# A falling object

1. Apr 18, 2008

### Bose

My limited understanding of relativity tells me that "gravitational forces" are really just due to the curvature of spacetime. That is, the reason why the earth orbits the sun is because the earth is following a "straight" line but the actual spacetime is curved so the "straight" line appears curved. Like an ant on a sphere, it walks straight but the sphere is curved so the path is curved. Anyways, when you hold an object in the air and then let go, it falls. Why? What's giving it the "push" to fall?

2. Apr 19, 2008

### JesseM

It falls because that's its natural geodesic path, the "straight line" through spacetime--it is before it falls that some force must be acting on it to keep it from following the geodesic (specifically, the electromagnetic force between the atoms in your hand and the atoms in the object which allows you to keep it in your grip).

Imagine I'm standing on a platform accelerating upwards in empty space at 1G. If I hold an object, the platform is accelerating my feet, which accelerates my body, which accelerates my hand, which accelerates the object I'm holding, so we all accelerate together. If I let go of the object, then now there is no force acting to accelerate it, so it will just move inertially in a straight line at constant velocity, the same velocity it had at the instant I let go--and since the platform is continuing to accelerate upward, the platform's velocity in this direction will increase and so it will eventually catch up with the object, which from my perspective on the platform makes it appear the object has "fallen" from my hand to the ground. Take a look at the nice little animation here for help visualizing this (especially the second part of the animation, with the caption 'in the frame of reference of the stars').

Last edited: Apr 19, 2008
3. Apr 19, 2008

### Antenna Guy

Let's say I have two equal masses free-falling side-by-side in a gravitational field.

Why do their "natural geodesic paths" converge?

If I limit the gravitational field to that of the two equal masses, why do the paths still tend to converge?

Regards,

Bill

4. Apr 19, 2008

### HallsofIvy

Staff Emeritus
"Why" geodesic paths are what they are involves solving a differential equation derived from the metric tensor of the space. In particular, if you have a space consisting of a single mass point, there exist "radial" geodesics so a "test mass" (assumed to have mass much lower than the original point mass and not used in calculating the field) will, if it has no initial velocity with respect to the point mass, move directly toward the mass point. Two such "test masses" will converge because both their radial geodesic converge to the point mass. If you want to include the masses of the two objects in calculating the field, that is much harder.

If your space consists entirely of two equal masses, their paths converge because there exist a geodesic from directly one to the other.

5. Apr 19, 2008

### Antenna Guy

Let's look at that particular geodesic ("from directly one to the other").

Let each mass have the same net charge (supporting a 'static" electric field about each).

At a point directly between the two masses, the electric fields cancel since they are equal in amplitude, and in opposite directions.

Consider that the field strength about either charged mass falls off with $\frac{1}{r^2}$, and note that r is in the direction of the respective geodesics (time). Would the magnetic fields associated with the superposition of the two $\frac{\delta E}{\delta r}$ terms cancel or add at the same mid-point?

Regards,

Bill

6. Apr 19, 2008

### LURCH

Don't know if this helps any, but:

You already seem to understand the geometry that causes orbits. This is the knowledge you should apply to the question you have asked. When you hold an object in the air and let it go, it is released and becomes free to travel its own "strait line" through curved space. As soon as the object is free, it begins to "orbit" the center of the earth (mind you, that's the center of Earth's gravitational field). The only difference ebtween the object orbiting Earth and the earth orbiting the sun, is that the object's orbital path is much lower. It is trying to follow an orbit that would take it within a few hundred yards of the Earth's center, but it can't travel that path very far without hitting an obstruction; the Earth's surface.

7. Apr 19, 2008

### Bose

Do you agree that an ant on a sphere has an option of not moving forward but an object in the air does not have this option once released?

8. Apr 19, 2008

### Crazy Tosser

An object has a choice, because you can give it an initial velocity.
If the initial velocity is great enough (and the angle is right), the object will go into orbit.

9. Apr 19, 2008

### Bose

No, I don't think an object has a choice when released free without a force act on it

10. Apr 19, 2008

### aaj

To that extent you are right. An object must always keep moving in a geodesic through spacetime. It cannot sit still in spacetime. For instance, in the absence of a gravitational field i.e in flat spacetime, an object is still moving through time. In other words, its worldline is still a straight vertical line in the fabric of spacetime.

In the presence of a gravitational field i.e curved spacetime, this worldline (geodesic) will deviate from the original vertical worldline. This deviation from the original vertical worldline manifests in the form of the object moving from its original spatial coordinates.

Pros, please correct me if my wording in the last para is slightly off the mark.

11. Apr 19, 2008

### MeJennifer

Objects, or more exact test objects, are not points but lines in spacetime.

12. Apr 19, 2008

### phyti

JesseM:
You gave him an analogy, or are you saying the earth is expanding?

Bose:
Why does an object move toward the center of mass?

13. Apr 19, 2008

### Antenna Guy

What direction is time?

When you move from one inertial frame to another, does the (reference) direction of time change? I think it does...

How else would you know which spatial dimension of an object at speed is Lorentz contracted within a particular inertial frame?

Regards,

Bill

14. Apr 19, 2008

### MeJennifer

The direction of time depends on the observer's chart of spacetime.

Out of curiosity did anyone prove under GR that wordlines must merge in all configurations of two test objects with not necessarily equal momenta and masses? It would be particularly interesting in cases that seem to have CTCs.

Anyone knows?

15. Apr 19, 2008

### Antenna Guy

Exactly.

Given that an infinite number of inertial observers could be constructed about a "test object", time (with respect to the "test object") must flow in all directions to account for any particular observer.

Is that a line - or an area?

Entropy issues with the above aside, have you thought about the superposition of magnetic fields that I mentioned earlier?

Pardon my ignorance, but what is a CTC?

Regards,

Bill

16. Apr 19, 2008

### Staff: Mentor

Even if you are at rest in space you are still moving through time. Everything always travels at c through spacetime. If you are moving inertially then your path through spacetime is a geodesic, but inertial or not nothing can "stop" in spacetime.

17. Apr 19, 2008

### Antenna Guy

Except the progression of time for a photon...

Regards,

Bill

18. Apr 19, 2008

### Crazy Tosser

If you are talking about no force acting on it, than ant on a sphere has no choice either, because every move it makes is a force that acts on it with nothing related to physics laws whatsoever.

A rock dropped on a spehere doesn't have much choice

19. Apr 20, 2008

### LURCH

No, a photon travels at c through space, but does not progress through time. Therefore, its total progression through space time is c.

Bose, has your original question that answered? A rock fallen to the ground is simply in a very low orbit around the center of the earth, and falls for the exact same reason that the Earth orbits the sun and the moon orbits the Earth.

20. Apr 20, 2008

### Antenna Guy

Why do you say I am wrong, and then tell me that a photon does not progress through time (essentially what I said in the first place)?

Regards,

Bill