Wave equation invariance under Galilean transformations
xray4abc
I have stated in an earlier thread in sci.physics.relativity
that Galilean transformation do preserve the wave equation,
contrary to current belief in physics.
That statement was based on my work on the subject
which you can access using the link below:
http://laszlo.lemhenyi.com/Galilei_transformations_and_wave_equation.pdf
Since then, I have made a demonstration in conventional style,
for plane waves only for now,
which I would like to bring to your attention
[and I am sure your critique as well :-) ].
The link for this second work is:
http://laszlo.lemhenyi.com/Wave_equation_invariance_in_classical_kinematics.pdf
Best regards, LL
Daryl McCullough
>
>Hi, guys!
>I have stated in an earlier thread in sci.physics.relativity
>that Galilean transformation do preserve the wave equation,
>contrary to current belief in physics.
It's not a matter of belief alone. It's a matter of mathematical
proof.
Let's assume that we have some kind of wave that in frame F
obeys the wave equation: (in one spatial dimension for simplicity)
(d/dx)^2 Phi - 1/c^2 (d/dt)^2 Phi = 0
If we let f(x) be any differentiable function that is spiky:
It's large and positive at x=0, and is small when |x| is big.
Then two solutions to the wave equation are:
Phi_1 = f(x-ct)
Phi_2 = f(x+ct)
The first solution is a pulse traveling at speed c in the +x direction.
The second solution is a pulse traveling at speed c in the -x direction.
Note: the speed in the +x direction is equal to the speed in the -x
direction.
Now, transform to a new frame F' using the Galilean transform:
x' = x - vt
t' = t
with the inverse
x = x' + vt'
t = t'
In frame F', the solutions look like this:
Phi_1' = f(x' - (c-v)t')
Phi_2' = f(x' + (c+v)t')
The first solution is a pulse traveling at speed (c-v) in the +x direction.
The second solution is a pulse traveling at speed (c+v) in the -x direction.
The speeds of the pulses in the two directions are *not* the same. So
it is clearly impossible to have a wave equation in frame F' of the
form
(d/dx')^2 Phi' - 1/c'^2 (d/dt')^2 Phi' = 0
such that Phi_1' and Phi_2' are solutions. We would need for c'
to simultaneously satisfy:
c' = c-v
c' = c+v
which is impossible. Your derivation in your paper
http://laszlo.lemhenyi.com/Galilei_transformations_and_wave_equation.pdf
only considered the case c' = c-v.
--
Daryl McCullough
Ithaca, NY
Juan R.
Any textbook in relativity shows (i) that Galilean transformations are a
limiting case of the Lorentz transformation when velocity of the moving
frame is very very small (v/c --> 0) and (ii) the wave equation is
invariant to Lorentz transformations.
It follows that the current 'belief' in physics is that the EM wave
equation is Galilean invariant *in the limit* when the Lorentz
transformations reduce to Galilean ones.
Moreover, your pdfs show that you do not understand the difference between
covariance and *invariance*, you do not know that c is a constant, you do
not know that the "well-known law of compounded speed in classical
mechanics" applies only to low speed phenomena, that energy in different
Galilean frames are not related by terms linear in v...
xray4abc
wrote:
> only considered the case c' = c-v.
>
> --
> Daryl McCullough
> Ithaca, NY
Thanks for your pertinent comments!
I will answer the issues you have raised. I just need a bit of
time to formulate a well enough documented answer.
Best regards, LL
xray4abc
> only considered the case c' = c-v.
>
> --
> Daryl McCullough
> Ithaca, NY
You got a good point here, but....there is always a but :-)
You see, I might rewrite my paper a bit, in order to be more specific.
And that, because in fact I have omitted to highlight the fact
that my relation (XXIV)
c=c'+v
* does not refer to the absolute values of corresponding velocities!*
It has to be written
C(x) =C'(x) + V(x)
where (x) is an *index* showing that we are dealing with
the scalar components of the implied velocities with respect of
coordinate axis OX.
(I am not sure if I am using here the best way to write indices)
For simplicity, OX and O'X' may be considered of the same orientation.
Of course, even in the Galilei transform x'=x-vt , v can be positive
or can be
negative depending on the situation.
(For readers not very familiar with the subject, I would mention that
the sign *minus* in the relation from above, has nothing to do
with the orientation of the relative movement of the 2 inertial
reference
frames. The above relation is simply a scalar "transcription" of the
relation between position-vectors in the 2 frames.)
That means , instead of v we shall understand V(x), in frame F
when writing the Galilei transforms.
In frame F' the movement of frame F is represented by V'(x')
of the same absolute value as V(x) , based on the reciprocity
principle.)
In conclusion, if we are careful with our notations, we realize that
C(x)=C'(x) + V(x)
*which appears in the transformed wave equation*
could mean *in absolute values*
c'=c + v or
c'=c - v
,depending on the case, *per se* is not likely to
contradict anything, even if for one wave leads to *value* c'=c+v
while for another wave we get c'=c+v.
In both cases *the relation itself*
that is
C(x)= C'(x) + V(x)
stays the very same!
Best regards, LL
juanR...@canonicalscience.com
> xray4abc says...
>>
>>Hi, guys!
>>I have stated in an earlier thread in sci.physics.relativity that
>>Galilean transformation do preserve the wave equation, contrary to
>>current belief in physics.
>
> It's not a matter of belief alone. It's a matter of mathematical proof.
