When to square a sum and when not to

[mess1n]

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' + (m_ec²)/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 = (m_ec²)/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

 

[tiny-tim]

Welcome to PF!

Hey Andrew! Welcome to PF!

(have a square-root: √ )

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.

(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 )

 

[mess1n]

Cheers for the welcome, and for the answer!

 

[Nick89]

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:
$$(a + b)^2 = a^2 + b^2 + 2ab$$

 

[CellCoree]

Hi Andrew,

Thank you for your question. In mathematics and science, there are certain rules and conventions that we follow when dealing with equations and calculations. In this case, the reason for squaring the sum instead of the individual components has to do with the properties of square roots and how they interact with operations like addition and multiplication.

First, let's understand what is happening in the equation you mentioned. The left-hand side (LHS) of the equation is the sum of three terms: v, -v', and (mec^2)/h. The right-hand side (RHS) is the square root of "something else". In order to solve for the variable, we need to isolate it on one side of the equation, which is why your lecturer squared both sides.

Now, let's look at the specific terms being squared. The reason for squaring the sum (a+b)^2 instead of the individual components a^2 + b^2 has to do with how square roots interact with addition. When we take the square root of a sum, we cannot simply take the square root of each individual term and add them together. This is because the square root of a sum is not equal to the sum of the square roots. In other words, √(a+b) ≠ √a + √b.

To demonstrate this, let's take a simple example: √(4+9) vs. √4 + √9. The first one can be simplified to √13, while the second one can be simplified to 2+3, which is equal to 5. As you can see, these two values are not equal. This is because the square root operation is not distributive over addition.

On the other hand, when we square a sum, we can distribute the operation and get the correct result. For example, (a+b)^2 = a^2 + 2ab + b^2. So when we square the LHS of your equation, we get (v-v'+(mec^2)/h)^2 = v^2 + 2vv' + (mec^2)^2/h^2. This allows us to simplify the equation and solve for the variable.

In conclusion, the reason for squaring the sum instead of the individual components has to do with the properties of square roots and how they interact with operations like addition. I hope this explanation helps and clears up