When did Photon Become Massless?
Logical Pike
carries mass. Max Born argued the same as well as others. When did this
particle become massless and under what circumstances? Why? What led to
the change? I'm interested in much feedback. Hopefully, there will be a
minimum of psychoanalysis.
Sent via Deja.com http://www.deja.com/
Before you buy.
DLZC
The photon may carry enough energy to create mass. It will of course, carry
away the energy from some sort of destroyed mass.
The legs of whatever 'mass table' quantum mechanics requires are folded up in
transit.
As far as the referent to either Bohr or Einstein, I cannot help. What were
their exact quotes?
David A. Smith
Village Idiot
and...@attglobal.net
>
> Einstein and Bohr, in their discussions, both argued that the photon
> carries mass. Max Born argued the same as well as others.
When? Cite some references.
> When did this
> particle become massless and under what circumstances? Why? What led to
> the change? I'm interested in much feedback. Hopefully, there will be a
> minimum of psychoanalysis.
>
In SR, E^2 = (p*c)^2 + (m*c^2)^2
For a photon, E = p*c. Guess where that leaves
m in the above equation.
John Anderson
Jackie & Barry
and...@attglobal.net wrote:
> In SR, E^2 = (p*c)^2 + (m*c^2)^2
> For a photon, E = p*c. Guess where that leaves
> m in the above equation.
That is why my equation is more consistent, (when used with the force
law F = mc*omega),
(mc^2)^2 = (mc^2/gamma)^2 + (pc)^2
In this formulation, which is fully consistent with observation, p can
continue to represent mv, since, for c = v,
(mc^2)^2 = 0 + (pc)^2 = (mc^2)^2
It is also fully consistent with Einstein's SR postulates, except that
the simplistic F = ma is dropped, as Einstein seems to have suggested
might be necessary.
Barry
Bill Rowe
<logic...@my-deja.com> wrote:
>Einstein and Bohr, in their discussions, both argued that the photon
>carries mass. Max Born argued the same as well as others. When did this
>particle become massless and under what circumstances? Why? What led to
>the change?
There really hasn't been a change. Whether you regard a photon as
massless or not depends precisely on how you define mass.
The normal definition is a quantity of matter that I can measure on a
balance pan by comparing it to a known mass. If you use this definition
then the relationship between mass, momentum and energy that holds for
all particles is
E^2 = (mc^2)^2 + (pc)^2
Also, for photons it is known E = pc. Substituting this into the above
relationship gives m = 0 for photons.
However, you could substitute mv for p in the above relationship. Then
with a little algebra the terms could be rearranged to give
E = g(v)mc^2 where I've used g(v) to be the usual gamma factor. Note
g(v) is dimensionless. So, you could replace the product g(v)m with
another variable which would have units of mass.
So you could define E/c^2 to be the "mass" of a photon.
The problem with this definition is this mass is neither invariant nor
scalar. Also, this "mass" is no longer simply related to a quanity of
matter. So, using this definition in a consistent manner will make mean
more complex expressions for relationships involving mass. Consequently,
there is a strong preference for keeping photons as massless.
--
-
PGPKey fingerprint: 6DA1 E71F EDFC 7601 0201 9243 E02A C9FD EF09 EAE5
and...@attglobal.net
>
> and...@attglobal.net wrote:
>
> > In SR, E^2 = (p*c)^2 + (m*c^2)^2
>
> > For a photon, E = p*c. Guess where that leaves
> > m in the above equation.
>
> That is why my equation is more consistent,
More consistent with what? Your speculation of
the month or years of real physics research?
John Anderson
Jackie & Barry
and...@attglobal.net wrote:
> Barry wrote:
> > That is why my equation is more consistent,
>
> More consistent with what? Your speculation of
> the month or years of real physics research?
As I wrote, in the portion of my post beneath the line which you have
quoted, which portion you apparently did not read:
"In this formulation, which is fully consistent with observation.... "
So, firstly, it's consistent with observation
"p can continue to represent mv, since, for c = v,
(mc^2)^2 = 0 + (pc)^2 = (mc^2)^2"
Secondly, it's consistent across all velocities, apparently "massless"
photons are treated in a fashion consistent with "massive" particles.
"It is also fully consistent with Einstein's SR postulates, except that
the simplistic F = ma is dropped, as Einstein seems to have suggested
might be necessary."
Thirdly, it's consistent with the postulates of SR, except the one which
Einstein himself seems to have doubted, that F = ma.
And fourthly, it's self consistent.
Barry
GLOBARR
Bill Rowe <bjr...@earthlink.net> wrote:
In <8oh9af$8sl$1...@nnrp1.deja.com>,
> Einstein and Bohr, in their discussions, both argued that the
> photon carries mass. Max Born argued the same as well as
> others. When did this particle become massless and under
> what circumstances? Why? What led to the change?
Bill Rowe wrote:
There really hasn't been a change. Whether you regard a
photon as massless or not depends precisely on how you define
mass.
The normal definition is a quantity of matter that I can
measure on a balance pan by comparing it to a known mass. If
you use this definition then the relationship between mass,
momentum and energy that holds for all particles is
E^2 = (mc^2)^2 + (pc)^2
Also, for photons it is known E = pc. Substituting this into
the above relationship gives m = 0 for photons.
However, you could substitute mv for p in the above
relationship. Then with a little algebra the terms could be
rearranged to give E = g(v)mc^2 where I've used g(v) to be the
usual gamma factor. Note g(v) is dimensionless. So, you could
replace the product g(v)m with another variable which would
have units of mass.
So you could define E/c^2 to be the "mass" of a photon.
The problem with this definition is this mass is neither
invariant nor scalar. Also, this "mass" is no longer simply
related to a quanity of matter. So, using this definition in a
consistent manner will make mean more complex expressions for
relationships involving mass. Consequently, there is a strong
preference for keeping photons as massless.
Gerald L. O'Barr (Globarr) comments:
What a refreshing post! Bill said, `There really hasn't
been a change.' And this is correct. There has not been any
change at all in the physics. There has not been any change in
the math. There has been no change in the concepts. The
photon today has just as much mass as it ever had, no more, no
less. Its relativistic mass is E/c^2. This mass gives the
photon momentum, and it gives it energy. This mass gives the
photon the power to move light sails, and to turn a radiometer.
What was done was an effort of mind control, where the SR
experts wanted to prevent any person from saying that a
particle moving at c could have mass. This violated several of
their arguments. And so they changed the meaning of the word
`mass' to prevent anyone from being able to say that a photon
had mass. But they cannot change the physics. A photon has
all the characteristics of mass, and thus no one should
question that a photon has mass. The rest mass of a photon
might be zero, but since the rest frame of a photon cannot
physically exist, the rest mass of a photon will never be
scientifically measured. If something cannot be measured, the
SR experts tell you that it cannot be science. So go figure!
The SR experts were wrong to do what they did, and I will not
rest until this error is corrected!
A question to Bill: You said that the correct measurement
of mass was to measure it on a balance pan by comparing it to a
known mass. What would you get if you measured 5 identical
photons on one side with 10 other identical photons on the
other side? If photons had no mass, the pan certainly would
remain level. But if the pan did not remain level, what would
that really indicate????? Or were you just kidding about this
being a meaningful measurement?
Gerald L. O'Barr glo...@yahoo.com
Please Read: http://www.uc-online.com/absolute
And Jan 99 issue of Physics Today about the ether!
(We need to improve the SR FAQ)
Henry Wilson
>Logical Pike wrote:
>>
>> Einstein and Bohr, in their discussions, both argued that the photon
>> carries mass. Max Born argued the same as well as others.
>
>When? Cite some references.
>
>> When did this
>> particle become massless and under what circumstances? Why? What led to
>> the change? I'm interested in much feedback. Hopefully, there will be a
>> minimum of psychoanalysis.
>>
>
>In SR, E^2 = (p*c)^2 + (m*c^2)^2
>
>For a photon, E = p*c. Guess where that leaves
>m in the above equation.
How do you know photon energy doesn't equal sqrt((p*c)^2 + (m*c^2)^2),
where m is known to be < a certain value.
All you have said is that "because a photon doesn't have mass, its
energy is p*c and because its energy is p*c it doesn't have mass".
That is not very convincing, John.
>John Anderson
Bilge
>browe said some stuff about
> The normal definition is a quantity of matter that I can
>measure on a balance pan by comparing it to a known mass. If
>you use this definition then the relationship between mass,
>momentum and energy that holds for all particles is
>
>E^2 = (mc^2)^2 + (pc)^2
>
That is incorrect. You cannot place a massless particle on
a scale to weigh it - Any massless particle is constrained
to move at c. The only comparison would be to dropping
a rock on the pan and removing the part that you weigh with
the rock stationary. Th photon that corresponds to:
p^2 = E^2 - m^2
for the rock. Obviously, if the rock just sits on the scale, this
cannot be a photon with a non-zero momentum.
Bill Rowe
dav...@david15.dallas.nationwide.net wrote:
>>browe said some stuff about
>
> > The normal definition is a quantity of matter that I can
> >measure on a balance pan by comparing it to a known mass. If
> >you use this definition then the relationship between mass,
> >momentum and energy that holds for all particles is
> >
> >E^2 = (mc^2)^2 + (pc)^2
> That is incorrect.
