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How big is an inertial frame?

Consider a huge rod which is rotating about a fixed point in a plane, its length is 1 light year.

Thus light from its end closer to the fixed point to the end farther from the fixed point takes one year to reach.

Now the angular velocity of the point closer to the fixed point is much slower than the angular velocity of the farther point.

Thus the end closer to the fixed point has a relative velocity to the end farther away.

At some point in time, the clocks at the two ends are synchronized by sending a light signal to both ends from exactly from the centre of the rod (half a light year from any point).

My question is, are the two ends of the rod in the same inertial frame? Is it possible for 1 object to be in two inertial frames simultaneously?

A simpler question. A rod is 1 light year long and travelling at a a velocity c/2 along its length. There will be length contraction of the rod by gamma. Now imagine a light source at one end of the rod. The light from one end takes one year to reach the other end (as measured by either end), when the rod is stationary.

Now since the distance is contracted when the rod is moving, so must the time be different (as compared to an observer looking at this rod) for the two ends of the rod as light must be measured with the same velocity on both ends. The time must be same at both ends by symmetry, so they are indeed in the same inertial frame.

However, if there is length contraction, they will not measure one light year, they should each measure less than one light year.

An external observer observing this rod would see the distance as less than one year. But to measure the same velocity of light, must also measure the same time as the ends of the rod, thus implying the observer is in the same frame as the rod?

The second part of my question is this. In zero gravity, a person holds a ball in his hand. Thus the person and the ball are in the same inertial frame. Then the person "throws" the ball and as a result, both and ball and the person now acquire a relative velocity to each other (action and reaction). Are the two in the same inertial frame?

The third part of my question is this. In zero gravity, can a ship and an observer on the ship be in the same inertial frame? That is, do they share acceleration? Assuming the ship is a plane, when the plane accelerates, the person is floating around, hence the person and the ship cannot share acceleration unless the person is strapped to the ship.Thus in zero gravity, every object must have its own inertial frame?

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A simpler question. A rod is 1 light year long and travelling at a a velocity c/2 along its length. There will be length contraction of the rod by gamma. Now imagine a light source at one end of the rod. The light from one end takes one year to reach the other end. Now since the distance is contracted, so must the time be different for the two ends of the rod as light must be measured with the same velocity on both ends. So the two ends must be in different inertial frames with different time. So, where does the frame of one side of the rod end and the frame of the other side start –  Physics n00b Mar 21 '13 at 11:07
    
To suggest that the ends of a rotating rod share a inertia frame means that you have misunderstood the term. From the POV of one end the other one is under acceleration. –  dmckee Mar 21 '13 at 12:48
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2 Answers

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General relativity complicates things a bit, but in special relativity an inertial frame is a time independent co-ordinate system that is homogenous and isotropic. The time independence means that everything in that frame is at rest wrt everything else, and the homogenous and isotropic bit means the co-ordinate system has no curvature.

Take your question 1: you could define a rotating frame that rotates along with your rod, but this frame would not be isotropic (except at the pivot point) because there is an acceleration acting towards the pivot point and this picks out a particular direction. It's also not homogeneous because the acceleration varies with position.

Incidentally, your example of the rod is basically the same as the rotating disk I mentioned in my answer to your other question. You can treat the rod as a radial strip in the disk.

Re your other questions, can you ask these separately. As it is, the answer is heading towards essay size!

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I agree with your take on question 1 and how it can be treated as a radial disk. –  Physics n00b Mar 21 '13 at 17:55
    
The time independent bit, I am still unclear about. Would you consider a moving photon in the inertial frame at rest? What I mean is, in the original time dilation experiment, the photon in the light clock is not at rest? –  Physics n00b Mar 21 '13 at 18:20
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Suppose the co-ordinates of some particle in your inertial frame are $(x, y, x)$, then time independence means these co-ordinates don't change in time, in other words the particle is stationary wrt you. If the particle is moving that doesn't mean it's not in an inertial frame, but it does mean it's not in the same inertial frame as you are. The photon is a special case because there is no frame in which a photon is stationary. Every observer in every physically possible frame will observe the photon to be moving (at the speed of light). –  John Rennie Mar 22 '13 at 14:20
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An inertial frame is a local property in general relativity. There is no frame that is inertial with respect to both ends of the rod in the first scenario you mention. The frame that includes both ends of the rod is rotating and therefore is equivalent to a frame with a curved space-time (wherein the force of "gravity" points outward).

Your example regarding a ball being thrown is different. After the ball is thrown the ball and person are simply moving with fixed relative velocity and therefore there any inertial frame with respect to one is also an inertial frame with respect to the other.

I'm afraid I don't really follow the last scenario you describe.

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Technical point: the spacetime is still flat, even in the rotating frame. A freely moving body accelerates in the rotating frame, but there are no tidal forces, therefore no curvature. –  Michael Brown Mar 21 '13 at 11:19
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