Science > Physics > The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case
| Topic: |
Science > Physics |
| User: |
"Y.Porat" |
| Date: |
09 May 2006 02:29:12 AM |
| Object: |
The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon
case !!
Copyright Y.porat 9/05-2006
explanations -- if asked for ..
ATB and TIA
Y. PORAT
-------------------------------
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| User: "Harry" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 06:14:43 AM |
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"Y.Porat" <maporat@012.net.il> wrote in message
news:1147159752.890429.289700@i39g2000cwa.googlegroups.com...
The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon
case !!
Work-around: try with m=0
That should work ;-)
Harald
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| User: "Y.Porat" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 06:37:16 AM |
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No Sir !
if you put in m=0
you still got E=pc
and there is some shocking news for you and others
pc is mc^2 as well !!!
but that is even not all the story !! it is only part of the beginning
...... (:-)
TIA
Y.Porat
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| User: "Harry" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 06:56:53 AM |
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"Y.Porat" <maporat@012.net.il> wrote in message
news:1147174636.905030.34270@u72g2000cwu.googlegroups.com...
No Sir !
if you put in m=0
you still got E=pc
and there is some shocking news for you and others
pc is mc^2 as well !!!
but that is even not all the story !! it is only part of the beginning
..... (:-)
It's not at all shocking that pc=Mc^2, if that's what you mean (with
M=equivalent relativistic mass).
Harald
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| User: "tomgee" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
10 May 2006 12:39:03 AM |
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Harry wrote:
"Y.Porat" <maporat@012.net.il> wrote in message
news:1147174636.905030.34270@u72g2000cwu.googlegroups.com...
No Sir !
if you put in m=0
you still got E=pc
and there is some shocking news for you and others
pc is mc^2 as well !!!
but that is even not all the story !! it is only part of the beginning
..... (:-)
It's not at all shocking that pc=Mc^2, if that's what you mean (with
M=equivalent relativistic mass).
Harald
Harald, I agree. That is exactly my argument - that the massless
photon exists only as M, an equivalent relativistic mass, but not
as m, a real mass having the energy of motion more commonly
known as momentum, p. M has momentum but no mass. The
mass m is real mass as opposed to relativistic mass, therefore,
it has momentum.
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| User: "Golden Boar" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 10:52:10 AM |
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Harry wrote:
"Y.Porat" <maporat@012.net.il> wrote in message
news:1147174636.905030.34270@u72g2000cwu.googlegroups.com...
No Sir !
if you put in m=0
you still got E=pc
and there is some shocking news for you and others
pc is mc^2 as well !!!
but that is even not all the story !! it is only part of the beginning
..... (:-)
It's not at all shocking that pc=Mc^2, if that's what you mean (with
M=equivalent relativistic mass).
Harald
Ok, look at it this way.
m is rest mass
M is relativistic mass
M = gamma.m
pc = M.c^2
E^2 = m^2 * c^4 + p^2 * c^2
E^2 = 0^2 * c^4 + M^2 * c^4
E^2 = 0^2 * c^4 + gamma2 * m^2 *c^4
E^2 = 0^2 * c^4 + gamma2 * 0 *c^4
E^2 = 0 + 0 = 0
What a surprise.
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| User: "The TimeLord" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
10 May 2006 01:56:35 AM |
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On Tue, 09 May 2006 08:52:10 -0700, "Golden Boar" <goldenboar@hotmail.com>
wrote in <1147189930.344764.91980@i40g2000cwc.googlegroups.com>:
Harry wrote:
"Y.Porat" <maporat@012.net.il> wrote in message
news:1147174636.905030.34270@u72g2000cwu.googlegroups.com...
[...]
It's not at all shocking that pc=Mc^2, if that's what you mean (with
It isn't.
M=equivalent relativistic mass).
Harald
Ok, look at it this way.
m is rest mass
M is relativistic mass
M = gamma.m
pc = M.c^2
That is incorrect. E=p*c and M=m=0 since m is invariant
under coordinate transformation. Remember, there is no
such thing as relativistic mass.
E^2 = m^2 * c^4 + p^2 * c^2
E^2 = 0^2 * c^4 + M^2 * c^4
E^2 = 0^2 * c^4 + gamma2 * m^2 *c^4
E^2 = 0^2 * c^4 + gamma2 * 0 *c^4
E^2 = 0 + 0 = 0
What a surprise.
Indeed! Since you have derived this for all photons
and we know photons have non-zero energy unless the
frequency is zero, that shows that this is non-sense.