Given what I've seen of your postings I am very surprised you are saying
what I wrote above is incorrect. Could you be more specific?
> You cannot place a massless particle on a scale to weigh it - Any
> massless particle is constrained to move at c.
Agreed.
> The only comparison would be to dropping
> a rock on the pan and removing the part that you weigh with
> the rock stationary. Th photon that corresponds to:
> p^2 = E^2 - m^2
And this is exactly the same equation I wrote in units where c = 1.
Where is the disagreement? It applies to photons as well since setting m
to zero gives E = pc.
> for the rock. Obviously, if the rock just sits on the scale, this
> cannot be a photon with a non-zero momentum.
No argument. But where is the diffence between what I've wrtitten and
what you seem to be saying?
Jackie & Barry
Bill Rowe wrote:
> E^2 = (mc^2)^2 + (pc)^2 (1)
...
> Also, for photons it is known E = pc. Substituting this into the above
> relationship gives m = 0 for photons.
...
> However, you could substitute mv for p in the above relationship.
....
> So you could define E/c^2 to be the "mass" of a photon.
Correct, but the following problems are not, in fact, problems.
> The problem with this definition is this mass is neither invariant nor
> scalar. Also, this "mass" is no longer simply related to a quanity of
> matter. So, using this definition in a consistent manner will make mean
> more complex expressions for relationships involving mass. Consequently,
> there is a strong preference for keeping photons as massless.
That doesn't necessarily follow, we can define Kinetic Mass Energy to be
mvc (pc).
We can differentiate mv wrt distance, and get the result mc*omega (in
Newtonian mechanics, we get ma).
For v << c, mc*omega ~ ma and the two expressions are observationally
almost indistinguishable.
Thus we can retain all the postulate of SR, changing only the Newtonian
F = ma to F = mc*omega, to obtain a model which is as consistent with
observation as is SR.
In particular, c is the maximum velocity wrt an inertial frame, inertial
rest mass energy is still invariant
and equation (1) becomes:
E^2 = (mc^2)^2 = [(mc^2)/gamma]^2 + (pc)^2
R^2 = I^2 + K^2
(Rest Mass Energy)^2 = (Inertial Mass Energy)^2 + (Kinetic Mass
Energy)^2
As a bonus, relating the frequency of a particle to its Inertial Mass
Energy gives a physical cause for time dilation.
[f(R)]^2 = [f(I)]^2 + [f(K)]^2
Very simple, very much observationally verifiable, very simple and very
self consistent.
Barry
Paul Alan Cardinale
>
<snip>
> The SR experts were wrong to do what they did [redefine the word "mass"), and I will not
> rest until this error is corrected!
>
Besides, its not an error, its a preference.
Your lack of understanding of semantics will drive you crazy (I suspect
that will be a short trip).
Paul Cardinale
and...@attglobal.net
It is when you realize that measured photon energies
and momenta are related by E = p*c. If you take that
data, it ought to be pretty obvious that m = 0.
John Anderson
Bilge
Re: When did Photon Become Massless? to usenet:
>In article <slrn8qvan...@radioactivex.lebesque-al.net>,
>dav...@david15.dallas.nationwide.net wrote:
>
>>>browe said some stuff about
>>
>> > The normal definition is a quantity of matter that I can
>> >measure on a balance pan by comparing it to a known mass. If
>> >you use this definition then the relationship between mass,
>> >momentum and energy that holds for all particles is
>> >
>> >E^2 = (mc^2)^2 + (pc)^2
>
>> That is incorrect.
>
>Given what I've seen of your postings I am very surprised you are saying
>what I wrote above is incorrect. Could you be more specific?
>
break in and make that comment. My reference was regarding
the way you equated the momentum of a moving massive object to a
photon mass via the m = E/c^2. In order to do that the photon needs
to have a velocity other than c to go with it. Using your scales
example, for instance, you can measure a value if you throw the
rock against the scale. You can measure a value if a photon strikes
the scale. If you cannot measure a velocity such that you may transform
one of the two to zero so that it may be weighed, then you cannot
define a mass. So it really isn't fair to define a mass for something
which cannot stop moving since the ability to stop something from
moving and weigh it in your rest frame is what made it a mass in the
first place.
>
>No argument. But where is the diffence between what I've wrtitten and
>what you seem to be saying?
>
unless I misattributed that line.
Bilge
>
>Thus we can retain all the postulate of SR, changing only the Newtonian
>F = ma to F = mc*omega, to obtain a model which is as consistent with
>observation as is SR.
>
>In particular, c is the maximum velocity wrt an inertial frame, inertial
>rest mass energy is still invariant
>and equation (1) becomes:
>
> E^2 = (mc^2)^2 = [(mc^2)/gamma]^2 + (pc)^2
>
> R^2 = I^2 + K^2
>
>(Rest Mass Energy)^2 = (Inertial Mass Energy)^2 + (Kinetic Mass
>Energy)^2
>
Allow me to simplify your expression:
(1) Factor out c^2
(mc)^2 = (mc/\gamma)^2 + p^2
(2) Mutiply by \gamma^2 and subtract (mc)^2 from both sides:
(mc)^2 [\gamma^2 - 1] = (p\gamma)^2
(3) Divide through by \gamma^2 and note that \gamma^ = 1/[ 1-\beta^2 ]
(mc)^2 [\gamma^2 - 1]/\gamma^2 = p^2
|
+-> [\gamma^2 - 1]/\gamma^2 = \beta^2
=> p^2 = (mc\beta)^2
(4) So far, I haven't changed anything you've defined and the above
gives me:
p = mc\beta If I now use the standard definition of \beta,
in which I'll denote the velocity as v' I have:
p = mv' Obviously, v' can't be the velocity. In fact,
guess what it is and guess what you are missing.
>As a bonus, relating the frequency of a particle to its Inertial Mass
>Energy gives a physical cause for time dilation.
>
> [f(R)]^2 = [f(I)]^2 + [f(K)]^2
>
How so? By implicitly calling your velocity the 4-velocity
and having no expression for the energy?
>Very simple, very much observationally verifiable, very simple and very
>self consistent.
I see nothing simple about it. Do some kinematics with it. For
example, try compton scattering. I get the following for the
invariants SR defines, i.e., p^up_u . I'll supress the indicies
for readability where it's obvious I need an upper and lower
and use p,q for initial and p',q' for final momenta:
(p + q)^2 = (p' + q')^2 = -m^2
Since p^2 = -m2 for the electron in any frame, then the following
is also true (and is further useful since we can ignore the final
electron momentum and work in the lab frame):
(p + q - q')^2 = -p'^2 = -m^2
=> p^2 + 2pq - 2pq' - 2qq' = -m2
=> p^2 = -m^2
So I get: pq - pq' - qq' = 0
In the lab frame (the electron rest frame), or any other, I can take
whatever axes I wish to use to define q as (hv,hv,0,0) so I can
call it (hv, hv n) and q' = (hv', hv'n') with n and n' being unit
(3) vectors in the direction of q and q' (q^ = q'^2 is still 0).
The factor q_{0_q^{0} is h^vv' and q_{i}q'^{i}, i=1,3 is then
just h^2 vv'(n.n') or h^2vv'cos(x):
mhv - mhv' - h^2 vv' - h^2 vv'cos(x) = 0
1-cos(x) = (m/h)vv'/(v-v')
I don't see this as being particularly simple via your redefined
version where p^2 isn't E^2 - m^2.
Jackie & Barry
Bilge wrote:
> Barry said
> > E^2 = (mc^2)^2 = [(mc^2)/gamma]^2 + (pc)^2 (1)
> > R^2 = I^2 + K^2 (2)
> >(Rest Mass Energy)^2 = (Inertial Mass Energy)^2 + (Kinetic Mass
> >Energy)^2
> Allow me to simplify your expression:
<several lines deleted>
> p = mc\beta If I now use the standard definition of \beta,
> in which I'll denote the velocity as v' I have:
> p = mv' Obviously, v' can't be the velocity. In fact,
> guess what it is and guess what you are missing.
Actually, that was rather clumsy, you can get there from (1) far more
quickly by
(pc)^2 = (mc^2)^2( 1 - 1/gamma)^2
p = (mc)(1 - 1/gamma)
Which means that p is at a maximum, wrt any inertial frame, when v = c
in that frame.
I'm glad you're at least trying the maths.
Now, what do think it physically *means*?
And what experiment falsifies it, that would not also falsify SR?
> >As a bonus, relating the frequency of a particle to its Inertial Mass
> >Energy gives a physical cause for time dilation.
> > [f(R)]^2 = [f(I)]^2 + [f(K)]^2 (3)
> How so? By implicitly calling your velocity the 4-velocity
> and having no expression for the energy?
Because f(R) = R/h, f(I) = I/h, f(K) = K/h and, using (2), we get
equation (3)
[f(R)]^2 = [f(I)]^2 + [f(K)]^2 (3)
> >Very simple, very much observationally verifiable, very simple and very
> >self consistent.