(...)
> The speeds of the pulses in the two directions are *not* the same. So it
> is clearly impossible to have a wave equation in frame F' of the form
>
> (d/dx')^2 Phi' - 1/c'^2 (d/dt')^2 Phi' = 0
>
> such that Phi_1' and Phi_2' are solutions. We would need for c' to
> simultaneously satisfy:
>
> c' = c-v
> c' = c+v
>
> which is impossible.
No Daryl, it is *possible*. Divide all by c
c'/c = 1-v/c
c'/c = 1+v/c
The system is trivially satisfied in the *limit* when v/c = 0, which is
precisely the limit when the Lorentz transformations reduce to Galilean
ones:
x' = gamma (x-vt) --> (x-vt)
t' = gamma (t-vx/c^2) --> t
Since Galilean invariance is a limiting case of Lorentz invariance,
Lorentz invariant wave equations are Galilean invariant in the limit,
when the limit properly understood and applied.
Of course, his claim that the wave equation is Galilean invariant for
arbitrary velocities is incorrect.
Tom Roberts
> Since Galilean invariance is a limiting case of Lorentz invariance,
> Lorentz invariant wave equations are Galilean invariant in the limit,
> when the limit properly understood and applied.
Not true. You are being inconsistent in the way you take that limit.
The correct way to take the limit of a quantity K' related to K by a
Lorentz transform, in the limit v/c => 0 of the Lorentz transform, is to
compute K' from K using the Lorentz transform and then take the limit.
One does NOT take the limit and then compute K' from K using the
Galilean formula (which is essentially what you did, for K'=c' and K=c).
Using the Lorentz transform it's clear that c' = c INDEPENDENT OF v/c,
so c' is unchanged in the limit v/c => 0. That is the correct way to
take the limit of c' as v/c => 0, and it does NOT yield Galilean
relativity for light.
[I am treating c' as a variable representing the speed of
light in the moving frame, and c as a constant which also
represents the speed of light in the initial (non-moving)
frame.]
This also applies to the wave equation -- the wave equation is not
Galilean invariant because that would require c' to differ from c.
Tom Roberts
Gordon Stangler
IF v/c = 0, then the frames F and F' are equal Otherwise, there is a
difference between 1+ v/c and 1 - v/c.
Don't take the limit when claiming two things are invariant. Plug in
v/c > 0, and you will get two different answers, when you claim we
should have one identical answer.
Juan R.
> juanR...@canonicalscience.com wrote:
>> Since Galilean invariance is a limiting case of Lorentz invariance,
>> Lorentz invariant wave equations are Galilean invariant in the limit,
>> when the limit properly understood and applied.
>
> Not true. You are being inconsistent in the way you take that limit.
Tom, you are the one who start from constant c' and finish with variable
c' for low v, because you did not properly understood and applied the
limit v/c --> 0 to expressions like (c'/c) = (1-v/c).
> The correct way to take the limit of a quantity K' related to K by a
> Lorentz transform, in the limit v/c => 0 of the Lorentz transform, is to
> compute K' from K using the Lorentz transform and then take the limit.
> One does NOT take the limit and then compute K' from K using the
> Galilean formula (which is essentially what you did, for K'=c' and K=c).
It is not a fair practice to snip all the parts that you are commenting.
> Using the Lorentz transform it's clear that c' = c INDEPENDENT OF v/c,
> so c' is unchanged in the limit v/c => 0. That is the correct way to
> take the limit of c' as v/c => 0, and it does NOT yield Galilean
> relativity for light.
>
> [I am treating c' as a variable representing the speed of
> light in the moving frame, and c as a constant which also
> represents
> the speed of light in the initial (non-moving) frame.]
Your treatment of c' as a variable may be related to your previous
nonsensical claim that "c is a parameter in GR". As then was said to you
Tom, c is not a parameter in GR but an universal constant. Your response
was "call them as you want":
http://groups.google.com/group/sci.physics/msg/94c04b345f963eac
> This also applies to the wave equation -- the wave equation is not
> Galilean invariant because that would require c' to differ from c.
The wave equation is Lorentz invariant for *arbitrary* values of v and
Galilean invariant when v/c --> 0.
It is a trivial thing to check that wave equation is invariant to
the special transformation x'=x and t'=t, which is *both* Galilean and
Lorentzian.
xray4abc
Yes, the Galilei transformations give c� <> c
I�d like to mention here that, even in the domain of Newtonian
physics,
c�=c could make sense in a certain situation.
This would be the situation where the coordinate transformation is
interpreted as taking place between 2 different Cartesian coordinate
systems
defined * inside the same inertial frame*.
One of coordinate systems must be considered having a fixed reference
point O,
while the other one is taken as having a mobile /moving reference
point O�
having speed v relative to the fixed point O.
This is an unusual way of considering coordinate systems, yet it is
workable.
According to this point of view, the wave-speed c, is considered
relative to the
*frame itself* and, as such, it can be the same in the different
coordinate systems
defined in the given frame.
The coordinate transformation which results from the above
supposition,
has more to do with the Lorentz transformations than with the Galilei
transformations,
as one can easily imagine.
Of course, it raises other important issues as well.
Best regards,
Laszlo Lemhenyi
Juan R.
Plain wrong. They can be two frames separated some distance D.
> Otherwise, there is a
> difference between 1+ v/c and 1 - v/c. Don't take the limit when
> claiming two things are invariant. Plug in v/c > 0, and you will get
> two different answers, when you claim we should have one identical
> answer.
