Severian

Ok, let me be more explicit. Lets assume A is not zero for a moment. The first equation [math]\frac{\hbar^2 k^2}{2m}=0[/math] implies that [math]k=0[/math] (since the other variables are just numbers). The one at pi/2 implies w=0. So putting this back in the original equation [math]\Psi_I = A \cos (kx - \omega t) = A \cos 0 = A[/math] So the equation is only a solution if it is a constant (the case A=0 is a subset of this too). This shows why the question isn't a very good one because both answers are correct. [math]\Psi_I[/math] is a solution of the Schroedinger equation if k=w=0. But, I am sure that this is not the answer intended by the question. This solution is a particle with zero momentum and zero frequency (infinite wavelength) which is non-physical. (There would be a problem with normalization too, but this is true also for a plane wave.) Think about it also in tems of eigenstates: Let's write [math]\Psi_I = \frac{A}{2}\left( \psi_+ + \psi_- \right)[/math] with [math]\psi_\pm = e^{\pm (kx-\omega t)}[/math] Putting [math]\psi_+[/math] in the equation gives [math]\frac{\hbar^2 k^2}{2m} = \hbar \omega[/math] so it is a valid solution as long as this holds. It is an energy eigenstate with [math]E= \frac{\hbar^2 k^2}{2m}[/math] Putting [math]\psi_-[/math] in the equation gives [math]\frac{\hbar^2 k^2}{2m} = - \hbar \omega[/math]. [math]\psi_-[/math] is an energy eigenstate with [math]E= -\frac{\hbar^2 k^2}{2m}[/math]. But this cannot hold if the earlier one did (unless [math]k=0[/math]). The state cannot be in two (non-degenerate) eigenstates at once. (I think that was the point of the question).

You made an assumption in your first line. (If A=0 you can't divide it out.) But lets go with the assumption. You have shown that for a non-zero solution (ie with A not zero) you need [math]\frac{\hbar^2 k^2}{2m}=0[/math]. But this is a necessary condition - not a sufficient one. What happens for values of [math]kx-\omega t[/math] of [math]\pi/2[/math] and [math]\pi/4[/math]? (Remember the equations need to work for all values of x and t.) Then think about how this answers the original question.

So to work you need LHS=RHS for all values of x and t. Pick one x and t and see what happens, eg. x=0, t=0.

I rather like bacon and brie myself. In fact, I often have a bacon and brie (with red onions too) toastie at O'briens for lunch. Did you ask for extra fecal matter on your sandwich that day?

So when will they find the 'penchant for high heels and stockings gene'? I think I may have that one....

I don't think that is fair. The themes were very different. In fact, John Savage's problems in Brave New World had nothing to do with his genetics. I have no problem with screening for genetic disorders, but one should not make aesthetic judgements such as choosing the color of the the yes, shape of the nose etc. (I have actually been genetically tested and have no known 'defects', not even recessive ones.)

m(34) pro-abortion, but with restrictions (like date from conception) And for rape cases, the rules should be the same.

I am sorry, but this is such a stupid statement. You may as well disagree with the statement "a dog is only a dog" or "a law is only a law". It is quite clear that "a theory is only a theory" is 100% correct. Furthermore, theories (in the scientific sense) are not always applicable. The theory of special relativity for example is not valid in accelerating frames of reference.

You can see from here, that the most constraining limit on the photon mass is by Ryutov et al, using observations of the magnetohydrodynamics of the solar wind. http://pdg.lbl.gov/2005/listings/s000.pdf

I am actually an author of the TESLA Technical Design Report http://tesla.desy.de/new_pages/TDR_CD/start.html but I have no idea how to go about builing a linac on a tabletop!

In this context they mean 'energy'. So the electroweak scale for example is the energy at which electroweak symmetry is broken; the Planck scale is the energy at which gravity becomes a strong force etc.

It is because quantum condensates form when there is a potential well away from zero field. By that I mean that there is a 'valley' in the potential at some non-zero field configuration with a barrier between it and the usual vacuum state. If the temperature is higher than the height of the barrier, the field will just jump over the barrier as if it isn't there, and there will be no quantum condensate.

I have met, the Queen, the BeeGees, Angela Merkle, Douglas Adams and a few Nobel prize winners.

So its ether one or the other?

