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Pentcho Valev
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 Posted: Sun Jul 13, 2008 2:04 pm Post subject: Re: massless or massive photon? |
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On Jul 11, 3:54 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote in
sci.physics.relativity:
| Quote: |
The REAL argument for massless photons is that when one treats the mass
of the photon as a free parameter and fits the theory to experimental
observations, one finds that the mass must be an INCREDIBLY small value,
and is consistent with zero. The PDG lists the current upper bound on
the photon mass as 6*10^-17 eV/c^2.
To put that in perspective, the lightest other particle
known (except neutrinos whose masses are not known) is the
electron. If an electron were scaled up to have the mass
of the earth, then the upper bound on the photon mass would
be scaled up to the mass of a half cubic meter of water.
|
But, Honest Roberts, no matter how small the mass of the photon is,
the dependence of its speed on the speed of the light source, v,
remains the same, c'=c+v:
http://books.google.com/books?id=JokgnS1JtmMC
"Relativity and Its Roots" By Banesh Hoffmann
p.92: "There are various remarks to be made about this second
principle. For instance, if it is so obvious, how could it turn out to
be part of a revolution - especially when the first principle is also
a natural one? Moreover, if light consists of particles, as Einstein
had suggested in his paper submitted just thirteen weeks before this
one, the second principle seems absurd: A stone thrown from a speeding
train can do far more damage than one thrown from a train at rest; the
speed of the particle is not independent of the motion of the object
emitting it. And if we take light to consist of particles and assume
that these particles obey Newton's laws, they will conform to
Newtonian relativity and thus automatically account for the null
result of the Michelson-Morley experiment without recourse to
contracting lengths, local time, or Lorentz transformations. Yet, as
we have seen, Einstein resisted the temptation to account for the null
result in terms of particles of light and simple, familiar Newtonian
ideas, and introduced as his second postulate something that was more
or less obvious when thought of in terms of waves in an ether. If it
was so obvious, though, why did he need to state it as a principle?
Because, having taken from the idea of light waves in the ether the
one aspect that he needed, he declared early in his paper, to quote
his own words, that "the introduction of a 'luminiferous ether' will
prove to be superfluous."
http://philsci-archive.pitt.edu/archive/00001743/02/Norton.pdf
John Norton: "Einstein regarded the Michelson-Morley experiment as
evidence for the principle of relativity, whereas later writers almost
universally use it as support for the light postulate of special
relativity......THE MICHELSON-MORLEY EXPERIMENT IS FULLY COMPATIBLE
WITH AN EMISSION THEORY OF LIGHT THAT CONTRADICTS THE LIGHT
POSTULATE."
Equivalently (that is, as one applies Einstein's equivalence
principle), no matter how small the mass of the photon is, the
dependence of the speed of the photon on the gravitational potential,
V, remains the same, c'=c(1+V/c^2):
http://groups.google.ca/group/sci.physics.relativity/msg/2d2a006c7d508022
Pentcho Valev: CAN THE SPEED OF LIGHT EXCEED 300000 km/s IN A
GRAVITATIONAL FIELD? Tom Roberts: Sure, depending on the physical
conditions of the measurement. It can also be less than "300000 km/
s" (by which I assume you really mean the standard value for c). And
this can happen even for an accelerated observer in a region without
any significant gravitation (e.g. in Minkowski spacetime).
Pentcho Valev
pvalev@yahoo.com |
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Dono
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| Posted: Sun Jul 13, 2008 2:11 pm Post subject: Re: massless or massive photon? |
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On Jul 13, 6:01 am, "Pmb" <physics_wo...@yahoo.com> wrote:
| Quote: |
"Dono" <sa...@comcast.net> wrote in message
news:38ce6321-c3e0-43aa-b647-5881231c41c3@c58g2000hsc.googlegroups.com...
On Jul 11, 4:49 am, "JuanShit R." González-Álvarez
juanREM...@canonicalscience.com> wrote:
The third mistake being his confusion between the Lagrangian and the
energy in special relativity. Note that Eric *multiplies* by the factor
(1 - (v^2/ c^2))
instead dividing by it. According to *Eric* when a particle travel to
energies close to c its energies vanishes...