This is nonsensical. If you do not take the limit v/c --> 0 then
gamma =/= 1 and (vx/c^2) =/= 0 and you are not more working in the
Galilean limit of the transformations.
Tom Roberts
please respond to what I said:
> Tom Roberts said:
>> The correct way to take the limit of a quantity K' related to K by a
>> Lorentz transform, in the limit v/c => 0 of the Lorentz transform, is to
>> compute K' from K using the Lorentz transform and then take the limit.
>> One does NOT take the limit and then compute K' from K using the
>> Galilean formula (which is essentially what you did, for K'=c' and K=c).
>> Using the Lorentz transform it's clear that c' = c INDEPENDENT OF v/c,
>> so c' is unchanged in the limit v/c => 0. That is the correct way to
>> take the limit of c' as v/c => 0, and it does NOT yield Galilean
>> relativity for light.
This directly contradicts your claim "The wave equation is Lorentz
invariant for *arbitrary* values of v and Galilean invariant when v/c
--> 0." Indeed, that claim of yours is wrong.
>> [I am treating c' as a variable representing the speed of
>> light in the moving frame, and c as a constant which also
>> represents
>> the speed of light in the initial (non-moving) frame.]
>
> Your treatment of c' as a variable may be related to your previous
> nonsensical claim that "c is a parameter in GR". As then was said to you
> Tom, c is not a parameter in GR but an universal constant.
First I'll discuss this, then your attempt at a diversion.
I treated c' "as a variable" because that is INHERENT in the question,
"what is the speed of light c' in the moving frame?". In SR, of course,
c'=c independent of v/c, AS I SAID.
And please note that this is THEORY DEPENDENT. In
Newtonian mechanics c is not at all a "universal constant".
With c as a "universal constant" and representing the
vacuum speed of light, the above question CANNOT BE ASKED
-- the speed of light in the moving frame is c. Period.
That's what "universal constant" means.
But read on for what c actually means.
In physics today, the symbol "c" has three distinct meanings:
A) the vacuum speed of light, measured in a locally-inertial frame.
B) the invariant speed of the Lorentz transform.
C) the universal constant defined by ISO in 1983 as part of the
re-definition of the meter.
The fact that the same symbol has three different meanings, all with the
same value, has confused you: c is a universal constant only in usage
(C), but c appears in GR as usage (B); it is a parameter of GR, because:
* it is a symbol appearing in its equations
* it is not determined or defined by the theory
* it has no dependency on anything else in the theory
That's in essence what we mean by "parameter". Its value can only be
determined by experiment, or by fiat.
In 1983 the ISO decided that defining c by fiat had
advantages over other approaches, and they did so in
a way designed to be consistent with the experiments
that had established its universal nature. This is
more a statement about units (as befits ISO) than it
is about the structure of physical theories -- if in
the future we discover and validate a theory in which
the vacuum speed of light varies, the c of usage (C)
will be unaffected.
> It is a trivial thing to check that wave equation is invariant to
> the special transformation x'=x and t'=t, which is *both* Galilean and
> Lorentzian.
But that is NOT the limit v/c->0, that is SETTING v/c to zero. Take the
limit PROPERLY, don't set v/c to zero -- a PROPER treatment of the limit
gets as different answer: c'=c independent of v/c, which is most
definitely not Galilean invariant.
Tom Roberts
Rock Brentwood
> I have stated in an earlier thread in sci.physics.relativity
> that Galilean transformation do preserve the wave equation,
> contrary to current belief in physics.
It's appropriate that my earlier reply should also be seen in this
venue, since it touches on some fundamental issue and answers your
question in a comprehensive fashion.
> Is the equation of an electromagnetic wave in vacuum
> invariant under Galilei transformations?
If the "vacuum" is defined as any medium which is invariant with
respect to rotational and boost symmetry, where "boost" symmetry is
that defined by the Galilean boost (e.g. (x,y,z,t) -> (x+vt,y,z,t))
then, clearly the answer is no -- there is no finite invariant speed.
If, you (instead) ask whether there is a set of equations that are CO-
variant under the Galilean transformations, then the answer is yes and
was provided by Maxwell himself (modulo a minor correction later noted
by Thomson). The Maxwell equations are given by:
div D = rho; curl H - @D/@t = J
B = curl A; E = -grad phi - @A/@t
(where @ denotes partial derivative operators)
>From the former two is derived the continuity equation
div J + @rho/@t = 0
and from the latter two is derived the "homogeneous Maxwell equations"
div B = 0; curl E + @B/@t = 0
This is invariant under the full diffeomorphism group (and a larger
group even than that) so is COMPLETELY independent of what signature
the underlying 4-dimensional continuum has (it applied, for instance,
not just to the Galilean signature of non-relativistic spacetimes or
the Lorentzian signature of relativistic spacetime; but even the
Archimedean signature of Hellenistic-era spacetime and the Euclidean
signature of 4-dimensional Euclidean space).
If the constitutive law is given by the "Lorentz relations"
D = epsilon E; B = mu H
this generates a wave equation with an invariant velocity. But it is
NOT invariant under the Galilean transformation. Rather, it selects
out a particular frame of reference. Maxwell identified this frame and
included another field (which he called G) to designate the velocity
of the observer relative to that distinguished frame.
