New answers tagged vectors
0
Definition of average acceleration:Change in velocity /time elapsed
$$\frac{\Delta v}{\Delta t}$$
Note that it is change ,$v_f-v_i$,Final velocity -intial velocity.
Now, do you see why we subtract?
0
By definition, acceleration is $d\vec{v}/dt$. In turn, by the definition of a derivative, $d\vec{v}/dt = (v_f - v_i)/dt$, where $v_i$ is the initial velocity and $v_f$ is the velocity at a short time $dt$ afterwards.
Since it's $v_f -v_i$ (instead of $v_f + v_i$), you subtract the vectors.
0
Any small segment of an area of any shape (in 3D) can be represented by a vector perpendicular to the surface. It may or may not be convenient to do this with a cross product. (The two vectors you start with must lie in the surface.) A dot product does not yield a vector result. If you are working with a closed surface, it may be customary to define the ...
0
A vector drawn from the origin to a point within any coordinate system (fixed or not) can be defined as the position vector for that point in that system. The coordinates for the point depend on how the system is defined. If you are going from a (2D) polar to a rectangular Cartesian system, then x = R cos(θ) and y = R sin(θ), where θ is measured from the x ...
0
You see the addition of velocities is done when you are changing reference frames, but in this case the velocity of the mass is the upward velocity of the point to which it is attached. Hope this explains your doubt
0
Forces can't like vectors however the velocity does not.
The velocity of the block will be the same as the velocity of the piece of rope attached to it. The velocity of the vertically hanging strand is equal to the rate at which the whole rope overhanging a pulley is moving up.
If the $ 2v \cos \theta$ was the velocity of the block, then it would mean that ...
-1
This kind of diagram is often used to study forces. Forces do add like you are thinking.
But this is velocity. Consider the case where $\Theta = 0$, and the pulleys are right next to each other. If P and Q drop as distance, you would expect M to rise the same distance because the string stays the same length.
0
In this problem, you are not concerned with the vector sum of the two tensions, only with the relative velocities. For that you can work with just one side.
1
In general, limits on an integral over a subset of $\Bbb R^n$ implicitly take care of integration direction. (The case $n=1$ is familiar; if $a<b$ the directions for $\int_a^bfdx,\,\int_b^afdx$ are obvious.) It doesn't matter whether what's integrated is a univariate dot product, a $3$-dimensional cross product or in general a $k$-dimensional integrand, ...
1
Actually, your matrix can be greatly simplified as
$$
M =
\begin{bmatrix}
\frac{1}{\sqrt{a^2 - p^2}} a & \frac{p}{\sqrt{a^2 - p^2}} & 0 & 0 \\
\frac{p}{\sqrt{a^2 - p^2}} & \frac{a}{\sqrt{a^2 - p^2}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
since $(\mathbf{e}, \mathbf{q}, \mathbf{r})$ forms ...
0
Let f be the magnitude of the force exerted by the particle at the origin on the particle being moved.
If the force exerted by the particle at the origin on the particle being moved is attractive, then you need to exert a force in the positive radial direction to move it +dr, and the work you expend is +fdr (and the work the particle does on you is -fdr).
If ...
0
There is nothing wrong!
The force is radially outward. If the particle travels from some point to infinite, then the work turns out to be positive. If the particle travel from infinite to some finite point the work is done by force will be negative. This can be understood from direction of force and displacement as you did. You can equally understand it as ...
1
As Charlie explained, one way to treat Minkowski spacetime is as manifold, which is what one does when he computes metric components as function of coordinates or when one uses the transformation formula
$$V'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}V^\nu.$$
Another way is to treat it as affine space, which is what you are trying to do in your question. ...
0
First of all keep this thing in mind that Dot product and cross products are a tool.
They are defined so as to ease the calculations.
Now when you go for calculation of flux you define it as a number of field lines crossing an area perpendicularly. Now for slanted field lines, we take the area perpendicular to the field lines and we find that we have to take ...
