Discussion in 'Physics & Math' started by Prosoothus, Apr 4, 2006.
Because you started a spelling cockfight. How's your 'schedule'?
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It takes two to fight. JamesR is not an innocent little sweety pie. Neither are you.
But this is all for now between you and me unless you want to discuss something important in physics (a first for you).
We're waiting for your relativity killer. In your own time, of course.
If you want to discuss something while taking a break from your analysis, there are unfinished discussions in the other threads. You have yet to respond to this post and a number of posts in [post=1042196]this thread[/post].
This would be a really good experiment. I agree that you would need an onboard clock whose motion you could control and not a distant clock like a pulsar. If they didn't keep synchronized then it would clearly indicate that the first postulate is wrong. I personally hope it gets done sometime, but it will obviously require a really good mechanical clock.
Any Rolex could be chosen at random and serve the purpose. For that matter, a hundred could be chosen at random at shot into orbit for a fraction of the cost of the average orbital Relativity experiment.
And do you remember that when Einstein promoted the time dilation, a Rolex would have been the only type of clock in existence? Or did you ever know?
Excuse me. There were pendulum clocks and hourglass clocks and the like. What happens to a pendulum clock in low gee?
Don't be silly CANGAS. This has nothing to do with what kinds of clocks were available in Einstein's day. It is all about precision and experimental sensitivity. If you have an imprecise clock then you must be able to achieve much higher relative velocities before the relativistic effects rise out of the noise and become measurable.
Einstein's thoughts about clocks had nothing to do with the only clocks he possibly knew about?
Now tell me the one about Goldilocks and the three bears. You know some good fairy tales, you silly goose.
I don't know about his thoughts, but his theory had nothing to do with the exact mechanism of a clock. The only times he even mentioned the construction of a clock was to say that identically constructed clocks would tick at identical rates. This idea applies equally well to clocks of any mechanism.
Let's assume that space is the photon's medium of travel, but gravitational fields are its medium of propulsion. If this was the case, then a photon entering a moving uniform gravitational field would have its speed altered (to match the moving gravitational field), but not its path (because the space is not moving). Only in a non-uniform gravitational field would a photon's path change due to the photon rotating to point towards the stronger part of the field.
Interesting! Could you please explain how gravitational fields could be a medium (means?) of propulsion for the light?
More importantly (from a practical standpoint), what's the local speed of gravity? What does it mean for a uniform gravitational field to have a speed? Is this concept being introduced merely to explain the behaviour of light near large masses?
Well, first let's assume that the gravitational interaction has two poles, positive and negative, just like the electrostatic interaction. However, unlike the electrostatic interaction, in the gravitational interaction like poles attract while opposite poles repel. Because of this difference, over billions of years the universe has seperated into large regions of positive gravitational areas and large regions of negative gravitational areas. For the sake of argument, let's assume that our galaxy is a large positive gravitational area that pushed all the negative gravitational matter out of it a long time ago.
Now, let's say that you have a piece of positive gravitational matter sitting in a uniform positive gravitational field. Standard matter generates a gravitational field that is equal at all sides of it, so a piece of positive gravitational matter sitting in a uniform positive gravitational field is equally attracted at all sides by the field. Because of this, the net forces on that piece of matter are equal to zero, which is why the piece of matter remains stationairy.
However, not all of the matter in the universe is the standard type. I believe that photons, and other light-speed particles, generate a non-uniform, or even a dipolar, gravitational field. In other words, the gravitational field on one side of this type of particle is larger than, or is the opposite of, the field on the other side. When this type of particle is placed in a uniform positive gravitational field, the field attracts the side of the particle that is gravitationally positive, and pushes against the other side of the particle that is gravitationally negative. This creates a net force on the particle that causes it to accelerate in the direction of its positive side (at least in our galaxy). Finally, when the particle reaches c its stops accelerating because c is the speed of the gravitational interaction.