The equations become CO-variant, when this extra field is explicitly
included. Hence, the constitutive law generalized to the following
form for "moving frames":
The Galilean Constitutive Lawsa
D = epsilon (E + G x B)
B = mu (H - G x D)
(The inclusion of the -G x D term is the one Thomson corrected Maxwell
on in the late 1880's and in the 1891 edition of Maxwell's treatise).
These equations are CO-variant under the Galilean transformation. They
remain the same under transformation, because G soaks up what had
formerly been the non-invariance of the constitutive law.
The wave equation assumes a form involving "covariant derivative"
operators --
del remains intact
@/@t becomes @/@t - G.del
Thus, the wave equation becomes
(del^2 - mu epsilon (@/@t - G.del)^2) (E/B/H/D/A/phi) = 0
for homogeneous fields (i.e. where J = 0 and rho = 0).
In this sense, the wave equation is "invariant" (i.e. covariant). But
in no other sense.
Notice by the way that because the first set of equations are
universal to ALL 4-dimensional signatures; then so are the equations
that include the G field -- except that they assume a slightly
different form for the other signatures:
D + alpha G x H = epsilon (E + G x B)
B - alpha G x E = mu (H - G x D)
where alpha = 0 for Galilean signature, alpha = (1/c)^2 for
Lorentzian
signature; and alpha < 0 for Euclidean signature (the Archimedean form
I haven't included here, and I'm not entirely sure there is one, in
fact).
Juan R.
> Rather than complain about irrelevancies and ignoring my main point,
> please respond to what I said:
Responses are on the parts of my messages that either you refuse to
read or you snipped off :-D
>> Tom Roberts said:
>>> The correct way to take the limit of a quantity K' related to K by a
>>> Lorentz transform, in the limit v/c => 0 of the Lorentz transform, is
>>> to compute K' from K using the Lorentz transform and then take the
>>> limit. One does NOT take the limit and then compute K' from K using
>>> the Galilean formula (which is essentially what you did, for K'=c' and
>>> K=c). Using the Lorentz transform it's clear that c' = c INDEPENDENT
>>> OF v/c, so c' is unchanged in the limit v/c => 0. That is the correct
>>> way to take the limit of c' as v/c => 0, and it does NOT yield
>>> Galilean relativity for light.
>
> This directly contradicts your claim "The wave equation is Lorentz
> invariant for *arbitrary* values of v and Galilean invariant when v/c
> --> 0." Indeed, that claim of yours is wrong.
The original poster states that the Galilean invariance applies at any
velocity v. You state that it never applies. Both are wrong. The truth is
in somehow the middle. Lorentz invariance is valid for arbitrary v and
Galilean invariance is valid in the limit v/c --> 0.
>>> [I am treating c' as a variable representing the speed of
>>> light in the moving frame, and c as a constant which also
>>> represents
>>> the speed of light in the initial (non-moving) frame.]
>>
>> Your treatment of c' as a variable may be related to your previous
>> nonsensical claim that "c is a parameter in GR". As then was said to
>> you Tom, c is not a parameter in GR but an universal constant.
>
> First I'll discuss this, then your attempt at a diversion.
>
> I treated c' "as a variable" because that is INHERENT in the question,
> "what is the speed of light c' in the moving frame?". In SR, of course,
> c'=c independent of v/c, AS I SAID.
Yes, this is the main question.
I will reply the equations that I wrote and *you sniped off*
Galilean invariance arises as the (v/c --> 0) limit of Lorentz
x' = gamma (x-vt) --> (x-vt)
t' = gamma (t-vx/c^2) --> t
The RHS expressions are obtained *iff* when terms of order v/c and
higher vanish.
*Your* pretense is to work with the INCONSISTENT system of equations
x' = (x-vt) (1)
t' = t (2)
c'/c = 1-v/c < 1 (3)
Take the first expression (1) and divide by the second (2)
this gives
c' = (c-v)
Or
c'/c = (1-v/c) = 1 (4)
This expression was obtained from (1) plus (2). Both of which are only
valid when (v/c --> 0). Thus trivially follows c'/c = 1.
This is the correct way to take the Galilean limit. You cannot use (3) and
pretend that magically c' is a variable. You cannot ignore term of order
(v/c) in some expressions you want and not in others.
(...)
> In physics today, the symbol "c" has three distinct meanings:
> A) the vacuum speed of light, measured in a locally-inertial frame.
B) the invariant speed of the Lorentz transform. C) the universal
> constant defined by ISO in 1983 as part of the
> re-definition of the meter.
>
> The fact that the same symbol has three different meanings, all with the
> same value, has confused you:
You have sniped the link to *your* own words containing *your* invalid
explanation that c and G are parameters GR.
You repeat now your incorrect assertion about c being a parameter, but the
sniped link contained my remark that c is a universal constant.
http://physics.nist.gov/cgi-bin/cuu/Value?c
I will treat to clarify all this again for you.
> c is a universal constant only in usage (C),
You start ignoring the history of physics.
c was an universal constant *much* before 1983. In that year, they only
gave us a new operational procedure to set the *value* of the constant.
See the entry "value" in above NIST link. Due to (i) the limitations to
define meter in a precise and exact way, (ii) progress made in the
stabilization of lasers and clocks, and (iii) the goal in maintaining
unchanged the value of the speed of light recommended in 1975 by the 15th
CGPM in its Resolution 2 (c = 299 792 458 m/s) for astronomy and geodesy,
the Resolution 1 of the 17th meeting of the CGPM (1983), decided to set
the *value* of the constant in its current one.