Bah, just accidentally deleted my response! Lets try again..... Firstly I was intending events A and B to be at the same space point but at different times. So for example, plane guy sets off at event A: Earth N:52`32, W:002`04, Alt:492ft at 18:20 March 2, 2006, and arrives back at point B: Earth N:52`32, W:002`04, Alt:492ft at 18:20 March 16, 2006 (two weeks later). You seem to be thinking of him setting off at A, travelling to B, where he turns round and comes back to arrive home at C. Notice that C is not equal to A because they are at different times. With the God thing there is a problem. The whole reason that your clocks are travelling at different speeds is that you are defining your time diferently. Neither is wrong - they are just different. So when you say 'instantly' you have to state which time definition you mean (because whether or not events are 'simultaneous' depends on your frame). I suspect from your wording that you want God to use the time direction as defined by the guy in the plane(?) In that case, he would read the guy on the Earth's clock as being slow. So it is actually the other way round from what you said. Plane guy's clock reads 12:00 while Earth guy reads 11:59. Also, the trip back will have as much effect as the trip out, so when he gets home his clock will read 12:10 (if it took 10 minutes) while the guy on Earth's will read 12:08. At this point it looks like the Earth guy's clock is slow! Actually there is no need for god-like powers - all you need is lots of little helpers standing stationary with respect to you at the turnaround point with a stopwatch.

You clock is measuring the time you have taken to get from event A to event B. That is a different time-length (if you see what I mean) for your journey than the time between A and B for the stationary person's journey. You have both travelled through space-time at 'c', but he used all of that velocity to travel from A to B along a direct route (not moving at all in space) while you used some of the 'c' moving through space so you weren't able to move as far in time as he was. Your clock is measuring the time you have spent moving between the two events, but his clock has measured the time he spent moving between them. Since you moved less (in the time direction) your clock should read less, even when you decelerate back into his frame.

No time is lost. The time between the two events (synchronising clocks at the start and comparing them at the end) is genuinely different in the different frames. Just as flying from Glasgow to Warsaw with a change of plane in London is travelling a different distance from a flight from Glasgow to Warsaw via Paris.

It was running correctly. It just looks to the (unaware) observer that it was running slow, because it does not match his own clock. Let me put it a little differently. Imagine you want to measure the distance between two points. You take a very long tape measure and stretch it between the points and have your answer. But your friend, whose tape measure is too short, has to measure it in lots of little steps instead of one big one. It is difficult to align his tape measure with the endpoints, so sometimes his tape measure is pointing in a slightly wrong direction and he measures a longer distance than you did. He has measured a distance just as you did, but you measured different distances so got different answers. The clocks are similar to the short tapemeasure. They measure very short timesteps (with their tick-tock), and we want to measure the time between the plane leaving and coming back in the Earth's rest frame.The guy on the ground's clock is automatically 'pointing' in the right 'direction' so he meausures the correct time displacement (in the Earth's rest frame). The guy in the plane's clock is pointing in the wrong direction so its steps are slightly out. He still measures a time displacement, but it is not the same displacement as the guy on the ground. they have measured different things, so we should not be surprised that they get different answers.

Imagine that the world had lines painted on it, some going east-west and others going north-south. (And imagine that your measurements of distance are small enough that we can neglect the curvature of the Earth.) Now let's say we measured any north-south distance in metres while we measured any east-west distance in feet. Everything would be consistant, and we would have no problems - but we would need to be careful when measuring a distance that we look at its orientation. Now some bright spark will realize that an object which is length x metres when laid north-south is a length y feet when laid east-west where y=a*x (a is about 3, I forget exactly) and this holds for any size object! He will then go on to hyposesize that the laws of physics are independent of which direction you are facing. He will suggest not using separate distance scales but measuring east-west distances in metres too! (And for the next hundred years, spotty teenagers and sad pseudoscientists will post nonsensicle arguments in web fora claiming he was wrong...) Time and distance are much like this. Einstein realized that the laws of physics are the same in any spac-time direction. You can rotate away from the time axis and into the space axis (this is a Lorentz transformation). The factor 'c' is just to compensate for our stupid units (like 'a' was in my analogy).

It is not a case of 'snapping back', it is a case of time not being the same in different reference frames. So after you accelerate/decelerate, the measurement of time you are making is a different quantity to what you were measuring before. It is no different (in principle) to measuring a relative velocity. Driving in your car, you may measure the relative velocity of an overpass as 50mph. If you stop the car, the relative velocity become 0mph. It doesn't snap-back - it is a different measurement. So in your plane example, the pilot's would actually see things on the Earth as running slow, becaue he is comparing their clocks with his clocks. The clocks are measuring different quantities. When he decellerates to the Earth's rest frame, his clock will now be in the same frame as an Earth clock, so it will be measuring the same quantity again.

Yes, I know it is a unique username. What I meant is that if you were called 'xyz8hd' then a search for 'Severian' in author and 'xyz8hd' in keyword would reveal the posts you are looking for because people usually quote the previous poster (so the post would contain "[ quote]xyz8hd.....". But with your username that tactic doesn't work. Hmm... would searching for a keyword "[ quote]Atheist" work....? Edit: No it doesn't Edit2: In fact, I don't think it includes anything which is in a quote. Can anyone confirm this?

You don't help yourself by having a name like 'Atheist' since including 'atheist' in the search key-words won't narrow things down much on these boards!