# Say I have my pen on my desk; does it describe a geodesic?

1. Aug 20, 2013

### Dreak

Let's say I have my pen on my desk; does it describe a geodesic.?

Or not because there is the normalforce working on it.

2. Aug 20, 2013

### WannabeNewton

It is certainly not a geodesic for exactly the reason you stated.

3. Aug 20, 2013

### Dreak

And what about someone falling from the sky? (let's assume that there is no fraction)?
Because I'm not certain about someone in freefall?

Is it ok if I write down my own thinking and you correct it?

If someone stands on the ground; we can go to a reference system in freefall by x'=x-1/2gt². So someone in freefall (falling from the sky, even if there is no fraction) also doesn't describe a geodesic.
The reason is because in a gravitational field; the metric is only local Minkowskian and thus not an inertial frame of reference?

Only outside a gravitation field, with no extern forces working on you (no rocket engines, no graf vield, no EM force...), I describe a geodesic?

4. Aug 20, 2013

### WannabeNewton

A test particle in free fall is described by a geodesic in space-time, by definition. A person falling in the Earth's atmosphere under influence of the Earth's interior gravitational field alone will of course be described by a geodesic. Why would you suspect otherwise?

5. Aug 20, 2013

### Dreak

quote from my course:

"A direct result of the universal movement of objects in a gravitational field is that a constant gravitation field g can always be transformed away by going to an other reference system.
Indeed, if we go over to reference system S' by coordinationtrasformation: x' = x - 1/2gt², we find that a = 0. We say that reference system S is in freefall."

But the pen on my desk also has a = 0, so it's also in freefall?

Or am I messing 2 things up?

6. Aug 20, 2013

### WannabeNewton

The pen on the desk doesn't have $a = 0$. It has $a = g$ because there is a normal reaction force from the desk on the pencil. If you instead dropped the pen towards the floor then $a = 0$ (ignoring air resistance). The second scenario describes a freely falling pen; the first scenario describes an accelerating pen.

Remember the Einstein elevator thought experiments: an object at rest in a uniform (constant) gravitational field $g$ is equivalent to an object accelerating in free space with magnitude $g$ whereas an object in free fall in a uniform gravitational field $g$ is equivalent to an object freely floating in free space. In our case, the pen on the desk is at rest in the uniform gravitational field of the Earth hence it is accelerating according to general relativity whereas the pen in free fall in the Earth's uniform gravitational field is actually inertial according to general relativity.

7. Aug 20, 2013

### Dreak

Of course! Thanks, can't believe I messed that up.

8. Aug 20, 2013

### WannabeNewton

No problem! More generally, if we have a particle described by some curve $\gamma$ in space-time with 4-velocity $u^a$, the 4-acceleration of the particle (i.e. the acceleration of $\gamma$) is given by $a^b = u^a \nabla_a u^b$ and $a^b =0$ if and only if $\gamma$ is a geodesic i.e. the particle is in free fall.

9. Aug 20, 2013

### Staff: Mentor

In GR geodesic worldlines are easy to identify. They are the ones where accelerometers read 0. An accelerometer strapped to the pen on your desk reads 1 g upwards.

10. Aug 21, 2013

### A.T.

You are confusing coordinate acceleration (dv/dt) and proper acceleration (what an acceleratometer measures). In General Relativity a geodesic world line corresponds to zero proper acceleration:
http://en.wikipedia.org/wiki/Proper_acceleration

In Newtons theory gravity is a real force that cancels the normal force, so the net force is zero for the pen on the table, and the world line is straight in undistorted space-time. In General Relativity there is only the normal force, which accelerates the pen, so it's world line is not geodesic (locally straight). See the apple hanging on the branch in this animation: