# Tag Info

22

This is not really an answer to your question, essentially because there isn't (currently) a question in your post, but it is too long for a comment. Your statement that A co-ordinate transformation is linear map from a vector to itself with a change of basis. is muddled and ultimately incorrect. Take some vector space $V$ and two bases $\beta$ and ...

18

I get the physical significance of vector addition & subtraction. But I don't understand what do dot & cross products mean? Perhaps you would find the geometric interpretations of the dot and cross products more intuitive: The dot product of A and B is the length of the projection of A onto B multiplied by the length of B (or the other way ...

11

A slow-motion video of an ollie makes the physics clear. The main idea is that the skateboarder pushes down hard on one side out past the wheels, torquing the board up into a hop, then pushes down with the other foot to level the board out and make it appear to stick to the feet. The skateboarder never lifts the board - the only necessary upward force in ...

11

This might be more of a math question. This is a peculiar thing about three-dimensional space. Note that in three dimensions, an area such as a plane is a two dimensional subspace. On a sheet of paper you only need two numbers to unambiguously denote a point. Now imagine standing on the sheet of paper, the direction your head points to will always be a way ...

10

With suitable boundary conditions, the decomposition is unique. Without them, it's not. Suppose that $(\phi,{\bf G})$ and $(\phi',{\bf G}')$ are two different decompositions for the same function. Then $$\nabla(\phi-\phi')+\nabla\times({\bf G}-{\bf G}')=0.$$ Take the divergence of both sides to find that $$\nabla^2(\phi-\phi')=0. ... 10 The object you're talking about is called, in mathematics, a Clifford algebra. The case when the algebra is over the complex field in general has a significantly different structure from the case when the algebra is over the real field, which is important in Physics. In Physics, in the specific case of 4 dimensions, using the Minkowski metric as you have in ... 10 It's just funny. Note that your equation doesn't actually use any single general quaternion. You only use the i,j,k imaginary units in an ad hoc way to get three minus signs whenever you need them. If you were using an actual quaternion$$ q = t + xi + yj + zk,$$then the only semi-natural real bilinear invariant you may construct out of it is$$ q\bar q ...

9

This is a note on why angular velocities are vectors, to complement Matt and David's excellent explanations of why rotations are not. When we say something has a certain angular velocity $\vec{\omega_1}$, we mean that each part of the thing has a position-dependent velocity $\vec{v_1}(\vec{r}) = \vec{\omega_1} \times \vec{r}$. We might consider another ...

9

The main regime of use is when an area is infinitesimally small, like one would use in an integral. In that case, we can easily see that it is flat, and the shape doesn't really matter. In which case, we can encode the information as a vector, with the magnitude representing the (scalar) area; the choice (as you noticed) of pointing out of any given side is ...

9

Vectors are probably the most important tool to learn in all of physics and engineering. Some random examples: Classical Mechanics: Block sliding down a ramp: You need to calculate the force of gravity (a vector down), the normal force (a vector perpendicular to the ramp), and a friction force (a vector opposite the direction of motion). E&M: Electric ...