> I see nothing simple about it.
Perhaps you can't see the wood for the trees.
Equation (3) clearly predicts time dilation, as a *direct consequence*
of Quantum Theory.
Barry
Bilge
>
>
Personally, I think it means gibberish.
>And what experiment falsifies it, that would not also falsify SR?
>
I can rearrange an equation any number of ways. There are usually
a few useful ways. Why don't you do as I asked and get something
useful from it?
>> >As a bonus, relating the frequency of a particle to its Inertial Mass
>> >Energy gives a physical cause for time dilation.
>
>> > [f(R)]^2 = [f(I)]^2 + [f(K)]^2 (3)
>
>
>> How so? By implicitly calling your velocity the 4-velocity
>> and having no expression for the energy?
>
>Because f(R) = R/h, f(I) = I/h, f(K) = K/h and, using (2), we get
>equation (3)
>
> [f(R)]^2 = [f(I)]^2 + [f(K)]^2 (3)
>
>
>> >Very simple, very much observationally verifiable, very simple and very
>> >self consistent.
>
>> I see nothing simple about it.
>
>Perhaps you can't see the wood for the trees.
Perhaps you could do as I asked and show me how much simpler you
can get a physical result like I got for compton scattering rather
than just asserting that you can.
>Equation (3) clearly predicts time dilation, as a *direct consequence*
>of Quantum Theory.
>
Nonsense. You have no quantum theory at all in anything. All
you have is a algebraic manipulations of a simple eqn for
no particular reason.
Jackie & Barry
Bilge wrote:
> Jackie & Barry said some stuff about
> >Now, what do think it physically *means*?
> I'm asking you. I have no idea what you think your variables represent.
I'm truly sorry that you don't understand what my variables represent.
The equation was:
E^2 = (mc^2)^2 = [(mc^2)/gamma]^2 + (pc)^2
Perhaps you can read a text book to find out what the symbols E, m, v,
gamma and p might represent.
Did I have any other variables in there?
How can you possibly understand any physics, if you don't understand
what the fundamental symbols represent?
> Personally, I think it means gibberish.
Did you study SR, GR and QM, and did you have lots of help - text books,
professors, fellow students, faith?
On your own, no formal text, no professor, no fellow students, no faith,
might they all have seemed gibberish to you?
> >And what experiment falsifies it, that would not also falsify SR?
> I can rearrange an equation any number of ways. There are usually
> a few useful ways. Why don't you do as I asked and get something
> useful from it?
I've listed several "predictions", each of which has been observed.
In the last 27 months, count them, 27, not one experimental observation
has been put forward which falsifies the above equation or any of my
predictions.
Do you think that a physical cause for time dilation is a useful
prediction?
> >> >As a bonus, relating the frequency of a particle to its Inertial Mass
> >> >Energy gives a physical cause for time dilation.
> >> > [f(R)]^2 = [f(I)]^2 + [f(K)]^2 (3)
> >> How so? By implicitly calling your velocity the 4-velocity
> >> and having no expression for the energy?
> >Because f(R) = R/h, f(I) = I/h, f(K) = K/h and, using (2), we get
> >equation (3)
> > [f(R)]^2 = [f(I)]^2 + [f(K)]^2 (3)
> >> >Very simple, very much observationally verifiable, very simple and very
> >> >self consistent.
> >> I see nothing simple about it.
> >Perhaps you can't see the wood for the trees.
> Perhaps you could do as I asked and show me how much simpler you
> can get a physical result like I got for compton scattering rather
> than just asserting that you can.
Time dilation isn't good enough for you?
Can you name any other model that gives a physical cause for time
dilation?
> >Equation (3) clearly predicts time dilation, as a *direct consequence*
> >of Quantum Theory.
> Nonsense. You have no quantum theory at all in anything. All
> you have is a algebraic manipulations of a simple eqn for
> no particular reason.
Are you saying that the equation f = E/h is wrong?
Have you ever seen it before?
If you have seen it before, where did you see it, if not in quantum
theory?
Barry
Jackie & Barry
Bilge wrote:
> Barry said
> > E^2 = (mc^2)^2 = [(mc^2)/gamma]^2 + (pc)^2 (1)
> I see nothing simple about it. Do some kinematics with it. For
> example, try compton scattering. I get the following for the
> invariants SR defines, i.e., p^up_u . I'll supress the indicies
> for readability where it's obvious I need an upper and lower
> and use p,q for initial and p',q' for final momenta:
> (p + q)^2 = (p' + q')^2 = -m^2
> Since p^2 = -m2 for the electron in any frame,
p^2 is the same for an electron in all frames?
Where does that come from?
p^2 = -m2?
That equation isn't even dimensionally correct.
surely the correct equation is
p^2 = (mv)^2
which is certainly not the same in all inertial frames.
Momentum isn't Lorentz invariant, even for photons, even in SR.
Since what followed seems to be based on the above, I won't respond to
the rest.
Barry
Luc Bourhis
You are still after that !!! One month ago I have shown to you that
your ideas are badly refuted by every experiment studying the motion of
a relativistic charge particle in an electromagnetic field. You have
never answered my last post.
> That doesn't necessarily follow, we can define Kinetic Mass Energy to be
> mvc (pc).
>
> We can differentiate mv wrt distance, and get the result mc*omega (in
> Newtonian mechanics, we get ma).
What do you mean by differentiating mv wrt distance ? That's even less
clear than last time. Let's start a seance of maieutics:
1) What is the energy E of a particle of velocity v ? By energy I mean
the quantity I can measure with a calorimeter in which the particle has
been stopped.
2) What is the equation of motion for a charged particle in an
electrical field ?
> [...] very much observationally verifiable
Fine ! Then you can answer my questions .....
--
Luc Bourhis
and...@attglobal.net
You wrote
(mc^2)^2 = (mc^2/gamma)^2 + (pc)^2
I assume that you mean are using gamma = 1/sqrt(1 - (v/c)^2)
What is the value of gamma when v = c?
John Anderson
Bill Rowe
dav...@david15.dallas.nationwide.net wrote:
[bunch of stuff skipped]
> Just the temptation to define an "equivalent photon mass" as E/c^2,
> unless I misattributed that line.
I thought when I made this comment I also made it clear from context
this is something that could be defined, not something that should be
defined.
Frank Wappler
> Let's start a seance of maieutics:
> 1) What is the energy E of a particle of velocity v ?
> By energy I mean the quantity I can measure with a calorimeter
> in which the particle has been stopped.
What is "velocity v", i.e.
how (and, if applicable wrt. whom) should "velocity v of a particle"
be measured?
Perhaps related to that is the question how any two
should measure whether they have "stopped" wrt. each other.
The corresponding measurement procedure might also be useful
for defining coordinate relations (such as pairwise distances)
in terms of which you might describe just what you mean by
"a calorimeter" ...
> 2) What is the equation of motion for a charged particle in an
> electrical field ?
... and how to characterize (the distribution of)
"an electrical field" in the first place.
Best regards, Frank W ~@) R
Bilge
>
>
>
>> Barry said
>
>> > E^2 = (mc^2)^2 = [(mc^2)/gamma]^2 + (pc)^2 (1)
>
>> I see nothing simple about it. Do some kinematics with it. For
>> example, try compton scattering. I get the following for the
>> invariants SR defines, i.e., p^up_u . I'll supress the indicies
>> for readability where it's obvious I need an upper and lower
>> and use p,q for initial and p',q' for final momenta:
>
>> (p + q)^2 = (p' + q')^2 = -m^2
>
>> Since p^2 = -m2 for the electron in any frame,
>
>
>p^2 is the same for an electron in all frames?
>
>Where does that come from?
>
>p^2 = -m2?
>
Rer^H^H^HRead the paragraph above. It's an ancient family
recipe.
>
>surely the correct equation is
>
>p^2 = (mv)^2
>
>which is certainly not the same in all inertial frames.
>
>Momentum isn't Lorentz invariant, even for photons, even in SR.
>
p^2 = -m^2 is the invariant 4 momentum which is
E^2 - p.p, from the eqn you seem
determined to convolute.
>Since what followed seems to be based on the above, I won't respond to
>the rest.
>
OK. Don't let the fact that it's correct stop
you though. Need another example?
Bilge
>
>Please show me your derivation.
>
Did you READ the part you have now deleted? No? Ok. I'll fill in
the blanks: p^{u} = (-E, p_x, p_y, p_z) or (E, -px, -py, -pz) depending
upon whther you choose -+++ or +--- for your metric.
p^{u}p_{u} = -E^{2} + \vec{p}\cdot\vec{p}
since p^{u}p_{u} is an invariant, I can evaluate it at v=0:
p^2 == p^{u}p_{u} = -m^2, with p^2 just the indicies suppressed
version AS IT WAS NOTED in the section
I suggested to read before replying.
>
>> >Momentum isn't Lorentz invariant, even for photons, even in SR.
>
>> q^2 = 0. For all (real) photons.
>> p^2 = -m^2 is the invariant 4 momentum which is
>> E^2 - p.p, from the eqn you seem
>> determined to convolute.
