New answers tagged

0

It is a scalar. Consider generic case - some $p$-surface. It's $p$-volume is something like $$ \int d^{p} \xi \sqrt{\det g (x(\xi))} $$ Where $g$ is a induced metric on the surface, $\xi$ are the coordinates on hypersurface, and the integration is over the hypersurface. In the case $p = 1$ you have some curve in the space, and for $p = 2$ some membrane, ...


0

I think you are referring to the I.E.IRODOV (Problems in General Physics) question number 1.6 (KINEMATICS) South-Eastern winds don't go towards South-East, in fact it's the complete opposite, they originate from South-East and go towards North-West. So, if you were to reverse the direction of the wind vector, the solution would be correct and so will be ...


5

If a map preserves the length $\|{\bf a}\|$ of vectors, then it also preserves the inner product because $$ {\bf a}\cdot {\bf b}= \frac 12 (\|{\bf a}+{\bf b}\|^2- \|{\bf a}\|^2- \|{|\bf b}\|^2). $$ This identity is true for both the usual Euclidean length and inner product, and also the spacetime interval and Lorentz inner product.


6

But the inner product is more general than norm of difference, Actually it's not. If you have a norm, then you automatically get an inner product for free, through linearity, because $(u+v)\cdot(u+v)=u\cdot u+v\cdot v+2u\cdot v$. Alternatively I could phrase my question as follows: is there a proof that a Lorentz transformation is equivalent to a change ...


1

There is a general argument that is beautifully simple. The (squared) speed is given by $\vec{v}\cdot\vec{v}$. Now, let's consider the rate of change of (squared) speed: \begin{align} \frac{d}{dt} (\vec{v}\cdot\vec{v})&= \frac{d\vec{v}}{dt}\cdot\vec{v}+ \vec{v}\cdot\frac{d\vec{v}}{dt} \\ &= 2 \frac{d\vec{v}}{dt}\cdot\vec{v}\\ &= 2 \vec{a}\cdot\...


1

It's very easy to do: \begin{align} |\mathbf{v}(t+dt)|^2&= \left|\mathbf{v}(t)+\frac{d\mathbf{v}}{dt}dt\right|^2\\ &=\left|v(t)\boldsymbol{\hat\theta}+\mathbf{a}(t)dt\right|^2\\ &=\left|v(t)\boldsymbol{\hat\theta}+a(t)dt\mathbf{\hat r}\right|^2\\ &=v^2(t)+2v(t)a(t)dt \boldsymbol{\hat\theta}\cdot\mathbf{\hat r}+a^2(t)(dt)^2 \end{align} Since $\...


0

Phasors are complex quantities used to partially represent real quantities that vary sinusoidally in time and perhaps in space. All phasors are time-independent. They represent partially the real quantity, and not completely, because they don't have information about the frequency. To say that phasors are like vectors, is to say that complex numbers are ...


1

We say that angular momentum is perpendicular to the plane of rotation (in a simple case like a rotating disk) only because Josiah Willard Gibbs and Oliver Heaviside popularized vector algebra (including a vector product using the right-hand rule) and vector calculus in the early 1900’s. There are more modern formalisms — not generally taught in high school ...


1

1) If the body is moving in a plane then it's position and velocity/momentum vectors are in that plane too. This is easy to verify. 2)Angular momentum is a cross product of position and momentum 3)A cross product is always perpendicular to the plane in which the vectors to be crossed are. 4) We conclude that the angular momentum is perpendicular to the ...


0

The set of points unaffected by rotation forms a line, the axis of rotation, which is perpendicular to the plane of rotation. One can take this axis as defining the direction of angular momentum. However, AM, is a pseudovector . It is in reality an antisymmetric tensor and such objects have 3 components in 3D. Coincidentally this can be treated as something ...


2

As with all mathematical conventions in physics, the reason why we represent something a certain way is because it is useful. Angular velocity is a pseudovector, so its direction is defined by the axis about which an object rotates, which makes the angular velocity vector normal to the plane of rotation. Angular momentum is defined by $$\vec L = I\vec\omega$...