> but c appears in GR as usage (B); it is a parameter of GR, because:
> * it is a symbol appearing in its equations * it is not determined
> or defined by the theory * it has no dependency on anything else in the
> theory
No. c is *not* a parameter in GR. c is an universal constant for this and
other theories, a constant whose *value* is taken from that given by the
operational procedure mentioned above.
Once again you confound the *theoretical* concept of c with the specific
*value* given to c in a specific set of units. In fact, it is usual in
research literature to choose a systems of units where c=1. This do many
computations easier, but does not mean that c was a parameter you can
change. It continues being a universal constant in GR.
Also, it is usual to take the limit (c --> infinity), when studying
non-relativistic limits. This does not mean that c can be changed and done
infinity in this universe at own desire.
This limit is a mathematical 'trick' to guarantee that the limits of a set
of expressions are taken in a consistent form at once. But again c
continues being a universal constant in GR.
This 'trick' is very much related to your misconceptions about Galilean
relativity showed in this thread. We can compute the limit (v/c) as was
done above or we can directly apply the 'trick' (c --> infinity).
If you apply the 'trick' to relativistic expressions you can see we obtain
(1), (2), and (4); we do not obtain your INVALID (3).
Once more again neither c nor G are parameters of GR. They are universal
constants, as said to you before
http://groups.google.com/group/sci.physics/msg/94c04b345f963eac
This is the reason which Clifford M Will in his excellent "The
Confrontation between General Relativity and Experiment" writes:
"Furthermore, the predictions of general relativity are fixed; the
theory contains no adjustable constants so nothing can be changed."
Of course, "adjustable constants" is a synonym for "parameters".
Let me finish recommending you the link
http://physics.nist.gov/cuu/Constants/introduction.html
which gives an excellent introduction to general subject of fundamental
physical constants, explaining his role in physical theory. As you can see
here c is often used as example of universal constant.
> That's in essence what we mean by "parameter". Its value can only be
> determined by experiment, or by fiat.
>
> In 1983 the ISO decided that defining c by fiat had advantages
> over
> other approaches, and they did so in a way designed to be
> consistent
> with the experiments that had established its universal nature.
> This is
> more a statement about units (as befits ISO) than it is about the
> structure of physical theories -- if in the future we discover and
> validate a theory in which the vacuum speed of light varies, the
> c of
> usage (C) will be unaffected.
>
>
>> It is a trivial thing to check that wave equation is invariant to the
>> special transformation x'=x and t'=t, which is *both* Galilean and
>> Lorentzian.
>
> But that is NOT the limit v/c->0, that is SETTING v/c to zero. Take the
> limit PROPERLY, don't set v/c to zero -- a PROPER treatment of the limit
> gets as different answer: c'=c independent of v/c, which is most
> definitely not Galilean invariant.
Once again, you have sniped the part of *your wrong* message that I was
replying. This is unfair!
What you wrote and sniped was:
>>> This also applies to the wave equation -- the wave equation is not
>>> Galilean invariant because that would require c' to differ from c.
Since you have difficulties to take the limit correctly, my response was:
>> The wave equation is Lorentz invariant for *arbitrary* values of v and
>> Galilean invariant when v/c --> 0.
>> It is a trivial thing to check that wave equation is invariant to the
>> special transformation x'=x and t'=t, which is *both* Galilean and
>> Lorentzian.
Which of course is true. It is trivial to check that in the special case
v=0 the expression c'=c holds and the Galilean invariance of the equation
is trivial to verify. The existence of a single case already invalidates
your *incorrect* statement of above.
Regards
Andreas Most
> Galilean invariance arises as the (v/c --> 0) limit of Lorentz
>
> x' = gamma (x-vt) --> (x-vt)
>
> t' = gamma (t-vx/c^2) --> t
>
> The RHS expressions are obtained *iff* when terms of order v/c and
> higher vanish.
Quite strange way of obtaining the Galilean transformation.
It is
x' = gamma (x - vt) = (x - vt) + O((v/c)^2)
t' = gamma (t - vx/c^2) = (t - vx/c^2) + O((v/c)^2)
If you actually take only the limit v/c << 1, i.e. dropping quadratic
and higher order terms in v/c, you are left with the transformation law
x' = x - vt
t' = t - v/c * x/c
There is in additional term proportional to v/c that needs another
limit to take into account. That is the spatial distances must be
small compared to the time scales:
Delta x/c << Delta t
Only with this additional limit it is possible to use the Galilean
transformation. Obviously, this doesn't apply to the wave equation where
you look at ranges of the order
Delta x/c = Delta t
Regards,
Andreas.
--
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Juan R. González-Álvarez
> "Juan R." González-Álvarez <juanR...@canonicalscience.com> writes:
>
>> Galilean invariance arises as the (v/c --> 0) limit of Lorentz
>>
>> x' = gamma (x-vt) --> (x-vt)
>>
>> t' = gamma (t-vx/c^2) --> t
>>
>> The RHS expressions are obtained *iff* when terms of order v/c and
>> higher vanish.
>
> Quite strange way of obtaining the Galilean transformation. It is
>
> x' = gamma (x - vt) = (x - vt) + O((v/c)^2) t' = gamma (t - vx/c^2)
> = (t - vx/c^2) + O((v/c)^2)
>
> If you actually take only the limit v/c << 1, i.e. dropping quadratic
> and higher order terms in v/c, you are left with the transformation law
>
> x' = x - vt
> t' = t - v/c * x/c
>
> There is in additional term proportional to v/c that needs another limit
> to take into account.
