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41

A unit vector has magnitude $1$ - as in, the dimensionless number $1$. Not $1\ \mathrm{cm}$ or $1\ \mathrm{kg}$ or $1\ \mathrm{N}$ or $1\ \mathrm{J}$. It's also not hard to show that for any vector $\vec A$, the dimensions of $\vec A$ and $\vert \vec A \vert$ are the same.


16

I think that your confusion lies in the word "unit". In the definition of unit vector, "A unit vector is every vector whose magnitude is 1 unit," the word does not really refer to units like meter and second. It rather means just '1' and could be skipped. A true unit vector has no physical dimension (like force) but only a direction. If $\...


9

$(1,0,0)={(2,0,0)\over2}$. According to your naming scheme, the left hand side is not a unit vector but the right hand side is. This seems to be a problem. For that matter, whenever $(x,y,z)$ has length $1$, then $(x,y,z)={(x,y,z)\over 1}$ both is and is not a unit vector. Oops!


5

The vector form of the kinematic equation is $$ 2\boldsymbol{a}\cdot\left(\boldsymbol{x}_{1}-\boldsymbol{x}_{0}\right) = \boldsymbol{v}_{1}\cdot\boldsymbol{v}_{1}-\boldsymbol{v}_{0}\cdot\boldsymbol{v}_{0} $$ where in bold are vector quantities, and $\cdot$ is the dot product. Proof Take the vector form of the change in velocity $$ \boldsymbol{v}_{1}-\...


4

The kinematic equations are scalar equations in general. They only include the magnitudes of vector quantities, not the vectors themselves. They are equations that apply along a bound path or direction given a sign so that the directionality is covered for before applying them. Otherwise not only the $v$'s but also $a$ and $\Delta x$ ought to have been ...


4

In many physics textbooks it is given the following definition of unit vector: "A unit vector is every vector whose magnitude is 1 unit". Yes, that is the definition of unit vector. I cannot remember seeing any other definitions. Is correct my definition of unit vector? You did not actually give a definition, unless you are referring to this ...


2

There are three common ways of understanding vectors: Algebraic Geometric Transformational The algebraic definition is common in mathematics where vectors are defined as those objects satisfying a short list of axioms. This is a coordinate-free definition. The geometric definition is common in physics, and in particular, in mechanics, where a vector is ...


2

The concept of a "magnetic circuit" exploits a one-to-one correspondence between the equations of the magnetic field in an unsaturated ferromagnetic material to that of an electrical circuit. Using this concept the magnetic fields of complex devices such as transformers can be quickly solved using the methods and techniques developed for electrical ...


2

Exactly as @Jacob1729 says. Strange though it may seem, positions are not vectors. Positioning in this or that coordinate system is just tagging points on a manifold. The definition of vectors essentially depends on a concept of tangency. And for that you need a derivative. Stop worrying about $x^i$ and $x'^i$. Those are just parametrisations of your ...


2

UPDATE: For clarity, here are some references relevant to the OP's question. There may be differences in conventions among the references. (Readers unfamiliar with this spacetime viewpoint may misinterpret the title of the OP's question.) Robert Geroch's General Relativity 1972 Lecture Notes (p. 53-54): An electromagnetic field is a (smooth) antisymmetric ...


2

By definition, if the acceleration points towards the origin it is centripetal. So if $\ddot{r} - r \dot{θ}^2 - r \sin^2 θ \dot{\phi}^2<0$ then the radial component is centripetal. You are most likely getting confused with introductory systems of uniform circular motion where we say "centripetal acceleration is $r\dot\theta^2$", but that is ...


2

I think the notation $ \frac{\partial}{\partial{\vec{x_k}}} $ stand for the gradient $ \vec{\nabla} $. The gradient operator act on scalar funtion like $V$ and gives a vector : $$\vec{\nabla} V= \begin{pmatrix} \frac{\partial}{\partial_{x_1}}V\\ \dots \\ \frac{\partial}{\partial_{x_n}}V \end{pmatrix} $$ Here you describe $V$ as a function of multiple ...


2

You can understand this with some examples. In three-dimensional space you can define a displacement vector with components $(\delta x,\delta y,\delta z)$. You can have other vectors such as velocity, momentum, force, acceleration. To understand the difference between a vector and a non-vector it helps to have some examples of non-vectors. Mass $m$, charge $...


2

You're in the right ball-park. One thing that is not explained well in my opinion is the distinction between covariant and contravariant components, and as you point out, how this relies upon the metric. If we fix a basis $e=(e_\alpha)$ in a vector space $V$, then for every vector $v \in V$, we have its components, $v^\alpha$. So far so standard. What's not ...


