Prove the following vector relation:
V * (pvv) = pv * vV + vV * (pv)
Note: Multiplication sign is supposed to be a dot, both v's are vectors, and V is the del operator
To prove the vector relation, we need to expand both sides of the equation using the properties of the dot product and the del operator.
Let's start with the left-hand side (LHS) of the equation:
LHS = V * (pvv)
Using the distributive property of the dot product, we can rewrite this as:
LHS = (V * pv) * v + (V * v) * v
Now, let's expand the right-hand side (RHS) of the equation:
RHS = pv * vV + vV * (pv)
Using the commutative property of the dot product, we can rewrite this as:
RHS = pv * vV + (pv) * vV
Now, let's expand each term separately:
RHS = pv * vV + (pv) * vV
Next, we can use the associative property of the dot product to rewrite the second term:
RHS = pv * vV + pv * (vV)
Finally, let's use the commutative property of the dot product to interchange the vector order in the second term:
RHS = pv * vV + pv * (Vv)
Now, we can see that both the LHS and RHS have a common term pv * vV. So, we can rewrite the RHS as:
RHS = pv * vV + pv * (Vv)
Combining like terms, we get:
RHS = (pv + pv) * vV
Simplifying further, we have:
RHS = 2pv * vV
Therefore, we have proven that:
LHS = V * (pvv) = RHS = 2pv * vV
To prove the vector relation V * (pvv) = pv * vV + vV * (pv), we will use the properties of the dot product and the cross product of vectors.
Let's break down each term in the equation:
1. V * (pvv):
Here, V is the del operator (also known as the nabla operator). The del operator applied to a vector gives a vector with partial derivatives. The dot product of V with (pvv) implies that we take the dot product of the del operator with the result of the cross product of vectors pv and v.
2. pv * vV:
In this term, pv is a vector, and vV is the cross product of vectors v and V. The dot product of pv and vV implies that we take the dot product of vector pv with the cross product of vectors v and V.
3. vV * (pv):
In this term, vV is the cross product of vectors v and V, and (pv) is a vector. The dot product of vV and (pv) implies that we take the dot product of the cross product of vectors v and V with vector pv.
Now, let's prove the relation step by step:
Starting with the left-hand side of the equation, V * (pvv):
The dot product of the del operator (V) with the result of the cross product (pvv) can be rewritten using the triple scalar product identity:
V * (pvv) = (V · pv) × v + (V · v) × pv
Next, let's simplify each term on the right-hand side of the equation:
pv * vV:
The cross product of vectors v and V can be written as a negative cross product of V and v. So, we can rewrite this term using the antisymmetric property of the cross product:
pv * vV = -(v * V) × pv
vV * (pv):
The cross product of vectors v and V can be written as a negative cross product of V and v. So, we can rewrite this term using the antisymmetric property of the cross product:
vV * (pv) = (pv) × -(V * v) = - (pv) × (V * v)
Now, let's substitute these simplified terms back into the original equation:
V * (pvv) = (V · pv) × v + (V · v) × pv
= -(v * V) × pv + (V · v) × pv
= (V · v) × pv - (v * V) × pv
= ((V · v) - (v * V)) × pv
We can see that ((V · v) - (v * V)) is zero, as it represents the dot product of the del operator V with a vector v, subtracted from the dot product of a cross product v * V with a vector v. This difference is zero by the properties of the del operator and the dot product.
Therefore, we have:
V * (pvv) = 0, which implies that the vector relation V * (pvv) = pv * vV + vV * (pv) is proven.
Note: The above proof assumes that V, pv, and v are vectors in three dimensions. The same process can be applied for vectors in higher dimensions, but the algebraic representation and properties may differ slightly.