# Re Speed of light straight from the classical wave

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```					                      Re: Speed of light straight from the classical wave equation ?

Re: Speed of light straight from the classical wave
equation ?

Source: http://sci.tech−archive.net/Archive/sci.math/2008−01/msg05038.html

• From: srp2inc@xxxxxxxxx
• Date: Wed, 30 Jan 2008 08:21:37 −0800 (PST)

On 30 jan, 10:05, Randy Poe <poespam−t...@xxxxxxxxx> wrote:

On Jan 30, 9:46 am, srp2...@xxxxxxxxx wrote:

On 30 jan, 09:14, Randy Poe <poespam−t...@xxxxxxxxx> wrote:

On Jan 30, 8:41 am, srp2...@xxxxxxxxx wrote:

Out of curiosity, I did a dimensional analysis
of the very
straightforward wave equation
describing a moving transverse wave on an
elastic string.

On the hyperphysics page,

v = sqrt[ T/(m/L) ]

Yes.

where m is mass of string, T is tension, and L is
length of string.

Re: Speed of light straight from the classical wave equation ?                         1
Re: Speed of light straight from the classical wave equation ?

http://hyperphysics.phy−astr.gsu.edu/Hbase/waves/string.html

It would be more comprehensible to actually state

The plane 2nd derivative equation you give below is fine.

From any intro ref on moving transverse waves on
solid strings.

you have mu which is in kg/m

Not necessarily, but I can imagine that there would
be an equation where the symbol mu was used to
represent mass per unit length.

It is in elementary intros to moving transverse wave on
elastic strings.

Also used in Griffiths intro to electrodynamics, which
is a popular ref.

A standard scalar wave equation is:

@^2 u/@t^2 = v^2 * del^2 u

The same, but with derivative on position at the left.

Re: Speed of light straight from the classical wave equation ?                       2
Re: Speed of light straight from the classical wave equation ?

where "@" indicates partial derivative,

Ok thanks. Didn't know which symbol to use here.

u is the quantity
which is oscillating, del^2 means the Laplacian, and
the parameter v turns out to be the propagation
speed.

For EM radiation, u is a field strength. For
sound waves, u is pressure. For your oscillating
string, u is transverse distance. All of these
things have different units.

Different "macro" units, yes. But at the end,
they all simplify to velocity squared, of course.

you have T which is in Newton

you have (2nd p derivative on position) f =
mu/T (2nd p derivative on
time) f

I take it f is the symbol for displacement here.

Yes. Transverse displacement

So mu/T = 1/v^2, or v = sqrt( T/mu ) as I said above.

Now, from E=mc^2 mass is really E/c^2,
which is Joules/c^2

Re: Speed of light straight from the classical wave equation ?                      3
Re: Speed of light straight from the classical wave equation ?

Yes, Joules/c^2 are a mass unit. I don't know if
I'd say mass is "really E/c^2". But that's one
possible system of units.

which gives mu the dimensions Joules/(c^2
m)

Newton is, after resolving to its most
fundamental dimensions, Joules/
meter

What does "most fundamental" mean? Yes, a Newton
is a Joule/meter. It's also a kg*m/sec^2, which
is probably more fundamental since those are the
fundamental mks units.

Agreed, but you don't see the energy component then (joules).

which results in the dimensions of mu/T
simplifying as follows

mu/T = [J/(c^2 m)] (m/J) = 1/c^2

Yes mu/T = 1/v^2 has the units of inverse velocity
squared.

More than that I think. If you introduce joules as a
unit, this 1/v^2 can be nothing but 1/c^2 by very definition
of mass as being E/c^2, directly from re−arranging E=mc^2.

Conclusion:

Re: Speed of light straight from the classical wave equation ?                      4
Re: Speed of light straight from the classical wave equation ?

There is not even need to go to derivations
Maxwell's forth equation to get it.

No, there is no need to go to Faraday and Maxwell
to conclude that velocity of a wave on a string
has the units of velocity.

Not what I meant. I referred to getting c^2 from
the elementary wave equation.

Nor does anything in Faraday or Maxwell have anything
to say about what the velocity on a string is.

Nothing but the conclusion that by similarity with
the classical wave equation for moving wave on a
string, the product eps0 mu0 of the 2nd derivative
was equated to the inverse of a velocity squared,
which was the speed of light.

I sort of wonder what the meaning of this is.

I wonder what you think "this" is.

You seem to have concluded that velocity has
units of velocity (already known), that mu/T has
units of inverse velocity−squared (already known,
standard equation for speed of waves on a string),
and that you don't need Maxwell's equations to
recognize what parameter represents speed in a
general wave equation (also already known).

Known stuff of course, but not really my conclusion.

Re: Speed of light straight from the classical wave equation ?                      5
Re: Speed of light straight from the classical wave equation ?

My conclusion is as stated. We can get c as a fixed
speed for energy from the straight classical wave
equation by simple dimensional analysis if we simplify
the standard units down to basic dimensions involving
energy.

But that's not what you calculated. You
calculated the DIMENSIONS of mu/T.

Well, I analyzed the dimensions, one aspect
of which turns out to be the fixed speed of
light squared by definition.

If you thought that this step

which results in the dimensions of mu/T
simplifying as follows

mu/T = [J/(c^2 m)] (m/J) = 1/c^2

actually was saying that mu/T = 1/c^2 numerically
rather than dimensionally, you would have just
concluded that waves on a string propagate
at the speed of light.

hmm... not really. I concluded that apparently,
when energy is being dealt with with the classical
wave equation, the only propagation speed that can
be had is the speed of light, by definition.

That is untrue. As you started out saying, you
were just doing a dimensional analysis.

Yes. But the "just" is your contribution, not mine.

Here's
how the numerical calculation would work out
in those units:

Re: Speed of light straight from the classical wave equation ?                         6
Re: Speed of light straight from the classical wave equation ?

First we need to know how to convert between
cgs or mks units and E/c^2 mass units. An
electron has a mass of 511 keV/c^2, or
511e3 * 1.602e−19 = 8.19e−14 J/c^2. It also
has a mass of 9.11e−31 kg.

So 1 kg = 8.99e16 J/c^2.

Suppose I have a piece of string of mass 10 g
(0.01 kg, 8.99e14 J/c^2) and length 1 m. So
mu in these units is 8.99e14 (J/c^2)/m.

Suppose the tension on this string is 100 N,
or 100 J/m.

Then mu/T = (8.99e14 J/m.c^2) / (100 J/m)

= 8.99e12/c^2

No sweat up to here.

As your dimensional analysis correctly determined,
mu/T has the dimensions of 1/c^2, i.e. it
can be expressed as a dimensionless constant
times 1/c^2.

Not really here, since the tention you chose is
arbitrary. See below.

Thus the speed of wave motion in this string,
sqrt(T/mu), is sqrt[c^2/8.99e12] = c/3e6,
which you can express in any units you like.
If we express c in m/sec, this gives us
that sqrt(T/mu) = 100 m/sec.

Calculation ok with the tension you chose, but
you can't chose an arbitrary tension in the present
context since the 1/c^2 conclusion already
integrates by definition the force related to
the energy you wish to use. It is the whole
of mu/T that equals 1/c^2.

So you have to use the precise force that
relates to the energy involved.

Re: Speed of light straight from the classical wave equation ?                          7
Re: Speed of light straight from the classical wave equation ?
But anyhow, if a mathematician has some input
to contribute on this, it would be welcome.

André Michaud
.

Re: Speed of light straight from the classical wave equation ?                       8

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