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
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| User: "Y.Porat" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 11:56:40 AM |
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and if you get a nonsero value for E waht will you say then ??
as in the link i brought above ?
TIA
Y.Porat
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| User: "Golden Boar" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 02:52:07 PM |
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Y.Porat wrote:
and if you get a nonsero value for E waht will you say then ??
as in the link i brought above ?
TIA
Y.Porat
----------------
If you take pc to mean M.c^2, where M is the relativistic mass, and you
let m=0, you cannot obtain a no-zero value for E.
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| User: "Y.Porat" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 09:10:31 AM |
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so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
E=mc^2 even if you put m=0 you got it back
at the back door !!!
only that this time you enjoy the croocked situation
that you dont have mass
AND YOU DO HAVE MASS !!
AND YOU JUST CALLED IT 'RELATIVISTIC MASS !!
TIA
Y.Porat
-----------------------
see alter
TIA
Y.Porat
----------------
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| User: "Harry" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
10 May 2006 07:24:32 AM |
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"Y.Porat" <maporat@012.net.il> wrote in message
news:1147183831.818066.258030@j33g2000cwa.googlegroups.com...
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
E=mc^2 even if you put m=0 you got it back
at the back door !!!
only that this time you enjoy the croocked situation
that you dont have mass
AND YOU DO HAVE MASS !!
AND YOU JUST CALLED IT 'RELATIVISTIC MASS !!
I can't imagine why you find that exiting, it's well known. Just a matter of
phrasing.
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| User: "The TimeLord" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 10:53:19 PM |
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On Tue, 09 May 2006 07:10:31 -0700, "Y.Porat" <maporat@012.net.il> wrote
in <1147183831.818066.258030@j33g2000cwa.googlegroups.com>:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!! E=mc^2 even if
However, E=m*c^2 is only valid if v=0. Since for photons,
v=c, the equation E=m*c^2 is not valid. Instead, E=p*c
for photons, which is consistent with m=0.
you put m=0 you got it back at the back door !!!
Not sure what you mean by this.
[...]
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
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| User: "Randy Poe" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 09:17:01 AM |
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Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
E=mc^2 even if you put m=0 you got it back
at the back door !!!
Nope. If you put m=0 into E=mc^2 you get E=0
(which is incorrect, because E = mc^2 is incorrect).
only that this time you enjoy the croocked situation
that you dont have mass
AND YOU DO HAVE MASS !!
AND YOU JUST CALLED IT 'RELATIVISTIC MASS !!
Nope. That term is not used in practice, though it remains
in some texts.
0 for 3.
- Randy
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| User: "Golden Boar" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 12:07:54 PM |
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Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E, and E = hf = hc /
lambda.
So pc = hc / lambda.
We now have,
E^2 = 0^2 + h^2.f^2, or
E = hf
E=mc^2 even if you put m=0 you got it back
at the back door !!!
Nope. If you put m=0 into E=mc^2 you get E=0
(which is incorrect, because E = mc^2 is incorrect).
only that this time you enjoy the croocked situation
that you dont have mass
AND YOU DO HAVE MASS !!
AND YOU JUST CALLED IT 'RELATIVISTIC MASS !!
Nope. That term is not used in practice, though it remains
in some texts.
0 for 3.
- Randy
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| User: "Randy Poe" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 02:18:19 PM |
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Golden Boar wrote:
Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E,
Incorrect.
For objects with nonzero mass p = gamma*m*v. For
v close to c, gamma > 1, that means p ~ gamma*m*c.
- Randy
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| User: "Dirk Van de moortel" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 02:20:27 PM |
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"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202299.823410.158160@j33g2000cwa.googlegroups.com...
Golden Boar wrote:
Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E,
Incorrect.
For objects with nonzero mass p = gamma*m*v. For
v close to c, gamma > 1, that means p ~ gamma*m*c.
I hope you realize you're talking to Androcles?
Dirk Vdm
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| User: "Randy Poe" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 02:27:19 PM |
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Dirk Van de moortel wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202299.823410.158160@j33g2000cwa.googlegroups.com...
Golden Boar wrote:
Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E,
Incorrect.
For objects with nonzero mass p = gamma*m*v. For
v close to c, gamma > 1, that means p ~ gamma*m*c.
I hope you realize you're talking to Androcles?
I didn't, no. I thought we had a new crank on the block. Dang,
that's twice in recent memory he's gotten one name-change
ahead of me.
