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The dose-depth curve of protons and photons can be seen in the image below:

Dose-depth curve

Now, what I've heard is, that in some cases, proton therapy is advantageous compared to photons, and of course the other way around in some other cases. I can see, that the Bragg peak of the proton curve is very sharp, which means, that in order to hit a tumour, with some size, you have to make a spread Bragg peak resulting in some higher dose in the plateau region, than then ~20% it is now. But I'm not sure that even that would result in a larger dose in that area, than if you were to use photons ?

So why is it, that in some cases protons are better than photons, and the other way around ? As far as I understand, the proton radiation borders, when it hits the desired target, is very sharp, and therefore very precise - but still not very good if you have to hit a tumour that is VERY close to tissue of high risk. And therefore you often use photons for that. But it just seems that photons are much more spread out, so I can't really see why they should be less of a concern to tissue very close to the irradiated area ?

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    $\begingroup$ Well, you want to radiate the tumor, and not healthy tissue around it. If you can range out protons in the tumor, you aren't putting much dose in the healthy tissue nearer to the surface. $\endgroup$
    – Jon Custer
    Mar 11, 2015 at 13:09
  • $\begingroup$ A non-helpful answer is at campaigns.scripps.org/services/cancer-care__proton-therapy/…. They say its complicated. Call us and we will decide which is best for you. $\endgroup$
    – mmesser314
    Mar 11, 2015 at 13:24

2 Answers 2

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Some practical information first - the charged particle beam passing through the matter suffers from energy (and also angular) straggling. That means that even if an ideally monoenergetic beam is used, there will be always a finite volume with Bragg peak losses. The bigger initial energy, the bigger is the volume. Protons stopping at 40 mm have straggling less than 1 mm, at 125mm have 1.5 mm.

1st question - Compare the dose at Bragg peak : in your picture - you need multiply the gamma curve by two to have the similar dose for protons and gamma at 150 mm. To compensate the unwanted irradiation in plateau, you can irradiate the patient from several angles, keeping the Bragg peak always in the same place. Which also can/should be done for gamma irradiation.

2nd question - charged particles are definitely less damaging the surrounding tissue than gamma. Brain and near-eye tumors are best candidates for this kind of strictly localized treatment. However gamma is much less expensive and it is said that studies on treated patients show, that statistical results are similar for gamma and protons.

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Although there is already an accepted answer, I want to give some further ideas:

While the Bragg-peak is the "perfect" solution to get a high percentage of effective dose to the desired tissue, the exact positioning of this peak in the tumor tissue can be challenging due to misalignments of CT/MRI data and the application gantry and furthermore due to motion of the patient. The importance of exact positioning becomes larger as soon as high risk tissue is present near the target or when the target region is very small.

Proton therapy is very expensive and only possible in distinctive medical centers. Since the effective dose calculated to destroy the tumor is the same for protons and gammas, in principle there should be no difference in the obtained results of radiation.

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    $\begingroup$ Don't overlook the need for precise control of the beam energy among the difficulties of using massive beams for this application. I say massive beams, because some facilities are fooling around with still heavier beam components (up to $^{12}\mathrm{C}$!). $\endgroup$
    – dmckee --- ex-moderator kitten
    Mar 14, 2015 at 19:08
  • $\begingroup$ You are right of course, energy control to even be able to control the location of the bragg peak is nowhere near "easy". And all that while moving the beamline around the patient to be able to irradiate from different directions. $\endgroup$
    – Dux
    Mar 15, 2015 at 16:14

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