9

Consider an $n$-dimensional space (two dimension in the picture), and let $f(\vec x)$ be a non-constant scalar function, like a temperature distribution in your case. Let $\vec y(t)$ be any curve in the space such that the function $f(\vec y(t))=c$ is constant along that trajectory (the colored lines). Now compute the scalar product $\left\langle ., ... 7 There are actually several different ways to interpret that question, depending on what you mean by "vector" and "rotation". But here's a sense that I've often wondered about myself: in introductory physics, the velocity vector is defined as the time derivative of the position vector (relative to some fixed point). Why is the same not true of angular ... 7 Your confusion comes from the difference between special and general relativity. In special relativity, the space-time manifold is assumed to carry the structure of 4-dimensional Minkowski space, which has the nice property that it is canonically identified with its own tangent space at the origin (since it is a vector space). So in special relativity you ... 7 When studying angular things - torque, angular velocity, angular momentum, etc. - physicists do a clever thing to avoid having to describe curves. You see, you might be tempted to draw a curved arrow for a torque, indicating that you are twisting something around in a circular-ish way. But then when you try to add two such arrows together, all of a sudden ... 7 The mistake you made is in the way you stated Coloumb's law. It's either $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}}$$ OR $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}}$$ but definitely NOT $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}}$$ 7 In terms of velocitiy in classical mechanics the only difference between the dimensions is what kind of objects$v$,$u$and$a$are. While$t$should always be a real number$v$,$u$and$a$should be vectors in$\mathbb{R}^n$where$n$represents the dimension you are talking about. If this does not help, please clarify. 7 In this equation$F_N$is the magnitude of the normal force, and$W$is the magnitude of the weight. The forces are in opposite directions, yes, but their magnitudes are equal. It would be correct to write this: $$\vec{F}_N = -\vec{W}$$ because$\vec{F}_N$refers to the full force vector, including its direction, not just the magnitude. (And similarly for ... 7 There is a identity for the derivative of the cross-product of two vector functions$\mathbf A(t)$and$\mathbf B(t); \begin{align} \frac{d}{dt} (\mathbf A \times \mathbf B) = \frac{d\mathbf A}{dt}\times \mathbf B + \mathbf A\times \frac{d\mathbf B}{dt} \end{align} Using this rule with the computation you're considering, we obtain \begin{align} ... 6 Scalar field is a field of scalars (don't blame me, you invited Captain Obvious yourself). In other words, it's just a function on the manifold. If the manifold were the surface of the Earth (I hope you don't mind I move to 2+0 dimensions for a bit), it could e.g. be the ground temperature in certain moment. Moving up to 2+1 dimensions, it can again be a ... 6 Defining properties of vectors are that you can add them and multiply them by constants. These both make sense for angular velocities. On the other hand, adding rotations doesn't make sense. What you can do with two rotations is compose them: first rotate one way, then rotate another. This operation doesn't look like addition of any sort. For one thing, it ... 6 You made a mistake in assuming that the angular acceleration (\alpha$) is equal to$v^2/r$which actually is the centripetal acceleration. In simple words, angular acceleration is the rate of change of angular velocity, which further is the rate of change of the angle$\theta$. This is very similar to how the linear acceleration is defined. ... 6 The equation${\rm d}T~=~ \nabla T \cdot {\rm d}{\bf r}$, says that the change in T, namely${\rm d}T$, is the scalar product of 2 vectors,$\nabla T$and${\rm d}{\bf r}$, which can also be written as the magnitude of the 1st vector times the magnitude of the 2nd vector times cosine the angle between them.${\rm d}T~=~ |\nabla T| |{\rm d}{\bf ...

6

In component notation, 3d and 4d vectors are usually distinguished using latin and greek letters respecitively, e.g. $u_i$ and $u_\mu$. Moreover, four-vectors without indices are usually just written as $u$, whereas three-vectors are denoted $\vec u$, as you say. You'll hardly find $\vec u$ denoting a four-vector. The option $\underline{u}$ is also ...

6

Length and distance are not vector quantities (they are scalar quantities), but position and displacement are vector quantities (at least according to common terminological conventions). Here is how all of these are defined. Note that I am restricting the discussion here to vectors in three-dimensional Euclidean space $\mathbb R^3$. Every point in ...

6

Both sides of the equation given are vectors and so represent 3 equations which are, on a Cartesian basis: $E_x + \frac{\partial A_x}{\partial t} = -\frac{\partial V}{\partial x}$ $E_y + \frac{\partial A_y}{\partial t} = -\frac{\partial V}{\partial y}$ $E_z + \frac{\partial A_z}{\partial t} = -\frac{\partial V}{\partial z}$

6

I'll risk answering a homework problem since I think there's a principle that we experimental physicists take for granted, but which may not be obvious to budding physicists. When doing experiments it's not uncommon for the experimenter to have no way of knowing if a result is correct, but you can usually work out whether it's reasonable. If you look at the ...

6

Because direction cosines are, unlike sines and tans, even functions of the angle which makes the sign of the angle irrelevant and that's a good thing. More importantly, the direction cosines of a unit vector $\vec v$ end up being the coordinates $v_x,v_y,v_z$, respectively, so the direction cosines obey $$\cos^2 a+\cos^2 b+\cos^2 c = 1$$ which is nice. ...

5

There's no wonder you're confused - the author obviously was as well. First, the operations he's talking about are direct sum $U\oplus V$ and tensor product $U\otimes V$ of vector spaces. This has nothing to do with the vector product (an ambiguous term which most often denotes the cross product you probably know from school). Both are two different ways ...

5

There was a rather lengthy discussion about whether force is naturally a vector or a covector over at physicsforums: http://www.physicsforums.com/showthread.php?t=666861 . If you define momentum as "that which is conjugate to position," then momentum is a covector. I.e. if you have a Lagrangian, then: $$p_\mu =\frac{\partial L}{\partial \dot{x}^\mu}$$ ...

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