(...)
Well I computed the non-relativistic limit (v/c --> 0). I also said in
words that "terms of order v/c and higher vanish". This derivation of
Galilean expressions from Lorentz ones is standard and reproduced in any
treatise of relativity I know.
*You* are retaining terms of order v/c, but this is *another* limit and
then, as you say, *you* get an extra term. As a consequence, your
t' = t - v/c * x/c
is not Galilean.
As explained in my response to Tom, one can also apply the 'trick'
c --> infinity
to get Galilean expressions directly.
If you apply this trick you will get the correct limiting transformation
t' = t
Regards.
xray4abc
> On Jul 9, 8:59 am, xray4abc <lemhen...@yahoo.ca> wrote:
>
> > that Galilean transformation do preserve the wave equation,
> > contrary to current belief in physics.
>
Hi, Brent
Thanks for your comments!
Invariance versus covariance deserves a bit more attention.
It was maybe not the best idea to consider Maxwell's equations
as an example to address the above issue!
I would rather consider now a very simple example:
applying coordinate transformations to
the law of linear uniform motion of a point-like
object, in order to make *my* point, regarding the physical
meaning of coordinate transformations.
I do this because I disagree with the
widely accepted statement, that changing
reference frames and changing coordinate systems is
one and the same thing.
Let us consider 2 different inertial reference frames (IRF)
moving with relative speed v, each having its fixed coordinate
axis OX and O'X' respectively, along the direction of the
movement of the object.
In frame S, the law will be
X=Xo +u(t-to) (I)
where u is the velocity of the object in IRF S.
In frame S', the law will be
X'=X'o + u'(t'-t'o) (II)
where u' is the velocity of the object in IRF S'.
The 2 laws have the very same form but they are not
identical equations. For that, we would need to have u=u' !
They are covariant equations, as applying the
Galilei transformations to (I) we will get (II) and vice versa.
There may be out there other transformations
(like X’=AX + B
t’=At + D )
, under which
2 equations, like the above, would transform identically
In a case like this
from (I) we would get
X'=X'o + u (t'-t'o) (III)
which is called the motion-law of the object in frame S' as well....!
Then we could say that, the law of linear and uniform motion
is invariant under these new transformations linking 2 IRFs.
The problem with the new transformations arises the
moment we compare (III) to (I) !
And this is because we can get (III) in a different way, simply
choosing in frame S another coordinate system setup :
reference point O'<>O and another reference time value t'o<>to.
The motion law we get for our object is in this case (in frame S
using the same notations as in S' !) is
X'=X'o + u(t'-t'o) (IV)
which is identical with (III) !
THEN, obviously, one can hardly call (III) the motion law *in
frame S' * any more!!
More than that, we have to conclude that
our NEW transformations, in fact operate within the
very same IRF, linking data not between 2 different IRFs but between
just 2 different coordinate systems of ONLY ONE IRF !
Right now, I am inclined to think that the Lorentz
transformations may be exactly
like the *new* transformations described above.
(The bases vector components are defined in the very same manifold
as you well know, old coordinates and new coordinates as well.)
I have thought quite a bit on this matter. Some of the results of
this thinking one can read in my pdf.file which can be found at
http://laszlo.lemhenyi.com/The_duality_of_Lorentz_transformation.pdf
I would appreciate your or the other guys' comments on the ideas
in the above pdf file.
Best regards,
Laszlo Lemhenyi
Andreas Most
> Andreas Most wrote on Tue, 21 Jul 2009 22:34:41 -0400:
>
>> "Juan R. González-Álvarez" writes:
>>
>>> Galilean invariance arises as the (v/c --> 0) limit of Lorentz
>>>
>>> x' = gamma (x-vt) --> (x-vt)
>>>
>>> t' = gamma (t-vx/c^2) --> t
>>>
>>> The RHS expressions are obtained *iff* when terms of order v/c and
>>> higher vanish.
>>
>> Quite strange way of obtaining the Galilean transformation. It is
>>
>> x' = gamma (x - vt) = (x - vt) + O((v/c)^2) t' = gamma (t - vx/c^2)
>> = (t - vx/c^2) + O((v/c)^2)
>>
>> If you actually take only the limit v/c << 1, i.e. dropping quadratic
>> and higher order terms in v/c, you are left with the transformation law
>>
>> x' = x - vt
>> t' = t - v/c * x/c
>>
>> There is in additional term proportional to v/c that needs another limit
>> to take into account.
>
> (...)
>
> Well I computed the non-relativistic limit (v/c --> 0). I also said in
> words that "terms of order v/c and higher vanish". This derivation of
> Galilean expressions from Lorentz ones is standard and reproduced in any
> treatise of relativity I know.
It is better to take the limit c -> oo.
I don't know how "everyday" text books do it.
> *You* are retaining terms of order v/c, but this is *another* limit and
> then, as you say, *you* get an extra term.
Yes, it is another limit. You demanded to drop also terms of order v/c
in which case you would also have to drop the term vt (= v/c * ct).
Then only x'=x and t'=t is left, which is of no use at all.
> As a consequence, your
>
> t' = t - v/c * x/c
>
> is not Galilean.
