Aeschylus

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

no it wont, if you speed up, light still travels at the same speed, your speed does not effect the speed of light around you.

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) this is impossible

(b) however, if you were, theoretically, light would not stop, it would still travel at C, its just you would be travelling at C + 1, therefore it would be like overtaking a car;

if car A is doing 70mph and i overtake them at 90mph, car A doesnt stop, it keeps going, just slower than me.

it is the same if you went faster than light, which would result in you going back in time. [remember post (a) though!]

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.

unless of course the theories are wrong causing the quantum mechanics to appear not to make sense. My quantum mechanics make sense' date=' but I do not agree with everything that I read.

 

Pincho.[/quote']

 

Special relatvity and quantum mechanics are comp0letely compatible and I do not believe anyoone has ever thought otherwise. Schroedinger formulated the fundmantal equation of quantum mechnaics in 1925, only one year later in 1926 was the Klein-Gordon equation formulated (initially rejected due to negative proabilty densities, but it didn't take long for phycists to relaize that tyey could be interpreted in a physical meaningful way) and three years later in 1938 the Dirac equation was formulated (which can be seen as perhaps the true birth of relatvstic 'quantum mechanics'). So you see that quantum mechnaics and relativstic 'quantum mechanics' have been around for just about the same period of time, which is not suprsing given that QM is partly based on SR.

I meant that from the photon/electrons point of view the journey is instant and therefore light would not see other light travel.

Photons do not have refrence frames, we cannot talk about what a photon sees.

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.

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.

I was under the impression they were a required consequence of said theory.

They're not required, infact I think it's highly, highly unlikely that they exist. Infact they violate some formulations of special relativty.

The velocity is the derivative with respect to time of displacement. Wht you mentioned is average velocity not instantaneous velocity.

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]

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.

0^0 is the same as 0/0, that's why x^0, where x is not 0 is x/x=1

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.

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

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.

as in the "collapsing" of a light wave into a photon

. Throughout space and time, that is what they are doing, quantum states and "probability waves" collapsing willy-nilly every time energy is exchanged. They behave no differently when we observe them doing so, or when we don't. Otherwise the universe would fall apart!

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?

okay then' date=' that makes sense.

 

One thing i want to understand:

 

A observer in motion at a fraction of c will witness the world around him age much faster than him.

An observer in motion at minute fractions of c will be witnessed to travel much faster than the 1st observer in the eyes of the 1st observer?

 

Observer 1 : Large fraction of c

Observer 2 : minute fraction of c

 

Obvr 1 travels so and so distance in so and so time witnessed by Obvr 1, he views Obvr 2 travel so and so distance much faster relative to Obvr 1 witnessing time according to Obvr 1.

 

does Obvr 2 witness Obvr 1 traveling at a large fraction of c or at minute speeds because his time rate of change (dt) is less than Obvr 2's dt?

 

I must be missing something.[/quote']

 

There's no such thing as absolute speed (except for photons I suppose), velocity is relative, i.e. it can be defined with refernce to some other object.

 

Lets say that the first observer is travelling at a large fraction of c relative to the Earth/faraway stars/CMBR frame and the second observer is travelling at a small fraction of c relative to what we have defined as the 'staionary' frame (though I stress that it's completely arbiatry anyinertial frame in SR can equally be considered a stationary frame).

 

Assuming constant velocity the first observer does not see the world (i.e. objects in the staionary frame) age (i.e. clocks or other measurments of time as compared to the proper time in his frame) faster he sees it age slower, just as an obsrever in the stiaonry frame would see observer 1 age slower too.

 

Observer two observes observer one travelling at a large fraction of c still.

i seem to understand it is irrelavent and mostly already stated was incorrect in this scenario.

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.

no, it isn't. the radius DOES NOT SHRINK, so it wouldn't be a point. did you not read the last few pages of this thread?

Have you?

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

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

remember radius doesn't shrink.

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.

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

my question was if the measure of the angle changes. if it does, then you were wrong about always using 3.14.... for pi

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

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

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]

but, pi isn't pi in this case, why would you use 3.14..........?

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.

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).

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.

It's not a very good critque as the author is not entirely famlair with the subject matter.