# How Calculate Coriolis aceleration from Christoffel Symbols?

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1. Aug 19, 2017

### alejandromeira

1. The problem statement, all variables and given/known data
Hi, We are trying to calculate the Coriolis acceleration from the Cristoffel symbols in spherical coordinates for the flat space. I think this problem is interesting because, maybe it's a good way if we want to do the calculations with a computer.
We start whit the Christoffel symbols.
(r->1; theta->2; varphi->3).
Theta is the angle whit Z-axis (from 0 to pi), and varphi is the angle with the X-axis (from 0 to 2pi)

$\Gamma^{1}_{22}=-R$
$\Gamma^{1}_{33}=-R·(Sin\theta)^{2}$
$\Gamma^{2}_{12}=\Gamma^{2}_{21}=\Gamma^{3}_{13}=\Gamma^{3}_{31}=1/R$
$\Gamma^{2}_{33}=-sin\theta·cos\theta$
$\Gamma^{3}_{23}=\Gamma^{3}_{32}=\frac {cos\theta}{sin\theta}$
All the others Christoffel symbols are zero.

2. Relevant equations
To do the calculation I have used equation, with Einstein summation criterium:
$$a_{k}=\frac{\partial^{2}x_{k}}{\partial{t^{2}}}+\Gamma^{k}_{ij}·\frac{dx_{i}}{dt}·\frac{dx_{j}}{dt}$$
First of all, I don't know if we can use this equation for solve the problem

3. The attempt at a solution
Ok. We have obtained the following solutions (we start only the calculations of $a_{r}$, the rest is very mechanical)
$a_{r}=\frac{\partial^{2}r}{\partial{t^{2}}}+\Gamma^{1}_{11}·\frac{dr}{dt}·\frac{dr}{dt}+\Gamma^{1}_{12}·\frac{dr}{dt}·\frac{d\theta}{dt}+\Gamma^{1}_{13}·\frac{dr}{dt}·\frac{d\varphi}{dt}+...$(1+9 therms in total)
The results we have obtained, in this way, for the accelerations are:

$$a_{r}=\frac{\partial^{2}r}{\partial{t^{2}}}-R·[\dot\theta]^{2}-R·(sin\theta)^{2}·[\dot\varphi]^{2};m/s^{2}$$$$a_{\theta}=\frac{\partial^{2}\theta}{\partial{t^{2}}}+\frac{2}{R}·\dot{R}·\dot\theta-sen\theta·cos\theta·[\dot\varphi]^{2};s^{-2}$$$$a_{\varphi}=\frac{\partial^{2}\varphi}{\partial{t^{2}}}+\frac{2}{R}·\dot{R}·\dot\varphi+2\frac{cos\theta}{sin\theta}·\dot\theta·\dot\varphi;s^{-2}$$
The problem we have is that we don't know how to interpret these result. We suppose that in these results they are: the centrifugal acceleration, the acceleration of Coriolis, and the acceleration of drag, but I don't know how to separate one from each other.
We also don't know if the left side of the equation is the aceleration in the inertial reference frame or not.
Ok we need a lot of help, you can see!!!

2. Aug 19, 2017

### Orodruin

Staff Emeritus
What you have written down are just the acceleration equations in spherical coordinates. In order to find the centrifugal and Coriolis forces you have to go to an actually rotating frame. How can you relate fixed spherical coordinates to those of a rotating frame?

Edit: Also, you should make a habit of writing the coordinate indices as contravariant indices, i.e., $x^i$ instead of $x_i$. This will help you when you check that your expressions make sense.

3. Aug 20, 2017

### alejandromeira

This is my challenge. I think that we can obtain these acelerations from the equations of acceleration shown above.
I think the term of centrifugal forces is in $a^{r}$, but there are two terms because the rotation axis does not have to coincide with a Cartesian axis.

