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1 vote

Magnitude of Acceleration Vector when Speed is Constant

"But acceleration is defined by the time derivative of velocity (which is a vector quantity) and thus by the vector difference - or am I wrong?" You are not wrong. If the velocity changes ...
Philip Wood's user avatar
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4 votes
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Magnitude of Acceleration Vector when Speed is Constant

The acceleration vector, any instant of the motion, regardless of whether or not the magnitude of the velocity is altered, is always perpendicular to the instantaneous velocity vector. You can work ...
Albertus Magnus's user avatar
1 vote

Instantaneous speed x instantaneous velocity

The direction of the tangent to the path of the particle in space at any point is the direction in which the particle covers the distance in an infinitesimal amount of time. To obtain the direction of ...
Piyush Lath's user avatar
1 vote
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Instantaneous speed x instantaneous velocity

If A and B are two points with a finite separation on the path of a moving particle, the segment of path between these points need not be straight, so, for this segment: $$\text{distance gone}\geq \...
Philip Wood's user avatar
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1 vote

Physical significance of $\vec{w}$ $\times$ $($curl $\vec{v})$

If it helps, the following identity holds (in index notation, where repeated indices are summed over): $$[\vec w\times(\vec \nabla\times\vec v)]_\mu = (\partial_\mu v_\nu - \partial_\nu v_\mu)w_\nu. $$...
0 votes

Scalar Flow Across a Small Area Element

I am reading the same book as well. Here is my attempt at a solution. Please let me know if I am thinking about this correctly because I am not sure it is correct, but hopefully on the correct tract. ...
Alexander Savadelis's user avatar
3 votes

Can we call numbers unidirectional vectors?

Good morning, what you're saying is interesting. Of course, real numbers are vectors, belonging to $(\mathbb{R},+,*)$, which is an $\mathbf{R}$-vector space. It's a one-dimensional version of the ...
Gorga's user avatar
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1 vote

Could you calculate the force between two NON-PARALLEL, straight current carrying wires?

Assuming we already know how to calculate the $B$-field of one infinite straight wire, we only need the integral of its force over the second wire. Without loss of generality we can assume that the ...
Jos Bergervoet's user avatar
0 votes

When we apply these concepts to physics, where do we put the UNITS in vector spaces and manifolds? Do units have a clear mathematical meaning?

It is good that you are already familiar with the concepts introduced by GR. Then I can just directly tell you that we are supposed to simply lay any coördinates (that work), and forget about trying ...
naturallyInconsistent's user avatar
0 votes

Comparing vectors on a sphere

Maybe you can try something like this: Reinterpret the two points $p_1$ and $p_2$ on the surface of the sphere as 3D points, which means they correspond to two unit vectors $ \vec{p}_1$ and $ \vec{p}...
Futurologist's user avatar
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1 vote

Proving a given vector to be a four-vector

We know, the four-position $$x^\mu=(ct,\vec{r})$$ is a four-vector. We also know, for a plane wave its phase at any certain point in spacetime $$\phi=-\omega t+\vec{k}\cdot\vec{r},$$ must be a four-...
Thomas Fritsch's user avatar
2 votes

Comparing vectors on a sphere

The two-sphere, i.e. $S^2$ is an example of a smooth manifold and the thing about smooth manifolds is that the usual notions of vectors are naturally lacking because of the decidedly non-linear ...
Albertus Magnus's user avatar
-1 votes
Accepted

Finding radial acceleration from $xy$ vector cordinate

The radial acceleration is not $\ddot A + \ddot B$. To see this, consider the case where $A(t) = t^2$ and $B(t) = t^2$, so $\vec {R(t)} = t^2 \vec i + t^2 \vec j$ and we have $r(t) = (\sqrt 2)t^2$ and ...
gandalf61's user avatar
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0 votes

Electromagnetic duality and vector/pseudovector transformation properties

Both are solutions. If $(\mathbf{E},\mathbf{H})$ is a solution, then so is $(-\mathbf{E},-\mathbf{H})$. Thus, $\mathbf{E}'=-Z_0\mathbf{H}$, $\mathbf{H}'=Z_0^{-1}\mathbf{E}$ is also a solution. On your ...
Nullius in Verba's user avatar
1 vote

Isomorphism of the tangent space and the space of directional derivatives

In the following answer I will assume you are familiar with germs of functions and derivations at a point. I will provide an argument starting with the definition of tangent space of a manifolds in ...
Sir Crackpot's user avatar
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4 votes
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A confusion about the inner product keeps disturbing me

But very often, I hear that people [cannot give a reference, :-(] say it is an operation between a vector from $\mathbb{V}$ and a dual vector from $\mathbb{V}^*$? If somebody could explain this simply ...
peek-a-boo's user avatar
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1 vote

A confusion about the inner product keeps disturbing me

Note: this is not a math answer, go to math.stackexchange.com for that. This is a physics answer. So my absolute favorite definition of the inner product is from a Kip Thorne lecture on "...
JEB's user avatar
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1 vote

