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J.C.MacSwell

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Posts posted by J.C.MacSwell

  1. I have no idea what this is suppposed to mean. Care to try again?

     

    10% of the radiation incident on the earth is from the sun? (I wouldn't think this would be accurate)

     

    How about 10% from "other than the sun"? Or is that still way too high?

  2.  

     

    If you insist on using a title' date=' please us the correct one, (Dr) though addressing me by my user name is perfectly fine.[/quote']

     

    Can we call you "Doc"?

  3. So no physical object in the universe is moving away from any other physical object at c? Now from a third party, can those two physical objects be seen to be moving away from one another at c[/b']?

     

     

    Yes, up to but not including, 2c. (locally)

     

    (first question is no anyway due to Hubble but yes locally in either objects rest frame)

  4. Exactly. I know its short (certainly way shorter than the grass where I spend most of my round.)' date=' but it still has mobility. The individual blades can be pushed closer together by the weight of the ball. That could be just enough to for the effect we are discussing.

    On practice putting surfaces I have observed the blades spring back after the ball has 'fallen' into the hole. Now, you make a very good point that the greens at Augusta are cut very close - closer for sure than a practice green - but I still think its a plausible explanation.[/quote']

     

    I agree with your assessment. The grass can "buckle" and then recover, partially or otherwise and always with some energy loss. The ball looked to be still poised to go downhill when it "stopped" but with the grass underneath straining/slowly collapsing under the weight. Crowd noise "plausibly" or butterfly wings flapping in Australia a week earlier "possibly" could have "tipped the scale".

  5. The simplest way to beat gravity is the "bootstrap" method. You need good balance and very strong arms and laces (regular shoe laces will probably break). You lean over so that your center of gravity is directly over your laces and pull straight up. If you pull hard enough (the force must exceed your weight plus the tension on your laces) you will rise and "hover" for a few seconds before falling over. Good balance is the key. Once you leave the ground it is like balancing a pencil on it's point so a couple of seconds is extremely good.

  6. Definition: Let (x1' date='y1,z1), (x2,y2,z2) denote the coordinates of a point in some frame. Let D be defined as follows:

     

    [math'] D = \sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2} [/math]

     

    The axes are rigid if and only if d(D) = 0

     

    Ok, I'm not going to let JC or Swansont confuse me. Here is the deal.

     

    I have an actual meterstick. This one single meterstick is used to define distances throughout the entire universe.

     

    So if I want to know the distance from my current location to the center of this universe, what I want to know is how many metersticks I would have to place in a striaght line, to reach from here to there.

     

    If I want to know the distance from here to the sun, again, I want to know how many metersticks I need.

     

    If I want to know the distance from the sun to some other star, again, I want to know how many metersticks I need.

     

    So the Pythagorean theorem is irrelevent.

     

    Again, if I want to know the distance from my current location to the sun, I want to know how many meter sticks i need, to rightnow simultaneously place in a straight line pointing right at the sun. In other words I want to know how many metersticks I can fit in between my current location, and the sun.

     

    So for example, here is the current approximate distance from the earth to the sun:

     

     

    Astronomical Unit

     

     

    So there it is.

     

    150 million kilometers

     

    And since 1000 meters = 1 kilometer

     

    The approximate distance from the earth to the sun is

     

    [math] 1.5 \times 10^8 Km \frac{1000 m}{Km} = 1.5 \times 10^{11} m [/math]

     

    So 150 billion metersticks would be needed. And this gives the mind a sense of the distance.

     

    So this is why we need rigid axes.

     

    Real rulers must be substitutable for the axes.

     

    But not rulers that can expand or contract though.

     

    When we were interested in the distance from here to the sun, we used rigid rulers. And the length of a real ruler doesn't vary much over large amounts of time. And any change in its length is certainly imperceptible.

     

    So this is the empirical thought behind rigid axes.

     

    In order to say this logically, it suffices to say that the distance between any two points on an axis is constant in time.

     

    So given any frame, we are given three coordinate axes.

     

    The distance between any two points on an axis is found by subtracting the lesser coordinate from the greater.