0

I would proceed as follows \begin{align} \hat{k} =& k_x\hat{x} + k_y\hat{y} + k_z\hat{z}\\ \hat{e} =& e_x\hat{x} + e_y\hat{y} + e_z\hat{z} \end{align} \begin{align} \hat{e}\cdot\hat{e} =& e_x^2+e_y^2+e_z^2 = 1 \\ \hat{k}\cdot\hat{e} =& k_xe_x+k_y e_y+k_z e_z = 0 \end{align} We are always free to arbitrarily choose $e_y=0$ for one of the ...


0

We have, for a free space plane wave, that $\hat{\boldsymbol{e}} \cdot \hat{\boldsymbol{k}} = 0$. This tells us that $\hat{\boldsymbol{e}}$ lies in the plane perpendicular to $\hat{\boldsymbol{k}}$. You are correct that in a fully general situation there is nothing else which determines the direction for $\hat{\boldsymbol{e}}$ within this plane. However, ...


2

$$ \frac12 n (n + 1) $$ The easiest way to see it is to consider the number of independent elements of the $n\times n$ symmetric matrix $m_{ij} \equiv p^\mu_i p_\mu^j$. The entries of the matrix clearly exhaust all possible Lorentz contractions. The number of independent elements is $1 + 2 + \cdots + n = \frac12 n (n + 1)$, which can easily be seen by ...


0

The scalars can be constructed by taking products of the 4-vectors taken two at a time. So $p_i^\mu p_{j \mu}$ constitute all the Lorentz scalars possible. The number of these scalars are therefore $^nC_2 + ^nC_1$ represent all the possibilities since There are $^nC_2$ ways to make a product $p_i^\mu p_{j \mu}$ where $i \neq j$. There are $^nC_1$ ways to ...


2

There are 6 ways to pick two things out of 4, and 4 ways to pick one thing out of 4. So 10 altogether: p1 p2 p1 p1' p1 p2' p2 p1' p2 p2' p1' p2' p1 p1 p2 p2 p1' p1' p2' p2' Hence the number of easily available scalar invariants here is 10 not 8 as you guessed. This is just me making up what seems to me to be a reasonable answer. If it is wrong ...


1

Consistent notation is needed for clear communication, but there is no one "correct" notation for anything. If you can remember what things mean in your own work, you can do whatever you want. If you are writing something that others must read too, like a graded homework assignment, it is important to define any notation. That being said if you clearly ...


0

Sorry for answering my own question, i just noticed a really simple solution for my question. The answer is not about the physics itself, but absence of mind , The current indeed decides the direction of magnetic field I still have a doubt on the conclusion that whether the Area vector convention is strictly necessary for the thing which prof. Lewin ...


6

I am a bit confused with your notation so I chose my notation. Assume the components of the position vector $\vec{R}=[x_1,x_2,\ldots,x_{n_R}]^T $are function of the generalized coordinates $q_1,q_2,\ldots,q_{n_Q}$ thus: $x_j=x_j(q_i)$ where $j=1,(1),n_R$ and $i=1,(1),n_Q\quad ,n_Q \le n_R$ We want to obtain the velocity vector $\vec{v}=\frac{d\vec{R}}{dt}...


10

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{dx}{dt}\frac{\partial\mathbf{v}}{\partial x}+\frac{dy}{dt}\frac{\partial\mathbf{v}}{\partial y}+\frac{dz}{dt}\frac{\partial\mathbf{v}}{\partial z}=(\mathbf{v}\cdot\nabla)\mathbf{v}$$


1

Just use Pythagoras to write squared vector magnitudes in terms of components $$u^2 = |u|^2 = u_x^2 + u_y^2$$ $$v^2 = |v|^2 = v_x^2 + v_y^2$$ and note that there is no change in the horizontal component of velocity.


1

You have a lot of text here that I will not parse through, but I get your confusion I believe. In the definition of kinetic energy $K=\frac12mv^2$, $v$ is the speed of the object. i.e. $v=\sqrt{\mathbf v\cdot\mathbf v}=\sqrt{v_x^2+v_y^2}$ in two dimensions. So another way to write out the kinetic energy is $K=\frac12m(v_x^2+v_y^2)$. This is not breaking up ...


