# Tag Info

17

It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. ...

6

It's not possible to derive the orbital angular momentum $L = r \times p$ from the $\mathfrak{so}(3)$ commutation relations alone, since the spin operator $S$ also fulfills the same commutation relations, but certainly is different from $r \times p$.

6

I have read somewhere that commutation relations of the form $$[a_i,b_j]=\epsilon_{ijk} c_k$$ admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is: $$... 6 Force is indeed a vector. Technically you should write |\overrightarrow{F}| = 30N, however there is usually context given that let you omit this. If you are working in one dimension, then the vector-like direction is all encapsulated in the sign once you've defined your coordinate system (e.g. -30N is 30N downwards.) Beyond that, it is typically just a ... 3 Some people use \mathbf{F} instead of \vec{F} or even \overrightarrow{F}. I agree that often F=\| \vec{F} \| is a convenient shortcut. So for example A force \mathbf{F}=(10 \mbox{ N},0,0) has magnitude \|\mathbf{F}\|=10 \mbox{ N}. The components of \mathbf{F} are F_x = 10\mbox{ N}, F_y=0 and F_z=0 So the subscript is used to ... 3 In mechanics the vector cross product is used to transform a force at a distance into a moment, and a rotation about an axis into linear velocity$$ \mathbf{M} = \mathbf{r} \times \mathbf{F} \\ \mathbf{v} = \mathbf{r} \times \mathbf{\omega} $$If you project those vectors into any plane, you will see that each component of force (or rotation) is multiplied ... 3 If you 'carry' an object 10 meters in one direction then return it back 10 meters from where you started the work done on the object is not the force you expended times distance walked. The formula you write is often misunderstood and misused. In your example, when you lift the object in a gravitational field, the work being done on the object is its weight ... 2 A negatively charged particle has an electric field,$$\mathbf{E} =-\vert\mathbf{E}\vert \, \hat{\mathbf{r}} =\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}Gauss's law gives, \begin{align*} \int \mathbf{E}\cdot d\mathbf{A} &= \int \Big(\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}\Big) \cdot (r^2 \sin \theta d\theta d\phi \, \hat{\mathbf{r}}) = ... 2 No, you should not write "\left|\vec{F}\right| = 30\textrm{N}", because it's no better than "F=30\textrm{N}" Since force is a vector, you could write out the list of components, either as a parenthetical list or a column vector:\vec{F} = \left(30\textrm{N}\right) = \left[ 30\textrm{N}\right]$$You could also write the one component as a scalar: ... 2 Everything depends on how your fields (vectors and spinors are fields in the classical theory, and when you quantize in QFT, they become operator-valued fields) transform when you make a Lorentz transform: An scalar is a field that doesn't change at all: \phi'(x') = \phi(x). Examples are the Higgs and pions. A vector field is a field that transform like ... 2 A vector product of two vectors is also called a cross product. As you may know, one way of describing a vector, as opposed to defining a vector is "a quantity which has magnitude and direction." In 3-dimensional space, one can specify a vector as the sum of x, y, and z components of parts. We can write$$\vec{A} = (A_x, A_y, A_z) \text{ or } ...

2

Electric flux is given by: Here, S (vector area is same) while the strength of the electric field E is more in A (length of arrows). $\theta=0$ in both cases. Hence flux is not same. A > B

2

Basically, vectors are called contravariant because their components transform oppositely to the basis vectors: if our change of coordinates is such that $$\frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j}$$ then if we have a vector $\mathbf{V}$, its components $V^i_x$ in the $x$ coordinates are related to its ...

1

This can be understood in terms of vector differentiation and the dot product. Take the example that $v \perp r$. The change in the square of the displacement is $$\frac{d}{dt}r^2$$ $$=2r \cdot v$$, and if they are perpendicular, the dot product is zero.

1

Here we are talking about instantaneous velocity. So,its -20 m/s. And the velocity will be always tangent to the circular track. So,it will be 20 m/s in magnitude every second. But the direction will be different so different values in x & y - axes

1

The dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. This operation can be defined either algebraically or geometrically. The cross product or vector product is a binary operation on two vectors in three-dimensional space and is denoted by the symbol ×. The cross product a ...

1

In short: The dot product gives you the multiplication of the parallel components. Example: The work expression $W=\vec F \cdot \vec r$, where only the force component parallel to the direction (or likewise, the position component parallel to the force) is wanted. The (magnetude of the) cross product gives you the multiplication of the perpendicular ...

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