| Thread Closed |
A falling object |
Share Thread |
| Apr18-08, 11:51 PM | #1 |
|
|
A falling object
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?
|
| Apr19-08, 01:24 AM | #2 |
|
Recognitions:
|
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'). |
| Apr19-08, 05:56 AM | #3 |
|
|
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 |
| Apr19-08, 08:44 AM | #4 |
|
|
A falling object
"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. |
| Apr19-08, 10:27 AM | #5 |
|
|
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 [itex]\frac{1}{r^2}[/itex], and note that r is in the direction of the respective geodesics (time). Would the magnetic fields associated with the superposition of the two [itex]\frac{\delta E}{\delta r}[/itex] terms cancel or add at the same mid-point? Regards, Bill |
| Apr19-08, 02:04 PM | #6 |
|
|
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. |
| Apr19-08, 02:30 PM | #7 |
|
|
|
| Apr19-08, 02:43 PM | #8 |
|
|
If the initial velocity is great enough (and the angle is right), the object will go into orbit. |
| Apr19-08, 03:05 PM | #9 |
|
|
|
| Apr19-08, 03:45 PM | #10 |
|
|
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. |
| Apr19-08, 04:55 PM | #11 |
|
|
Objects, or more exact test objects, are not points but lines in spacetime.
|
| Apr19-08, 06:22 PM | #12 |
|
|
JesseM:
Bose: |
| Apr19-08, 06:33 PM | #13 |
|
|
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 |
| Apr19-08, 06:53 PM | #14 |
|
|
Anyone knows? |
| Apr19-08, 07:22 PM | #15 |
|
|
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? Regards, Bill |
| Apr19-08, 08:00 PM | #16 |
|
Mentor
|
|
| Apr19-08, 08:11 PM | #17 |
|
|
Regards, Bill |
| Thread Closed |
Similar discussions for: A falling object
|
||||
| Thread | Forum | Replies | ||
| Falling Object | Introductory Physics Homework | 20 | ||
| A falling object | Special & General Relativity | 16 | ||
| Falling Object | Introductory Physics Homework | 5 | ||
| Falling object at c | General Astronomy | 5 | ||