Calculating Norm of a Vector with Two Vectors

[physguy09]

Ok, so I have no idea how to take the norm of a vector composed of two vectors. I have
$$\vec{q}$$=$$\vec{pi}$$ - $$\vec{pf}$$we are given:
|$$\vec{pi}$$|=|$$\vec{pf}$$|=|$$\vec{p}$$|

so i know that
|$$\vec{q}$$| $$\neq$$ 0, that would be too easy, and it doesn't make sense.

now, is the following right? it just doesn't seem to be

|$$\vec{q}$$|=$$\sqrt{2}$$*|$$\vec{p}$$|

 

[CompuChip]

You can try working it out explicitly through the inner product.
Note that
$$|\vec q|^2 = \vec q \cdot \vec q$$
so
$$(\vec p_i - \vec p_f) \cdot (\vec p_i - \vec p_f) = \ldots?$$

 

[physguy09]

ok so that gets me |$$\vec{q}$$|=$$\sqrt{|\vec{p_{i}^2}|+|\vec{p_{f}^2}|}$$,
would that then be $$\sqrt{2}$$p?

 

[quantumdude]

ok so that gets me |$$\vec{q}$$|=$$\sqrt{|\vec{p_{i}^2}|+|\vec{p_{f}^2}|}$$,

No, it doesn't get you that. Not unless $p_i$ is orthogonal to $p_f$, and the problem statement (as you gave it) doesn't say that.

 

[CompuChip]

Please stop guessing and work it out properly.

$$(\vec p_i - \vec p_f) \cdot (\vec p_i - \vec p_f) = (\vec p_i - \vec p_f) \cdot \vec p_i - (\vec p_i - \vec p_f) \cdot \vec p_f = \vec p_i \cdot \vec p_i + \cdots$$

 

[physguy09]

Please stop guessing and work it out properly.

$$(\vec p_i - \vec p_f) \cdot (\vec p_i - \vec p_f) = (\vec p_i - \vec p_f) \cdot \vec p_i - (\vec p_i - \vec p_f) \cdot \vec p_f = \vec p_i \cdot \vec p_i + \cdots$$

not guessing, just not typing it out entirely. i wrongly assumed p_i and p_f were orthogonal, but i was trying to make use of the fact that
|p|=|p_i|= |p_f|.

i know |q|=$$\sqrt{\vec{p_{i}^2} + \vec{p_{f}^2 }-2*\vec{p_i}*\vec{p_f}}$$

but i know |p|=|p_i|= |p_f|, so |p|$$^2$$ must equal |p|²=|p_i|² = |p_f|², hence, possibly with faulty reasoning,
|q|=$$\sqrt{2*p^2-2*\vec{p_i}*\vec{p_f}}$$

 

[CompuChip]

You are getting closer, it's just not entirely clear to me what
$2 * \vec p_i * \vec p_f$
means. In particular, what do the asterisks mean?

After clearing that up, can you write that expression in terms of p as well?

 

[physguy09]

apologies, I began to get sloppy, as I'm still trying to get used to Latex, it should be
|q|=$$\sqrt{2 p^2-2 \vec{p_i}\cdot\vec{p_f}}$$. I would not know how to write it out in terms of p, and it might be useless in this case, as we are supposed to plug in a known value for p. $$\vec{q}$$ was mainly momentum transfer, and we were given that for this case $$\vec{q} = \vec{p_i} - \vec{p_f}$$. i already turned in the assignment, just wanted to reach the answer for the sake of learning.

 

[CompuChip]

apologies, I began to get sloppy, as I'm still trying to get used to Latex, it should be
|q|=$$\sqrt{2 p^2-2 \vec{p_i}\cdot\vec{p_f}}$$. I would not know how to write it out in terms of p, and it might be useless in this case, as we are supposed to plug in a known value for p. $$\vec{q}$$ was mainly momentum transfer, and we were given that for this case $$\vec{q} = \vec{p_i} - \vec{p_f}$$. i already turned in the assignment, just wanted to reach the answer for the sake of learning.

Yes, at first you will work more slowly when you have to do the math and learn how to write it out in $\LaTeX$ - compliments for doing it anyway :)

What I was hinting at was to define $\theta$ as the angle between the ingoing and outgoing momentum, and writing
$$\vec p_i \cdot \vec p_f = |\vec p_i| |\vec p_f| \cos\theta = 2 |\vec p| \cos\theta$$.

Then you can take $\sqrt{2}p$ outside the square root, if you want to write it more beautifully.