>
>Please show me your derivation.
For a photon (one more time) q = (-hv, hv, 0, 0)
q^2 = q^{u}q_{u} = (-hv,hv,0,0)(hv,hv,0,0) -(hv)^2 + (hv)^2 = 0.
>
>Please show me your derivation.
Save this one so I can just refer to it next time.
Jackie & Barry
Bilge wrote:
> Barry said
> >Please show me your derivation.
> >
>
> Did you READ the part you have now deleted? No? Ok. I'll fill in
> the blanks: p^{u} = (-E, p_x, p_y, p_z) or (E, -px, -py, -pz) depending
> upon whther you choose -+++ or +--- for your metric.
>
> p^{u}p_{u} = -E^{2} + \vec{p}\cdot\vec{p}
>
> since p^{u}p_{u} is an invariant, I can evaluate it at v=0:
>
> p^2 == p^{u}p_{u} = -m^2, with p^2 just the indicies suppressed
> version AS IT WAS NOTED in the section
> I suggested to read before replying.
That doesn't seem to have anything to do with my model.
How do you get from *my* equation
(mc^2)^2 = (mc^2/gamma)^2 + (pc)^2
to p^2 = -m^2?
> >> >Momentum isn't Lorentz invariant, even for photons, even in SR.
> >
> >> q^2 = 0. For all (real) photons.
> >> p^2 = -m^2 is the invariant 4 momentum which is
> >> E^2 - p.p, from the eqn you seem
> >> determined to convolute.
> >
> >Please show me your derivation.
>
> For a photon (one more time) q = (-hv, hv, 0, 0)
> q^2 = q^{u}q_{u} = (-hv,hv,0,0)(hv,hv,0,0) -(hv)^2 + (hv)^2 = 0.
Very pretty, but nothing to do with my equation.
> >Please show me your derivation.
> Save this one so I can just refer to it next time.
But it has nothing to do with my equation
(mc^2)^2 = (mc^2/gamma)^2 + (pc)^2
in which p^2 certainly does not equal -m^2
You seem to be very confused.
Barry
Luc Bourhis
> Luc Bourhis wrote:
>> Let's start a seance of maieutic:
>
>> 1) What is the energy E of a particle of velocity v ?
>> By energy I mean the quantity I can measure with a calorimeter
>> in which the particle has been stopped.
>
> What is "velocity v", i.e.
> how (and, if applicable wrt. whom) should "velocity v of a particle"
> be measured?
1) Choose a procedure to synchronize clocks. Use it to build an network
of synchronized clocks.
2) Choose a procedure to measure distances.
3) The speed of a point moving from A to B is the distance between A
and B divided by the difference between the instant of arrival
according to the clock at B and the instant of departure according to
the clock at A.
And please do not translate that in your pairwise notation nobody
understands.
> Perhaps related to that is the question how any two
> should measure whether they have "stopped" wrt. each other.
The particle is stopped in the calorimeter if it enters it but does not
go out of it.
> The corresponding measurement procedure might also be useful
> for defining coordinate relations (such as pairwise distances)
> in terms of which you might describe just what you mean by
> "a calorimeter" ...
The actual devices are complicated apparatus whose calibration relies
on model which relies on electrodynamics. In this discussion just
consider a block of matter whose variation of temperature is monitored
so as to deduce the energy brought by the particle in its bulk.
>> 2) What is the equation of motion for a charged particle in an
>> electrical field ?
>
> ... and how to characterize (the distribution of)
> "an electrical field" in the first place.
1) choose a method to measure a difference of potential between two
points A and B
2) A being fixed moves B so as to maximize the difference of potential.
construct the vector whose modulus is this maximum and which points
from A to B
3) redo step 2) with distances between A and B decreasing toward 0
4) the limit of the vector when the distance goes to zero is the
electrical field at A
--
Luc Bourhis
Luc Bourhis
> Luc Bourhis wrote:
>
>> Barry wrote:
>
>>> We can differentiate mvc wrt distance, and get the result mc*omega (in
>>> Newtonian mechanics, we get ma).
>>
>> What do you mean by differentiating mv wrt distance ? That's even less
>> clear than last time. Let's start a seance of maieutics:
>
> Of course, that was a typo, I meant mvc, which is Kinetic Mass Energy.
How does this answer my question ? Since we can take c=1 if that's suit
us that's a very irrelevant detail.
>> 1) What is the energy E of a particle of velocity v ? By energy I mean
>> the quantity I can measure with a calorimeter in which the particle has
>> been stopped.
>
> The energy of a particle is invariant in this model, it is always mc^2.
>
> v don't enter into it.
>
> Your question is meaningless in this model
It is meaningless to ask what is the prediction of the outcome of such
a fundamental experiment !?!?!? Are you kidding ? If it is so you can
not even claim you can compete with SR for the latter answer this
question unambiguously, and from the very beginning.
> , as you would know if you actually tried to *understand*.
Why should I try to understand your theory when you admit implicitely
that it is vastly inferior to SR ? Newcomers have to prove their value.
>> 2) What is the equation of motion for a charged particle in an
>> electrical field ?
>
> That's another question that shows that you don't understand the model.
>
> If you *understood* the equation, which is *simple*, you would see that
> the answer is trivial.
>
> Hint: what are the two relevant Force Laws.
>
> Hint: equate the two for a charged particle.
>
> It's algebra for 10 year olds.
So your model can predict the motion of charged particle whereas my
first question was meaningless. Are you kidding ? Anyway if the
calculations are so simple you should have no difficulties showing
them. Right ? Will you take up the challenge or will you continue to
behave as an ostrich ?
--
Luc Bourhis
Henry Wilson
'Measured' with what acuracy? Not to 1 part in 10^40 surely.
and...@attglobal.net
>
> > I assume that you mean are using gamma = 1/sqrt(1 - (v/c)^2)
>
> > What is the value of gamma when v = c?
>
>
> Which is 0, when v = c.
>
> I wrote it as /gamma for ease of understanding.
>
> Equation (1) can be written as:
>
> (mc^2*X)^2 = ( mc^2)^2 - (pc)^2
>
> = (mc^2)^2 - (mc^2*v/c)^2
>
> WE can solve for X, as follows:
>
> Dividing by (mc^2)^2,
>
> X^2 = 1 - (v/c)^2
>
> X = [(1 -(v/c)^2]^0.5
>
>
> which is equivalent to /gamma, except that the equation doesn't "blow
> up" at v = c.
>
> and the equation
>
> (mc^2)^2 = (mc^2/gamma)^2 + (pc)^2
>
>
> i.e. it is consistent with SR.
>
> And, if you examine the equation from a "physical" viewpoint, certain
> factors like time dilation "fall out" naturally, and the Kinetic Mass
> Energy for all particles has the same value (pc).
>
> In particular, it is consistent with my starting point,
>
> F = mc*omega.
>
> since if we integrate that Force, over distance we get pc for Kinetic
> Mass Energy.
>
> AS I've said umpteen times now, it's a kind of renormalization of SR.
>
In SR, for a particle with v < c (and in units in which c = 1),
E = gamma*m, p = gamma*m*v where gamma = 1/sqrt(1 - v^2)
and m is the mass of the particle.
m^2 = E^2 - p^2
= gamma^2*(m^2 - (m*v)^2)
is a mathematical identity that is true for m > 0, m = 0
or even m < 0. In other words, it doesn't tell you what
m is. It's also only defined for v <> 1, since both
E and p are infinite in that limit.
If I multiply through by 1/gamma, I get your equation
except what you call p is m*v.
That's still a mathematical identity as long as v <> 1.
What I get is
m^2*(1 - v^2) = m^2 - (m*v)^2
ie m^2 = m^2
it doesn't define the mass for v <> 1 or even for v = 1.
For v = 1, it especially doesn't determine the mass since
both sides of the equation are identically zero.
The other problem with what you're trying to do
is that you can't play around with equality
when you're multiplying and dividing by zero.
x*0 = 0
y*0 = 0
Therefore, x*0 = y*0
so if I divide both sides by 0, x = y, no matter what
x or y are. You can't get rid of problems like this
by multiplying through or dividing through by 0.
This invalidates your claim that your formula is equivalent
to m^2 = E^2 - p^2 because you wrote down the valid formulas
for v <> c and then extrapolated them to v = c which produces
results that are valid only if you assume that division
by 0 produces meaningful numbers.
John Anderson
Luc Bourhis
> Luc Bourhis wrote:
>> 1) Choose a procedure to synchronize clocks.
>
> [...]
>
> Which _explicit_ procedure/thought_experiment do you suggest
> as definite, at least in principle?
Examples are
- Esynch or its generalized version: A sends a light signal to B which
sends it back to A which measures the round-trip time T. At t=0
according to its clock A resends a light signal to B which sets its
clock at t=u*T where u is an arbitrary numerical constant (Esynch
corresponds to u = 1/2).
- slow clock transport
>> 2) Choose a procedure to measure distances.
>
> Which _explicit_ procedure/thought_experiment do you suggest
> as definite, at least in principle ?