That is actually the point I was trying to make: Only taking linear
terms of v/c into account for v << c does not give the Galilean
transformation. This modified version is the transformation law for the
wave equation in the low velocity limit (where terms O((v/c)^2) have to
be dropped) but not the Galilean.
> As explained in my response to Tom, one can also apply the 'trick'
>
> c --> infinity
>
> to get Galilean expressions directly.
>
> If you apply this trick you will get the correct limiting transformation
>
> t' = t
Sure. But in this case the wave equation is meaningless because the
constant c in it is infinity.
Andreas.
Juan R. González-Álvarez
I have one more comment to add. Consider momentum
p = m gamma v + eA
the limit (v/c --> 0) gives the Galilean expression
p = m v
used, for instance, in the non-relativistic equation of motion
d/dt p = -grad V
your limit gives the non-Galilean
p = m v + eA
because A is of order (v'/c) for source particle. You would finally
consider a limit like (v'/c) --> 0 to get the correct final response.
Juan R.
I did not demand "to drop also". I took the limit v/c --> 0.
You demanded (v/c) << 1 and demanded also x/c --> 0.
> in which case you would also have to drop the term vt (= v/c * ct).
That is so incorrect like if I said to you that you have to drop the term
x because x (= x/c * c).
Note that your (v/c * ct) can be rewritten like (v/c * c/v * vt), which
evidently do not vanish in the limit v/c --> 0.
> Then
> only x'=x and t'=t is left, which is of no use at all.
It is the special case when the relative velocity between frames is zero.
That is when frames are separated by some constant distance D.
Invariance of laws for frames verifying v=0 is the reason that labs at
different places on Earth apply the same laws. I find that very useful.
>> As a consequence, your
>>
>> t' = t - v/c * x/c
>>
>> is not Galilean.
>
> That is actually the point I was trying to make: Only taking linear
> terms of v/c into account for v << c does not give the Galilean
> transformation. This modified version is the transformation law for the
> wave equation in the low velocity limit (where terms O((v/c)^2) have to
> be dropped) but not the Galilean.
>
>> As explained in my response to Tom, one can also apply the 'trick'
>>
>> c --> infinity
>>
>> to get Galilean expressions directly.
>>
>> If you apply this trick you will get the correct limiting
>> transformation
>>
>> t' = t
>
> Sure. But in this case the wave equation is meaningless because the
> constant c in it is infinity.
It is a *trick*. If c was infinity then energy would be infinity
E = mc^2 + 1/2 mv^2 + ... = mc^2 = infinity.
You would get difficulties giving meaning to (ct,x,y,z), etc.
Andreas Most
> I have one more comment to add. Consider momentum
>
> p = m gamma v + eA
>
> the limit (v/c --> 0) gives the Galilean expression
I am sure you are aware of the difference in the notation between
x --> 0 and x << 1.
> p = m v
With the notation v/c --> 0 you would have p = 0.
> used, for instance, in the non-relativistic equation of motion
>
> d/dt p = -grad V
Even with v/c --> 0, how do you extract this equation from p = mv?
> your limit gives the non-Galilean
>
> p = m v + eA
Actually, one would replace
p^m -> p^m + eA^m
in the relativistic Lagrangian of a free particle. This is called the
minimal coupling. I am not sure whether it is possible to get the
correct non-relativistic expression by first taking the Galilean limit
for the Lagrangian. Probably, it would be necessary to take terms of
order (v/c)^2 in the Lagrangian into account.
Usually one would calculate the (relativistic) equation of motion from
the Lagrangian
L = 1/2 (p + eA)^2
where terms O(e^2) are dropped because otherwise it is necessary to
consider the interaction of the particle with its own field.
Variation yields
dp_m/d tau = e u^m F_mn
(BTW, p^m is *not* "m gamma v + eA" in this equation as you seem to have
implied by your above statement)
For v/c << 1 this gives the well known result
d vec(p) / dt = e ( vec(E) + vec(v) x vec(B))
> because A is of order (v'/c) for source particle. You would finally
> consider a limit like (v'/c) --> 0 to get the correct final response.
I am not sure what you mean by this. We are talking here about a slow
moving, charged particle in an external field. Why would I need to apply
a low velocity limit to the external field? Even when considering slowly
moving electrons in a wire (as the source of the external field) it is
not possible to apply a simple non-relativistic limit because the
magnetic field is still a relativistic effect due to the huge number of
electrons in the wire.
Andreas.
Andreas Most
> Andreas Most wrote on Fri, 24 Jul 2009 21:49:32 -0400:
>
>> "Juan R. Gonz�lez-�lvarez" <email....@not.supplied> writes:
>>> *You* are retaining terms of order v/c, but this is *another* limit and
>>> then, as you say, *you* get an extra term.
>>
>> Yes, it is another limit. You demanded to drop also terms of order v/c
>
> I did not demand "to drop also". I took the limit v/c --> 0.
That is essentially the same
lim_{v/c --> 0} v/c = 0
That means the term vt is also 0 in this limit.
> You demanded (v/c) << 1 and demanded also x/c --> 0.
I have not demanded x/c --> 0.
I said that when looking at small velocities the Lorentz transformation
does not reduce to the Galilean transformation because of the additional
term xv/c^2.
Usually it is not so interesting to look at the "absolute" coordinate.
One is more interested in time differences (t2-t1) and spatial
differences (x2-x1). When looking at time differences (t2-t1) which are
much larger than (x2-x1)/c one can neglect the term xv/c^2 because
(x2-x1)v/c^2 << (t2-t1)
in this case.