A confusion about the inner product keeps disturbing me

There are differences in real vectors and imaginary vectors. In particular, real vectors in a measure space (the dot product establishes a measure) are combined in a dot product by multiplication of ...
Whit3rd's user avatar
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0 votes

Tensions in the chains

Here a 2D moodel and the internal actions And here some details of the computations
basics's user avatar
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4 votes
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Tensions in the chains

The answer is very simple. As can be seen in the photo, there is no tension in the chains on the left because they are visibly slack. There is some tension in the chains on the right, but this is only ...
KDP's user avatar
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3 votes

Is the electromagnetic 4-potential a Lorentz 4-vector in the Coulomb gauge?

but I don't understand how this can say that $A$ isn't a 4-vector Because the four-tuple is defined, in all frames, in such a way that it obeys $$ \nabla\cdot \mathbf A = 0, \varphi=0 $$ (in all ...
Ján Lalinský's user avatar
6 votes

Is the electromagnetic 4-potential a Lorentz 4-vector in the Coulomb gauge?

It's rather that the Coulomb gauge is not Lorentz invariant. If you fix the gauge in a given frame, then if you boost to another frame, the new 4-potential obtained by the Lorentz transformation does ...
LPZ's user avatar
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1 vote

Calculating the magnitude of the average acceleration of a clock hand

My initial thought was to calculate the average velocity of the minute hand during this time period. This isn't the correct approach. We're interested in the acceleration, meaning how the velocity ...
The Photon's user avatar
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2 votes
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Pseudo forces in accelerating reference frames

It's because in case of rotational frames of referance at each position the translational acceleration(a) is different and also there will be a psudo force the centrifugal force . This centrifugal ...
user140490's user avatar
0 votes

When we apply these concepts to physics, where do we put the UNITS in vector spaces and manifolds? Do units have a clear mathematical meaning?

A similar question was asked on Maths Stack Exchange, and the top answer was that dimensional object live in a tensor product of 1-dimensional graded vector spaces. This builds on an essay by Terence ...
Anders Sandberg's user avatar
0 votes

When we apply these concepts to physics, where do we put the UNITS in vector spaces and manifolds? Do units have a clear mathematical meaning?

I think taking the formal tensor product of (say) $\mathbb{R}\otimes m$ for a length (meter) is good enough, in analogy how one can complexify vector spaces. Now use the natural convention of $(\...
Confuse-ray30's user avatar
0 votes

Avg. velocity in plane polar coordinates

If you insist on starting with polar coordinates, you'd do, $$\vec v _\mathrm{avg} = \dfrac 1{t_2-t_1} \int_{t_1}^{t_2} \vec a \, \mathrm dt = \dfrac{\vec v_2 - \vec v_1}{t_2-t_1}$$ and you express ...
user256872's user avatar
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1 vote

Lorentz covariant but Lorentz inconsistent $4$-forces?

This is a comment working out knzhou's example of coupling to a rank-2 tensor field $g$. I'll use primes to denote differentiation with respect to $\lambda$. Consider a parametrization-invariant ...
user196574's user avatar
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0 votes

Signs terms including acceleration

More or less. Velocity, accelerations are vector quantities. You can get a scalar quantity related to a direction taking the dot product of the vector quantity $\mathbf{v}$ of interest and a unit-...
basics's user avatar
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1 vote

Integral involving unit vectors in spherical coordinates

Any integral of this form is zero by symmetry (assuming it converges). Consider $$ \mathbf{I} = \int_{\mathbb{R}^3} dv\, \mathbf{r} f(r)$$ Now, rotate $\mathbf{I}$ with an arbitrary matrix $M$ in $O(3)...
lcv's user avatar
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1 vote
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Lorentz boost that result in $\vec{E}$ parallel to $\vec{B}$

We have $$ \vec{E'} = \gamma(\vec{E}+\vec{v'}\times\vec{B}) = (1 - \alpha B^2) \vec{E}+\alpha (\vec{E} \cdot \vec{B}) \vec{B} \\ \vec{B'} = \gamma(\vec{B}-\vec{v'}\times\vec{E}) = (1 - \alpha E^2) \...
Michael Seifert's user avatar
1 vote

Newton's Law of Gravitation Explanation?

$\vec r$ is the position vector of the planet relative to the Sun. $r$ is the magnitude of $\vec r$ (in other words, $r = |\vec r|$). So $\frac {\vec r} {r}$ is a unit vector in the direction of $\vec ...
gandalf61's user avatar
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0 votes

Calculating the magnitude of the average acceleration of a clock hand

The average acceleration is: $$ \langle\vec a\rangle = \frac{\vec v_f - \vec v_i}{\Delta t} $$ Since: $$ \vec v_i = \vec v_f = 0 $$ it is zero. Now if you just consider the average tip speed of $\bar ...
JEB's user avatar
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