The problem statement, all variables and given/known data
To beguin we will consider the surface of the earth and we will forget about gravity.
We will assume that the Z axis is the axis of rotation of the earth. The earth rotates with constant angular velocity omega. In this way we have: $$\dot\theta=0$$$$\dot\varphi=\omega$$
Initially the mobile is in $(R, \theta, \varphi)$

The attempt at a solution

ATTENTION I DO NOT KNOW IF THE FOLLOWING CONCLUSIONS ARE OK OR NOT​

A) DISCUSSION ABOUT $a^{r}$
With these assumptions, the acceleration equation for $a^{r}$ we have calculated is: $$a^{r}=\dfrac{d^{2}r}{dt^{2}}+R·\omega^{2}·(sin\theta)^{2}$$ I think that the centrifugal acceleration is: $R·\omega^{2}·sin\theta=\rho·\omega^{2}$. Also the radial component is $R·\omega^{2}·(sin\theta)^{2}$ and vanishes a $\theta=0$, and it is maximal at the equator when $\theta=\pi/2$.
We have made a figure with pencil and paper, and centrifugal acceleration has a radial and tangential components. We assume that the tangential component is that which appears in $a^{\theta}$. (I don't know if this is correct, but looking at the figure it seems to me that yes.

B) DISCUSSION ABOUT $a^{\theta}$
If we assume $\dot\theta=0$, whe have obtained for $a^{\theta}$ the equation: $$a^{\theta}=\dfrac{d^{2}\theta}{dt^{2}}-\omega^{2}\sin\theta·\cos\theta$$It is an angular acceleration. If we multiply this equation by R for obtain te lineal acceleration it is clear that the second term is the tangencial component of centrifugal force.

C) DISCUSSION ABOUT $a^{\varphi}$
I have obtained for $a^{\varphi}$: $$a^{\varphi}=\dfrac{d^{2}\varphi}{dt^{2}}+\frac{2}{R}·\dot{R}·\omega$$ It is also an angular acceleration. Then if we multiply the equation by R we can obtain the Coriolis term: $2·\dot{R}·\omega$

Questions

1) I do not know if I have interpreted the problem correctly.
2) In equation of $a^{\varphi}$ and the Coriolis term only appears $\dot{R}$. I know that if the particle moves from the equator to the pole, by a meridian, ar R=const, coriolis acceleration appears. Is this described in the above equations?
3) What is the drag acceleration?
4) How can we introduce the acceleration of gravity into the equation of $a^{r}$?

Ok, I have a lot of doubts. All help is welcome.

Last edited: Aug 20, 2017
4. Aug 20, 2017

### Orodruin

Staff Emeritus
You cannot put $\dot \theta=0$ and $\dot \varphi = \omega$ and expect to recover the Coriolis force since it appears due to motion in the rotating frame. Your assumptions would describe an object at a fixed direction in the rotating frame.

5. Aug 20, 2017

### alejandromeira

Thank you. I am very disoriented.
Do you know any reference where I can study the problem?

6. Aug 20, 2017

### Orodruin

Staff Emeritus
Just think about how the coordinates in a rotating spherical coordinate system (preferably around the z axis) relate to a fixed spherical coordinate system. What coordinates change between them?

7. Aug 21, 2017

### alejandromeira

Ok, I already found what I was looking for:
https://www.maplesoft.com/applications/view.aspx?SID=4892&view=html
Although in the previous webpage the $\theta$ and $\varphi$ coordinates are changed.
I need to multiply by R the equation of $a^{\theta}$, and by $R·sin\theta$ the equation of $a^{\varphi}$ to obtain accelerations in $m/s^{2}$

I only need to know how to introduce the acceleration of gravity into equation.

8. Aug 23, 2017

### alejandromeira

Hello we have been working on the subject. To introduce a constant gravitational field we have tried to use common sense. The gravitational field g is radial and must have negative sign. Then the radial component of the centrifugal force must have positive sign. There are no gravitational field on $\theta$ and $\varphi$ directions. Using these assumptions we have: $$-g=\ddot{R}-R·[\dot\theta]^{2}-R·(sin\theta)^{2}·[\dot\varphi]^{2}$$ $$0=R·\ddot\theta + 2·\dot{R}·\dot\theta - R·sin\theta·cos\theta·[\dot\varphi]^{2}$$ $$0=R·sin\theta ·\ddot\varphi + 2sin\theta · \dot{R}·\dot\varphi + 2R·cos\theta ·\dot\theta · \dot\varphi$$ We can then clear, in these equations, the accelerations in m/s², for each component, we get: $$\ddot{R}=-g+R·[\dot\theta]^{2}+R·(sin\theta)^{2}·[\dot\varphi]^{2}$$ $$R·\ddot\theta = -2·\dot{R}·\dot\theta + R·sen\theta·cos\theta·[\dot\varphi]^{2}$$ $$R·sin\theta ·\ddot\varphi = -2sen\theta · \dot{R}·\dot\varphi - 2R·cos\theta ·\dot\theta · \dot\varphi$$ The cosine function makes these terms change sign at the equator, as it should be. In addition we have verified that when we move on a meridian, from the north pole to the equator, an acceleration towards the west appears as one would expect. This acceleration is contrary beyond the equator due to the cosine sign change. If we move from the equator to the north pole, the sign of the velocity is opposite, and the acceleration appears towards the east, as expected.