0

For this problem, I will think from the mathematical point of view. Magnetic field, current flow and position vector are all vector quantities. Hence, mathematically, it must be done in "vectorial" way. Vectors multiplication that result a vector is cross product. And mathematically, cross product is proportional to the $sine$ of the angle between the two ...


1

Answer: Equality: $\frac{\vec r}{r^3}= \frac{1}{r^2}$ is not true Explanation: A vector quantity (having both magnitude & direction) & a scalar quantity (having only magnitude) can never be equated. Now, $\frac{\vec r}{r^3}$ is a vector quantity while $\frac{1}{r^2}$ is a scalar quantity hence $$\therefore \frac{\vec r}{r^3}\ne \frac{1}{r^2}$$ ...


4

There is no requirement that potentials must be zero at infinity, nor is there a requirement that we need to be able to achieve zero by a gauge transformation. When the potential does not have a finite value at spatial infinity, this just means you cannot turn a bound state of the potential into a free state by endowing it with a finite amount of energy. ...


1

The component of $\mathbf{E}$ along the normal $\mathbf{\hat{n}}$ is given by $(\mathbf{E} \cdot \mathbf{\hat{n}}) \mathbf{\hat{n}}$. So the component of $\mathbf{E}$ along the plane is simply $\mathbf{E} - (\mathbf{E} \cdot \mathbf{\hat{n}}) \mathbf{\hat{n}}$. The point $\mathbf{p}$ is irrelevant.


1

You are perfectly free to assign units of length to $dx$ and units of inverse length to $\partial_x$. This doesn't change any of the conclusion of dimensional analysis, as we can see from several examples. First, this assignment gives additional units of $(\text{length})^{n-m}$ to rank $(m, n)$ tensors, i.e. those with $n$ covariant indices and $m$ ...


1

Angular momentum is given as $\textbf{L}=\textbf{r}\times\textbf{p}$. Thus, we obtain that torque, defined as $\frac{d\textbf{L}}{dt}$ is $\textbf{r}\times\frac{d\textbf{p}}{dt}=\textbf{r}\times\textbf{F}$. As you know, the $\frac{d\textbf{r}}{dt}\times \textbf{p}$ factor vanishes because it is the cross product of two parallel vectors. Now, the reason as ...


5

The formula for torque is ultimately set by an arbitrary choice in the way that we define angular velocity. We have chosen to define angular velocity according to the right-hand rule - in other words, we have arbitrarily chosen that counterclockwise motion corresponds to an angular velocity vector pointing upward. Defining angular velocity this way means ...


0

For vector analysis try to use vector algebra. The resultant vector for $\vec{P}$ and $\vec{Q}$ is given by $\vec{F}_{total}=\vec{P}+\vec{Q}$. So the magnitude of resultant force/vector will be $|\vec{F}_{total}|=\sqrt{\vec{F}_{total}^2}$ $$|\vec{F}_{total}|=\sqrt{\vec{F}_{total}\cdot\vec{F}_{total}}=\sqrt{(\vec{P}+\vec{Q}).(\vec{P}+\vec{Q})}$$ $$|\vec{F}_{...


2

Why is there no tangential acceleration in uniform circular motion? Because, then we wouldn't call it uniform. Circular motion can have both tangential $a_\parallel$ and centripetal $a_\perp$ acceleration. They are then the components of an acceleration vector $\vec a=(a_\parallel,a_\perp)$ pointing at an angle. Those special cases where there is no ...


3

The velocity is changing only the direction, which is why there exists only a centripetal acceleration. If the velocity would have been changing its magnitude (in other words, if the speed was changing) then you'd see that tangential acceleration would exist. This can also be seen by the relevent formula of a body's acceleration in polar coordinates: $$\...


0

The centripetal acceleration (CA) is not $\vec \omega \times \vec v$, as it does not contain any spatial deriative. Furthermore the CA is fundamentally different from the nonlinear term $(\vec v \cdot \vec \nabla) \vec v$. The CA and Coriolis acceleration originate in a frame transformation of the total time derivative of any vector, while the nonlinear term ...