Just as mass moves around in our universe, the mass drags the gravitational fields that it generates around with it. When a light-speed particle exits one gravitational field that is moving at one speed and enters another gravitational field moving at another speed, the speed of the particle will gradually change so that it's speed is equal to c relative to the new field it is passing through.
No. There are gravitational fields everywhere in the universe, and no matter how weak they are in certain areas, photons, and other light-speed particles, will travel at c relative to those fields. It just may take a longer time for a photon to accelerate, or decelerate, to c in a weaker field than in a stronger one.
This is a very basic question, but I seem to be lost on old ground.. Can someone clarify if it is the relative velocity of the photon clock pair that determines time dilation, or whether the direction of that velocity (towards or away from each other) matters? Before I post the follow up question, I just want to make sure I am perfectly straight on that one. thx.
If you really want to do it correctly you should always use the full Lorentz transform. That will always evaluate all three relativistic effects (time dilation, length contraction, relativity of simultaneity) correctly. Time dilation is isotropic, but length contraction is not, so sometimes the direction of the velocity matters.
Thanks Dale.. Here's what precipitated the question. If you take your two sets of parallel mirrors (A/B and C/D) and start them off with the equivalent light particle bouncing up and down and start the mirror pairs off at rest to each other, they tick in unison. the first question is how do they observe the other's tick rate.. and the answer has to be by "watching" ie observing the photons and that can't happen unless photons are being sent from one to the other so simplify it.. Imagine that on the bottom mirror of both pairs is a sensor and each time the bouncing photon strikes the bottom mirror a photon emitter sends a pulse (photon) horizontally to the other mirror set (assuming A and C are the bottom mirrors of each set). So there is an emitter and a sensor on the inward edge of A and C facing each other. when both sets are at rest relative to each other, both send and recieve the horizontal photon at the same exact time no matter how far apart they are and the tick rate remains consistently in step. ok.. now assume mirror set A/B starts moving at 100,000 miles per second horizontally away from mirror set C/D, in this case.. assuming that each vertical bounce takes one second (in the rest frame) each time a horizontal communication photon gets emitted the mirror sets will be 100,000 miles further apart so both mirrors will see the tick rate of the other set slowed by the time it takes the photon to cover the additional 100,000 miles between them. Now assume that both A and C keep track of the number of vertical photon bounces in their own frame and the number of horizontal photons recieved from the other set. If we now apply a force on C/D in the direction of A/B until both mirrors are once again at rest to each other, won't both A and C show that the other set had less ticks recieved by from the other set while the same number of vertical ticks?
Additionally, if C/D was accelerated to a faster relative velocity toward A/B so that it would catch up and pass A/B, wouldn't it observe A/B's tick rate as faster than its own?
Yes, identically constructed clocks at inertial rest wrt each other will tick at identical rates.
This is true, but it is irrelevant to understanding time dilation. In other words, even when the time it takes the photon to cover the additional distance is taken into account you are still left with an "extra" slow down. This extra bit is the time dilation, not the communication delay.
Lets think of a slightly different scenario. Instead of having A/B and C/D lets say that you have A1/B1, A2/B2, A3/B3, ..., and C/D. All of them are initially at rest and synchronized. All of the An/Bn clocks are spaced at 100,000 mile intervals. Now as C/D travels past them at 100,000 miles per second then at the n'th second it can compare its clock with the An/Bn clock with no communication delays involved. Since all of the An/Bn clocks are synchronized in their rest frame then comparing the time of the "local" one is the same as comparing the time with the original one (accounting for the distance).
Here is where the full Lorentz transform will be helpful. It does not matter how you arrange the clock's paths through spacetime, both clocks will agree on what the other clock will say (the proper time is frame invariant). In other words, if clock A/B travels along some path between events 1 and 2 and records a time t, then C/D will agree that A/B records that time t even though C/D recorded a different time during the measurement.
This is indeed possible. Although inertial observers will always observe relatively moving clocks to tick slower a non-inertial observer can observe relatively moving clocks to tick faster.
