I Vector math (small angle approximation)

[Boltzman Oscillation]

Given the following vectors:

how can i determine that Θ = Δp/p ?
I can understand that p + Δp = p' but nothing arrives from this. Any help is welcome!

 

[anorlunda]

Sin Φ ≈ ∅ for small angles.

(Different) Mentor edit: The above should be $\sin(\theta) \approx \theta$, for $\theta$ in radians.

 

[Delta2]

Sin Φ ≈ ∅ for small angles.

You meant to say that $\sin\theta\approx \theta$ for small angles.

 

[Boltzman Oscillation]

Sin Φ ≈ ∅ for small angles.

wait wait what? How is this true? I've never known this!

 

[Boltzman Oscillation]

You meant to say that $\sin\theta\approx \theta$ for small angles.

Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.

 

[jedishrfu]

To be clear, the angle is in radian measure.

 

[A.T.]

Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.

You also see that from the fact that sin(0) = 0, sin'(0) = 1 and sin''(0) = 0. So sin is equivalent to the identity function up to the 2nd derivative around 0.

 

[RPinPA]

Ah I guess I could see that being true since the taylor expansion of sin is theta - theta^3/3! +theta^5/5! so a small theta would cause the terms after the first to be significantly small.

It's instructive to take $\sin(\theta)$ for some small values of $\theta$ (always in radians) to see just how good the approximation is. Even at $\theta = 0.1$, which is a little larger than what we usually consider "small compared to 1", it's a pretty good approximation.

 

[sophiecentaur]

how can i determine that Θ = Δp/p ?

Are you looking for more than just basic Trigonometry here?

 

[Boltzman Oscillation]

Are you looking for more than just basic Trigonometry here?

Well the other gentlemen helped me understand better now but maybe you can provide more insight. So, yes.

 

[Boltzman Oscillation]

It's instructive to take $\sin(\theta)$ for some small values of $\theta$ (always in radians) to see just how good the approximation is. Even at $\theta = 0.1$, which is a little larger than what we usually consider "small compared to 1", it's a pretty good approximation.

okay i will try some small numbers. Thank you sir.