# Tag Info

34

[Disclaimer: I'm not providing an argument where the distinction would be useful. I am providing an argument that pseudovectors and vectors describe intrinsically different geometrical concepts, and should, for clarify of argument, never be conflated just because they look so similar] The point is that pseudovectors, by their very nature, are not the same ...

26

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 ...

19

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 ...

18

The best way is to ignore the garbage authors put in elementary physics books, and define it with tensors. A tensor is an object which transforms as a product of vectors under rotations. Equivalently, it can be defined by linear functions of (sets of vectors) and (linear functions of sets of vectors), all this is described on Wikipedia. There are exactly ...

18

No, in general you cannot divide one vector by another. It is possible to prove that no vector multiplication on three dimensions will be well-behaved enough to have division as we understand it. Regarding force, area and pressure, the most fruitful way is to say that force is area times pressure: $$\vec F=P\cdot \vec A.$$ As it turns out, pressure is not ...

17

There are some quantities that only come with one sign, for example mass/energy. So in any process if you start with some mass/energy $m$ then whatever happens you can't end up with less than $m$. But other quantities come in both positive and negative magnitudes, and these will cancel. If you add a velocity of 10 m/s North to a velocity of 10 m/sec South ...

12

It comes from the cross product. Every situation in which you have to use the right-hand rule corresponds to some mathematical equation that involves a cross product. In this case, the relevant equation is the Biot-Savart law, $$\vec{B} = \frac{\mu_0}{4\pi}\int \frac{I\,\mathrm{d}\vec{l}\times\vec{r}}{r^3}$$ If you use the right-hand rule - the version ...

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

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 $$q\bar q ... 10 The object you're talking about is called, in mathematics, a Clifford algebra. The case when the algebra is over the complex field in general has a significantly different structure from the case when the algebra is over the real field, which is important in Physics. In Physics, in the specific case of 4 dimensions, using the Minkowski metric as you have in ... 10 With suitable boundary conditions, the decomposition is unique. Without them, it's not. Suppose that (\phi,{\bf G}) and (\phi',{\bf G}') are two different decompositions for the same function. Then$$ \nabla(\phi-\phi')+\nabla\times({\bf G}-{\bf G}')=0. $$Take the divergence of both sides to find that$$ \nabla^2(\phi-\phi')=0. ...

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

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

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 ., ... 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 In a class I'm lecturing, I mention to my students (in a very, very elementary way) that vectors and covectors do not live in the same space. It's a typical school phrase... "Do not add apples and pears", and it's true! If you keep in mind the custom column and row representation of a vector, you can prove that both of them (by themselves) satisfy the ... 9 The gradient of a scalar is again vector. 8 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 $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}}$$ 8 There's absolutely nothing wrong with a vector having all zero components. But a vector like that does have an undefined direction. So the definition of a vector as "something which has a magnitude and a direction" can throw you off here a little. 8 Not only can you do physics "in a mirror", but I've been part of an experiment involving exactly that. The weak interaction is, well, weak. And that makes it very hard to get access to in any physical process which can also proceed through other interactions. So, you can see the weak interaction at work in beta decay, but to leading order you can't see it ... 7 Since torque is defined as the rate of change of angular momentum, the more fundamental question would be whether angular momentum is a vector in SR. The answer is no, because there is no vector cross product in four dimensions. Angular momentum is a rank-2 tensor. 7 The issue here is that$\dot{r}$is not the magnitude of$\dot{\vec{r}}$, but rather the rate of change of$r$, the magnitude of$\vec{r}$. Think about a particle moving in a circle. Since$r$is constant,$\dot{r}=0$, but the velocity is certainly not zero! The formula$\vec{r}\cdot\dot{\vec{r}} = r\dot{r}$makes perfect sense, because$\dot{r}$is the ... 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 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 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 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: $$p_\mu =\frac{\partial L}{\partial \dot{x}^\mu}$$ ... 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} ...

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