Examples are
(1) choose a rod and put two marks on it. Mark the middle point between
them by using the well known construction with a compass. Iterate the
processus until there are enough marks on the rod to measure lengths
with a given precision.
(2) decide that the length between A and B is the round-trip time of
light A->B->A multiplied by an arbitrary numerical constant.
> Note the _order_ in which you've asked for choices to be made,
> and in which the corresponding explicit definitions are to be given:
>
> the procedure addressing (2) may use and refer to results/values
> obtained by conducting the procedure that would have been defined
> in addressing (1),
Not necessarily so as shown by my example (1).
> but vice versa, the procedure addressing (1) may _not_ use and refer
> to results/values that are not yet definite, but only promised
> to be defined later, in addressing (2).
Easily satisfied as shown by my examples.
>> 3) The speed of a point moving from A to B is the distance between A
>> and B divided by the difference between the instant of arrival
>> according to the clock at B and the instant of departure according
>> to the clock at A.
>
> Sure - but keep in mind that in order to take differences between
> various labels of individual instants/states/readings/proper_times,
> they must have been _mapped_ to each other by the procedure
> that would have been defined in addressing (1);
which is why my 1) came before this 3)
> and in order to divide a distance value by an interval,
> any distance value measured through the procedure
> that would have been defined in addressing (2)
> should be obtained (presumably) as a product,
> with (at least) one factor being an interval.
Why ? Both numerators and denominators are real number in the above
definition of speed. That's enough to make this division a rigorous
mathematical operation.
> Arithmetic is done with _numbers_,
> not with incidental labels/marks/symbols.
They are number.
>>>> By energy I mean the quantity I can measure with a calorimeter
>>>> in which the particle has been stopped.]
>
>>> Perhaps related to that is the question how any two
>>> should measure whether they have "stopped" wrt. each other.
>
> [...]
>
>> The particle is stopped in the calorimeter if it enters it
>> but does not go out of it.
>
> I guess that "entering" and "going out" are defined in terms
> of distance values that are measured by _several distinct_
> observers/constituents of the calorimeter wrt. the particle.
Place a layer of detectors just outside the boundary of the
calorimeter. The particle is stopped inside the latter if and only if
one and only one detector is activated.
>>> The corresponding measurement procedure might also be useful
>>> for defining coordinate relations (such as pairwise distances)
>>> in terms of which you might describe just what you mean by
>>> "a calorimeter" ...
>
>> The actual devices are complicated apparatus
>
> Unless you call _every_ complicated apparatus "a calorimeter" [...]
which is why I did not give any definition since I did not want to
consider a real calorimeter.
>> whose calibration relies on model which relies on electrodynamics.
> (See below.)
>> In this discussion just consider a block
>
> ... procedures (1) and/or (2) will surely allow a definition
> of just what you mean by "a block" ...
A connected object is all I need. This topological concept is
independent of the notion of distance as everybody knows. And this has
nothing to do with time at all.
>> of matter
>
> ... would that _exclude_ any particular observers/things
> who otherwise had succeeded in determining that they
> constituted "a block" wrt. each other?
Try as many materials as you want until you manage to stop the particle
inside the calorimeter.
>> whose variation of temperature is monitored
>
> What do you mean by/how do you suggest to reproducibly
> measure/monitor "temperature"?
First we need a thermomether. There are many different sort of such
devices as you know. Some rely on a measurement of length. Other are
electronic devices which produces an electrical tension as an
intermediate result, which is then digitalized. What matters is of
course that the measurement of energy does not depend on our choice.
See later ....
Then put a thermometer T1 at the surface of the calorimeter, a layer of
heat insulator around it, and another thermother T2 ouside as close as
possible to T1 (the precise measurement of the distance between T1 and
T2 is not needed). Start noting temperatures of T1 and T2 long before
the processus producing the particle has been activated, which means
that the precise timing of the instant of production does not matter.
When T1 becomes constant after having increasing for some time check
that T2 has kept its initial value, up to a given accuracy. If not
choose another insulator. If so note the constant reading T of T1.
So now we can have a curve T = f(v) where v is the speed of the
particle, measured independently, _using Esynch_ .
>> so as to deduce the energy brought by the particle in its bulk.
>
> (Note that you haven't concluded a reproducible definition
> and measurement procedure of what you mean by "energy" yet.)
For a large variety of materials used to build the calorimeter and a
large variety of thermometers one will find the relation
T(v) = a/(1-v^2)^(1/2) + b
By computing the ratio of pair of [T(v)-b]'s obtained with pair of
different materials and pair of different thermometers one can check
that the corresponding ratio of a's do not depend on the particle type.
Once we know these latter ratios we can deduce that, up to an unknown
constant,
T(v) = r E(v) + b
where E(v) is independent of the calorimeter material and of the
thermometer. By making measurements for different particles one gets
E(v) = m/(1-v^2)^(1/2)
The exact value of m depends on a choice of units, i.e. it is defined
up to a constant multiplicative factor.
>>>> 2) What is the equation of motion for a charged particle in an
>>>> electrical field ?
>
>>> ... and how to characterize (the distribution of)
>>> "an electrical field" in the first place.
>
>> 1) choose a method to measure a difference of potential between two
>> points A and B
>
> Please specify such a method.
Again there are plenty of possible method, as you might know. Some rely
on measurement of angles or distances (the old devices which measured
in fact a force between two plate charged because of the difference of
potential). Modern electronic devices do not rely on the measurement of
times or distances but I do not know precisely how they work.
Anyway this does not really matter for what I wanted to do. Choose many
such different devices. Apply my algorithm at the same point A.
Normalize all the resultant outcomes to the outcome obtained with one
of them. Check that my algorithm gives the same result with all devices
at other points A, up to a given accuracy. Then one knows the
electrical field up to a multiplicative constant factor, which
corresponds to a choice of unit. If one of this device gives a result
too different from the others, don't use it. We have plenty of
experimental proofs that a large set of apparatus meets this criteria.
> Also, if that's relevant, please define/specify a procedure by which
> to determine whether or not some particular A and B are "points".
Point is the usual geometrical concept. What this idealization
approximates depends on the device used to measure the difference of
potential. So the outcome of my algorithm depends only on this device,
which is ultimately removed as explained above.
>> 2) A being fixed moves B so as to maximize the difference of
>> potential.
>
> Therefore it is necessary (but not sufficient) that the values of
> potential differences obtained above are non-negative real numbers.
No. The usual convention is that 2 > -5.
> But I may not be able to identify B for which that difference
> is _the_ maximum. Perhaps I may find that all of sufficiently
> many values are below some particular bound, and I may use this
> bound to _estimate a probable maximum value_, for some
> particular B_stat.
If the set of points B for which the difference of potential is
indistiguishingly maximum is not reduce to one point, just do the
procedure I described later for all points of this set. This will give
a set of vectors at each steps. One can then check that this set
converges to a single point.
>> Construct the vector whose modulus is this maximum
>> and which points from A to B
>
> Above it should have been clarified how A and B measure
> their _distance_ wrt. each other.
I do not really need to know the distance between A and B. Any
Euclidian distance will do it, since the notion of A converging toward
B is independent of the distance used on a finite dimensional metric
topological space as R^3 which I use implicitely in my algorithm.
> Now, what do you mean by "vector, which points from A to B" ?
The usual geometrical concept in R^3.
>> 3) redo step 2) with distances between A and B decreasing toward 0
>> 4) the limit of the vector when the distance goes to zero is the
>> electrical field at A
>
> Can you prove that such limits exist?
No. But I can check that the difference between two vectors of the
above sequence is smaller than some given precision after some point in
the sequence. If this never happens I decide that the method failed and
that I cannot say anything about the electrical field at A. If this
happens I have a strong evidence of the existence of a limit.
So no I can't prove that it exists but my algorithm gives unambiguously
a result when it succeeds. One can hardly ask for more than that.
> Also, unless you call _every_ such limit, wrt. _every_ particular B,
> "the electrical field at A", among _all_ such limits, wrt. any
> particular B there may surely be some which then differ more or less
> strongly from what you understand to be "the electrical field at A",
> in dependence on the particular B, or that may gravitate to entirely
> different values altogether.
I disagree. The only ambiguity in the above algorithm is not associated
with B but with the method chosen to measure a difference of potential.
To remove that, try several different ones and check that the algorithm
produces values whose differences are smaller than some given
precision.
--
Luc Bourhis
Jackie & Barry
and...@attglobal.net wrote:
> Barry wrote:
> > > (mc^2)^2 = (mc^2/gamma)^2 + (pc)^2 (1)
> > In particular, it is consistent with my starting point,
> > F = mc*omega.
> > since if we integrate that Force, over distance we get pc for Kinetic
> > Mass Energy.
> If I multiply through by 1/gamma, I get your equation
> except what you call p is m*v.
Like I said, equation (1) doesn't "blow up" at v = c.
> This invalidates your claim that your formula is equivalent
> to m^2 = E^2 - p^2 because you wrote down the valid formulas
> for v <> c and then extrapolated them to v = c which produces
> results that are valid only if you assume that division
> by 0 produces meaningful numbers.
