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Aeschylus

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Everything posted by Aeschylus

  1. It's an operator so you've gotta be very careful about what the index actually means. There is such a thing as Del^4 which is called the bilapacian operator: http://icl.pku.edu.cn/yujs/MathWorld/math/b/b194.htm
  2. No light always travels at c to all inertial observers so whether you speed up or slow down light is still travelling at c in your MCIF (momentarily comoving inertial frame). Remeber- relativty is all about, well, relativty- there is no absolute frame of refernce to define an absolute speed. Of course the light could be moving towards you at c tho'. a) correct, all objects with real mass must travel at speeds < c in any inertial frame and objects moving at c do not have MCIFs so we cannot talkabout thier refernce frames. b) You cannot overtake light, because at whatever speed your travelling in some inertial frmae, in your own MCIF light is always travelling at c. Also we cannot say theoretically what would happen as the barrier totravelling at light speed and defing MCIFs for objects moving at lightspeed is theoretical itself.
  3. Photons do not have refrence frames, we cannot talk about what a photon sees.
  4. To paraphrase Bernard Schultz in his book 'A First Course in General Relativty': Special relativity is probably the most tested theories in physics, as it is tested everyday under the most rigorous scientific conditons in particle accelerators around the world.
  5. When I was a child going down to the River Meadway in Gillingham, I seem to remeber that they had concrete boats beached near the harbour down there.
  6. They're not required, infact I think it's highly, highly unlikely that they exist. Infact they violate some formulations of special relativty.
  7. I think what he was trying to say is that velocity is given by: [math]\lim_{\Delta t \rightarrow 0} \frac{\Delta r}{\Delta t}[/math]
  8. an n dimensional vector needs n mubers to speficy it. For example a 3-dimensional vector can be expressed in terms of three numbers v_1, v_2 and v_3 [math] \vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}[/math] Where the i, j and k are some (usually orthogonal, but not necessarily) unit vectors or simply by a matrix: [math] \vec{v} = \left(\begin{array}{c}v_1\\v_2\\v_3\end{array} \right) [/math] So the difference between speed and velcotiy is importnat, because velocity (in three dimensions requires 3 numbers to specify it, whereas speed only rrequires one number.
  9. What you're arguing from is the fact that that x^n/x^m generally equals x^(n-m), but it dods not neccessarily follow that x^0 = 0/0. You should be very, very careful about equating any intdeterminate or undwefined values.
  10. Hmm I've never been clear on this: is indeterminate value (such as 0/0) also undefined? Often 0^0 is simply defined as 1.Any way read this: http://mathworld.wolfram.com/Indeterminate.html and this: http://mathforum.org/dr.math/faq/faq.0.to.0.power.html
  11. The author is claiming to of soilved the measuremnt problem, which is ridculous and the following statemnts show that the author is noit cloompletely famlair with the subject matter. Now if (though it doesn't seem to me that the he is) the author is talking about decoherence it's not so ridculous, but it's still incorrect. And how can you even talk about an objective reality in QM without even discussing Bell's theorum?
  12. Aeschylus

    Circle.

    No what you don't understand is that angles in such spaces cannot be defined by arbitary circles centred on the vertex like they are in plane geomerty. A full rotation is still 2pi radians.
  13. Aeschylus

    Circle.

    Have you? We're talking about angles between lines which are defined by the point where they intersect not by a circle.
  14. Aeschylus

    Circle.

    I answered your question, cos the 'circle' that we are considering is actually a point.
  15. Aeschylus

    Circle.

    I think your getting confused, I'm not talking about the specific example of the relatvistic disc, I'm talking about circles in spatial slices of the Lorentzian metrics of GR. For an equation describing the circle as the parameter r (the radius) tends to zero, the ratio between the circle's diameter to it's circumference tends to pi.
  16. Aeschylus

    Circle.

    rember as r -> 0 the ratio tends to pi.
  17. Aeschylus

    Circle.

    The way angles are measured doesn't change. The kind of spaces we are talking about are Rimeannian manifolds and they are 'locally' Euclidean.
  18. Aeschylus

    Circle.

    I'm not talking about radians I'm talking about pi in general. Though we still use radians in GR.
  19. Aeschylus

    Circle.

    But the aplicvations of pi go beyond geometry. Is the identity below related to whetehr you do your maths in Euclidean space?: [math]e^{\pi i} + 1 = 0[/math]
  20. Aeschylus

    Circle.

    because pi as in 3.14... is an important mathematical constant whichever way you look at it and pi has uses far beyond describing the ratio of the diameter of a circle to it's circunference.
  21. Aeschylus

    Circle.

    It's a matter of defintion, read these (notice the firstone delibartely avoids saying thta pi is not constant): http://www.google.com/url?sa=U&start=1&q=http://mathforum.org/library/drmath/view/55021.html&e=747 http://mathforum.org/library/drmath/view/58292.html If you ever see the number pi in any equation, even one that descirbes a non-Euclidea space you can be sure that it will be the pi that we're famlair wtih (i.e. the one defined by plane geometry).
  22. The Aspect experiment was slightly more complex than this, but they're communicating with each other. I just shows the inconsistineceies that appear when you try to interpret the wavefunction in a realist manner.
  23. It's not a very good critque as the author is not entirely famlair with the subject matter.
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