- #1

foges

- 53

- 0

You should upgrade or use an alternative browser.

- Thread starter foges
- Start date

- #1

foges

- 53

- 0

- #2

haushofer

Science Advisor

- 2,640

- 1,107

What I never understood is that when viewing a sattelite photo of the earth, people tend to say that the curly movement of the clouds is due to the Coriolis force. The Coriolis force emerges because one goes to a non-inertial frame, so on the earth one would experience the Coriolis force, while in outer space, one would NOT experience something like a Coriolis force.

Maybe this video helps,

The mathematical easiest way to see the emergence of inertial forces is, I think, the following: consider Newton's law in Cartesian coordinates, which holds in inertial frames:

[tex]

\frac{d^2 x^i}{dt^2} = 0

[/tex]

Now one goes (transforms) to a rotating frame. This can be achieved by

[tex]

x^i \rightarrow x^{'i} = R^i_{\ j}(t) x^j \ \ ,

[/tex]

where R is an element of SO(3) (which describes rotations), but in which the angles can be arbitrary functions of time t. Plugging this in Newton's law gives

[tex]

\ddot{R}^i_{\ j}(t)x^j + 2 \dot{R}^i_{\ j}(t)\dot{x}^j + R^i_{\ j}(t)\ddot{x}^j = 0

[/tex]

Using the orthogonality of the rotation matrices R (which, ofcourse still holds if you make the angles time dependent) you see the centrifugal force (first term) and Coriolis force (second term) arising. If you take an explicit example, like a rotation around the z-axis, you can explicitly calculate the corresponding R and see that it coincides with the expressions you learn in classical mechanics.

Maybe this video helps,

The mathematical easiest way to see the emergence of inertial forces is, I think, the following: consider Newton's law in Cartesian coordinates, which holds in inertial frames:

[tex]

\frac{d^2 x^i}{dt^2} = 0

[/tex]

Now one goes (transforms) to a rotating frame. This can be achieved by

[tex]

x^i \rightarrow x^{'i} = R^i_{\ j}(t) x^j \ \ ,

[/tex]

where R is an element of SO(3) (which describes rotations), but in which the angles can be arbitrary functions of time t. Plugging this in Newton's law gives

[tex]

\ddot{R}^i_{\ j}(t)x^j + 2 \dot{R}^i_{\ j}(t)\dot{x}^j + R^i_{\ j}(t)\ddot{x}^j = 0

[/tex]

Using the orthogonality of the rotation matrices R (which, ofcourse still holds if you make the angles time dependent) you see the centrifugal force (first term) and Coriolis force (second term) arising. If you take an explicit example, like a rotation around the z-axis, you can explicitly calculate the corresponding R and see that it coincides with the expressions you learn in classical mechanics.

Last edited by a moderator:

- #3

haushofer

Science Advisor

- 2,640

- 1,107

- #4

foges

- 53

- 0

Thanks for the replies haushofer, yes newtons bucket (or rather two stones and a string) seems to be exactly in line with my confusion (glad to see i wasn't the only one at odd with this concept, haha). Not quite sure I fully grasp it yet, but I'll read a bit more about it.

While on the topic can you maybe help me with Foucault pendulum, not really about the Coriolis force, but more about the experiments assumptions. Take a Cartesian coordinate system and make the axis of the earth's rotation in line with the z-axis. The coordinate system is stationary, respectively in it the earth makes one full revolution every 24hrs. Further assume we are standing on the true north pole, have attached a pendulum to the z-axis and we are holding the mass at some distance r from (perpendicular to) the z-axis. Since we are holding this pendulum it must also rotate with a period of 24hrs (in our coordinate system). When we let the pendulum go it clearly accelerates in the direction of the true north pole, however it still has some component of angular velocity (still with a period of 24hrs). Foucault's pendulum experiment on the other hand requires this angular velocity to be zero, does it not?

While on the topic can you maybe help me with Foucault pendulum, not really about the Coriolis force, but more about the experiments assumptions. Take a Cartesian coordinate system and make the axis of the earth's rotation in line with the z-axis. The coordinate system is stationary, respectively in it the earth makes one full revolution every 24hrs. Further assume we are standing on the true north pole, have attached a pendulum to the z-axis and we are holding the mass at some distance r from (perpendicular to) the z-axis. Since we are holding this pendulum it must also rotate with a period of 24hrs (in our coordinate system). When we let the pendulum go it clearly accelerates in the direction of the true north pole, however it still has some component of angular velocity (still with a period of 24hrs). Foucault's pendulum experiment on the other hand requires this angular velocity to be zero, does it not?

Last edited:

- #5

cjl

Science Advisor

- 1,961

- 543

What I never understood is that when viewing a sattelite photo of the earth, people tend to say that the curly movement of the clouds is due to the Coriolis force. The Coriolis force emerges because one goes to a non-inertial frame, so on the earth one would experience the Coriolis force, while in outer space, one would NOT experience something like a Coriolis force.

To respond to this: the satellite might be in outer space, but the clouds are on earth. Thus, the clouds can in fact experience a Coriolis effect, and it is indeed the source of much of the large-scale vorticity seen in the Earth's atmosphere.

- #6

foges

- 53

- 0

Anyone else on the non-zero angular velocity?

- #7

Subductionzon

- 172

- 2

If you have a point where this spinning space idea of yours worked then the only other points that would feel no force from this spin would be on the axis of that spin. That is not what is observed in the universe. Rotation is always observed to be local. Whether is is the rotation of the Earth, the Moon about the Earth, the Earth about the Sun, or the Sun around the Milky Way. Those different local rotations could not coexist with your idea of spatial rotation foges.

Share:

- Replies
- 5

- Views
- 3K

- Last Post
- Mechanics

- Replies
- 7

- Views
- 1K

- Replies
- 14

- Views
- 2K

- Last Post
- Mechanics

- Replies
- 4

- Views
- 3K

- Last Post

- Replies
- 16

- Views
- 5K

- Last Post

- Replies
- 3

- Views
- 5K

- Last Post

- Replies
- 8

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post
- Mechanics

- Replies
- 3

- Views
- 2K