# Using dummy variable to integrate (formal solution of rad. trans eqn)

1. Apr 7, 2013

### ck99

1. The problem statement, all variables and given/known data

(Everything here is frequency dependant, but I have left off the v from all the variables to minimise chance of typos - hopefully that is ok but please let me know if not and I will try and retype!)

The equation of radiative transfer is

dI/dτ = -I + S

We guess that I(τ) = f(τ)e

Taking the derivative of this gives

dI/dτ = fbe + edf/dτ

which is just the same as

dI/dτ = bI + edf/dτ

Substituing this back into the first equation gives

bI + edf/dτ = -I + S

So b = -1 and

S = edf/dτ

I have been able to follow all this but the next step is where I get lost - my lecture notes go straight to

f = ∫Setdt + c0 (Integral limits from 0 to τ)

"where t is a dummy variable".

2. Relevant equations

All given above I think

3. The attempt at a solution

My approach was to go from

S = edf/dτ

df/dτ = Seτ

Separate variables

df = Seτ

Then integrate from 0 to f and from 0 to τ

f = ∫Seτ

But I'm not sure how to do that integral and (according to my notes) it is the wrong approach anyway! If someone could fill in the gaps and take me step-by-step through the method used in the notes it would be really helpful. I don't get how this other variable t is allowable or useful? Or where the c0 comes from?

2. Apr 7, 2013

### CompuChip

You can't just choose different integration limits on both sides of the equation. You are allowed to integrate both sides but you should choose consistent limits, so let's choose 0 to τ:

$$\int_0^\tau \frac{df(t)}{dt} \, dt = \int_0^\tau S e^t \, dt$$

You will see that I picked your equation which I marked by (*), as I'm not too fond of writing df = ... dt.

As τ is an integration limit, you need another variable in the integration, if the use of t and τ confuses you, you are also allowed to call it x:
$$\int_0^\tau \frac{df(x)}{dx} \, dx = \int_0^\tau S e^x \, dx$$

Or you can give the integration limit a different name, e.g. x:
$$\int_0^x \frac{df(\tau)}{d\tau} \, d\tau = \int_0^x S e^\tau \, d\tau$$

Let's use the first equation. I hope you're still with me.
On the left hand side, we have the fundamental theorem of calculus saying we will get f(τ) - f(0).
On the right hand side, well, we can't really do much with that integral. So now we have
$$f(\tau) - f(0) = \int_0^\tau S e^t \, dt$$
so you can pull f(0) over to the right hand side and call it c0.