Why is a Linear Approximation Used for Momentum in a Coordinate System?

In summary, the conversation discusses a coordinate system where the momentum in the x direction is approximated by px' where x' is the horizontal displacement of the beam from the ideal path. The confusion lies in how this linear approximation applies to the momentum. However, it is later realized that x' can be defined using the relation x' = dx/ds, allowing for the use of a linear approximation for the momentum.
  • #1
DukeLuke
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0
I'm looking at a coordinate system where [itex] \vec{s} [/itex] is tangential to the ideal beam path [itex] \vec{S}(s) [/itex], and x and y are the horizontal and vertical displacement of the beam from this path s. I'm not understanding why a linear approximation of the momentum in the x direction would be [itex] p_x \approx px' [/itex] where [itex] x' = \frac{dx}{ds} [/itex]. I know in general what a linear approximation is (from the Taylor Series), but I can't seem to understand how it applies to the momentum in this coordinate system. If someone could point me in the right direction it would really help me out. One confusing thing is that it seems to me like x' should be undefined because s and x are perpendicular coordinates.

Edit: Never mind, I figured it out.
 
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  • #2
The momentum in the x direction is given by p_x = \frac{dE}{dx'} , and since x' = \frac{dx}{ds} , we can rewrite this as p_x = \frac{dE}{dx}\cdot \frac{ds}{dx} . Since we are approximating dE/dx to be a linear function, we can then use the linear approximation p_x \approx px' , where x' = \frac{dx}{ds} .
 

Related to Why is a Linear Approximation Used for Momentum in a Coordinate System?

1. What is linear transport question?

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