# Tag Info

17

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 ...

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 $$q\bar q ... 10 This might be more of a math question. This is a peculiar thing about three-dimensional space. Note that in three dimensions, an area such as a plane is a two dimensional subspace. On a sheet of paper you only need two numbers to unambiguously denote a point. Now imagine standing on the sheet of paper, the direction your head points to will always be a way ... 9 With suitable boundary conditions, the decomposition is unique. Without them, it's not. Suppose that (\phi,{\bf G}) and (\phi',{\bf G}') are two different decompositions for the same function. Then$$ \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

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 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 $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}}$$ 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 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 ... 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 ...

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: $$p_\mu =\frac{\partial L}{\partial \dot{x}^\mu}$$ ...

5

Yes, they're representations of $SO(8)$, more precisely $Spin(8)$ which is an "improvement" of $SO(8)$ that allows the rotation by 360 degrees to be represented by a matrix different from the unit matrix, namely minus unit matrix. ${\bf 8}_v$ transforms normally as $$v\mapsto M v$$ where $MM^T=1$ is the $8\times 8$ real orthogonal $SO(8)$ matrix. The ...

5

It's defined as a dot-product (or scalar product) of force and displacement, both of which are vectors. A scalar product of two vectors gives a scalar result (aptly named!). $$dW = \vec{F}\cdot\vec{S} = |F||S|\cos\theta$$ ($\theta$ being the angle between the vectors). No direction, only magnitude. Thinking logically, what would be the direction of ...

4

In a certain sense, 3-d vectors can be divided in physics. To define the division, you begin by defining multiplication using the cross product. I.e. if $\vec{u} \times \vec{v} = \vec{w}$ then $\vec{w}/\vec{v} = \vec{u}$. This is a definition of division that is non commutative. It is also not well defined (as written) because you can always add multiples of ...

4

Can vectors in physics be represented by complex numbers? Absolutely. There exists a direct isomorphism between the 2D Euclidean vector space and the Argand plane, for a start. In fact, it is possible to talk of mathematical objects called quaternions and use quaternion algebra analogously to vector algebra. Historically quaternions were used to represent ...

4

This is the law of cosines! The Pythagorean theorem is just a special case of this more general result - in the Pythagorean theorem we are dealing with a right triangle only, and $C=90$ degrees. This equation is true in the more general case of any triangle.

4

I think the real question is actually posed most directly in your comment (so you might want to consider editing some of this into the original question): I'm more concerned about understanding whats going on and making sure that I know how to do it. I've heard that you shouldn't worry about significant figure rules until you have your final answer. How ...

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