- Randy
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| User: "Dirk Van de moortel" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 02:33:23 PM |
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"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202839.571668.164850@i40g2000cwc.googlegroups.com...
Dirk Van de moortel wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202299.823410.158160@j33g2000cwa.googlegroups.com...
Golden Boar wrote:
Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E,
Incorrect.
For objects with nonzero mass p = gamma*m*v. For
v close to c, gamma > 1, that means p ~ gamma*m*c.
I hope you realize you're talking to Androcles?
I didn't, no. I thought we had a new crank on the block. Dang,
that's twice in recent memory he's gotten one name-change
ahead of me.
:-))
Dirk Vdm
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| User: "Eric Gisse" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 05:44:41 PM |
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Dirk Van de moortel wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202839.571668.164850@i40g2000cwc.googlegroups.com...
Dirk Van de moortel wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202299.823410.158160@j33g2000cwa.googlegroups.com...
Golden Boar wrote:
Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E,
Incorrect.
For objects with nonzero mass p = gamma*m*v. For
v close to c, gamma > 1, that means p ~ gamma*m*c.
I hope you realize you're talking to Androcles?
I didn't, no. I thought we had a new crank on the block. Dang,
that's twice in recent memory he's gotten one name-change
ahead of me.
:-))
Dirk Vdm
You sure?
They are both dumb as hell but at least Golden Boar can construct a
paragraph without using "*****", and lacks the abject psychosis of
Androcles.
...they also have different IP addresses. They aren't both on the same
class C, or even B.
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| User: "Dirk Van de moortel" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
10 May 2006 03:30:35 AM |
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"Eric Gisse" <jowr.pi@gmail.com> wrote in message news:1147214681.011393.168610@i39g2000cwa.googlegroups.com...
Dirk Van de moortel wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202839.571668.164850@i40g2000cwc.googlegroups.com...
Dirk Van de moortel wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202299.823410.158160@j33g2000cwa.googlegroups.com...
Golden Boar wrote:
Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E,
Incorrect.
For objects with nonzero mass p = gamma*m*v. For
v close to c, gamma > 1, that means p ~ gamma*m*c.
I hope you realize you're talking to Androcles?
I didn't, no. I thought we had a new crank on the block. Dang,
that's twice in recent memory he's gotten one name-change
ahead of me.
:-))
Dirk Vdm
You sure?
Not 100%.
They are both dumb as hell but at least Golden Boar can construct a
paragraph without using "*****", and lacks the abject psychosis of
Androcles.
..they also have different IP addresses. They aren't both on the same
class C, or even B.
Yes I know. But that doesn't prove anything.
Let's see how "they" evolve :-)
Dirk Vdm
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| User: "Golden Boar" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 02:47:17 PM |
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Randy Poe wrote:
Dirk Van de moortel wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message news:1147202299.823410.158160@j33g2000cwa.googlegroups.com...
Golden Boar wrote:
Randy Poe wrote:
Y.Porat wrote:
so we have at least one step forwards
pc is actually mc^2 th e famous old one again !!!
Nope. pc and mc^2 are not in general the same.
If pc is not mc^2, then it is either hc/lambda or E,
Incorrect.
For objects with nonzero mass p = gamma*m*v. For
v close to c, gamma > 1, that means p ~ gamma*m*c.
I hope you realize you're talking to Androcles?
I didn't, no. I thought we had a new crank on the block. Dang,
that's twice in recent memory he's gotten one name-change
ahead of me.
- Randy
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
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| User: "Randy Poe" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 03:01:11 PM |
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Golden Boar wrote:
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
Maybe you are. But Porat is obsessed with "proving" that
the photon mass is not zero, and when he emits a statement
like "E = mc^2, even if you put m=0..." or "pc = mc^2" then
it's pretty clear he's not limiting to the case of m=0.
- Randy
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| User: "The TimeLord" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
10 May 2006 01:05:32 AM |
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On Tue, 09 May 2006 13:01:11 -0700, "Randy Poe" <poespam-trap@yahoo.com>
wrote in <1147204871.030616.47610@j33g2000cwa.googlegroups.com>:
Golden Boar wrote:
[...]
Maybe you are. But Porat is obsessed with "proving" that the photon mass
is not zero, and when he emits a statement like "E = mc^2, even if you
I thought he was talking about his claim that
E^2=p^2*c^2+m^2*c^4
does not apply to photons. By the way, that is what *we* were
talking about.
[...]