As I wrote, I used 1/gamma to make the equation easier to "relate" to.
The "correct" form, which does *not* involve division by zero is:
(mc^2)^2 = [(mc^2)(1 - v^2/c^2)^0.5]^2 + (mvc)^2
Indeed, it is Classical SR that leads to division by zero, since
p = gamma*mv
which "blows up" at v = c.
If you look at the derivations in SR, Classically we have:
F = m dp/dt ---> p = gamma*mv.
[Or, conversely, some texts work backwards from the definition of
momentum:
p = gamma*mv ----> F = m dp/dt]
In my model:
p = mv -----> F = mc*omega
[or alternatively
F = mc*omega ----> p = mv]
It's a choice between retaining the definition of momentum as mv and
retaining Newton's F = dp/dt.
In my opinion, the best (simplest) choice was not made in 1905.
Barry
and...@attglobal.net
>
> and...@attglobal.net wrote:
>
> > Barry wrote:
>
> > > > (mc^2)^2 = (mc^2/gamma)^2 + (pc)^2 (1)
>
> > > In particular, it is consistent with my starting point,
>
> > > F = mc*omega.
>
> > > since if we integrate that Force, over distance we get pc for Kinetic
> > > Mass Energy.
>
> > If I multiply through by 1/gamma, I get your equation
> > except what you call p is m*v.
>
> Like I said, equation (1) doesn't "blow up" at v = c.
>
That is part of my point. If you multiply through by
gamma, everything is fine for v < c. If you then set
v = c, then the equations before and after your multiplication
don't have to be equivalent.
Saying that your result doesn't "blow up" is fine, but for
v < c, it's a mathematical identity that is true for any value
of m. It can't be used to calculate m if you set v = c.
> > This invalidates your claim that your formula is equivalent
> > to m^2 = E^2 - p^2 because you wrote down the valid formulas
> > for v <> c and then extrapolated them to v = c which produces
> > results that are valid only if you assume that division
> > by 0 produces meaningful numbers.
>
> As I wrote, I used 1/gamma to make the equation easier to "relate" to.
>
> The "correct" form, which does *not* involve division by zero is:
>
> (mc^2)^2 = [(mc^2)(1 - v^2/c^2)^0.5]^2 + (mvc)^2
>
> Indeed, it is Classical SR that leads to division by zero, since
>
> p = gamma*mv
>
> which "blows up" at v = c.
>
Which it should. The energy and momentum of a massive object
diverge as v -> c. What you need to do is to write
E^2 = (p*c)^2 + (m*c^2)^2 (1)
For m <> 0 and v < c, this is a mathematical
identity if
E = (m*c^2)/sqrt(1 - (v/c)^2) (2)
p = (m*v)/sqrt(1 - (v/c)^2) (3)
Note that equation (1) doesn't contain any gamma's either
unless I explicitly make the substitutions (2) and (3).
It's true if I do make the substitutions, but it doesn't
contain gamma unless I do that.
But (1) is also true if I make the substitutions
E = p*c (3)
m = 0 (4)
If I insist on interpreting (1) using (2) and (3)
then I get no blow up if I let m -> 0 as v -> c.
But then I'm looking at 0/0. That is meaningless.
And all you're doing is to try to pretend that
you can take 0/0, multiply it by zero anc calculate
the mass. But if 0/0 is meaningful, then any
number is equal to any other number as I pointed
out in my posting.
But that's not what SR does. It uses (3) and (4).
And (3) is certainly true for EM radiation.
John Anderson
Jackie & Barry
and...@attglobal.net wrote:
>Barry wrote:
> > and...@attglobal.net wrote:
> > > Barry wrote:
> > > > > (mc^2)^2 = (mc^2/gamma)^2 + (pc)^2 (1)
> > > > In particular, it is consistent with my starting point,
> > > > F = mc*omega.
> > Like I said, equation (1) doesn't "blow up" at v = c.
> That is part of my point. If you multiply through by
> gamma, everything is fine for v < c. If you then set
> v = c, then the equations before and after your multiplication
> don't have to be equivalent.
> Saying that your result doesn't "blow up" is fine, but for
> v < c, it's a mathematical identity that is true for any value
> of m. It can't be used to calculate m if you set v = c.
> > As I wrote, I used 1/gamma to make the equation easier to "relate" to.
> > The "correct" form, which does *not* involve division by zero is:
> > (mc^2)^2 = [(mc^2)(1 - v^2/c^2)^0.5]^2 + (mvc)^2
> > Indeed, it is Classical SR that leads to division by zero, since
> > p = gamma*mv
> > which "blows up" at v = c.
> Which it should. The energy and momentum of a massive object
> diverge as v -> c.
If we take the postulates of SR PLUS Newton's F = ma, then energy and
momentum both diverge as v --> c.
If we take the postulates of SR PLUS my F = mc*omega, then energy and
momentum do not diverge as v --> c.
In this model, momentum (*defined* as mv) tends to mc, and Kinetic Mass
Energy (*defined* as pc, or mvc) tends to mc^2.
Rest Mass Eneregy (invariant R, |mc^2|) is represented by a vector.
That vector has component vectors - Kinetic Mass Energy (K, |mvc|) and
Inertial Mass Energy (I, |mc*(1 - v^2/c^2)^0.5|) in any given inertial
frame in which the particle has velocity = v.
In the particle's rest frame, R = I and K = 0
> What you need to do is to write
> E^2 = (p*c)^2 + (m*c^2)^2 (1)
No, what I Do write is:
R = I + K
--->
R.R = I.I + K.K
--->
(mc^2)^2 = [mc^2*(1 - v^2/c^2)^0.5]^2 + (mvc)^2
> For m <> 0 and v < c, this is a mathematical
> identity if
> E = (m*c^2)/sqrt(1 - (v/c)^2) (2)
> p = (m*v)/sqrt(1 - (v/c)^2) (3)
which both "blow up" for v = c.
> Note that equation (1) doesn't contain any gamma's either
> unless I explicitly make the substitutions (2) and (3).
> It's true if I do make the substitutions, but it doesn't
> contain gamma unless I do that.
> But (1) is also true if I make the substitutions
> E = p*c (3)
> m = 0 (4)
> If I insist on interpreting (1) using (2) and (3)
> then I get no blow up if I let m -> 0 as v -> c.
But m is invariant, it never ---> anything except m.
> But then I'm looking at 0/0.
But my model never divides by 0 anywhere.
> But that's not what SR does. It uses (3) and (4).
> And (3) is certainly true for EM radiation.
And it's true in my model too, if we retain the classical definition, p=
mv,
Since K = pc (= (mv)c = mc^2)
And my equation
(mc^2)^2 = [mc^2*(1 - v^2/c^2)^0.5]^2 + (mvc)^2
in which m is invariant, leads, in the frame in which v = 0 and K = 0,
to
Inertial Mass Energy = I = mc^2 = R
and leads, in the frame in which v = c and I = 0, to
Kinetic Mass Energy = K = mvc = pc = mc^2 = R
And c is the maximum velocity for any particle, in an inertial frame,
since Kinetic Mass Energy is at a maximum value, K = mc^2, when v = c,
and Inertial Mass Energy is at a minimum, 0.
Barry
Jackie & Barry
and...@attglobal.net wrote:
> Which it should (blow up). The energy and momentum of a massive object
> diverge as v -> c.
If we take the postulates of SR PLUS Newton's F = ma, then energy and
momentum both diverge as v --> c.
If we take the postulates of SR PLUS my F = mc*omega, then energy and
momentum do not diverge as v --> c.
In this model, momentum (*defined* as mv) tends to mc, and Kinetic Mass
Energy (*defined* as pc, or mvc) tends to mc^2.
Rest Mass Energy (invariant R, |mc^2|) is represented by a vector.
That vector has component vectors - Kinetic Mass Energy (K, |mvc|) and
Inertial Mass Energy (I, |mc*(1 - v^2/c^2)^0.5|) in any given inertial
frame in which the particle has velocity = v.
In the particle's rest frame, R = I and K = 0
> What you need to do is to write
> E^2 = (p*c)^2 + (m*c^2)^2 (1)
No, what I do write is:
R = I + K
--->
R.R = I.I + K.K
--->
(mc^2)^2 = [mc^2*(1 - v^2/c^2)^0.5]^2 + (mvc)^2
> For m <> 0 and v < c, this is a mathematical
> identity if
> E = (m*c^2)/sqrt(1 - (v/c)^2) (2)
> p = (m*v)/sqrt(1 - (v/c)^2) (3)
which both "blow up" for v = c.
> If I insist on interpreting (1) using (2) and (3)
> then I get no blow up if I let m -> 0 as v -> c.
But m is invariant, it never ---> anything except m.
> But then I'm looking at 0/0.
But my model never divides by 0 anywhere.
> But that's not what SR does. It uses (3) and (4).
> And (3) is certainly true for EM radiation.
And it's true in my model too, if we retain the definition, p= mv,
since K = pc (= (mv)c = mc^2)
Bilge
>
>If we take the postulates of SR PLUS Newton's F = ma, then energy and
>momentum both diverge as v --> c.
