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I feel this is not correct because if you work back from the tumor the skin will always see that same amount of radiation. Which way of thought is correct?

If anyone can follow that they are a guru of illogical thought at the very least.

INtj

- Thread starter INtj
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- #1

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I feel this is not correct because if you work back from the tumor the skin will always see that same amount of radiation. Which way of thought is correct?

If anyone can follow that they are a guru of illogical thought at the very least.

INtj

- #2

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[tex] P = \frac{k}{r^2}[/tex]

If so, consider a linear approximation of the difference at a distance r.

[tex] \frac{dP}{dr} = \frac{-k}{2r^3}[/tex]

[tex] dP = \frac{-k}{2r^3}dr[/tex]

[tex] \Delta P = \frac{-k}{2r^3}\Delta r[/tex]

Notice that the difference in power between the skin and the tumor decreases the larger the distance to the skin is, since the denominator increases as r increases.

--J

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INtj

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Suppose the distance to the skin is r and the distance from the skin to the tumor is h. Then the distance to the tumor is (r + h). Then, from the previous equations, we have

[tex]\Delta P = \frac{-k}{2r^3}((r+h) - r) = \frac{-k}{2r^3}h[/tex]

For there to be no difference in energy received by the skin and the tumor, this difference must be zero. There's no way that they can be zero unless k is zero (there is no radiation being emitted) or h is zero (the tumor is the same distance from the source as the skin).

Now, when is this difference small and when is it large? What does this tell us about how far away we should set the source?

--J

- #5

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if you are considering only inverse square losses in radiation intensity, then the relative intensity between the skin and tumour is

[tex]\frac{I_{tumour}}{I_{skin}}=(\frac{r-h}{r})^2[/tex]

where r is the source-tumour distance and h is the tumour depth (skin-tumour distance).

as r increases, [tex]r-h \rightarrow r[/tex], so this ratio approaches unity.

[tex]\frac{I_{tumour}}{I_{skin}}=(\frac{r-h}{r})^2[/tex]

where r is the source-tumour distance and h is the tumour depth (skin-tumour distance).

as r increases, [tex]r-h \rightarrow r[/tex], so this ratio approaches unity.

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I think this tells us the ratio will decrease but the absolute value of the radiation difference will remain the same if the tumor is required to get the same amount of rads.Justin Lazear said:[tex]\Delta P = \frac{-k}{2r^3}((r+h) - r) = \frac{-k}{2r^3}h[/tex]

--J

- #7

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then consider this.

if r = source-tumour distance, h = tumour depth (skin-tumour distance), then the radiation intensities at the skin and the tumour are proportional to the inverse square of their distances.

[tex]I_{tumour} \alpha \frac{1}{r^2}[/tex]

[tex]I_{skin} \alpha \frac{1}{(r-h)^2}[/tex]

The ratio of the two intensities approaches unity as r increases. The absolute difference between the two is

[tex]I_{skin} - I_{tumour} \alpha \frac{1}{(r-h)^2} - \frac{1}{r^2}[/tex]

as r increases, you again have [tex](r-h) \rightarrow r[/tex], so this difference approaches 0.

there will still be a difference, but as your prof said, the differences become less important as r increases.

plug in some numbers and see for yourself.

if r = source-tumour distance, h = tumour depth (skin-tumour distance), then the radiation intensities at the skin and the tumour are proportional to the inverse square of their distances.

[tex]I_{tumour} \alpha \frac{1}{r^2}[/tex]

[tex]I_{skin} \alpha \frac{1}{(r-h)^2}[/tex]

The ratio of the two intensities approaches unity as r increases. The absolute difference between the two is

[tex]I_{skin} - I_{tumour} \alpha \frac{1}{(r-h)^2} - \frac{1}{r^2}[/tex]

as r increases, you again have [tex](r-h) \rightarrow r[/tex], so this difference approaches 0.

there will still be a difference, but as your prof said, the differences become less important as r increases.

plug in some numbers and see for yourself.

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What? The ratio of the values is different, but the values themselves are the same?INtj said:I think this tells us the ratio will decrease but the absolute value of the radiation difference will remain the same if the tumor is required to get the same amount of rads.

2/5 = .4, and 5 - 2 = 3

4/5 = .8, and 5 - 4 = 1

5/5 = 1, and 5 - 5 = 0

but maybe

5/5 = .4, and 5 - 5 = 0.

and

4/5 = .4, and 5 - 4 = 1.

Plus, the equation doesn't tell you anything about the ratio. For a sufficiently smooth curve and a sufficiently small [itex]\Delta r[/itex], it approximates the difference quite well. You can even test it on functions you trust. Like 1/x^2. Or e^-x.

Are you trying to get somebody to just tell you you're right, or are you trying to understand why you're wrong?

--J

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