the earth will appear to tick slowly when seen
from the surface, and even more slowly from outer space. There's a
reciprocal blueshift of clocks higher up when seen from lower down, so it
makes sense to think of this as a "real" slowing of the clock.

Since the surface-with-pits drawings actually show the Newtonian potential,
a good rule of thumb is that clocks lower down in the pits tick slower --
it's height, rather than slope, that matters.

-- Ben

sue jahn wrote:
"Ben Rudiak-Gould" wrote in message ...
For roughly the same reason, a
photon emitted at the center of the earth (imagining the earth to be
transparent) will lose energy as it rises, which means that it will redshift

http://scitation.aip.org/getabs/serv...cvips&gifs=yes

You're right. It was a terrible explanation. I retract it.

What's true is that (in the linearized theory) there's a relationship
between gravitational potential and time dilation, and the "surface with
pits" is actually a plot of the gravitational potential (as opposed to a
spacetime embedding), so lower is slower.

-- Ben