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Particle on a curved trajectory

by Sunfire
Tags: curved, particle, trajectory
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Sunfire
#1
Sep8-12, 09:28 PM
P: 215
Hello

Particle with nonzero mass moves on a curved trajectory. Why does it feel centrifugal gravity? Does spacetime curve or...?

One answer might be that it feels gravity because the 1st law of Newton says so; but this wouldn't be good enough an answer :)

Thank you
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Matterwave
#2
Sep8-12, 10:01 PM
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What is centrifugal gravity? ...
tom.stoer
#3
Sep9-12, 01:35 AM
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even particles with zero rest mass move on such trajectories called geodesics

Sunfire
#4
Sep9-12, 07:17 AM
P: 215
Particle on a curved trajectory

Perhaps I didn't say something right... There is a moving frame of reference - a radius-vector drawn from the coordinate origin points to the particle. In the frame where this vector is motionless, fictitious force appears - the centrifugal force. It is a gravitational force, acting radially out. This frame of reference is non-inertial.

Why does gravity appear in such frame? Is there a spacetime explanation for it?
pervect
#5
Sep9-12, 09:20 AM
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The "centrifugal force" isn't due to space-time curvature, but something called the "connection coefficients" or "Christoffel symbols". I'm not sure how much help that will be, but I can't think of a less technical explanation at the moment. I'm not sure if Ill be able to come up with one when I'mmore awake - probably not, but there's some chance I might.

Because curvature is a tensor, if space-time is flat for the non-accelerating observer, i.e. if there aren't any large masses nearby, it's also flat for the accelerating one. The connection coefficients are NOTtensors, so they can and do vary with the state of motion of the observer.
Sunfire
#6
Sep9-12, 01:06 PM
P: 215
Dear pervect

thank you for your reply. I am not very familiar with Christoffel symbols, but I understand the explanation to be similar to what is happening in rotating polar coordinates - expressing the motion in them, "fictitious" forces appear (centrifugal/Coriolis) purely from the mathematical derivation itself. Do I understand you correctly that you are referring to something similar?
pervect
#7
Sep9-12, 09:41 PM
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Certain components of the Christoffel symbols can be closely identified with the various pseudoforces you mention. I dont think you can interpret ALL of the components in the symbols in this manner, though. If you think of the Christoffel symbols as a superset that includes information about the pseuo-forces due to acceleration and rotation , it will probably be the safest approach.
DrGreg
#8
Sep10-12, 08:04 AM
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When an apple falls from a tree, if you are stood on the Earth, the apple accelerates with g downwards, relative to you, and the Earth is stationary. However, from the apple's point of view, the apple is stationary and you are accelerating upwards with g.

General relativity works on the assumption that the apple's point of view is correct and your own view is "fictitious". According to the apple, no forces act on it, so it is stationary. You are accelerating upwards because the Earth is pushing you upwards with a force. According to you, you have to introduce a "fictitious force" to explain what pulls the apple downwards and acts on you opposing the Earth pushing you up.

General relativity provides a method for expressing the laws of physics in any frame you choose. In some frames there will be "fictitious forces" and these are calculated using the Christoffel symbols referred to above. When the frame is in local free-fall, like the apple's frame, all the local Christoffel symbols are zero.
Sunfire
#9
Sep10-12, 02:51 PM
P: 215
As I understand it then, gravity is not due to spacetime curvature; neither is it due to the presence of mass
Whovian
#10
Sep10-12, 03:01 PM
P: 642
Quote Quote by Sunfire View Post
As I understand it then, gravity is not due to spacetime curvature; neither is it due to the presence of mass
What?

Okay, let me explain the equivalence principle. Say I'm standing on Earth in a box. This is locally indistinguishable from being in a box in space accelerating at 9.8 m/s^2. By locally indistinguishable, that means that I can't detect the minor variations in the magnitude and direction of the gravitational field by moving a little in a certain direction.

If you're in an elevator in freefall, this is locally indistinguishable from being in an airtight elevator drifting through space.

Now, what depends on the presence of mass is how much acceleration someone standing on the surface of this mass "feels" upwards. That is, if the Earth were twice as massive with the same radius, we'd feel an upwards acceleration of about 20 m/s^2 (I'm too lazy to do 9.8*2 right now.) A good way to visualize this whole thing with curvature is to visualize a two-dimensional space with a third dimension representing time.
DrGreg
#11
Sep10-12, 03:09 PM
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Quote Quote by Sunfire View Post
As I understand it then, gravity is not due to spacetime curvature; neither is it due to the presence of mass
You might find this post helpful.
Sunfire
#12
Sep10-12, 03:17 PM
P: 215
Sorry, I was just about to edit my last post so as to appear less provocative :)

What I meant to say is, that it appears to me that centrifugal gravity is not a result of spacetime curvature (correct me if I am wrong).
Whovian
#13
Sep10-12, 03:24 PM
P: 642
Quote Quote by Sunfire View Post
Sorry, I was just about to edit my last post so as to appear less provocative :)

What I meant to say is, that it appears to me that centrifugal gravity is not a result of spacetime curvature (correct me if I am wrong).
Ah, correct. Sorry.
Austin0
#14
Sep10-12, 04:00 PM
P: 1,162
Quote Quote by DrGreg View Post
When an apple falls from a tree, if you are stood on the Earth, the apple accelerates with g downwards, relative to you, and the Earth is stationary. However, from the apple's point of view, the apple is stationary and you are accelerating upwards with g.

General relativity works on the assumption that the apple's point of view is correct and your own view is "fictitious". According to the apple, no forces act on it, so it is stationary. You are accelerating upwards because the Earth is pushing you upwards with a force. According to you, you have to introduce a "fictitious force" to explain what pulls the apple downwards and acts on you opposing the Earth pushing you up.

General relativity provides a method for expressing the laws of physics in any frame you choose. In some frames there will be "fictitious forces" and these are calculated using the Christoffel symbols referred to above. When the frame is in local free-fall, like the apple's frame, all the local Christoffel symbols are zero.
Hi DrGreg
It seems to me that the upward force in the cases of being static in gravity or accelerating in flat spacetime is equally fictitious .

In the gravity case it is essentially the same as the fictitious centrifugal force arising in a rotating system. Inertial momentum acting against a physical constraint (centripetal acceleration). Say a rotating space station.

In the accelerating system outside of gravity it is just due to the propagation of momentum acting on the inertia of a particular structure. There is nothing going on but conservation of momentum and inertia.

In fact, in the case of mechanical acceleration (direct contact) , the Newtonian force F itself, seems to be a fiction. A useful abstract variable that allows us to quantify in general terms what, in any specific instance, is nothing more than the transference and conservation of momentum. DO you see any meaning to the term distinct from this??

I have only a hazy concept of connection coefficients or Christoffel symbols but am curious to know if they change anything or add to the physics described by the basic Newtonian motion laws in these cases??
Those laws seem to completely explain and describe all these circumstances but I may be missing something. Thanks
Sunfire
#15
Sep10-12, 04:05 PM
P: 215
It also appears to me, that centrifugal gravity is not dependent on the presence of mass


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