>
>If we take the postulates of SR PLUS my F = mc*omega, then energy and
>momentum do not diverge as v --> c.
>
Since both postulates of special relativity explicitly state that
inertial frames are being considered, why do you think you should
be concerned with forces? The clocks of accelerated observers do
not remain synchronized.
Frank Wappler
. By energy I mean the quantity I can measure with a calorimeter
. in which the particle has been stopped.
> Frank Wappler wrote:
> > Luc Bourhis wrote:
> > > 1) Choose a procedure to synchronize clocks.
> > [Recognizing that any two distinct clocks cannot have
> > equal states (the no-cloning theorem),
> > it may be sufficient to choose a procedure to calibrate clocks.
> > (I.e. for the clocks and anyone else to determine and agree on
> > which state/reading/proper_time of one corresponds to
> > which state/reading/proper_time of another, if any.)]
> > Which _explicit_ procedure/thought_experiment do you suggest
> > as definite, at least in principle?
> Examples are
> - Esynch
AFAIK, that would be the calibration procedure suggested
by A. Einstein in 1916, as sketched in chapter viii,
"On the idea of time in physics" of "Relativity":
that pairs A and B calibrate their individual states
to signals from (or to) "the middle between A and B".
I suppose that you and Einstein meant by "the middle between A and B"
in any particular trial the observer M which satisfies that
- A finds two roundtrips to M same as one roundtrip to B,
- B finds two roundtrips to M same as one roundtrip to A,
- M finds the roundtrip to A same as the roundtrip to B
(perhaps aming other criteria).
> or its generalized version: A sends a light signal to B which
> sends it back to A which measures the round-trip time T.
... i.e. explicitly the interval
T == { A_sees_B_sees_A_signals, A_signals }.
Note that this is an ordered pair of symbols/labels,
_not_ a real number, in general.
> At t=0 according to its clock A resends a light signal to B
... surely B can observe transitions between A's states,
such as A's transition into state A_0.
> [B] sets its clock
_Which_ particular state of B's clock is to be reset/altered?
And since A surely must conduct the same/reproducible procedure,
which particular states of A's clock are being reset/altered
according to A's observations of transitions between B's states?
(A calibration procedure wouldn't require
"setting/altering/relabelling" of a state;
it merely establishes a relation/map _between_ states.)
> at t=u*T where u is an arbitrary numerical constant
AFAIU, that means B relabels some particular state B_&
to B_u*{ A_sees_B_sees_A_j, A_j }
for some particular A_j which B observed before B_$.
What calibration is thereby established?
Which particular state of A corresponds to B's state
B_$ == B_u*{ A_sees_B_sees_A_j, A_j };
and how should A determine and agree on that?
> (Esynch corresponds to u = 1/2).
Einstein's calibration procedure doesn't involve
"setting/altering/relabelling" states of any of the observers involved.
> - slow clock transport
What would you mean by "slow" _before and without_ having completed
the definition for
> > > 2) Choose a procedure to measure distances.
which would allow you to define
> > > 3) The speed of a point moving from A to B is the distance
> > > between A and B divided by the difference between
> > > the instant of arrival according to the clock at B
> > > and the instant of departure according to the clock at A.
??
> Which _explicit_ procedure/thought_experiment do you suggest
> as definite [for (2)], at least in principle ?
> Examples are
> [...] choose a rod and put two marks on it.
Fine - there exist at least two distinct marks. Surely even more.
> Mark the middle point between them by using the well known
> construction with a compass.
Please specify "the well known construction with a compass".
Note that any reproducible construction that is to be used
within a definition of the procedure for (2)
cannot reference any results of (2) in turn already;
specificly, that construction must not involve
measurements of and requirements/conditions on distance values.
> Iterate the processus until there are enough marks on the rod
> to measure lengths with a given precision.
Before and without having completed a definition of the procedure
for measuring distance values, there is no definite notion of
"given precision of distance values" either.
> [...] decide that the length between A and B is the round-trip
> time of light A->B->A multiplied by an arbitrary numerical constant.
Since individual states and therefore also intervals can be
calibrated, that appears to be a reproducible procedure.
With the symbolic/numeric constant "c/2"
it corresponds to Einstein's distance definition.
Note that an interval multiplied by an arbitrary numeric constant
is _not_ a real number, in general;
an interval multiplied by an arbitrary symbolic/numeric constant
is obviously _not_ a real number, in general, either.
> > Note the _order_ in which you've asked for choices to be made,
> > and in which the corresponding explicit definitions are to be given:
> > the procedure addressing (2) may use and refer to results/values
> > obtained by conducting the procedure that would have been defined
> > in addressing (1),
> Not necessarily so as shown by my example [...]
Please note again that "(2) _may_ use and refer to ...";
and that some of your examples failed to show anything definite yet.
> > but vice versa, the procedure addressing (1) may _not_ use
> > and refer to results/values that are not yet definite,
> > but only promised to be defined later, in addressing (2).
> Easily satisfied as shown by my examples.
Pending your definition of "slow".
> > and in order to divide a distance value by an interval,
> > any distance value measured through the procedure
> > that would have been defined in addressing (2)
> > should be obtained (presumably) as a product,
> > with (at least) one factor being an interval.
> Both numerators and denominators are real number in the above
> definition of speed.
Then that definition couldn't be used at all
to determine speed from the intervals and distances obtained above.
> > > The particle is stopped in the calorimeter if it enters it
> > > but does not go out of it.
> > I guess that "entering" and "going out" are defined in terms
> > of distance values that are measured by _several distinct_
> > observers/constituents of the calorimeter wrt. the particle.
> Place a layer of detectors
... that's apparently more than one, corresponding to my guess.
> just outside the boundary of the calorimeter.
(This involves measurements of distances already.)
> The particle is stopped inside the latter if and only if
> one and only one detector is activated.
> > [If so, then your notion of "being stopped" could be defined
> > even _stronger_, e.g. as equal ratios between those various
> > distance values, in all trials.]
> > > In this discussion just consider a block
> > ... procedures (1) and/or (2) will surely allow a definition
> > of just what you mean by "a block" ...
> A connected object is all I need. This topological concept is
> independent of the notion of distance as everybody knows.
AFAIK, the topological concept of connectedness is defined
in terms of membership in "neighborhoods", which in turn
is _defined_ through measured distance values
of member pairs wrt. each other, to be compared to each other.
How would you define "connectedness" without reference
to measured distance values?
> And this has nothing to do with time at all.
By the only definite distance definition we've identified so far
(namely Einstein's, above) distances values _are_ derived from
measured and calibrated intervals.
> Try as many materials as you want until you manage to stop
> the particle inside the calorimeter.
(A suitable choice of "detector layer" might do even without
_any_ "materials inside".)
> > > whose variation of temperature is monitored
> > What do you mean by/how do you suggest to reproducibly
> > measure/monitor "temperature"?
> First we need a thermometer. There are many different sort of such
> devices as you know.
I know that there are many things labelled/marked "thermometer";
just now the screen I'm looking at got so marked, too.
Which of them would constitute a thermometer in the context
of the measurement procedure/thought experiment
by which you define "energy"?
> Some rely on a measurement of length.
"Length" of _what_; and what's the correlation of various
measured length values with "temperature", in the absence
of any definition and measured values of "temperature"?
> Other are electronic devices which produces an electrical tension
> as an intermediate result, which is then digitalized. What matters
> is of course that the measurement of energy does not depend
> on our choice. See later ....
> Then put a thermometer T1 at the surface of the calorimeter,
> a layer of heat insulator around it
What's the procedure to decide whether or not some layer constitutes
"heat insulator", in any particular trial?
> and another thermometer T2 ouside as close as possible to T1 (the
> precise measurement of the distance between T1 and T2 is not needed).
(Then they might possibly be found even closer to each other,
in some trials.)
> Start noting temperatures of T1 and T2
I may be able to measure and note various length/distance values, or
values of "electrical tension" (subject to what you'll define below).
> long before the processus producing the particle has been activated,
> which means that the precise timing of the instant of production
> does not matter.
(Surely a corresponding calibration procedure may be conducted by
the "thermometers" and who/whatever "activates particle production".)
> When T1 becomes constant after having [been] increasing for some time
How do you suggest to compare two lenght/"temperature" values
_between distinct trials_, "initially" vs. "after activation" ??
I can only compare _ratios_ of distances (from the same one trial)
between several trials.
> check that T2 has kept its initial value, up to a given accuracy.
(Same question.)
> If not choose another insulator.
... i.e. discard this trial, conduct another.
> If so note the constant reading T of T1.
If there exist any trials in which were determined that
"T1 became constant, after having increased" and
"T2 kept its initial value, up to a given accuracy" at all,
which hasn't been established yet.
> So now we can have a curve T = f(v) where v is the speed of the
> particle, measured independently, _using Esynch_ .
> For a large variety of materials used to build the calorimeter
> and a large variety of thermometers one will find the relation
> T(v) = a/(1-v^2)^(1/2) + b
Surely the specifications given so far may allow to find
many other relations, too.