>> in which case you would also have to drop the term vt (= v/c * ct).
>
> That is so incorrect like if I said to you that you have to drop the term
> x because x (= x/c * c).
Obviously, "x" is not small compared to "x".
> Note that your (v/c * ct) can be rewritten like (v/c * c/v * vt), which
> evidently do not vanish in the limit v/c --> 0.
Inserting a "1" does not change the limit. If you claim c and t being
finite then
lim_{v/c --> 0} vt = 0
>> Then
>> only x'=x and t'=t is left, which is of no use at all.
>
> It is the special case when the relative velocity between frames is zero.
> That is when frames are separated by some constant distance D.
Why have you then written out the Galilean transformation in the first
place? Taking the limit v/c --> 0 for the Lorentz transformation simply
yields x'=x and t'=t.
> Invariance of laws for frames verifying v=0 is the reason that labs at
> different places on Earth apply the same laws. I find that very
> useful.
Sure, but that is not the the scope of the thread. This is about the
invariance of the wave equation under transformations with non-zero
velocities.
>>> As a consequence, your
>>>
>>> t' = t - v/c * x/c
>>>
>>> is not Galilean.
>>
>> That is actually the point I was trying to make: Only taking linear
>> terms of v/c into account for v << c does not give the Galilean
>> transformation. This modified version is the transformation law for the
>> wave equation in the low velocity limit (where terms O((v/c)^2) have to
>> be dropped) but not the Galilean.
>>
>>> As explained in my response to Tom, one can also apply the 'trick'
>>>
>>> c --> infinity
>>>
>>> to get Galilean expressions directly.
>>>
>>> If you apply this trick you will get the correct limiting
>>> transformation
>>>
>>> t' = t
>>
>> Sure. But in this case the wave equation is meaningless because the
>> constant c in it is infinity.
>
> It is a *trick*. If c was infinity then energy would be infinity
>
> E = mc^2 + 1/2 mv^2 + ... = mc^2 = infinity.
This is not an argument against this type of approximation.
Adding a constant to the energy does not change kinematics. For v<<c it
is actually
E_Galilean = E_Lorentz - mc^2
> You would get difficulties giving meaning to (ct,x,y,z), etc.
There is no use in the limit c --> oo to do the substitution
t -> ct
Juan R.
> "Juan R. González-Álvarez" <email....@not.supplied> writes:
(...)
>> used, for instance, in the non-relativistic equation of motion
>>
>> d/dt p = -grad V
>
> Even with v/c --> 0, how do you extract this equation from p = mv?
I did not say (or even suggested) in any part that the equation of motion
was extracted from the momentum.
(...)
> Usually one would calculate the (relativistic) equation of motion from
> the Lagrangian
>
> L = 1/2 (p + eA)^2
Apart from lacking the mass m and the potential V, yours is *not* the
relativistic Lagrangian. As a consequence, the equation of motion obtained
from your L is *not* valid for arbitrarily large velocities.
> where terms O(e^2) are dropped because otherwise it is necessary to
> consider the interaction of the particle with its own field. Variation
> yields
>
> dp_m/d tau = e u^m F_mn
>
> (BTW, p^m is *not* "m gamma v + eA" in this equation as you seem to have
> implied by your above statement)
Since you do not start from the relativistic Lagrangian, you do not get
the relativistic momentum p. Moreover, it is evident that you are using a
notation where your "e" is my "-e".
(...)
Pmb
message news:pan.2009.07...@canonicalscience.com...
>> but c appears in GR as usage (B); it is a parameter of GR, because:
>> * it is a symbol appearing in its equations * it is not determined
>> or defined by the theory * it has no dependency on anything else in the
>> theory
That is a misuse of the term "parameter". c is not a parameter in GR. It's a
constant. A parameter is a variable and can thus take on more than one
value. E.g. if the position of a particle is expressed as a function of time
then the proper time would be a parameter of the particles motion. It a
quantity is not a variable then it cannot be called a parameter.
> No. c is *not* a parameter in GR. c is an universal constant for this and
> other theories, a constant whose *value* is taken from that given by the
> operational procedure mentioned above.
That's correct.
Pete
Pmb
message news:pan.2009.07...@canonicalscience.com...
> Andreas Most wrote on Mon, 27 Jul 2009 12:41:25 -0400:
>
>
> (...)
>
>>> used, for instance, in the non-relativistic equation of motion
>>>
>>> d/dt p = -grad V
That is a relativistically correct expression. Why would someone call it a
"non-relativistic equation of motion"? E.g. in a frame of reference S in
which there is a static electric field and no magnetic field the force f =
dp/dt on a charged particle is related to the potential energy V of the
particle by f = -grad V = dp/dt. This is an exact expression.
Phillip Helbig
discussion should be taken to private email. -P.H.]
Pmb wrote on Wed, 29 Jul 2009 08:09:40 +0200:
>>>> d/dt p = -grad V
>
> That is a relativistically correct expression.
Of course.
> Why would someone call it
> a "non-relativistic equation of motion"?
This may be understood in context. Just before, in the sniped part, I had
took the nonrelativistic limit for which p = mv.
That is, the above equation really mean *there*
d/dt (mv) = -grad V
As you correctly point the equation d/dt p = -grad V is also valid for
relativistic regimes, but then p is not mv and V also includes
relativistic effects.
(...)