Loss of Energy

[frederico09]

We know that if we put a charge in a curved path with constant speed, this charge emite electromagnetic radiation while your path still a curve. In LHC the charges still in a circular motion and, therefore, is emitting radiation. There is a use for this radiation? Wich?

[chriwi]

This kind of radiation has even aspecial nam, its call syncoronradiation.
In the LHC the syncotronradiation is not used, but there are other smaller accelerators (f.e. in france only about 2h drive from CERN) which are specialiced for producin and working with syncotronradiation. It is used for material examinations and even for medicinal popuses.

[Kasuha]

Synchrotron radiation in LHC is used in some of BLMs and BPMs, i.e. to monitor beam intensity and position. It's not used for separate experiments, just to support the experiment itself.

[frederico09]

First, thanks for the answers.
Second, In my humble point of view, the synchrotron radiation lost in LHC can be used for other experiments so exciting as the colisions. A large number of countries would like to use this synchrotron radiation for materials analysis, for example. Certainly some person thought about this. Thereby another question arises: Why this radiation is not used for another research proposals?

[Kasuha]

I believe the main reason is that LHC is not designed for that. Its radiation is evenly distributed along the whole ring and relatively weak in any given spot. For LHC, synchrotron light is unwanted byproduct and is suppressed to necessary minimum. Also LHC produces synchrotron light with properties given by the accelerator needs, not given by needs of the eventual experiment using that light.

[oxodoes]

The energy loss per particle per turn in the machine is given by:
As one can easily see the loss in energy is strongly related not only to the energy but also to the mass of the particle. The lighter the particle the more energy is lost. This is why all synchrotron light sources use electrons. In addition the intensity of the particle beam is needed to be orders of magnitudes higher then at the LHC to get a brilliant light beam.

Edit: For those that can't seem to be able to see the image: The formula is given by:
delta E = (Z*e)^2 * E^4 / (epsilon0 * 3 * R * (m0 * c^2)^4)

[tomey36]

Speaking of electrons why is it that the LHC can’t collide them? Why do you need a linear accelerator to do that?

[Kasuha]

They already had electron-positron collider (LEP) in the same tunnel. They dismantled it because they exhausted its discovery capabilities.
By the way, synchrotron radiation was their main obstacle in reaching higher energies by then.