JuanShit,
You ARE an idiot, the relativistic Lagrangian is indeed:
L=-m_0*c^2 *(1-(v/c)^2)
MULTIPLICATION, Shito. Not division. PRETENDER.
Juan. You were correct.
Dono!! You are *such* a liar!!! You claimed that you didn't write the above
and its perfectly clear that you *did*.
You forgot that people use Oulook Express which gets messages from a
different source than google. The data is downloaded from the newsgroups
server and then loaded/saved in a file which is located on disk. While he
can destroy anything you want on google, which you *did*, you can't destroy
what is on my hard drive and in the newsgroups server.
You're such a pest. I had to unblock you, delete Outlook's newsgroup file,
and then reload from scratch. All so that I could see you *lying*. You sure
are a piece of work Dono. You are a sad specimen for sure.
Now
Pete
|
"Brown-nose" Pete
Still with your nose up Juanshito's ass?
Here is what I said:
http://groups.google.com/group/sci.physics.relativity/msg/396f4aa1a1171a7b
And you and your pal Juanshito keep eating shit. Bon apetit!  |
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Dono
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| Posted: Sun Jul 13, 2008 2:20 pm Post subject: Re: massless or massive photon? |
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On Jul 13, 5:50 am, stevendaryl3...@yahoo.com (Daryl McCullough)
wrote:
| Quote: |
Dono says...
How about if you showed how the relativistic Lagrangian is DERIVED
In Jackson's Electrodynamics, the following heuristic argument is
used:
We want the action A = Integral L dt to be a scalar, independent
of which coordinate system is used. However, the parameter t is
coordinate-dependent. So convert it to tau, the proper time:
dt = gamma d tau
Then the action becomes
Integral gamma L d tau
For this to be independent of coordinate systems, the quantity
(gamma L) must be a Lorentz scalar. The simplest choice is for
it to be a constant, K. With this heuristic,
gamma L = K
L = K/gamma = K square-root(1-(v/c)^2)
Working out the corresponding momentum gives:
p = @L/@v = -Kv/(c^2 square-root(1-(v/c)^2))
To give the usual expression for momentum, we let K = -mc^2.
So L = - mc^2 square-root(1-(v/c)^2)
Alternatively, you can just start with the equation
p = @L/@v = mv/square-root(1-(v/c)^2)
and integrate with respect to v:
L = m Integral of v/square-root(1-(v/c)^2) dv
= -mc^2 square-root(1-(v/c)^2)
--
Daryl McCullough
Ithaca, NY
|
Dang,
I was making Juanshito squirm for almost a year asking him to show
that he can derive it, you blew it :-)
Anyways, I prefer a different approach that I came up:
1. Start with the non-relativistic (classical regime) Lagrangian.
Show that
L=mv^2/2 +e(<A_vector,v_vector>-\Fi) satisfies the equation d/dt(dL/
dv)-dL/dr=0
2. Move to the relativistic regime:
p_vector=\gamma*m_0*v_vector
F_vector=dp_vector/dt=m_0*\gamma^3(\gamma^-2*dv_vector/dt+vv_vector/
c^2*dv/dt)
Look for a lagrangian of the form:
L=L'+e(<A_vector,v_vector>-\Fi)
Show that , in order to satisfy d/dt(dL/dv)-dL/dr=0 L' must be of the
form -m_0c^2/\gamma |
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Juan R." González-Álvarez
Guest
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| Posted: Sun Jul 13, 2008 2:33 pm Post subject: Re: massless or massive photon? |
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Pmb wrote on Sun, 13 Jul 2008 09:29:39 -0400:
| Quote: |
"Juan R. Gonzlez-lvarez" <juanREMOVE@canonicalscience.com> wrote in
message news:pan.2008.07.13.11.35.50@canonicalscience.com...
Pmb wrote on Sat, 12 Jul 2008 21:35:10 -0400:
Peter, on the other hand, sometimes makes a good point. Unlike
Koobee and Juan and Sue, he's not a crackpot, there are just some
(unfortunately, permanent) gaps in his understanding of physics.
Easy claims to make Daryl. Why not state these supposed gaps and I'll
show you where (1) you're wrong or (2) I realized it and posted a
correction.
Pete
As explained in future version of guidelines, crackpots claim for
mistakes in other's knowledge but don't say where. /Ad hominem/ attacks
is a usual tactic they use.