Wouldn't the net force be zero if the forces act in opposite directions? Unless the photon has a net gravitational 'charge'?
I don't see how this could work in a non-uniform field. Net pulls on dipoles depend on a difference in the electric field between the two poles.
But where does one field end and another begin? Presumably you have a gradual change in mind, rather than fixed borders between distinct fields.
Intuitively, if I wanted to associate a speed with the gravitational field at some point, I'd take an average of the velocities of the particles contributing to the field, weighted by the magnitude of their contributions to the field at that point. The problem with this approach is that (unless I've made a mistake somewhere) the field velocity near the surface of the Earth would be roughly equal to the velocity of its centre of gravity, and MM type experiments would be influenced by the Earth's daily rotation on it's axis.
Alternatively, maybe you're thinking of the gravitational field as if it were some kind of fluid, being dragged by something resembling friction near the surface of the planet. Are you claiming forces can pull other forces?
That's not the problem. The issue is that I've never heard of any force field vector, gravity or otherwise, having a speed associated with it. I've heard of propagation speeds, and I know that objects exerting a force can have a speed, but I've never heard anything about velocity being a component of forces or fields.
The forces don't act in opposite directions. The force in the back of the photon is pushing the photon from the back, and the force in the front of the photon is pulling the photon from the front. They're both causing the photon to move forward.
Let me just say that I'm not sure exactly how a dipole would react in a uniform unipolar field, but now I realise how it can be tested:
Let's say you take a large hollow empty sphere and you negatively charge it. Math would dictate that you would have created a uniform negative electric field everywhere inside that sphere. Next, let's say you take two pieces of metal and seperate them with an insulator. You charge one piece of metal negatively, and the other positively, thereby creating an electric dipole. Finally you place your dipole in the center of the hollow sphere. How will the dipole react to the uniform negative electric field inside the sphere? This experiment would show whether my model could work or not.
I'm not sure I understand your question. Are you asking the difference between a dipole in a non-uniform field as opposed to a dipole in a uniform field? If so, I would think that the dipole in the non-uniform field would accelerate just like a dipole in a uniform field, but the dipole in the non-uniform field would also tend to rotate so that it would face the strongest part of the field. This can also be tested using the same experiment I described above. All you would have to do is split the large hollow sphere in half, seperate the two halves with an insulator, and negatively charge both halves by different amounts.
There would be a gradual change. Both fields would be having a tug-of-war on the photon, each trying to make the photon travel at c relative to itself. Of course, the stronger field would have a larger effect on the photon than the weaker one.
Exactly what I was thinking.
Instead of imagining a gravitational field as a curvature of space-time, or as an exchange of particles, I imagine it as simply an extension of the mass that creates it. So a moving mass drags its gravitational field along with it, and a rotating mass rotates its field as well. So a person standing on the surface of the Earth would be stationairy in the Earth's gravitational field, because the field is rotating with the Earth.
However, you may have presented a problem in my theory. If mass can drag gravitational fields around, why can't gravitational fields drag mass around? I'll have to think about that. Please Register or Log in to view the hidden image!
Let's say you have a wind blowing at 100 mph. If you are stationairy, you feel the force of the wind. However, if your traveling at 100 mph in the same direction of the wind, you feel no force. In this example you can say that the force of the wind has a speed of 100 mph in a certain direction.
In another example, let's say that you have a fundamental interaction that is the result of an exchange of particles. Let's also say that those particles have a constant speed relative to a certain medium. You can then say that the speed of that interaction is equal to the speed of those particles. So if you were travelling at at the speed of those particles, you would cease to feel any force from that interaction.
As a third example, it's not hard to imagine that a local interaction may not be instantaneous, but instead that there may be a small delay between when the conditions of the interaction are met, and when the interaction actually occurs. So for example, if a particle with a gravitational field is moving at a fast enough speed through an external gravitational field, the particle might have flown past the parts of the external field that would have been responsible for the interation before the interaction could actually occur. This would also set a speed limit for that interaction, and the force generated by that interaction.