-2
AB cos¶ is a product of two vectors a and b, where a and b are vectors and A, B their magnitudes
but both a and b
are just representations, not the actual quantities they represent.
Area can't be an ordinary arrow vector nor can the magnetic field be an arrow vector.
They can both be the products of 2 arrow vectors. For example a magnetic field is a certain ...
2
I'm assuming here we treated the area as a vector but how can we treat the area as a vector?
Yes, area is considered as a vector in physics. You will come across another law called Gauss Law which also considers area as vector.
and how can the area decrease due to change in orientation?
The area does not change. I am assuming that you are not familiar ...
0
I am pretty sure your book explains how the vector associated with the area is defined. The direction of the vector is perpendicular to the surface of the given area. Changing the angle between the area and the magnetic field does not change the magnitude of the area. Same as when you rotate any other vector, you don't change the magnitude of the vector. ...
2
a) Yes, this equation is valid even when we are in a coordinate system in which the metric is not the Minkowski metric. In fact, since the inner product of two vectors is a scalar (and thus, a coordinate invariant quantity) you will get exactly the same valued in a different coordinate system.
b) You have to be careful here, Minkowski space can be treated ...
0
How do I now express cartesian unit vectors in terms of polar unit vectors to show that they are independent of the $r$ and $θ$?
If you express the cartesian unit vectors in terms of polar unit vectors that means that they are not independent of each other. The unit vectors $i$ and $j$ are independent and so is $\hat{r}$ and $\hat{\theta}$. But one set is ...
0
Just to flesh out @ZeroTheHero 's impeccable answer for you, with a hint of how to escape conceptual hash by small finite-dimensional matrices. I'll avoid addressing your unsound conjectures/probes, to protect your attention from expressions which are not even wrong, in favor of standard stuff.
The states are tensor product states,
$$|\mathbf r\rangle = |x\...
4
It’s a tensor product as the various kets you have live in distinct Hilbert spaces. In this space $\hat x$ really is $\hat x\otimes \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}$, $\hat y$ is formally $\hat{\mathbb{I}}\otimes \hat y\otimes \hat{\mathbb{I}}$ etc. Indeed $\vert x\rangle $ is formally $\vert x\rangle \otimes \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}...
2
Superposition principle and using vectors for modeling forces are entirely independent things.
Using vectors for modeling forces
In Classical Mechanics, forces either are defined through the Second Law ($m \mathbf{a} = \mathbf{F}$), or are introduced as primitive entities which satisfy all the axioms of a vector space over $\mathbb{R}$. In all cases, the ...
1
You're always going to use vectors for forces, because forces have magnitude and direction, so nothing else really models them well. Even if the superposition principle was not valid we would still have to use vectors. We would have to have to use some law $\vec{F} = \vec{F}(\vec{r},Q_1, \ldots , Q_n)$.
1
$${\bf \tau = r \times F }$$
Here "r" is actually the position vector(also called radius vector) i.e. the vector from the axis of rotation to where the force is applied.
Its magnitude actually does equal to the distance from the axis, but it's still a vector(having both magnitude and direction, as seen in the figure).
3
That is not distance, It is position vector.
There is no concept that for a vector quantity always a cross product of two vectors is taken.
In physics some quantities are even given direction so that you can take their cross product.
Cross product is a tool not a property.
1
The position of the ant can be calculated as a sum of the displacement vectors at each step (walk + turn):
$$
\mathbf{r} = \sum_{n=1}^{+\infty}\mathbf{r}_n,
$$
where every displacement vector is obtained as
$$
\mathbf{r}_n=\frac{1}{n}\hat{R}^{n-1}\mathbf{r}_1,
$$
where $\hat{R}$ is the rotation matrix by $60^\circ$.
It remains to write down the rotation ...
0
You can resolve the mg (100N force) vector into components either as in your second (components O and H) or third (components O and A) drawing; but O is not the same vector in the two drawings. Look at your second drawing; the direction of vector O is not totally along (parallel to) the plane and so it still contributes to the net force perpendicular to the ...