Notice also how crackpot Daryl confound the Lagrangian written in terms
of four velocities v^b with the Lagrangian written in terms of four
proper velocities u^b.
The terms "proper velocity" and "4-velocity" are most often used to
refer to exactly the same object.
|
That is right, it is usual, but if one calls to the proper frame
Lagrangian
L = - 1/2 m u^b u_b
The "Lagrangian in terms of four velocities", then how would one call to
L = - m c (\sqrt v^b v_b) ?
It is more precise to use the terms "four proper velocity" for the former
and "four velocity" for the latter.
Also by consistency of notation, the spatial projection of the "four
proper velocity" gives the "proper velocity" u and the projection of the
"four velocity" gives the usual "velocity" v.
| Quote: |
See also the difference in minus sign in his proper frame Lagrangian.
I am not sure if crackpot Daryl know why I wrote the minus :-)
The choice of the minus sign makes the energy come out with the right
sign. If not then its still an integral of motion anyway. No matter what
the minus sign is, the equations of motion are the same.
Pete
|
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org |
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Juan R." González-Álvarez
Guest
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| Posted: Sun Jul 13, 2008 2:34 pm Post subject: Re: massless or massive photon? |
|
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Dono wrote on Sun, 13 Jul 2008 07:11:17 -0700:
| Quote: |
On Jul 13, 6:01 am, "Pmb" <physics_wo...@yahoo.com> wrote:
"Dono" <sa...@comcast.net> wrote in message
news:38ce6321-c3e0-43aa-
b647-5881231c41c3@c58g2000hsc.googlegroups.com...
On Jul 11, 4:49 am, "JuanShit R." GonzĂĄlez-Ălvarez
juanREM...@canonicalscience.com> wrote:
The third mistake being his confusion between the Lagrangian and the
energy in special relativity. Note that Eric *multiplies* by the
factor
(1 - (v^2/ c^2))
instead dividing by it. According to *Eric* when a particle travel
to energies close to c its energies vanishes...
JuanShit,
You ARE an idiot, the relativistic Lagrangian is indeed:
L=-m_0*c^2 *(1-(v/c)^2)
MULTIPLICATION, Shito. Not division. PRETENDER.
Juan. You were correct.
Dono!! You are *such* a liar!!! You claimed that you didn't write the
above and its perfectly clear that you *did*.
You forgot that people use Oulook Express which gets messages from a
different source than google. The data is downloaded from the
newsgroups server and then loaded/saved in a file which is located on
disk. While he can destroy anything you want on google, which you
*did*, you can't destroy what is on my hard drive and in the newsgroups
server.
You're such a pest. I had to unblock you, delete Outlook's newsgroup
file, and then reload from scratch. All so that I could see you
*lying*. You sure are a piece of work Dono. You are a sad specimen for
sure.
Now
Pete
"Brown-nose" Pete
Still with your nose up Juanshito's ass? Here is what I said:
|
http://sci.tech-archive.net/Archive/sci.physics.relativity/2008-07/
msg00824.html
http://groups.google.com/group/sci.physics.relativity/msg/6db93e88919413fe
Try to delete above messages as you did with Google archive troll :-)
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org |
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Juan R." González-Álvarez
Guest
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| Posted: Sun Jul 13, 2008 2:39 pm Post subject: Re: massless or massive photon? |
|
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Pmb wrote on Sun, 13 Jul 2008 09:31:40 -0400:
| Quote: |
"Juan R. Gonzlez-lvarez" <juanREMOVE@canonicalscience.com> wrote in
message news:pan.2008.07.13.11.25.34@canonicalscience.com...
Pmb wrote on Sat, 12 Jul 2008 21:31:50 -0400:
Don't waste your time here Juan. Both Eric and Dono have no clue how
the relativistic Lagrangian is derived. At best they're searching for
a text which derives it and then they'll copy it here.
Pete
Yes, both copied from my blog article of past year
http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-
and-
limitations.html
But that is not a derivation Juan. You didn't derive the Lagrangian
there.
|
Agree I only gave it, but they even copied it wrong!
The Lagrangian is usually derived from the action A taking into account
the ds for spacetime...
