In heavy ion therapy ,how do you determined the exact location of the tumor?

  • Thread starter Thread starter naveen5626
  • Start date Start date
  • Tags Tags
    Ion Therapy
naveen5626
Messages
1
Reaction score
0
In heavy ion therapy ,how do you determined the exact location of the tumor?

1.How position of any object can be determined by a nuclear detector.
2.Instrumentation involved in a heavy ion accelerator for cancer therapy.
 
Physics news on Phys.org
Depends on whether you are talking about IGRT or not. In IGRT a CT or MRI is used in conjunction with the treatment planning software to increase the accuracy of the treatment location. As of now, there are no FDA approved devices that give real time images during treatment, so the image is taken beforehand and input into the software.
 
Here is a link to many excellent presentations from last summer's workshop in Erice. There are a few talks on real time dosimetry.

http://erice2011.na.infn.it/speakers0.htm
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top