I've used quite a simple algorithm to create the appearance of a jumbled up bunch of letters. These letters represent a song. What is the song title?
(Everyone should know this song, unless you live in a non-English speaking country.)
Answer
The answer is
"Twinkle Twinkle Little Star", each letter corresponds to the first letter of each word in the song.
So I understand how moments work with regards to basic examples like pushing a door, in that the further you are away from the hinges of the door, the greater the moment, which is like a turning force. I also know, though perhaps do not fully understand, the formula for calculating torque, $\tau=\mathbf r\times \mathbf F$ where $\mathbf F$ is the force applied, $\mathbf r$ is distance from centre, and $\times$ denotes cross product.
However, I don't really understand how it is connected to the centre of gravity (sometimes called centre of mass). For instance, a problem in my textbook addresses the issue of the centre of gravity of a triangular-shaped piece of thin card, with equal mass distribution. How do moments help me find the centre of gravity? What forces are being applied to the piece of card?
Answer
Now consider the exact same situation, just with gravity being the force instead of your push:
This point is the center of gravity. Let's call it CoM. That is all. You talk about the CoM in connection to where gravity pulls.
How do moments help me find the centre of gravity?
Only moments caused by gravity can help you, since only then does the moment have anything to do with the CoM.
Gravity will pull in the CoM as a force applied at a point. So as long as this CoM is not straight under the rotation point, gravity will try to rotate the object (it will cause a torque, since the distance $r$ is not zero in your formula, unless the CoM is exactly under the rotation point).
So when your object hangs still and doesn't rotate anymore, you know that the CoM must be somewhere directly below the rotation point. On the vertical line below.
If you now hang the object in another point so that you have a new rotation point, then when the object hangs still again, you again know that the CoM must be directly below somewhere on the vertical line under the new rotation point. These two lines have only one point in common. So this point must be the CoM, because it must be the same for both situations (since you didn't redistribute the mass).
What forces are being applied to the piece of card?
If it hangs freely in a hook for example, so it can rotate freely, then only two forces act on the card. Gravity from the CoM and the hook's normal force holding it.
Since the normal force by the hook works in the rotation point, this gives no torque (the distance $r$ is zero). Therefore only gravity is left to do a torque to make it rotate.
And if the CoM is vertically below the rotation point, then also gravity gives no torque as explained above. Therefore the above method to find the CoM will only work if no more forces than the gravity can cause a torque.
If you compress a large mass, on the order of a star or the Earth, into a very small space, you get a black hole. Even for very large masses, it is possible in principle for it to occupy a very small size, like that of a golf ball.
I started to think, how would matter react around this golf ball sized Earth? If I let go of a coffee mug next to it, it would go tumbling down toward the "golf ball". Isn't that exactly how magnets work, with paperclips for example?
Magnets are cool because they seem to defy the laws of gravity, on a scale that we can casually see. Clearly, the force carrier particles that produce electromagnetic attraction are stronger than gravity on this scale (or are at least on par: gravity plays some role in the paperclips path, but so does electromagnetism).
My question is, why do we try to consider gravity as anything different than magnetism? Perhaps "great mass" equates to a positively (or negatively) charged object. Pull so much matter in close and somewhere you've crossed the line between what we call electromagnetic force and gravity force. They are one in the same, no?
Answer
There are several qualitative and quantitative differences between gravity and magnetism.
When you attract 'neutral' bits of metal with a magnet, or attach it to something like a plate of metal, what's happening is that individual atoms of the metal react to the magnetic force. In a ferromagnetic metal, one with a similar electronic structure to Iron or Nickel, the individual atoms work like nanoscopic magnets; but they are very weak, and they are not lined up with one another, so that their fields cancel one another out over any macroscopic distance. But if you bring a "large" magnet (such as a fridge magnet) up to them, the field of the large magnet causes them to align with the field, so that they are pulled towards the magnet — and the magnet is pulled towards them. This is why some metal objects are attracted to magnets.
Other metals, such as aluminum or silver, also react to magnets, but much more weakly (and in some cases repulsively): the way that they react to magnetic fields is described as paramagnetism (for materials which align very weakly with magnetic fields) and diamagnetism (for materials which align very weakly against magnetic fields).
The very fact that different materials react differently to magnetic fields is something that sets magnetism apart from gravity. Gravitation works equally with masses of any sort, and is always attractive (as noted by Nic); magnetism can both attract and repel, and do so with different degrees of force, as between ferromagnetic, paramagnetic, and diamagnetic materials. But of course, quite famously, even a single object can be both attracted and repelled by magnetic forces: the north poles of two magnets repel each other, as do the south poles; only opposite poles attract each other. (This, of course, is the basis on which compasses work.)
The way that these forces operate over distance also varies. Gravity very famously (but only approximately) obeys an inverse-square law; the field far from a bar magnet, however, decreases like the inverse of the cube of the distance from the magnet.
Finally, moving electric charges produce magnetic forces; whereas they don't cause any gravitational forces which could not be accounted for just by the fact that the charged particles have mass (whether moving or at rest).
So, on both the macroscopic level and on the level of individual atoms, the forces of gravity and magnetism act quite differently.
A question on a practice test I'm taking is as follows:
By shaking one end of a stretched string, a single pulse is generated. The traveling pulse carries:
A. mass
B. energy
C. momentum
D. energy and momentum
E. mass, energy and momentum
How would one describe the momentum of a wave?