But the point is that once you start from a given Lagrangian you can
derive the corresponding Hamiltonian using the Legendre, and the result
is of course the Hamiltonian I wrote in the original poster. Not the
nonsense that Eric and Dono wrote.
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org |
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Dono
Guest
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| Posted: Sun Jul 13, 2008 2:43 pm Post subject: Re: massless or massive photon? |
|
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On Jul 13, 7:34 am, "JuanShito R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
<snip>
Keep it up, Shito
You love to be shown what an idiot you are:
http://groups.google.com/group/sci.physics.relativity/msg/396f4aa1a1171a7b
In case you can't read, here is what I wrote:
"JuanShit,
You ARE an idiot, the relativistic Lagrangian is indeed:
L= - m_0 * c^2 * sqrt((1-(v/c)^2))
MULTIPLICATION, Shito. Not division.
Square root
PRETENDER. "
See? Mutiplication, not Division
PRETENDER |
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Juan R." González-Álvarez
Guest
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| Posted: Sun Jul 13, 2008 2:55 pm Post subject: Re: massless or massive photon? |
|
|
Daryl McCullough wrote on Sun, 13 Jul 2008 05:50:00 -0700:
| Quote: |
Dono says...
How about if you showed how the relativistic Lagrangian is DERIVED
In Jackson's Electrodynamics, the following heuristic argument is used:
We want the action A = Integral L dt to be a scalar, independent of
which coordinate system is used. However, the parameter t is
coordinate-dependent. So convert it to tau, the proper time:
dt = gamma d tau
Then the action becomes
Integral gamma L d tau
For this to be independent of coordinate systems, the quantity (gamma L)
must be a Lorentz scalar. The simplest choice is for it to be a
constant, K. With this heuristic,
gamma L = K
L = K/gamma = K square-root(1-(v/c)^2)
Working out the corresponding momentum gives:
p = @L/@v = -Kv/(c^2 square-root(1-(v/c)^2))
To give the usual expression for momentum, we let K = -mc^2.
So L = - mc^2 square-root(1-(v/c)^2)
Alternatively, you can just start with the equation
p = @L/@v = mv/square-root(1-(v/c)^2)
and integrate with respect to v:
L = m Integral of v/square-root(1-(v/c)^2) dv
= -mc^2 square-root(1-(v/c)^2)
|
A less convoluted approach starts from arguments about worldlines for
free particles with invariant action
A = \Int k ds
where k is a constant which is found to be k = -mc
From the above action one can find the coordinate frame Lagrangian
A = \Int k ds = \Int L dt
in one much more easy and direct way.
--
Center for CANONICAL |SCIENCE)
http://canonicalscience.org |
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Juan R." González-Álvarez
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Dono
Guest
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| Posted: Sun Jul 13, 2008 3:01 pm Post subject: Re: massless or massive photon? |
|
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On Jul 13, 7:56 am, "JuanShito R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
<snip>
Shito,
I don't know what mental illness you suffer from but it is clear that
you love to be shown what an idiot you are. Preferably multiple times
a day. So, I oblige you :-)
http://groups.google.com/group/sci.physics.relativity/msg/396f4aa1a1171a7b
I think being proven an idiot is music to your ears |
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Dono
Guest
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| Posted: Sun Jul 13, 2008 3:02 pm Post subject: Re: massless or massive photon? |
|
|
On Jul 13, 7:55 am, "JuanShito R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
| Quote: |
A less convoluted approach starts from arguments about worldlines for
free particles with invariant action
A = \Int k ds
where k is a constant which is found to be k = -mc
From the above action one can find the coordinate frame Lagrangian
A = \Int k ds = \Int L dt
|
PRETENDER. |
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Dono
Guest
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| Posted: Sun Jul 13, 2008 3:30 pm Post subject: Re: massless or massive photon? |
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On Jul 13, 8:05 am, "Peter M. Brown-nose" <physics_wo...@yahoo.com>
wrote:
| Quote: |
"Brown-nose" Pete
Still with your nose up Juanshito's ass?
Sure. Go ahead. Keep flaming. It only demonstrates your true character. And
just because Juan hasn't flamed me and I thus have no problem talking to him
|
....you talk with each other as in PRETENDER to PRETENDER.
