Aeschylus

Neither a classical wave description or a classical particle description of light is fully adequate to describe light.

I may have incorrectly stated what I meant above. Also' date=' forget what I asked about continuity.

 

Suppose a function, f, is defined on only 3 values of x that are infinitely close to each other. My question is what is the limit as x approaches the x-value of the middle point. It seems to me that it should be f(the middle point), but according to the epsilon-delta definition of a limit, taking the limit only makes sense on an open interval. f, being defined on only 3 points infinitely close to each other appears to me to be defined on a closed interval. So, what is the limit as x approaches the middle point? If the answer is that the limit doesn't exist, what is the reason behind a limit existing only on an open interval?[/quote']

Yes but there's no such thing as 'infinitely close', in the reals if the values are not the same as each other than there are an infinite number of reals between them.

I would' date=' except when people describe light as exhibiting "wave like properties", which suggests to me that light is a function of something other than particles.

 

If light is made of particles.. then it is made of particles, but then why describe it as something other than photons moving in wave like patterns? Why describe light as a [i']wave[/i]?

Personally I think if your going to think of light as anything, then think of it as waves, with the possible exception of high frequency em radiation which exhibits particle-like behaviour that's not difficult to observe.

Of course for a fuller descripion you need to look at quantum mechanics, which shows that neither a classical particle nor a classical wave description of light is completely adequate.

Z-space: What exactly are you reffering to as pure noise. What would be the source of that noise?

 

Coming back to a topic of a discussion' date=' can anyone elaborate on the randomness in quantum physics mentioned by Aeschylus earlier?[/quote']

 

The collapse of the wavefucntion from a state of superpostion into a single eigenstate is inhererntly random.

Aeschylus: You are talking about a noisy signal. If you want to talk about a noisy signal' date=' then I absolutely agree with you that a noisy signal can be either completely deterministic or chaotic. The pattern in noise that you mention was also mentioned in my preceeding post (#3 in this thread).

 

In the post #7 in this thread, I am talking about pure noise (ie, no signal mixed in with the noise). Pure noise is a different thing altogether from a noisy signal, and its mathematical properties are also different.[/quote']

 

I've took noice to do with coding, etc. Are you talking about something like a noise sphere??

sinexec: Noise and chaos are 2 very different things. Noise can not be chaotic. Noise is not determinant.

 

The best way to differentiate noise from chaos is determine the Lyapunov exponent for the function. In practice' date=' this is of limited applicability because usually the function is not known.[/quote']

 

No noise just means unwanted information in a signal it can be completly deterministic and not even chaotic. Noise can either be patterned or random.

Not all maths is completely deterministic, sattistics is obviously not deterministic. Chaos theory is a just an branch of maths that is detreministic.

Noise just means unwanted informatioon in a signal and can be completely dedteriministic. If there's anythg random in nature it's quantum physics.

So in other words the direction most defintely does matter.

It's an absolute must for a vB science or maths forum for Html to be turned on in order to make equations more readable.

Remember:

x' = γ(x - u_xt)

y' = y

z' = z

t' = γ(1- βx/c)

edited to add: for the love of God turn Html on!!!!!!!!

Hmmm... so it would seem that the rate of acceleration only applies to things with Mass then?

and as a photon has no mass' date=' it can effectively "get away with " instant velocity?[/quote']

 

It's important to relaize that the phton never accelartes or deccelrate sit always has a velcoity of c.

 

Instant velcoity is not peculair to the photon, imagine an electron-posiron annihilation in some refernce frame which produces lets say produces two pions. These two pions will in most refence frames have some velocity when emitted without having to accelerate t reach this velcoty(of course unlike the photon they will have a rest frame), indeed if this was not the case 4-momentum could not be conserved.

so from Zero to C in no time?

No, because the photon is never at rest in any inertial frame, when your a photon your born at c and you die at c.

In all models of the universe the universe is unbounded and yes an infinite universe can expand, there's certianly nothing unphysical about an infinite expanding unievres.

Hugh Everret was the father of the many-worlds interpreation of quantum mechanics, I don't think it's that helpful to say that this means mutiple universe, all it means is that we and presuambly the universe are in a superpostion of states.

Here's nice little proof (well, it's not laid out formally and it uses inequalities that really should be proved first (though they're easy inequalties to prove), but I've gauged it exactly for the sort of person who will argue 0.99.... is not equal to 1. Though are alot better and more comphrehensive proofs, people who've I've shown this one seem to liek it) that I came up with:

(Firstly just about every crank whose claimed that 0.999.. is not equal to 1 claims that there are nbo numebrs inbetween 0.999.. and 1, this proof really sets out to deal with this claim)

if x = 0.999... then there are 3 possible relations that x and sqrt(x) could have

1. sqrt(x) > 1

Howver if this is the case then 0< x < 1 must be UNTRUE as for any 0 < x < 1, 0 < sqrt(x) < 1. If this is the case x > 1

2. sqrt(x) = 1

If this is the case then x = 1 is the only solution, meaning that 0.999 does indeed equal 1.

3. sqrt(x) < 1

If this is the case then there is a number between 1 and x (as if 0 < x < 1 then sqrt(x) > x); what is the decimal representation of this number? Of cousre at this point someone could argue that in this case sqrt(x) = x, but this means that x msut equal 1 or zero).

and you need distance to single out movement too, so why the bias? One is two temporal coordiantes, and one is two spatial coordiantes, and as GR shows us, space and time can be treated as equivalent.

Yes indeed velocity is dx/dt, so in order to have velocity we must 'have' dx and dt (i.e. there must be a change in both postion and time).

Well GR has some situations in which spacelike dimensions become timelike, but this is the exceptional case in places like the interior of a black hole. In relatviity even though the concept of a spacetime continuum is introduced, there is still a very definite differnce between the two. This is easiest to see in the simplest case of a spacetime continuum, as described by (a form of)the Minowski metric (taking c =1): ds² = dx² + dy² + dz² - dt², notice in this form that dt cannot really be considered equivalent to dx, dy or dz due it's minus sign.

Objects with real rest mass cannot travel at or above c, any object that travels at c MUST have a rest mass of zero.

The inertial mass of an object in some inertial refernce frame will tend infinity as that object's speed tends to c.

1/0 is undefined in the real numbers, but in the extended reals it is taken as infinity. Physicists don't use the extended reals, but they are lazy and they will often say that x is infinty at y = z if x tends to infinity as y tends to z. If you get 1/0 in an equation inn special relativity you should interpret the result as unphysical and hence referring to an impossible situation.