If I have time after lunch, I will write some particular problem of using these equations, for example that of an object that is still at a certain latitude, and also that of an object moving along a meridian from the pole to the equator (they are the ones that we have solved until now)

Last edited: Aug 23, 2017
9. Aug 24, 2017

### alejandromeira

Relevant equations
We will use these equations obtained above, to solve some examples. $$\ddot{R}=-g+R·[\dot\theta]^{2}+R·(sin\theta)^{2}·[\dot\varphi]^{2}$$ $$R·\ddot\theta = -2·\dot{R}·\dot\theta + R·sen\theta·cos\theta·[\dot\varphi]^{2}$$ $$R·sin\theta ·\ddot\varphi = -2sen\theta · \dot{R}·\dot\varphi - 2R·cos\theta ·\dot\theta · \dot\varphi$$ Left side are the accelerations in m/s².
Remember that the $\theta$ angle (0..pi) is positive from the north to the soud, and also this is the criterium for $\dot\theta$. The movement from the sud to the north have, then, negative sign.
********************************************************************************************************************************************************
EXAMPLE 1: A mass remains still on the surface of the Earth at a certain latidude. Calculate the acceleration perceived by the mass.

If the mass is still, then: $\dot{R}=0$ and $\dot\theta=0$, only varies $\dot\varphi=\omega$ due to the Earth rotation. Substituting these values in the previous equations, and calculating we have: $$\ddot{R}=-g+R·\omega^{2}·(sin\theta)^{2}$$ $$R·\ddot\theta=R·sin\theta·\omega^{2}·cos\theta$$ $$R·sin\theta·\ddot\varphi=0$$
$\ddot{R}$ indicates that the apparent weight is lower. $R\ddot\theta$ indicates that there is a tangential force towards the ecuator. (is great the cos function changing the sign a pi/2 in the equator and then changing the direction of the tangential force).
In this case only appears the real weigth an the centrifugal force. The centrifugal force is due only by $\omega$.
*******************************************************************************************************************************************************
EXAMPLE 2: An observer moves along a meridian, with angular velocity $\dot\theta=constant$, from the north pole to the equator. Calculate the acceleration perceived by the observer.
In this case $\dot{R}=0$; $\dot\theta=constant$ and $\dot\varphi=\omega$ Replacing in the equations and calculating we have: $$\ddot{R}=-g+R·[(\dot\theta)^{2}+\omega^{2}·(sin\theta)^{2}]$$ $$R·\ddot\theta=R·sin\theta·\omega^{2}·cos\theta$$ $$R·sin\theta·\ddot\varphi= -2R·\dot\theta·\omega·cos\theta$$ $\ddot{R}$ indicates that with respect to the previous case, appears another radial centrifugal component, due to the movement along the meridian.
$R\ddot\theta$ is the same as in the previous example. We have already discussed it above, and now:

$R·sin\theta·\ddot\varphi$ IS THE CORIOLIS ACCELERATION. Note that due to the minus sign, when going from the north pole to the equator, the observer drifts westward. If we move from the equator to the pole, then $\dot\theta$ is negative and then as $(-)·(-)=+$ the observer drifts eastward.

In addition, as you can see, the cosine function changes sign at the equator (pi/2), and therefore in the south hemisphere the shifts are inverse.

It is impressive how the cosine function works at the equator, to keep the correct N-S, E-W displacements.

Okay, I'd appreciate it if anyone observes any errors, let me know. Sorry for my little English I'm not native.