You keep your respective noses up to your respective asses. |
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Daryl McCullough
Guest
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| Posted: Sun Jul 13, 2008 5:50 pm Post subject: Re: massless or massive photon? |
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Dono says...
| Quote: |
How about if you showed how the relativistic Lagrangian is DERIVED
|
In Jackson's Electrodynamics, the following heuristic argument is
used:
We want the action A = Integral L dt to be a scalar, independent
of which coordinate system is used. However, the parameter t is
coordinate-dependent. So convert it to tau, the proper time:
dt = gamma d tau
Then the action becomes
Integral gamma L d tau
For this to be independent of coordinate systems, the quantity
(gamma L) must be a Lorentz scalar. The simplest choice is for
it to be a constant, K. With this heuristic,
gamma L = K
L = K/gamma = K square-root(1-(v/c)^2)
Working out the corresponding momentum gives:
p = @L/@v = -Kv/(c^2 square-root(1-(v/c)^2))
To give the usual expression for momentum, we let K = -mc^2.
So L = - mc^2 square-root(1-(v/c)^2)
Alternatively, you can just start with the equation
p = @L/@v = mv/square-root(1-(v/c)^2)
and integrate with respect to v:
L = m Integral of v/square-root(1-(v/c)^2) dv
= -mc^2 square-root(1-(v/c)^2)
--
Daryl McCullough
Ithaca, NY |
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Pmb
Guest
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| Posted: Sun Jul 13, 2008 6:01 pm Post subject: Re: massless or massive photon? |
|
|
"Dono" <sa_ge@comcast.net> wrote in message
news:38ce6321-c3e0-43aa-b647-5881231c41c3@c58g2000hsc.googlegroups.com...
On Jul 11, 4:49 am, "JuanShit R." González-Álvarez
<juanREM...@canonicalscience.com> wrote:
| Quote: |
The third mistake being his confusion between the Lagrangian and the
energy in special relativity. Note that Eric *multiplies* by the factor
(1 - (v^2/ c^2))
instead dividing by it. According to *Eric* when a particle travel to
energies close to c its energies vanishes...
JuanShit,
You ARE an idiot, the relativistic Lagrangian is indeed:
L=-m_0*c^2 *(1-(v/c)^2)
MULTIPLICATION, Shito. Not division. PRETENDER.
|
Juan. You were correct.
Dono!! You are *such* a liar!!! You claimed that you didn't write the above
and its perfectly clear that you *did*.
You forgot that people use Oulook Express which gets messages from a
different source than google. The data is downloaded from the newsgroups
server and then loaded/saved in a file which is located on disk. While he
can destroy anything you want on google, which you *did*, you can't destroy
what is on my hard drive and in the newsgroups server.
You're such a pest. I had to unblock you, delete Outlook's newsgroup file,
and then reload from scratch. All so that I could see you *lying*. You sure
are a piece of work Dono. You are a sad specimen for sure.
Now
Pete |
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Pmb
Guest
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| Posted: Sun Jul 13, 2008 6:29 pm Post subject: Re: massless or massive photon? |
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|
"Juan R. González-Álvarez" <juanREMOVE@canonicalscience.com> wrote in
message news:pan.2008.07.13.11.35.50@canonicalscience.com...
| Quote: |
Pmb wrote on Sat, 12 Jul 2008 21:35:10 -0400:
Peter, on the other hand, sometimes makes a good point. Unlike Koobee
and Juan and Sue, he's not a crackpot, there are just some
(unfortunately, permanent) gaps in his understanding of physics.
Easy claims to make Daryl. Why not state these supposed gaps and I'll
show you where (1) you're wrong or (2) I realized it and posted a
correction.
Pete
As explained in future version of guidelines, crackpots claim for
mistakes in other's knowledge but don't say where. /Ad hominem/ attacks
is a usual tactic they use.
Notice also how crackpot Daryl confound the Lagrangian written in terms
of four velocities v^b with the Lagrangian written in terms of four
proper velocities u^b.
|
The terms "proper velocity" and "4-velocity" are most often used to refer to
exactly the same object.
| Quote: |
See also the difference in minus sign in his
proper frame Lagrangian.
I am not sure if crackpot Daryl know why I wrote the minus
|
The choice of the minus sign makes the energy come out with the right sign.
If not then its still an integral of motion anyway. No matter what the minus
sign is, the equations of motion are the same.
Pete |
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