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
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| User: "Schoenfeld" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
09 May 2006 07:13:52 PM |
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Randy Poe wrote:
Golden Boar wrote:
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
Maybe you are. But Porat is obsessed with "proving" that
the photon mass is not zero, and when he emits a statement
like "E = mc^2, even if you put m=0..." or "pc = mc^2" then
it's pretty clear he's not limiting to the case of m=0.
- Randy
According to the laws of physics, a photons mass is not 0.
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| User: "Bilge" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
11 May 2006 12:53:23 AM |
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Schoenfeld:
Randy Poe wrote:
Golden Boar wrote:
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
Maybe you are. But Porat is obsessed with "proving" that
the photon mass is not zero, and when he emits a statement
like "E = mc^2, even if you put m=0..." or "pc = mc^2" then
it's pretty clear he's not limiting to the case of m=0.
- Randy
According to the laws of physics, a photons mass is not 0.
Prove that.
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| User: "Schoenfeld" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
11 May 2006 06:00:48 AM |
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Bilge wrote:
Schoenfeld:
Randy Poe wrote:
Golden Boar wrote:
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
Maybe you are. But Porat is obsessed with "proving" that
the photon mass is not zero, and when he emits a statement
like "E = mc^2, even if you put m=0..." or "pc = mc^2" then
it's pretty clear he's not limiting to the case of m=0.
- Randy
According to the laws of physics, a photons mass is not 0.
Prove that.
Proposition A: A photon's mass is not 0 = True
Proposition B: NOT A = True
Since B violates the uncertainty principle, B is false.
The principle of bivalence thus gives
NOT B = NOT (NOT A) = A = True
So, according to the laws of physics, a photon's mass is not 0.
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| User: "Bilge" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
12 May 2006 03:35:24 AM |
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Schoenfelch, corporate shill:
Bilge wrote:
Schoenfeld:
Randy Poe wrote:
Golden Boar wrote:
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
Maybe you are. But Porat is obsessed with "proving" that
the photon mass is not zero, and when he emits a statement
like "E = mc^2, even if you put m=0..." or "pc = mc^2" then
it's pretty clear he's not limiting to the case of m=0.
- Randy
According to the laws of physics, a photons mass is not 0.
Prove that.
Proposition A: A photon's mass is not 0 = True
Proposition B: NOT A = True
Since B violates the uncertainty principle, B is false.
No, it doesn't. You have no understanding of the uncertainty
relations.
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| User: "Y.Porat" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
12 May 2006 04:12:58 AM |
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1 a remark to Bilge:
diverting the thread to sci.trolls without noting it explicitly
is an act of a shamless crook !!
now the Nazi disturbed mokey Gisse =dda1 =dda2 = etc etc wiol jump
for more bananas
2 see later
Bilge wrote:
Schoenfelch, corporate shill:
Bilge wrote:
Schoenfeld:
Randy Poe wrote:
Golden Boar wrote:
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
Maybe you are. But Porat is obsessed with "proving" that
the photon mass is not zero, and when he emits a statement
like "E = mc^2, even if you put m=0..." or "pc = mc^2" then
it's pretty clear he's not limiting to the case of m=0.
- Randy
According to the laws of physics, a photons mass is not 0.
Prove that.
Proposition A: A photon's mass is not 0 = True
Proposition B: NOT A = True
Since B violates the uncertainty principle, B is false.
No, it doesn't. You have no understanding of the uncertainty
relations.
---------------------------------Let me make clear what was my meaning
in the OP post:
please note i referee only to photon cases :
so that the above formula defines actually
*two or more photons * because :
if you detect a photon by its energy (mc^2)
you cant detect it (THE SAME PHOTON !)again by its momentum!!
i think it is even against the HUP principle !
so the above formula deals actually with two or more photons !!
th conclusion out of it is :
if you what to measure A SINGLE PHOTON
you must do it either by its energy **or ** by its momentum!
you cant have that single case in two forms at he same equation
so we can say at least that the
above formula (with energy +momentum presentation)
DOES NOT APPLY TO A *SINGLE PHOTON **
may be that should be the way that i had to present my thread
in order to me more accurate and understandable !!
how about that ??
(not for dada1 dada 2 =Gisse
pleae bypass the moron cook
TIA
Y.Porat
----------------------
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| User: "Schoenfeld" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
12 May 2006 04:43:56 AM |
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Bilge wrote:
Schoenfelch, corporate shill:
Bilge wrote:
Schoenfeld:
Randy Poe wrote:
Golden Boar wrote:
Why don't you try reading the last few posts again, and you will find
that we were talking about objects with m=0.
