FrediFizzx wrote in message
...
"Bill Hobba" wrote in message
...
|
| | FrediFizzx wrote:
| | I was wondering if the electron compton wavelength divided by 2pi
is
| an
| | invariant quantity? A main result from compton scattering is,
| |
| | lambda_C/2pi = hbar/m_e*c
| |
| | Everything on the right hand side of the equation is invariant, so
why
| | wouldn't this length be invariant?
| |
|
| Bill Hobba wrote:
| | Are you sure the m is the rest mass?
|
|
| FrediFizzx wrote;
| Why wouldn't it be?
|
| No particular reason except it has been a while since I looked at
compton
| wavelength and thought it may have been an out for the invariance. I
cant
| recall if it is invariant or not. A quick look at the textbooks I had
left
| me none the wiser.

Eric says it is a definition, which seems to be right. We just take three
invariant constants and produce an invariant length. However, now that we
have this invariant length, what is its significance? This seems to be
somewhat mind boggling. Does this mean this length will always be the
same
no matter what frame you are looking from? Seems like it would mean that.

FrediFizzx

The significance is in the invariance of the inverse root-mean-square
difference between the electromagnet and de Broglie wavelengths.
That invariance results from conservation of energy and momentum
in Compton scattering between electron and photon. [Old Man]

FrediFizzx:
"Bill Hobba" wrote in message
...
|
| Regarding the Compton wavelength Bill Hobba wrote:|
| | No particular reason except it has been a while since I looked at
| compton
| | wavelength and thought it may have been an out for the invariance. I
| cant
| | recall if it is invariant or not. A quick look at the textbooks I had
| left
| | me none the wiser.
|
|
| FrediFizzx replied:
| Eric says it is a definition, which seems to be right. We just take
three
| invariant constants and produce an invariant length. However, now that
we
| have this invariant length, what is its significance? This seems to be
| somewhat mind boggling. Does this mean this length will always be the
| same
| no matter what frame you are looking from? Seems like it would mean
that.
|
| Not really because one would need to know the physical meaning in other
| reference frames. My recollection of the Compton wavelength is such that
I
| have no idea what that. Suppose you need to have a read in a textbook.
|
| Thanks
| Bill

If it is just a definition, which I am sure that it is, then maybe mass is
our way out of this as you originally proposed. We know that mass can be
converted completely to energy so to me that would mean that it can't be 100
percent invariant. It is maybe only "partially" invariant. Yeah, must hit
the books for this one. But maybe it is just due to the compton wavelength
applying to a photon and the mass is applying to an electron.


There is a simple relation that will answer your question:

E^2 = (pc)^2 + (mc^2)^2


The compton wavelength is \lambda_c = h/mc = 2\pi\hbar/mc.
so that,

mc = 2\pi\hbar/(lambda)

Now, plug the compton wavelength in for m and since p = \hbar k, you
get:

E^2 = (\hbar ck)^2 + (2\pi\hbar/lambda_c)^2

Now, k = 2\pi/(lambda) [not lambda_c!], so the expression is:

E^2 = (2\pi\hbar c)^2 [ (1/lambda)^2 + (1/lambda_c)^2 ]

lambda is the deBroglie wavelength. lambda_c is the compton wavelength.
Since lambda_c is equivalent to the mass, it's an invariant.

"Bilge" wrote in message
...
| FrediFizzx:
| "Bill Hobba" wrote in message
| ...
| |
| | Regarding the Compton wavelength Bill Hobba wrote:|
| | | No particular reason except it has been a while since I looked at
| | compton
| | | wavelength and thought it may have been an out for the invariance.
I
| | cant
| | | recall if it is invariant or not. A quick look at the textbooks I
had
| | left
| | | me none the wiser.
| |
| |
| | FrediFizzx replied:
| | Eric says it is a definition, which seems to be right. We just take
| three
| | invariant constants and produce an invariant length. However, now
that
| we
| | have this invariant length, what is its significance? This seems to
be
| | somewhat mind boggling. Does this mean this length will always be
the
| | same
| | no matter what frame you are looking from? Seems like it would mean
| that.
| |
| | Not really because one would need to know the physical meaning in
other
| | reference frames. My recollection of the Compton wavelength is such
that
| I
| | have no idea what that. Suppose you need to have a read in a
textbook.
| |
| | Thanks
| | Bill
|
| If it is just a definition, which I am sure that it is, then maybe mass
is
| our way out of this as you originally proposed. We know that mass can
be
| converted completely to energy so to me that would mean that it can't be
100
| percent invariant. It is maybe only "partially" invariant. Yeah, must
hit
| the books for this one. But maybe it is just due to the compton
wavelength
| applying to a photon and the mass is applying to an electron.
|
|
| There is a simple relation that will answer your question:
|
| E^2 = (pc)^2 + (mc^2)^2
|
|
| The compton wavelength is \lambda_c = h/mc = 2\pi\hbar/mc.
| so that,
|
| mc = 2\pi\hbar/(lambda)
|
| Now, plug the compton wavelength in for m and since p = \hbar k, you
| get:
|
| E^2 = (\hbar ck)^2 + (2\pi\hbar/lambda_c)^2
|
| Now, k = 2\pi/(lambda) [not lambda_c!], so the expression is:
|
| E^2 = (2\pi\hbar c)^2 [ (1/lambda)^2 + (1/lambda_c)^2 ]
|
| lambda is the deBroglie wavelength. lambda_c is the compton wavelength.
| Since lambda_c is equivalent to the mass, it's an invariant.

Thanks. Old Man already got me straight on that one. However, is it
lambda_C that is invariant or is it lambda_C/2pi? IOW, which one is more
correct, hbar/m_e*c or h/m_e*c? I thought it was the first one which
results in lambda_C/2pi that is equivalent to the mass.

FrediFizzx