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I'm reading some papers that compare different values for a materials opacity to a particular particle. The first is given as $\frac{dE}{dX}$, a single particles energy loss per unit column depth (X = $x\rho $) to a continuous process.

Makes sense. So then the author goes on to compare his value to that of other authors, who have their opacities for the process expressed in $\kappa $ given in $cm^{2}/g $.

They're both presented as measuring the same thing, but I usually think of $ \kappa $ as being an opacity for a large group of particles that are removed discretely from a beam, so that the total energy in the beam dies of exponentially (since energy loss per distance is proportional to the number of particles left in the beam). $\frac{dE}{dX}$ is due to a continuous process acting on all the particles, and so would have a roughly linear effect per distance.

So my question is how does one compare the two values. What does $ \kappa $ mean in a context of continuous energy loss.

Thanks

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I should add that the paper refers to $\frac{1.}{\kappa} $ as the "range". Perhaps that helps identify what they're talking about? –  simplicio Jun 12 '13 at 20:53
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Couldn't hurt to give a link to the paper. Non-paywalled or preprint versions preferable. –  Chris White Jun 13 '13 at 0:20

1 Answer 1

up vote 1 down vote accepted

What does $\kappa$ mean in a context of continuous energy loss

In reference to energy $E$ (reaching "[normalized] column depth" $X$) opacity $\kappa_E$ may simply be defined as

$\kappa_E := \frac{1}{E} \frac{dE}{dX}$.

(If the so defined "opacity" is constant wrt. "[normalized] column depth $X$" this describes "exponential loss" (as a function of $X$) instead of "linear dependence on $X$".)

but I usually think of $\kappa$ as being an opacity for a large group of particles that are removed discretely from a beam

That's rather opacity $\kappa_N$, in reference to the (discrete, natural) number $N$ of particles (reaching "[normalized] column depth" $X$):

$\kappa_N := \frac{1}{N} \frac{dN}{dX}$; or perhaps rather

$\kappa_N := \left\langle\frac{1}{N} \frac{\Delta N}{\Delta X}\right\rangle_{\text average}$.

Of course, the relation between $E$ and $N$, or correspondingly between $\kappa_E$ and $\kappa_N$ may be complicated ...

Also, instead of being defined in reference to extensive quantities $E$ or $N$, opacity may be defined in reference to the corresponding (average) intensities:

$I := \frac{\Delta E}{\Delta t \Delta A}$ or $I := \frac{\Delta N}{\Delta t \Delta A}$, as

$\kappa_I := \frac{1}{I} \frac{dI}{dX}$.

[Note on edited version:
Consistent with the "unit of opacity $\kappa$" given as "$cm^2/g$" is the definition, in reference to "energy" $E$, as

$\kappa_E := \frac{1}{E} \frac{dE}{dX}$, and not (as stated in the initial version) $\kappa_E := \frac{dE}{dX}$.

The same consideration applies to any definition in reference to intensity.

In reference to particle number $N$ I correspondingly changed the explicit definition of "opacity $\kappa_N$" as well, even though its "unit" (or "dimension") is not affected since the number $N$ is "dimensionless".]

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Thanks for the reply user12262. But the $ \kappa_E $ in the paper can't simply be $\frac{dE}{dX} $ since the former doesn't have energy in the units. I should add that the paper refers to $\frac{1.}{\kappa} $ as the "range". Perhaps that helps identify what they're talking about? –  simplicio Jun 12 '13 at 20:47
    
You're welcome ... But -- as you noticed -- I have to put a major correction to my answer: Consistent with the given unit of "opacity $\kappa$" is instead the definition $\kappa_E := \frac{1}{E} \frac{dE}{dX}$; or in reference to (any) intensity $I$ correspondingly: $\kappa_I := \frac{1}{I} \frac{dI}{dX}$. If the so defined "opacity $\kappa$" is constant wrt. "[normalized] column depth $X$", this describes "exponential loss" (as a function of $X$) instead of "linear dependence". –  user12262 Jun 13 '13 at 5:29
    
I think you're right. So if $ \kappa_E $ as you define it was constant, then its inverse would be the distance the particle would travel before its energy hits zero, hence the authors referring to it as a "range". Thanks for the help. –  simplicio Jun 14 '13 at 15:52

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