When to square a sum and when not to

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In Compton scattering calculations, when squaring an equation, it's essential to square the sum correctly. The left-hand side (LHS) should be squared as a whole, following the formula (a + b)² = a² + b² + 2ab, rather than squaring individual components separately. This is crucial because a² + b² does not account for the cross-term, which can affect the outcome. The distinction is particularly important in vector components, where perpendicular components can sometimes lead to the same result. Understanding when to apply each method is key to accurate calculations in physics.
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Hey, I've got a question which might be really simple, I'm not too sure yet!

Basically, I'm going over the compton scattering calculations, and there's a part where:

v - v' + (mec2)/h = \sqrt{something else}

Basically, the next step is to square both sides of the equation.

To do this, my lecturer squares the LHS as a sum (i.e. in the form (a+b)2 instead of doing a2 + b2... where a = (v - v') and b = (mec2)/h).

My question is... why do you in some instances take the square of the sum, and in other instances take the square of the individual components. I'm assuming there is a non-arbitary reason for this.. but I don't know about it!

Any help or pointers in the right direction would be much appreciated.

Cheers,
Andrew
 
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Welcome to PF!

Hey Andrew! Welcome to PF! :smile:

(have a square-root: √ :wink:)
mess1n said:
My question is... why do you in some instances take the square of the sum, and in other instances take the square of the individual components. I'm assuming there is a non-arbitary reason for this.. but I don't know about it!

You always take the square of the sum. :smile:

(Though there are a few cases where that is the same as taking the sum of the squares, for example if they are perpendicular components of vectors :wink:)
 
Cheers for the welcome, and for the answer!
 
If you got
x = \sqrt{y}
and you square it to
x^2 = y

If x happens to be a sum
x = a + b
then you get
(a + b)^2 = y
and not
a^2 + b^2 = y

The reason is simple. In general,
a^2 + b^2 \neq (a + b)^2
because there's also the crossterm:
(a + b)^2 = a^2 + b^2 + 2ab
 
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