Maybe you are. But Porat is obsessed with "proving" that
the photon mass is not zero, and when he emits a statement
like "E = mc^2, even if you put m=0..." or "pc = mc^2" then
it's pretty clear he's not limiting to the case of m=0.
- Randy
According to the laws of physics, a photons mass is not 0.
Prove that.
Proposition A: A photon's mass is not 0 = True
Proposition B: NOT A = True
Since B violates the uncertainty principle, B is false.
No, it doesn't.
Since _no_ experiment can show a photon's mass to be 0, it means it is
not 0. If you did measure the photons mass to be 0, it means there was
no uncertainty in the energy of that photon and that violates HUP.
Either HUP is true and photon mass is unobservable or HUP is false a
photon's mass is 0. In relativity, there is HUP, but a photon is not a
relativistic construct.
You have no understanding of the uncertainty
relations.
It's a pitty you can't show why it does not violate HUP. Face it Semon,
you're a fraud - you did not even understand why a Procrustean stretch
was a Lorentz transform and even now still deny this.
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| User: "Bilge" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
12 May 2006 09:20:44 PM |
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Schoenfelch, clueless corporate shill and perennial idiot:
Since _no_ experiment can show a photon's mass to be 0, it means it is
not 0.
Are you stoned or just incredily stupid?
If you did measure the photons mass to be 0, it means there was no
uncertainty in the energy of that photon and that violates HUP.
You are completely clueless and haven't got the slightest idea
of what the uncertainty relations mean.
Either HUP is true and photon mass is unobservable or HUP is false a
photon's mass is 0. In relativity, there is HUP, but a photon is not a
relativistic construct.
What planet are you from? Last time I checked, qed was a fully
relativistic quantum field theory. Do yourself a favor. Leave physics
to those of us who are physicists and pick a less mentally taxing
hobby for your entertainment.
You have no understanding of the uncertainty
relations.
It's a pitty you can't show why it does not violate HUP. Face it Semon,
you're a fraud - you did not even understand why a Procrustean stretch
was a Lorentz transform and even now still deny this.
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| User: "Schoenfeld" |
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| Title: Re: The formula E^2=(mc^2)^2 + ((pc)^2 is not valid for the photon case |
13 May 2006 03:49:10 AM |
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[malicious followup-to header removed]
Mark Semon wrote:
Schoenfelch, clueless corporate shill and perennial idiot:
Getting all upset over your confusions will not make them go away,
Semon.
You are completely clueless and haven't got the slightest idea
of what the uncertainty relations mean.
I told you what they meant in terms of physics and all the forms it can
be formalized mathematically. You deliberately snipped this and, in
your characteristic style, started insulting and going off on
irrelevant tangents.
Either HUP is true and photon mass is unobservable or HUP is false a
photon's mass is 0. In relativity, there is HUP, but a photon is not a
relativistic construct.
What planet are you from? Last time I checked, qed was a fully
relativistic quantum field theory.
Read properly. Relativity, by itself, does not admit photons, you need
to introduce the quantum physics to recover this notion. You are
deliberately obfuscating and misreading things.
Do yourself a favor. Leave physics
to those of us who are physicists and pick a less mentally taxing
hobby for your entertainment.
You aren't a physicist, you are a fraud - you did not even know, and
you still deny, that a Procustean stretch was a Lorentz transform
(which is undergrad mathematics). Besides, the assumption that
academia's physcists by and large understand physics is questionable,
since most of them (like you) advocate that buildings can in-place
freefall due to a couple of hours of minor fire damage and yet fail
miserably to understand the basic physics laws which are violated or
the massive contradictions with the evidence. Yet, they and you are
still going around making these claims and calling others who oppose
them with objective analysis 'crackpots' and 'idiots'.
Physics is not about calling others 'laymen', not answering questions,
getting basic things wrong and then denying them with insults - it is
not about the pursuit of ones vanity by deliberately obfuscating basic
mathematics hoping people think 'you are smart' and 'special'.
Now if you would like objective discussion, you can restore the parts
of the previous post which you deliberately snipped without indicating
so - the parts which answered each and every non-issue you pointlessly
re-raised in this post - and then provide a criticism that actually
contains some mathematics and/or physics. Otherwise, do not expect me
to respond anymore to these pointless emotional games.
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