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Re: Massless Particles

by Oh No
Tags: massless, particles
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Oh No
#1
Nov16-06, 05:00 AM
P: n/a
Thus spake Christophe de Dinechin <christophe@dinechin.org>
>Oh No wrote:
>> To understand the reasons for a maximal speed we have to look at the way
>> in which space-time co-ordinates in physics are really defined. We have
>> to challenge the assumption that space-time co-ordinates are pre-
>> existent, and then measured, but actually think in terms of the
>> measurement providing the definition.

>
>I second that, but then, don't these measurements themselves rely on
>electromagnetic waves?


Yes, indeed. The actual measurements rely on electromagnetic waves. But
deeper than that there is a point of principle. The notion of a
coordinate system depends on the maximal speed of information transfer.
If it turned out, for example, that the photon has a very tiny mass,
then the laws of electromagnetic radiation would be slightly different,
but the laws of special relativity would still obtain. Technically, I
believe this possibility always remains open; figures have been given on
spr in the past for just how small the mass the photon would have to be
to be undetectable at the current level of accuracy of measurement.

>> From that point one establishes the properties of vectors in inertial
>> coordinate systems, and finds that, corresponding to the Pythagorean
>> magnitude of a 3-vector, for space time a 4-vector (E,p) has a magnitude

>
>An essential point which you did not make explicit is that Einstein
>wanted laws of physics to be invariant by change of coordinates.
>Therefore, acceptable entities in general relativity had to be tensors.
>

Quite right. The original poster may need to be told that a tensor is a
straightforward generalisation of a vector. If you have two vectors, a
and b, with components a^i and b^i then the thing you get by multiplying
the components, c^ij = a^i * b^j is a tensor. If you have two tensors
c^ij and d^ij, then the thing you get by adding the corresponding
components together, c^ij+d^ij, is also a tensor. These are rank 2
tensors, but you can carry on the same idea to any order.

I was just trying to give a broad overview of the ideas. Of course
general relativity came a bit later, but his fundamental philosophical
idea even in special relativity was that there is no difference in
principle between one bit of matter and another, so that there is no
distinction in principle between one observer and another, or between
the coordinate systems established by two different observers. He
encapsulated this idea first in the principle of special relativity,
that laws of physics are the same for all inertial observers, and later,
and more accurately, in the principle of general relativity, that local
laws of physics are the same for all observers. After introducing
coordinate systems defined by observers, the way this principle is most
easily expressed mathematically is the principle of general covariance,
which you cite, that the laws of physics have tensorial form.

>It so happens that what we called "mass" is a rank 0 tensor which
>happens to be the norm of a rank 1 tensor containing of the
>measurements we traditionally called "energy" and momentum. That's
>where the relation below comes from.


>> m^2 = E^2 - p^2


Indeed. The original poster may need to be told that a rank 0 tensor is
simply a number, and that a rank 1 tensor is simply a vector. The norm
is basically just the magnitude.

>> where E is the time component of the 4-vector and p is a space 3-vector.
>> then the statement m^2=0 is equivalent to a statement that the vector is
>> directed in a light-like direction, representing something moving at the
>> speed of light

>
>Another important step: it's the contravariance relation between
>energy-momentum and "space-time displacement" (dx,dy,dz,dt) that
>implies that if one has zero norm, then so has the other. And that's
>how you deduce that m=0 implies a speed-of-light displacement for the
>corresponding entity.


Yes. But you can also do it quite simply even without mentioning stuff
like contravariance and using the notations of differential geometry. If
you start with a vector representing a stationary object, then it has a
component in the time direction and zero components in space directions.
If you look at how that appears to an observer moving in the x direction
at speed v you can quickly find the principle results of special
relativity from the k-calculus using simple algebra. From there you can
find the Lorentz transformation develop the more sophisticated stuff.

Regards

--
Charles Francis
substitute charles for NotI to email

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