Decomposition of linearly polarized field MRI

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SUMMARY

The discussion focuses on the decomposition of the linearly polarized B1 field in MRI, specifically transitioning from equation (3.48) to (3.49). The key insight is that the B1 field can be represented as a combination of clockwise and counterclockwise circularly polarized fields, achieved by introducing sine terms through a mathematical trick of adding and subtracting the same term. This method clarifies how a cosine-only term can be expanded into both cosine and sine components, essential for understanding the underlying physics of MRI signal generation.

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Jen2114
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Homework Statement


Hi, I am having trouble understanding how the B1 field as described by (3.48) in the image attached in MRI which is described as a linearly polarized field is decomposed into it's final two circularly polarized field as described by (3.49) in the image attached.

Homework Equations

The Attempt at a Solution


I understand that because it's 2B1 that you have one in the clockwise direction and one in the counterclockwise direction. That's why the first B1 has a negative in front of the sine term and the second doesn't. However, how do you get a cosine and sine term if in the first equation (3.48) if you only have a cosine to begin with. I know you describe a circle like the unit circle with cosine and sine but I still don't understand why a term with only cosine is expanded to cosine and sine. Clarification on how to go from 3.48 to 3.49 would be greatly appreciated.
 

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Jen2114 said:
However, how do you get a cosine and sine term if in the first equation (3.48) if you only have a cosine to begin with.
You simply add and subtract the same term:
$$
2a = 2a + b - b = (a+b) + (a-b)
$$
This is a common trick used when you want to introduce an additional term, just like multiplying by ##b/b##.
 
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DrClaude said:
You simply add and subtract the same term:
$$
2a = 2a + b - b = (a+b) + (a-b)
$$
This is a common trick used when you want to introduce an additional term, just like multiplying by ##b/b##.
Thank you ! I wasn't aware of this but now that I am it makes much more sense!
 

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