0
The matrix you wrote will take the standard basis $\{(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)\}$ into the basis constructed from $\mathbf{X}$. Therefore, to take $\mathbf{X}$ into something proportional to $(1,0,0,0)$, you need to use the inverse matrix.
0
The 3+1-Lorentz transformation is
\begin{align}
\mathbf{x}^{\boldsymbol{\prime}} & \boldsymbol{=} \mathbf{x}\boldsymbol{+} \dfrac{\gamma^2_{\mathrm u}}{c^2 \left(\gamma_{\mathrm u}\boldsymbol{+}1\right)}\left(\mathbf{u}\boldsymbol{\cdot} \mathbf{x}\right)\mathbf{u}\boldsymbol{-}\dfrac{\gamma_{\mathrm u}\mathbf{u}}{c}c\,t
\tag{01a}\...
0
When the rotating disk has friction, the friction between the coin and the disk is the centripetal force, it is what is opposing the coin's centrifugal force to keep it from sliding outwards. Your friction arrow should be labeled centrifugal force. If the disk has no friction it cannot affect the movement of the coin except for the normal force that keeps it ...
2
The major thing to understand here is that centripetal force is not a real force.
It is an effect of the resultant forces.It is not created on rotating the disk, but the forces acting towards the center provide the force.
Centripetal force is the effect not the cause. It is not a separate force but merely the sum of the components of all the forces acting ...
0
In the situation with no friction, the only forces acting on it are the normal force and gravity
which do not provide torque about the axis of the disc.
Also, the disc is in translational equilibrium.
Therefore the coin will stay in its original place and will not go towards the center.
The coin will stay in its original place or move with its initial ...
0
If $\vec c=\vec a + \vec b$ and $\vec a$ and $\vec b$ are at right angles to each other then the vectors $\vec a, \vec b,\vec c$ form a right angled triangle. From the properties of right angled triangles we know that if the angle between $\vec a$ and $\vec c$ is $\theta$ then
$|\vec a|=|\vec c|\cos \theta
\\|\vec b|=|\vec c|\sin \theta$
1
By the mathematical definition of a vector, it can be resolved as the addition of constituent vectors using trigonometry. Using a Cartesian coordinate system, we want to know the components of the vector along each of the three orthogonal axes and for that the trigonometric functions are used to resolve the vector along the axes. The resolved components add ...
0
Magnitude is always relative to the chosen measurement scale.
For example if my room is 5 m long and I have a pile of books on my desk which weighs 5 kg, it is wrong to say simply that they have the same magnitude. Measured in feet (16.4 ft) and pounds (11 lb), the magnitudes are not equal. One has to specify that the magnitude of the length in metres is the ...
4
"Suppose a body has mass 5 kg. Is the magnitude 5 or is the magnitude 5 kg?"
5 kg is the only sensible interpretation because 5 kg is the quantity. It is meaningless without its unit. [We'd call the 5 by itself the numerical value of the mass in kg.]
Some scalar quantities can't be negative, for example speed or, in normal use, mass. For such a ...
0
The answer to this may vary from level to level.
Physics is a quantitative science, based on the measurement of physical quantities. A physical quantity possesses at least two characteristics in common, one is the numerical magnitude and the other is the unit in which it is measured. Suppose you are measuring something, say the length of a rod, so you are ...
-1
I believe it would be roughly the same? If two points on Earth, wich in my humble opinion I experience as flat and motionless, Mars , one of the seven wondering stars, would roughly be the same distance. Peace and Love to you all!
0
REFERENCE : My answer here What is the drawing scheme of the parallel transport of a vector?.
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$
See a 3d view of Figure-01 here
Your path of parallel transport is arc of a ...
0
When you write
$$(L^†L)^† = (L^†)(L^†)^† = L^† L$$
you are implicitly using two facts, that $(AB)^† = B^† A^†$ and that $(A^†)^† = A$.
If you already accept those facts then the second version is fine. But if you want to prove those two facts you will need to consider the inner product.
You must consider the inner product because the adjoint is only defined ...
