## Converting Linear Accelerator To Circular One

Hi guys ,

Suppose there is a circular accelerator has the same inner structure of a linear accelerator in terms of the rings of acceleration ...

The angular frequency of accelerated particles is : ω = qB/m = 2πf , and this is also the supplied current frequency ...

The question is , The above formula of ω works in this case ?

Sincerely , ...
 Mentor Blog Entries: 27 Just for future references, "accelerator" topics do not belong in High Energy/Nuclear Physics. While accelerators ARE used in those fields, the area of accelerator physics deals mainly with classical E&M, and thus, belong in the Classical Physics forum, which is where this thread will be moved to. Zz.

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For slow particles, that is fine. For relativistic particles, you need an additional factor of γ: ##\omega = \frac{qB}{\gamma m}##
 and this is also the supplied current frequency ...
Current where?

## Converting Linear Accelerator To Circular One

 Quote by ZapperZ Just for future references, "accelerator" topics do not belong in High Energy/Nuclear Physics. While accelerators ARE used in those fields, the area of accelerator physics deals mainly with classical E&M, and thus, belong in the Classical Physics forum, which is where this thread will be moved to. Zz.
Okay , I'll be careful in the next time

 Quote by mfb For slow particles, that is fine. For relativistic particles, you need an additional factor of γ: ##\omega = \frac{qB}{\gamma m}## Current where?
The current that supplied to the accelerator

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 Quote by Mohamed Nedal Hi guys , Suppose there is a circular accelerator has the same inner structure of a linear accelerator in terms of the rings of acceleration ...
Not sure what you are imagining here:
I think of a linear accelerator like that (link) ... so the picture I get is somehow putting the rings in a circle. But this uses electric fields so...
 The angular frequency of accelerated particles is : ω = qB/m = 2πf , and this is also the frequency of the supplied current ...
... I'm not sure where the B comes in to it.

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 Quote by mfb For relativistic particles, you need an additional factor of γ: ##\omega = \frac{qB}{\gamma m}##
just curious, is the decrease in angular frequency due to time dilation or an increase in mass?

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 Quote by Mohamed Nedal The current that supplied to the accelerator
That is extremely vague.

The connection to the electrical grid? That can be any frequency (or even DC), it does not matter as long as you know it in advance.

The electricity for dipole magnets? That depends on the energy of the particles in the ring. In storage rings, it is constant. If the energy of the particles increases, the current has to increase in most setups, too.
The electricity for kicker magnets? Varies with the beam and the setup.
The electricity for other magnets? Depends on the magnets.

The electricity for radio-frequency cavities? Those cavities need radio frequency as input, where the frequency depends on the cavity geometry.

@greswd: It comes from the modified reaction of particles to forces: ##F_T=\gamma m a_T##. The mass of particles is constant, "relativistic mass" is something you find in outdated text books only.
 yes I mean the current ( electricity ) that I apply on the accelerator . According to what I understand , The angular frequency of moving particles is equal to the angular frequency of the current which is applied on the accelerator , is equal to : ω = q*B/m = 2*π*f , where ( f ) is the frequency of the current , right ? If it was right , It is for one particle , So for all particles .. should I multiply with ( N ) the total number of particles ?
 Mentor Blog Entries: 27 You do know that circular accelerators do not move electrons the way you are describing here's don't you? Zz.
 yes I know how cyclotron , synchrotron , betatron .. etc. works I suppose another design for circular accelerator

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 Quote by Mohamed Nedal yes I know how cyclotron , synchrotron , betatron .. etc. works I suppose another design for circular accelerator
I don't think you do, and you're making very little sense in this thread, but OK.

Zz.

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 Quote by Mohamed Nedal yes I mean the current ( electricity ) that I apply on the accelerator .
As mentioned before, that is not a useful description.

 I suppose another design for circular accelerator
In that case, you should describe it first.

I agree with ZapperZ.