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You'd only need to test it if you were proposing a theory that gave different predictions than currently established ones. The same field will exert opposite forces on opposite charges, unless your view of the gravitational field is different from the current view of the electric field (aside from like charges attracting instead of repelling).
Actually, the field is uniformly zero inside a charged sphere. If you want a non-zero uniform field, the usual way of creating one is to use two oppositely charged flat plates, placed parallel to one another some distance apart.
The field would try to push the positive charge one way, and the negative charge the other way. The forces would either stretch or compress your dipole, but they wouldn't move it. As I said, the only motivation for testing this is if you have a reason to doubt current electrostatic theory.
As I understand it, the forces on the positive and negative charges on the dipole would be of different strengths if the field were non-uniform, giving a net force. As you also noted, the dipole would orient itself to face the strongest part of the field, and would be pulled in that direction. Also, there's no need for lab apparatus to witness this: you can use electrically charged objects to pull polar molecules (like water) around, for instance. It doesn't matter whether the object is charged positively or negatively - the net force is always a pull.
I was considering a uniform stationary rotating ring of mass a few months ago. I did a quick integration for an arbitrary distance outside the circle, and got zero velocity for the field in the direction tangent to the circle. Since a sphere can be viewed as a bunch of concentric circles, you'd get the same thing for a uniform sphere of mass.
I wouldn't look at it that way. Your attributing the force to a speed difference between air and something moving through it, but does that give the force itself a speed? It just looks like a cause/effect relationship to me.
This is the propagation speed of the field, which isn't the same thing. It's how quickly changes to the field will spread out.
I suppose so. Note that in your three examples the speeds you mentioned aren't really analagous to one another. One's a moving source, one's a propagation speed, and one is, as you put it, a speed limit. I didn't say forces couldn't be affected by some parameters that happen to be velocities, or that you couldn't find different ways of associating a velocity with a force. As a simple example, you could consider pushing some object along a table top at a certain speed. You might feel inclined to say that the force was "moving along" with the object, but does this really say anything new and useful?
I don't really see any potential for generalizing this force-velocity concept, especially when considering fields. I've never heard of fields being considered to have a 'velocity' of any kind; they just vary as a function of time.
I disagree with you that the electric field is uniformly zero inside the sphere. I would say that the net force on a charged particle in that sphere would be zero, but that doesn't mean that the electric field strength is equal to zero. The electrons on the surface of the sphere are still all generating electric fields.
As for your example of the field between two charged plates, I consider that field different from one created by the charged sphere. It seems that the field inside the sphere has no direction but the one between the plates does.
How do you figure? Where would the field push the charges when the field inside the sphere is uniform? If you look at the dipole as two seperately charged unipoles, one could argue that the net force on them both by the sphere would be zero, so there wouldn't be any stretching, compressing, or motion on the dipole. But when the two unipoles are brought together to create a dipole, the electric fields around the unipoles are no longer symmetrical. This break in symmetry in the fields around the unipole causes the positive unipole to be more attracted to part of the sphere that is opposite the negative unipole, and the negative unipole to be pushed from the part of the sphere that is opposite the positive unipole. The combination of these two forces would cause the dipole to start moving in the direction that its positive pole is pointing.
Let me also say that two oppositely charged unipoles placed together in a charged hollow sphere may not react the same way as a true electrostatic dipole in the same situation.
Here's a question for you: If you were God, and you wanted to make a particle inside a charged hollow sphere move in a certain direction, what kind of electric field would you give that particle to achieve your goal?
If an observer was standing at a certain distance from the rotating ring, the velocity of the field from the side of the ring which is closest to the observer would be equal and opposite to the velocity of the field on the the other side of the ring, but the side of the ring closest to you would be generating the stronger field. You have to take not only the velocity of the field, but the field strength as well, into consideration to calculate the actually velocity of the field relative to the observer.
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