     

    So the distance between the point 4 on the x axis, and the point 7 on the x axis is simply 7-4=3. And this distance never changes. If this distance could change, then the axis wouldn't be rigid.

     

    The key thing here, is temporal constant.

     

    This is the concept to be developed.

     

    You are doing quite a job explaining Euclidean space. If you try real hard you can explain Euclidean space in a number of ways. You can come up with many more examples of Euclidean space. But it will still be Euclidean space. It will be a very good model of reality at low speeds and a very poor one when higher speeds are involved.

  7. Why do you have to go and complicate things. :)

     

    I don't want the axes to be able to expand and contract though' date=' because there was a unit of distance chosen. I want the axes rigid. :P

     

    Let matter expand and contract, not the axes.[/quote']

     

     

    If you find an absolute reference frame (if such a thing exists) it is unlikely to be Euclidean (or even simply Lorentzian).

  8. Then in the one frame, the statement that Jim turned 40 when John turned 31 will be true, and in the other frame, the statement that, "John turned 40 when Jim turned 31" will be true. This is a temporal paradox. And it is caused by assuming that the time dilation formula is true in both frames[/b'].

     

    Johnny, this is how these frames work by definition. If you somehow come up with absolute frames that work differently that would be great. I won't say it is impossible.

     

    But, the way these frames are defined, that is the way they work.

  9. You better explain this one more.

     

    Also' date=' in a previous post in this thread, you stated the speed of light isn't constant in the frame in question. Could you say why please?[/quote']

     

    You chose the Earth as the frame. The Earth is spinning. You therefore have to account for the centrifugal "force" on your pebble even at "rest". While low speed newtonian physics works great in this frame on or near the Earth "corrections" must be made for greater precision. Pseudo forces such as the centrifugal force or coriollis force can be used to balance things out or correct for the fact that this is not an inertial frame.

     

    The speed of light is not constant in this frame. This would seem obvious if the Earth was spinning at much higher speed. Since it is not, the discrepancy is small locally but more obvious and much larger if you extended this frame out beyond the solar system. Go far enough and all matter, all stars etc. are all going much faster than 300,000 km/s.

     

    Of course this is far beyond the normal use of this frame. It is a very useful and common frame to use but you have to be aware of the limitations. It is a poor choice of frame to use to try and prove that SR or GR or the equivalence principle are wrong. They all require "corrections" in this frame.

  10. When it is not an inertial frame the forces do not have to net to zero to have the particle/body remain stationary in that frame. It often requires a pseudo force (coriollis etc.) to accomplish this.

  11. Take an object' date=' like a pebble in your hand. Hold it at a constant distance above the center of the earth.

     

    Now, you have to choose a frame to define the acceleration of the object in.

     

    In the rest frame of the sun, for sure that stone has a nonzero acceleration.

     

    BUT, and here is the really important point. Frame switch to the rest frame of the earth.

     

    Somewhere inside the earth, is the center of mass of the earth. And let the pebble be roughly spherical, so that we can insure that it's center of mass is inside of it.

     

    Ok so now there is a nonzero distance between the center of inertia of the earth, and the center of inertia of the pebble.

     

    And since you aren't raising or lowering your hand in the earth frame, this center to center distance is constant in time (at least for now).

     

    Now switch back to the rest frame of the sun.

     

    In that frame, the earth is spinning, so that the radial vector from the center of the earth to the center of the pebble is rotating. And that means that the center of the pebble is accelerating in the sun frame.

     

    Now switch to the earth frame where the radial vector is an axis of the frame.

     

    Therefore, the location of the center of inertia of the pebble isn't moving in this frame. That means this now...

     

    That means that the pebble is at rest in this frame.

     

    Certainly the pebble could be spinning in your hand, but the center of mass of that pebble is at rest in this frame.

     

    And when something is at rest in a frame, then its speed in the frame is zero. And this pebble is remaining at rest in this frame, hence it's speed isn't changing. Now here is the definition of acceleration:

     

    [math'] \vec a = \frac{d\vec v}{dt} [/math]

     

    As you can see acceleration is a vector quantity. That means that it has magnitude, as well as direction.