1

To answer your questions directly: Squaring a vector is meaningless. However, in this case, the equation works if you interpret the squaring as a dot product. John Alexiou’s answer gives a nice proof of this.


1

There are a couple of subtleties that I'd like to supplement. First of all, there is a potential confusion between a (1) "flux" as an integrated amount of energy passing through some pre-specified surface and a (2) "flux density" as an continuous distribution that you are supposed to integrate. Usually the "density" bit is ...


1

Flux is in a direction relative to a normal to the surface area. Traditionally, the normal points outward from the control volume that is bounded by the area. Flux into the control volume is negative (opposing the normal) while the temperature gradient is positive (cold to hot as referenced by the direction of the normal to the area). Flux out of the control ...


1

Flux of heat has direction, not only magnitude. So this is why it's a vector. The unit area is represented by a unit vector that indicates the direction of the flow. Think about the Fourier's law: $\vec{J} = - k\nabla T$. The flux is determined by the gradient of a scalar function, so it's a vector.


1

So, when you applied the equation $B_{ij}=X_{ijkl}A_{kl}$ to get $X_{ijkl}' a_{kx}a_{ly} A_{xy} = a_{ip}a_{jq}X_{pqkl}A_{kl}$, note that in the expression $X_{ijkl}A_{kl}$, $k$ and $l$ are essentially dummy indices, and can be renamed to whatever we want. In particular, the next step is much clearer if we rename them to $x$ and $y$! Then we instead get $$X_{...


1

Any $4$-vector forms an irreducible representation of the Lorentz group, since any Lorentz transformation mixes all four components. But from the point of the $SO(3)$ subgroup it is reducible since spatial rotations do not mix the temporal component $V^0$ with the spatial components $V^i$ of the $4$-vector. Clearly the temporal component is invariant under ...


1

Just to expand on J. Murrays answer: Every unit vector is a vector, but not every vector is a unit vector. Hence, if you use the "hat" notation to indicate unit vectors its fine to write $ \hat u = \vec u = \vec e_u = \hat e_u = \ldots $. So, if we consider a force of magnitude 1N pointing in the $x$-direction, we write $\vec F = F \cdot \hat x = ...


1

If you're talking about newtonian mechanics, then $\mathbf{F} = - \nabla V$. I'm not sure why you're indexing $\mathbf{F}$, either it's a vector $\mathbf{F}$ or you're referring to the components $F_k$. The components are then $F_k = \left[-\nabla V \right]_k = -\frac{\partial V}{\partial x_k}$. In general if you have a scalar function $f: \mathbb{R}^n \to \...


1

Your link for the 3D case will work fine for a 2D case. The concept is the same, you just don’t have z components. Be careful though - note that all the equations in your link only address velocity, and mass is only noted in a caption on a diagram as a weighting factor (nice pun by the author). You are calculating the velocities, in which case, as Pim notes, ...


1

If $\:\mathbf{f}\:$ is a pure force 3-vector applied on a particle of velocity 3-vector $\:\mathbf{u}\:$, then the 4-dimensional vector \begin{equation} \mathbf{F}\boldsymbol{=}\gamma_{\mathrm u}\left(\mathbf{f}\:,\:\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c} \right) \tag{01}\label{01} \end{equation} is a Lorentz 4-vector. We use the term $''$pure ...


1

Yes. You can slide any force along its line of action (parallel line through the point of application) and it does not change the system. In your case $R$ is going to be along the line of action where it intersects the rod. As shown below: Note that $F$ and $R$ must have the same magnitude and direction.


1

The distance from $+q$ (at $\vec{d}/2$ ) to point $\vec{r}$ is $|\vec{r}- \vec{d}/2|$. And The distance from $-q$ (at $-\vec{d}/2$) to point $\vec{r}$ is $|\vec{r}+ \vec{d}/2|$. Thus the potential at $\vec{r}$ $$ V(\vec{r}) = \frac{q}{4 \pi \epsilon_o} \left( \frac{1}{|\vec{r} - \frac{\vec{d}}{2}|} - \frac{1}{|\vec{r} + \frac{\vec{d}}{2}|} \right) $$ for $d&...


1

Imagine you dropped a ball onto the wedge, it would bounce off at an angle and receive a horizontal impulse. The wedge then must receive a horizontal impulse in the other direction to ensure conservation of momentum. Similarly for a weight of mass $m$ sliding down a wedge, it would receive a horizontal component of velocity, the wedge then gets pushed in ...


1

What Griffiths is calling a vector is, strictly speaking, a Euclidean vector in three-dimensional space or a four-vector in four-dimensional spacetime. The components of a Euclidean vector (or a four-vector) are a way of representing it by giving its projections along three (or four) co-ordinate axes or the angles between the vector and certain axes. The ...


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