2
Since you have two vectors in the same direction with the length one ($\vec{e}_r$) they yield 1 when multiplied with each other.
$$
\vec{e}_{r}\cdot (dr\vec{e}_{r})= (\vec{e}_r\cdot \vec{e}_r)dr=dr
$$
So you are left with
$$
\int_{i}^{f}-\frac{dU}{dr}\cdot dr=-(U(f)-U(i))
$$
Here you integrate the derivative of U with respect to r (that is $dU/dr$) with ...
0
I think both ways are fine. But you have to keep in mind that by writing
$$
(LL^\dagger)^\dagger = (L^\dagger)^\dagger (L)^\dagger = LL^\dagger
$$
you implicitly mean what you wrote in your first version. The definition of an adjoint operator, and by this the definition for self-adjoint and Hermitian, are based on a scalar product (in our case the $L^2$ ...
0
As you move north or south the angle between Mars and the horizon (its altitude) will change because the plane in which your local horizon lies is tipping one way or the other. As you move east or west the angle between the direction of Mars and north (its azimuth) will change because your local direction of north is changing.
To see a one degree difference ...
0
Suppose your friend is nearly half way around the world from you at the same latitude.
You might see Mars on the south-eastern horizon while your friend will see it on the south-western horizon.
If you moved closer together, the directions would move more south and less east/west.
Keep moving closer and closer to each other and the directions will become ...
1
Example in 2D, but it's the same in any other number of dimensions.
Represent $\mathbf{v}$ in some basis as $(v_x,v_y)$, then
$$
A_x = \frac{v_x^2+v_y^2}{v_x} = {\bigg{|}} \frac{dx}{dt} {\bigg{|}} + {\left(
\frac{dy}{dt}
\right)}^2
{\left(
\frac{dx}{dt}
\right)}^{-1}
$$
and similarly for $A_y$. So, the answer is no, $A_x \neq v_x$.
1
Your derivation fails in the third equality, when you assume that $\frac{dX^a/dt}{dX^i/dt}=\frac{dX^a}{dX^i}$. For that to be possible, you would need $X^a(t)$ to be an invertible function, which is not possible for a curve in general. That is, you cannot write $t=t(X^a)$.
Consider for example a 2D circular motion where $\vec{X}=(\cos t,\sin t)$. Given an ...
1
The answer by PaulPhy basically says everything, but I thought that this image would be a nice addition:
Starting from your text, the $XY$ (or $x_1 x_2$) Plane is already given. So there is only one choice for the $Z$ ($x_3$) axis as it must be perpendicular to both the $X$ and $Y$ axis.
The only thing we can choose is, as described in your textbook, the ...
1
In such a case the only thing you can do is to differentiate vector to write them in terms of component derivative. Here what I mean
$$\frac{d}{dt}\mathbf{r}=\frac{dx}{dt}\hat{i}+10y\frac{dy}{dt}\hat{j}+20z\frac{dz}{dt}\hat{k}$$
or
$$\dot{\mathbf{r}}=\dot{x}\hat{i}+10y\dot{y}\hat{j}+20z\dot{z}\hat{k}$$
That's all you can do in such a cases unless you have ...
6
The argument Griffiths is making (I think. My copy of the text isn't with me right now), is not
$$
-\vec{E} \cdot d\vec{l} = \nabla V \cdot d\vec{l} \implies-\vec{E} = \nabla V
$$
Rather, he is arguing that if
$$
-\int_{\vec{a}}^{\vec{b}} \vec{E} \cdot d\vec{l} = \int_{\vec{a}}^{\vec{b}} \nabla V \cdot d\vec{l}
$$
for all $\vec{a},\vec{b} \in \mathbb{R}^3$ (...
2
It means that you have a choice about which direction is positive in the $z$ direction. If you have a sheet of paper, you can draw a line orthogonal to the plane (piece of paper). You have a choice to make about if coming above the paper is negative or positive. It's largely arbitrary, but usually for cross products, we have decided that $\hat{x}\times\hat{y}...
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