> By computing the ratio of pair of [T(v)-b]'s obtained with pair of
> different materials and pair of different thermometers
> one can check that the corresponding ratio of a's do not depend
> on the particle type.
Surely the specifications given so far may allow to find
many values for such ratios, for various sets of trials,
if "a" is defined there at all.
> Once we know these latter ratios we can deduce that,
> up to an unknown constant,
> T(v) = r E(v) + b
> where E(v) is independent of the calorimeter material and of the
> thermometer.
As far as the procedure to determine T( v ) were reproducible,
your definition of "E( v )" were, too.
> By making measurements for different particles one gets
> E(v) = m/(1-v^2)^(1/2)
> The exact value of m depends on a choice of units, i.e. it is defined
> up to a constant multiplicative factor.
This expression for E( v ) closely resembles the expression for
"energy, of B wrt. A" with the measurement procedure/definition
of "energy" as what's invariant for any particular calibrations:
E_A( B ) == hbar i d/dt_A( t_B ); for which
E_A( B ) =proportional_to= 1/sqrt( 1 - (v_A( B )/c)^2 ).
However, since the specifications for measuring "T( v )"
so far may allow many other relations, too,
this resemblance appears coincidental.
We're obviously using different notions/measurememnt procedures to
define "energy"; and yours is so far not even reproducibly stated.
. 2) What is the equation of motion for a charged particle in an
. electrical field ?
. ... and how to characterize (the distribution of)
. "an electrical field" in the first place.
> > > 1) choose a method to measure a difference of potential
> > > between two points A and B
> > Please specify such a method.
> Again there are plenty of possible methods, as you might know.
I know that if distinct reproducible methods are defined
nontrivially different from each other, then they can obtain
different measured results, from the same set of trials/observations;
and even the same/reproducible procedure may obtain in general
different results for the same A wrt. different B's,
and/or in different trials.
> Some rely on measurement of angles or distances
(Surely measurements of "angles" may be defined in terms of
sets of measured pairwise distances.)
> (the old devices which measured in fact a force between two
What's the definition/procedure for any two to measure
"force between" each other?
> plate charged because of the difference of potential).
> Modern electronic devices do not rely on the measurement of
> times or distances but I do not know precisely how they work.
> Anyway this does not really matter for what I wanted to do.
> Choose many such different devices. [...]
(See below.)
> Apply my algorithm at the same point A.
> > please define/specify a procedure by which to determine
> > whether or not some particular A and B are "points".
> Point is the usual geometrical concept.
How does it relate to real/physical observations and measurements?:
> What this idealization approximates depends on the device used
> to measure the difference of potential. So the outcome
> of my algorithm depends only on this device,
> which is ultimately removed [via]
> Normalize all the resultant outcomes to the outcome obtained
> with one of them.
If at least one outcome is obtained, and if a non-negative
real number is obtained from that outcome, then yes.
> Check that my algorithm gives the same result with all devices
> at other points A, up to a given accuracy.
It doesn't, in general;
indeed that's how it might be measured and distinguished
if and how much charge is "carried by" any particular point A,
in any particular trial.
> If one of [those devices] gives a result too different
> from the others, don't use it.
Surely there may exist results that are "too different"
for any particular finite notion of "too different".
Please state yours more specificly.
Also, I'm not certain if you meant not to use this particular result,
in this particular trial, or to discard every result obtained
through this particular device, ever.
Either way, you'll end up discarding results for trials/points
which have not been otherwise obtained.
> > But I may not be able to identify B for which that difference
> > is _the_ maximum. Perhaps I may find that all of sufficiently
> > many values are below some particular bound, and I may use this
> > bound to _estimate a probable maximum value_, for some
> > particular B_stat.
> If the set of points B for which the difference of potential is
> indistiguishingly maximum is not reduce to one point,
I'm not sure which condition you're describing; but I considered
the case that values of potential difference are not obtained
wrt. _all_ points B, in _all_ trials,
for instance because certain values would have been discarded
as you suggested above.
> > > A being fixed moves B so as to maximize the difference of
> > > potential.
> > Therefore it is necessary (but not sufficient) that the values of
> > potential differences obtained above are non-negative real numbers.
> No. The usual convention is that 2 > -5.
Sure (that is, if we agree on "-5 * -5 = 25").
But note that you've been using non-negative real numbers:
"2" and "5", along with the operator "-".
OTOH, I'd be unable to make a definite statement
about "t > f", or "t < f"
if "t" and "f" were given merely as (not necessarily positive)
numbers with t * t = 2 * 2 = -2 * -2, and f * f = 5 * 5 = -5 * -5.
> > > Construct the vector whose modulus is this maximum
> > > and which points from A to B
> > Above it should have been clarified how A and B measure
> > their _distance_ wrt. each other.
> I do not really need to know the distance between A and B.
> Any Euclidian distance will do it, since the notion of A converging
> toward B is independent of the distance used on a finite dimensional
> metric topological space as R^3 which I use implicitely [...].
How do you suggest to _determine whether or not_ all A and B are
contained in "a finite dimensional metric topological space as R^3",
at least while conducting their corresponding trials?
> > Now, what do you mean by "vector, which points from A to B" ?
> The usual geometrical concept in R^3.
AFAIK that can be represented by the ordered set { A, B, modulus }.
> > > 3) redo step 2) with distances between A and B decreasing toward 0
> > > 4) the limit of the vector when the distance goes to zero is the
> > > electrical field at A
> > Can you prove that such limits exist?
> No. But I can check that the difference between two vectors of the
> above sequence is smaller than some given precision after some point
> in the sequence. If this never happens I decide that the method failed
> and that I cannot say anything about the electrical field at A.
At least you could obtain the relation between "given precision"
and whether or not "the method failed"; perhaps even in dependence
of "some point in the sequence".
> So no I can't prove that it exists but my algorithm gives
> unambiguously a result when it succeeds.
> > Also, unless you call _every_ such limit, wrt. _every_ particular B,
> > "the electrical field at A", among _all_ such limits, wrt. any
> > particular B there may surely be some which then differ more or less
> > strongly from what you understand to be "the electrical field at A",
> > in dependence on the particular B, or that may gravitate to entirely
> > different values altogether.
(That's unless various corresponding results were discarded already.)
> I disagree. The only ambiguity in the above algorithm
> is not associated with B but with the method chosen
> to measure a difference of potential.
I'd expect that not all B's can conduct all different methods
in any one trial, and that distinct B's may derive
different results wrt. A from their individual observations,
even using the same/reproducible measurement procedure.
If you'll state the procedure you suggest more specificly,
then surely one may construct correspondingly different B's.
> To remove that, try several different ones
Please specify at least two reproducibly.
> and check that the algorithm produces values whose differences
> are smaller than some given precision.
I'd expect that for any given A and any finite value of
"given precision" this can depend on B.
and...@attglobal.net
>
> and...@attglobal.net wrote:
>
> > Which it should (blow up). The energy and momentum of a massive object
> > diverge as v -> c.
>
> If we take the postulates of SR PLUS Newton's F = ma, then energy and
> momentum both diverge as v --> c.
>
> If we take the postulates of SR PLUS my F = mc*omega, then energy and
> momentum do not diverge as v --> c.
>
Funny that this isn't reflected in the design of particle accelerators
that need to get this right to actually work. I guess Fermilab hasn't
really worked properly and we never noticed it.
John Anderson
Jackie & Barry
and...@attglobal.net wrote:
> Funny that this isn't reflected in the design of particle accelerators
> that need to get this right to actually work. I guess Fermilab hasn't
> really worked properly and we never noticed it.
That's the answer I always get, sooner or later.
It's disappointing that nobody ever understands the equation.
If the acceleration law is F = mc*omega, omega gets smaller, "m" doesn't
get bigger. "p" doesn't "blow up".
That's the whole point!
The equation *agrees* with experiment (including Fermilab), c is a
limiting velocity in any inertial frame.
Do you really think that I'm stupid?
Barry
Bennett Standeven
"Jackie & Barry" <here...@istar.ca> wrote in message
news:39BD9543...@istar.ca...
>
>
> and...@attglobal.net wrote:
>
> > Funny that this isn't reflected in the design of particle accelerators
> > that need to get this right to actually work. I guess Fermilab hasn't
> > really worked properly and we never noticed it.
>
> That's the answer I always get, sooner or later.
>
> It's disappointing that nobody ever understands the equation.
>
> If the acceleration law is F = mc*omega, omega gets smaller, "m" doesn't
> get bigger. "p" doesn't "blow up".
>
That's nice. Unfortunately, it _does_ "blow up" in the real world.
Jackie & Barry
Bennett Standeven wrote:
> Barry wrote
> > If the acceleration law is F = mc*omega, omega gets smaller, "m" doesn't
> > get bigger. "p" doesn't "blow up".
> That's nice. Unfortunately, it _does_ "blow up" in the real world.
I defined p to be mv.
I stated that m was invariant (and finite).
I showed that 0 < v < c if we use the proposed Force law.
It follows that, for any m, p has a maximum finite value of mv
In that case, p cannot "blow up"?
Can you cite any observation of a finite mass having unbounded p in the
"real world"?
Please show your calculations.
Barry