Relativistic electric field derivation

In summary, the conversation is about deriving the equation tan(phi) = γtan(θ) by finding the flux of E through two spherical caps, with the flux through each cap being equal. The first cap spans the angle θ and the second cap spans the angle phi. The field is constant and equal to the field of a stationary point charge for the first cap, while the field for the second cap is described by E = (Q/4∏εr^2)((1-β^2)(1-β^2sin^2(phi))^(3/2). The element of surface area for the first cap is 2∏r^2sinθdθ and for the second cap it is 2
  • #1
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Homework Statement



I am supposed to derive this equation: tan(phi) = γtan(θ) by performing an integration to find the flux of E through each of 2 spherical caps; the flux through each of these caps should be equal. The first cap spans the angle θ; the element of surface area may be taken as 2∏r^2sinθdθ and the field is constant and equal to the field of a stationary point charge. The second caps spans the angle phi, the field through which is described by E = (Q/4∏εr^2)((1-β^2)(1-β^2sin^2(phi))^(3/2) and the element of surface area is 2∏r^2sin(phi)d(phi). I've tried to do this integration but can't seem to get the end result right. If it helps this is from Purcell problems 5.11.

http://books.google.com/books?id=Z3bkNh6h4WEC&printsec=frontcover#v=onepage&q&f=false

Homework Equations





The Attempt at a Solution

 
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  • #2
I've tried doing this: E for the first cap: Q/4pi*epsilon r^2 (point charge at rest) times integral of the area element given = E = (Q/4∏εr^2)((1-β^2)(1-β^2sin^2(phi))^(3/2) times the area element given for the second cap...I don't see how this will end up giving me tangent on both sides and in effect it doesn't. Is there something else I have to add?
 

FAQ: Relativistic electric field derivation

1. What is a relativistic electric field derivation?

A relativistic electric field derivation is a mathematical process used to describe the behavior of electric fields at high speeds, approaching the speed of light. It takes into account the principles of relativity, which state that the laws of physics should be the same for all observers regardless of their relative motion.

2. Why is a relativistic electric field derivation important?

Relativistic electric field derivations are important because they allow us to accurately describe and predict the behavior of electric fields in situations where objects are moving at high speeds. This is crucial in fields such as particle physics and astrophysics, where objects can reach velocities close to the speed of light.

3. How is a relativistic electric field derivation different from a classical electric field derivation?

A classical electric field derivation follows the laws of classical physics, which do not take into account the effects of high speeds and relativity. A relativistic electric field derivation, on the other hand, incorporates the principles of relativity and gives more accurate results for objects moving at high speeds.

4. What are the key equations involved in a relativistic electric field derivation?

The key equations involved in a relativistic electric field derivation are the Lorentz transformations, which describe how physical quantities such as time, length, and velocity change between different frames of reference, and Maxwell's equations, which describe the behavior of electric and magnetic fields.

5. How is a relativistic electric field derivation used in practical applications?

A relativistic electric field derivation is used in many practical applications, including particle accelerators, nuclear reactors, and high-speed electronics. It is also important in understanding the behavior of electromagnetic waves, such as radio waves and light, which are crucial in communication and technology.

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