Time dilation in a gravitational field A friend and I were talking about relativity and we had different opinions about this situation: If I choose a frame of reference on the surface of a planet, would time "run" differently between the top of a tree and the ground? 
I think that a clock in the floor would run slower than one in the top of the tree due to the gravitational field of the planet, but my friend claims that both the floor and the tree are at rest in our frame of reference, so there would be no difference when measuring times.
Could you clarify who's right?
Thank you.
 A: You're correct. This is what was tested in the 1959 Pound-Rebka experiment.

my friend claims that both the floor and the tree are at rest in our frame of reference, so there would be no difference when measuring times.

In relativity, an inertial frame of reference is a free-falling frame. Therefore neither the bottom nor the top of the tree is an inertial frame. You can derive the same result as in Pound-Rebka if the whole experiment is done in an accelerating spaceship, far from any planet's gravitational field. This is the equivalence principle at work.
A: Yes, time does run differently on the ground an at the top of the tree.
To get measurable differences, you need to compare the clocks at the ground and at a satellite orbiting the planet.
Here on Earth, we now know that clocks on satellites run faster then clocks on the ground (if we disregard the movement of the satellite relative to the ground, as per the correct comment).
This is due to gravitational time dilation. Basically, the clock on the ground is closer to the center of gravity, and experiences more stress-energy.
It is a misconception that mass creates gravitational time dilation. It is stress-energy. The ground is deeper inside the gravitational field of the Earth, and experiences more stress-energy and the gravitational field slows the clocks more on the ground.
Basically, the idea wass originally that even small clocks are made of energy, and like photon clocks, are based on the distance light has to travel from one mirror to the other. 
Now deeper inside the gravitational field, spacetime is bent. Inside the photon clock, the space between the mirrors gets bent (gets longer compared to straight 3D) and light needs more time to travel from one mirror to another. One tick will then last longer. So the clock will tick slower, time will pass slower. This is only true if you compare it to the clock on the satellite. The clock on the satellite will seem to tick faster (if we disregard the movement of the satellite relative to the ground). 
A: According to special relativity, ignoring gravitation, your friend is right, at least if we ignore the fact that the earth is spinning. The factor that time is slowing down in a spherical symmetric gravitational field due to graviational time dilation as compared to the graviattional field not existing goes as:
$$\text{time elapsed if no gravity} \times {\sqrt{1-\frac{2GM}{rc^2}}} = \text{time elapsed}$$
So even if the top of the tree and the bottom of the tree do not move in relation to each other they are situated at different gravitational potentials and thus experiences different amounts of time passed. There are two different types of time dilations, both attributed to relativity, but the formula for velocity-related time dilation is much more known to the general public.
A: tl;dr: As far as the following is concerned:

...but my friend claims that both the floor and the tree are at rest in our frame of reference, so there would be no difference when measuring times.

the clock at the top of the tree will run about 1E-15 faster than the bottom of the tree, and this is due to a combination of both time dilation and gravitational shift.



*

*The Earth is rotating so it is not an inertial frame, and since one point is farther from the center of rotation than the other, they are at a fixed distance apart *but moving at different speeds.

*Even if the Earth weren't rotating, one position is in a deeper gravitational well than the other.


From here (or here if you are ambitious, also used here) the lowest order terms to the relativistic frequency shift of a clock in orbit around a gravitational body are:
$$ \frac{\Delta f}{f} \approx -\frac{\Phi}{c^2} - \frac{v^2}{2c^2} =  -\frac{GM}{r c^2} - \frac{v^2}{2c^2},$$ 
where the first term is the gravitational shift and the second is time dilation. 
Earth's standard gravitational parameter $GM$ is about 3.986E+14 m^3/s^2 A typical value for Earth's radius at the equator is 6,378,137 meters and it moves at about 465.1 m/s.
Let's say it's a 10 meter tall tree.
                            gravitational shift   time dilation        sum
On the Earth's surface        -6.9535914E-10      -1.203451E-12   -6.9656259E-10
In a 10 meter Tree            -6.9535805E-10,     -1.203454E-12   -6.9656150E-10
    Difference                                                     1.1E-15

So the clock at the top of the tree will go about 1E-15 faster than the bottom of the tree.
You'd be able to measure that with some atomic clocks, but it will take some effort!