2

Vectors are usually denoted on figures by an arrow. The length of the arrow indicates the magnitude of the vector and the tip of the arrow indicates the direction. The vector is labeled with an alphabetical letter with a line over the top to distinguish it from a scalar. We will denote the magnitude of the vector by the symbol |a|. The direction will be ...


2

There is nothing wrong with calling the axes as infinite vectors$^1$, however there is nothing useful about it as well. Why? Mostly because of the undefined (or infinite) magnitude of the vectors along the axes. Thus if we really need to show the directions of the axes using vectors then using vectors of finite magnitude would be far more helpful. This is ...


2

The canonical answer is no because $\infty$ is not a member of $\mathbb{R}$. The points in a plane can be mapped to the vector space $\mathbb{R}^2$ where each point $(x,y)$ is identified as a vector, but infinite points are specifically excluded. Now, you could artificially put infinite vectors into this vector space as long as you are being careful and ...


1

You can think in this case that $x$ and $y$ mark to independent coordinates of a point. But one can always argue that there is a position vector between the origin of the $\mathbb R^2$ space associated with that point, so there is map between points of $\mathbb R^2$ and position vectors in the same space and in that sense you may see $(x,y)$ as a couple ...


1

All of the answers about vectors and matrices seem technically correct, but I am not sure that that is what is being asked here. If I understand the question correctly, the short answer is "it is entirely arbitrary". However, there is a convention. If you only have one direction then it is conventionally labeled "the x axis". It is conventionally drawn ...


4

I think it would help to read up on the definition of a vector. Velocity, Acceleration, Force, and many other quantities in Physics are described by vectors. From a geometrical point of view most texts begin with the idea of a directed line segment. Just as we have the idea of a line segment from points P to Q, where only the length has meaning, we ...


5

The 3D Cartesian Axes are usually defined as Z 'upwards' as positive Z direction, Y 'right' as positive Y direction and X 'out-of-the-page (towards you)' as the positive X direction. 2D obviously the Y and Z axes swap, there is no Z axis, and the X axis 'becomes the 3D Y axis' in effect. Each axis is independent from each other so have their own positive ...


16

In general velocity is a vector. In 1D motion we can get away with saying the velocity is positive or negative by associating each sign with a direction. However, once you move in more dimensions you can't say the velocity is positive or negative. You just have the vector (in Cartesian coordinates) $$\mathbf v=v_x\,\hat x+v_y\,\hat y+v_z\,\hat z$$ Of ...


0

A vector, in its geometric interpretation, is not merely a tuple of numbers. As Feynman says, a tuple is a vector if it satisfies a precise transformation law, when the coordinate system is changed. So let us assume that $\mathbf a$ and $\mathbf b$ are vectors. Under a coordinate transformation we can find a matrix $\Lambda$ such that the new components of ...


3

We say something is a vector if its transformation follows certain rules. This implies that we need to define transformation before we define the vector. A Transformation transforms any three numbers from K to K' by such a rule that $T([x,y,z])=[x',y',z']$. We don't need the concept of vectors there. To say $a$ is a vector means by knowing its ...


0

I think Feynman is making very heavy weather of this. Vectors can be defined by the rules of addition and scalar multiplication (in linear algebra we might also want to prove that they satisfy the axioms of vector space, which make no specific mention of coordinates). It is then straightforward to show that coordinate transformations are obeyed. Physicists ...


0

This question is resolved in relativity. Galilean transformation applies to 3-vectors, in relativity we use 4-vectors and describe boosts with Lorentz transformation. In general relativity, physical laws written in terms of vectors and tensors are unchanged in all frames, not just in inertial frames.


2

In the reference frame of the airplane the sum of physical and fictitious forces has to be always equal to zero, thus we can readily tell that the second and third diagrams labelled "slipping turn" and "skidding turn" are incorrect. I think the missing component is the lateral drag in the horizontal direction. The value of this component of drag would ...


1

I love this question and was very curious about it, so I built on a 3blue1brown video to answer it with the video below. I’d argue one doesn’t fully understand/appreciate divergence, curl or Maxwell’s equations unless they get this. Here’s the nutshell version: This connection is difficult to conceptualize because like you said, the del operator, $\nabla ,$ ...


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