Consider an RLC circuit in series, of the form
If the source drives the circuit in AC at the resonance frequency $\omega =1/\sqrt{LC}$, the peak-to-peak voltages on the capacitor and the inductor, $$ V_C=\left|\frac{Z_C}{Z_\mathrm{tot}}\right|V_S=\frac{\frac{1}{\omega C}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}V_S \quad \text{and}\quad V_L=\left|\frac{Z_L}{Z_\mathrm{tot}}\right|V_S=\frac{\omega L}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}V_S ,$$ can both be larger than the peak-to-peak voltage $V_S$ of the source.
The math might say one thing, but this is till terribly counterintuitive. How can this be?
I got badly lost in derivation of relativistic formulas for energy and momentum.
I stumbled upon relativistic action as follows (which should explain relativistic motion of a classical particle):
$$ S = \int Cds=C\int_{t_i}^{t_f}\sqrt{c^2-(x')^2}dt $$
Where $C$ is some constant (depends on what kind of physics we put in equation) and $s$ is relativistic interval. Later on Lagrangian $$ L(x')\equiv C\sqrt{c^2-(x')^2} $$ is used in derving relativistic energy and momentum.
I am familiar with Lagrangians and symmetry rules which connect energy and momentum to Lagrangian formalism. What I do not understand is this action - weren't action sum over time? Why all of a sudden it is sum over relativistic interval?
Answer
The action is commonly written in terms of $ds$ because it is a Lorentz scalar. $dt$ is not a Lorentz scalar, but $dt \sqrt{1 - v^2} = dt / \gamma = ds$ is, so you can write the action as an integral over time if you want: $$ S = - m \int ds = - m \int dt \sqrt{1 - v^2} $$ We can check this in the non-relativistic limit: $$ S = - m \int dt \sqrt{1 - v^2} \approx \int dt \left(-m + \frac{1}{2} m v^2 \right) $$ The constant $-m$ does not affect the equations of motion, so it can be removed, and we are left with $L = m v^2 / 2$, as expected.
The speed of light $c=1$ above.
I was deriving the matrix form of Lorentz boosts and I came up with a doubt. I don't think I quite understand hyperbolic rotations.
The stadard basis of the Minkowski space is given by $\{e_0, e_1,e_2,e_3\}$, where the square norm of any vector $x=x_0 e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3$ is $|x|^2 = x_0 ^2 - x_1 ^2 - x_2 ^2 -x_3 ^2$. Note: $x_0$ is the time component ($e_0$ indicates time axis), $x_1, x_2, x_3$ is the spatial component ($e_1 , e_2, e_3$ indicate the $x, y,$ and $z$ axis).
Let $B_i$ be a Lorentz boost in the ith direction. This boost will only modify the time component and the $ith$ component, and like any other lorentz transformation, it will preserve the norm of any vector. Consider $B_i e_0 = a e_0 +b e_i = e_0 '$. Then,
$(B_i e_0)^2 = e_0 ^2 = 1$
$a^2 - b^2 = 1$
The solutions of this lie on a hyperbola along the $e_0$ axis. We can parametrize it and obtain:
$a=cosh \theta$ and $b=sinh\theta$
I do the same procedure for $B_i e_i = a e_0 +b e_i = e_i '$. $a$ and $b$ satisfy the equation: $b^2 -a^2 = 1$ which is a hyperbola along the $e_i$ axis.
My intuition: I imagine the $e_0$ as the horizontal axis in the x-y plane and the $e_i$ as the vertical axis. If I shift $e_0$ by and angle $\theta$ (counterclockwise) over the hyperbola, the resulting vector $e_0'$ will fall on the first quadrant. So $e_0 ' = cosh \theta e_0 + sinh \theta e_i $. Now I need to shift $e_i$ by the same angle (counterclockwise) over the corresponding hyperbola so that I can keep $e_i$ and $e_0$ orthogonal to each other. The resulting vector $e_i'$ will fall over the 2nd quadrant. So $e_i ' = -sinh \theta e_0 + cosh \theta e_i $
For $i=1$, I end up with a boost $B_i$ of the form: $$B_i= \left( \begin{matrix} \cosh \theta & -sinh \theta & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ when I should end up with:
$$B_i= \left( \begin{matrix} \cosh \theta & sinh \theta & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ Edit: The columns of this matrix are correct because the matrix satisfies Lorentz transformation condition $\eta = B_1 ^T \eta B_1$ (where $\eta$ is the minkowski metric tensor) while my boost matrix doesn't.
Question : Since taking the square might make me lose/add a negative sign, is there a better way to obtain $e_0'$ and $e_1'$ with all the necessary negative signs in its components? Right now I would just be using trial an error, until the boost matrix I obtain satisfies $\eta = B_1 ^T \eta B_1$ I appreciate any insights, thanks!
Answer
In your attempted construction, you first use the future unit-hyperbola for the tips of your unit-timelike vectors. Note that this hyperbola has spacelike tangents. In fact, following Minkowski's own construction from "Space and Time"
We decompose any vector, such as that from O to x, y, z, t into four components x, y, z, t. If the directions of two vectors are, respectively, that of a radius vector OR from O to one of the surfaces ∓F = 1, and that of a tangent RS at the point R on the same surface, the vectors are called normal to each other. Accordingly, $$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$$ is the condition for the vectors with components x, y, z, t and $x_1$, $y_1$, $z_1$, $t_1$ to be normal to each other.
If R is the tip of the unit-timelike vector OR from O, the tangent to the hyperbola at R is [Minkowski-]orthogonal to OR. The "intuition" to have is that the tangent to the "circle" in that geometry is orthogonal to the radius vector.
(You don't need to use the other hyperbola [with timelike tangents].)