     

    Now, dt is strictly positive, hence the only way the LHS (left hand side) of the equation above can be zero, is if the numerator of the RHS is zero.

     

    Now here is the numerator of the RHS:

     

    [math] d\vec v [/math]

     

    ok so

     

    That quantity is a differentital of a velocity vector.

     

    Velocity is speed in a frame, times direction of motion in the frame.

     

    Right now, we are in in a frame in which the center of mass of earth is at rest.

     

    Now, there are many reference frames in which the center of mass of earth is at rest.

     

    In some of them, a fixed point on the surface of earth may be orbiting the CM of earth. But other frames we can concieve of have fixed points on the surface being at rest, in addition to the CM of earth being at rest. And this is the type of frame that the pebble is in, because it is hovering at a height h over the the same spot on the ground.

     

    So the point is, that the speed of the center of inertia of the pebble in this frame is zero, and hence the product of the speed of the pebble in this frame, and it's direction of motion in this frame must be zero, since its speed is zero, regardless of what it's direction of motion was previously, or later (since it's speed is zero over some interval of time in this frame, as measured by a clock at rest in this frame).

     

    With all of this keenly in mind, it must be the case that the acceleration of the pebble in this frame of reference is zero.

     

    It is only when you release the pebble, that it begins to fall, or using other words... it is only when you release the pebble that it begins to accelerate in this reference frame.

     

    My thoughts on this matter are improving.

     

    Now, once you release it, it has a measurable acceleration of g.

     

    But whilst you are holding that pebble in your hand, it has NO acceleration in this frame.

     

    On the other hand, you can feel the weight of the pebble at all moments in time for which the pebble has no acceleration.

     

    This is a very commonly used frame, perhaps the most common. Be careful not to use this frame in extremes however. The speed of light is not constant in this frame.

  12. Huh? I'm in the 3rd dimension' date=' as are you ([i']in that I have height (x), length (y), width (z)[/i]) Why would "someone" disappear?

     

     

    .

     

    I think they meant if they were only 2D they would disappear to the others in 2D. Similarly if you are only 3D you would disappear into a fourth dimension.

  13. This is neat. For a conspiracy theory, this looks like full body(space suit)weathered by time and covered with dust. Up top right would be the head, a little lower you can see the arm sleeves as if someone laid down on their side and died. All the way to the bottom for the legs.

     

    It's a Bi-ped for sure, as the sea turtle entering the bottom of the picture can clearly see! Obviously some humanoid arrived first, followed by even more intelligent sea creatures that do not require spacesuits in that atmosphere.

     

    As for the colour of the surface, I don't think we should jump to any conclusions without some more data.

  14. ']How would the photons alone be responsible for the orbital paths?

     

    Photons have never been obvserved to have any attractive force' date=' that would be easily tested here on earth with a laser for example.

     

    Plus we know the moon orbits around the earth, but the earth certainly emits photons in comparitavely small amounts, [b']and only on its day side[/b].

     

    Has temperature/emits photons.

  15. That makes no sense.

     

    Here is the formula for the one' date=' in terms of the other:

     

    [math'] C = 2 \pi R [/math]

     

    If the circumference C increases, then the radius R increases, since the value of pi is constant in time.

     

    If the circumference C decreases, then the radius R decreases, since the value of pi is constant in time.

     

    The equation has to always be a true statement.

    If either C,R varied, but the other didn't, a false statement would be true.

     

    Regards

     

    I think you are right if your assumptions are right.

     

    However, I think you are not right, except when your "loop" is at rest (not spinning).

  16. If a spaceship travels at v=.9c toward a star 5.4 LY away' date=' it wll reach it in

    6 years. (5.4/.9)[/quote']

     

    Yes, but less than 3 in "Spaceship Standard Time".

     

    Of course, it will be less than 2.7 LY in "Spaceship Standard Distance".

  17. That's the way it is. You can see it clearly from the argument which is being developed here.

     

    Yep, Lorentz and Galileo. Clearly there ain't enough room in this town for the two of them.

     

    Now, a kinder, gentler town (read lower the relative speeds of them yardsticks). I can see maybe they might get along a little better there. (not perfectly though)

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