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4

$10^{-10}m$ actually. According to Wikipedia, it is metric but not SI. SI is a subset of the metric system. "The ångström is not a part of the SI system of units, but it can be considered part of the metric system." Like many units, it is named after a person and this person happened to be Swedish so it is not very surprising that his name contains ...

1

The SI system (aka metric system) has, by definition, seven "base units" that can be scaled by a standardized list of prefixes. The base unit for length is the meter. There is no $10^{-10}$ prefix, so you cannot express 1 angstrom as 1 of any such combination. You can convert it, of course, to any such combination, for example the 0.1 nm that you listed ...

0

Depends on how you are defining "weight". If you mean weight as in the vector or scalar of the gravity acting on an object, Marco's answer is what you want. There is a third meaning of weight. This is essentially "what a scale measures". This definition also fits with how people typically experience the effects of weight. If we use that definition, then ...

1

This is an arbitrary convention that was fixed historically around the time that Einstein published the general theory of relativity. It's similar to the right-hand rule for defining torque, or the convention that the charge of the electron is negative. Although it's arbitrary, it's fixed, so different authors do not use different conventions as a matter of ...

2

There are two objects in relativity that contain the same information, but are of different mathematical nature - vectors and covectors. Covector is dual vector to the original vector. That is, it is member of different vector space, but (in space endowed with metric) there exists natural one-to-one correspondence between vectors and covectors. Now, since ...

6

For all practical purposes the weight will remain 1000 pounds when you freeze it. In theory the mass of the water will reduce somewhat through cooling because it will contain less energy when frozen, but the effect would be utterly negligible. Although the weight will be the same, water expands in volume by about a tenth as it freezes (which is why, for ...

0

You can of course find this on wiki...or many other places. Today, charge is related to the elementary charge e The SI unit of charge, the coulomb, "is the quantity of electricity carried in 1 second by a current of 1 ampere A". Conversely, a current of one ampere is one coulomb C of charge going past a given point per second: $1 A = 1 \frac{C}{s}$ The ...

0

Written as scalars, the DE looks like: $$m\frac{\text{d}^2x}{\text{d}t^2}+k^2x=0$$ Or: $$x''(t)+\frac{k^2}{m}x(t)=0$$ Set: $$\omega^2=\frac{k^2}{m}$$ The DE then solves to: $$x(t)=A\cos(\omega t+\varphi)$$ Where $A$ and $\varphi$ are determined from the initial conditions (not specified here). Using this notation the angular velocity $\omega$ then ...

1

But how can energy be negative? Your question is not entirely clear to me, but is it correctly understood that you essentially are asking why electric potential energy can be negative? The answer is that the value of potential energy doesn't matter. Only the difference matters. If you compare the amount of stored potential energy with another amount of ...

0

It may depend on the metric convention in the papers. In the West Coast metric $g_{\mu\nu}= {\rm diag}(+,-,-,-)$ the flat space Dirac equation is usually written as $$(-i\gamma^\mu\partial_\mu +m)\psi=0.$$ In the East Coast metric $g_{\mu\nu}= {\rm diag}(-,+,+,+)$ it is $$(\gamma^\mu\partial_\mu+M)\psi=0.$$ In both cases $$\gamma^\mu\gamma^\nu+\... 3 check the definitions of \Gamma^{\mu}, D _{\mu}, and the signature of g^{\mu\nu}. I think you will find that the equations are actually the same, possibly modulo an overall sign. 1 A statvolt is a statcoulomb per centimeter, because the electrostatic potential of a point charge in Gaussian units is \varphi=q/r. 0 Conventional current_ Current flow from positive to negative termminal of a body due to flow of positive charge known sa conventional current. 0 Note that these are slightly modified conventions which I developed after practising a few questions. For mirrors, simply use coordinate system conventions (fixing the pole of the mirror as the origin), but for lenses, as far as I know, this doesn't hold so good. You will have to check the direction of incident light to the lens. In the direction of ... 0 Negative sign is used for virtual images. For mirror, if the reflected ray comes where the radius of the curvature is and forms a real image, give a plus sign before i. Otherwise, give minus sign. 0 Our universe seems to tell a story that is independent of the words in which we have always chosen to express it. – Kate Becker I like to shorten it down to: The world works the same, regardless of how we speak about it. Mathematical formulations also count as a way of speaking about it. Math is a human invention - it is a "language" - that we use to ... 1 We can say "light travels in water 1.333 times slower than in vacuum", OR "light travels in water at 0.75c". Both ways of describing speed of light in medium are valid. Besides more valid definitions of light speed in medium can be constructed, so what ? It changes nothing at all. EDIT Several alternative refractive index definitions:  n= \lambda_0 / \... 7 All one would have to do is replace n with 1\over n everywhere. Changing the definition will not affect the physics in any way. 2 What is the historical reason why this unit of measure was adopted as a submultiple of the International System? The SI prefixes denote factors of ten (deci, deka), hundred (centi, hecto), and thousand (milli, kilo). Beyond that, the prefixes go in steps of 1000. According to the wikipedia article, micro- (10^{-6}) dates to 1873; nano- and pico- (10^{-... 1 I'm getting the same result, which is not correct since this is a left-handed system. What am I missing here? You are not missing anything. It is correct, you should be getting the same result. You have changed both the handed ness of the coordinate system and the roles of the vectors. Those two changes cancel out so that in the end you get the same result. ... 1 In a left-handed system (which is the one on the right), the relation that connects your basis vectors \textbf{e}_1, \textbf{e}_2, \textbf{e}_3 (that signify \textbf{i}, \textbf{j} and \textbf{k} respectively) is:$$\textbf{e}_i \times \textbf{e}_j = \sum_{k=1}^3\epsilon_{ijk} \textbf{e}_k$$where \epsilon_{ijk} is the \textbf{Levi - Civita Symbol}... 0 In a left-handed system, positive rotation is clockwise about the axis of rotation. Did you take this into account properly? And did you use left hand? It seems to work for me. The method you sited is for a right-handed coordinate system. Its discussed in your reference but a specific method is not given for a left-handed system. 0 A wrong assumption was made in the question asked. An element A\in SL(2,\mathbb{C}) that induces the Lorentz transformation X \rightarrow A X A^\dagger is a Spinor transformation for a right-handed Spinor (not left-handed, as claimed). Rather, a left-handed spinor transformes with an element A\to SL(2,\mathbb{C}) that induces the Lorentz ... 0 As stated in some of the comments above, signs are a matter of convention, but they are not arbitrary ! Consistency is the key. Since e^{\pm i \frac{1}{2} \vec{\phi} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}} and e^{\pm i \frac{1}{2} \vec{\phi} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}} (same sign for the ... 0 The naming of the two types of electricities is not wrong and it cannot be a matter of human's convention at all. (in this answer I will use the words “plus” and “minus” instead of “positive” and “negative”) “Plus” is the effect towards outside (expansion, blowing, explosion, yang), “minus” is the effect towards inside (contraction, suctioning, implosion, ... 0 To be consistent with the vector notation, when r points to the center of mass from the center of rotation it is$$ v = \omega \, r$$in scalar form and$$ \vec{v} = \vec{\omega} \times \vec{r} $$in vector form where \times is the vector cross product. 6 If you're seeing web sites disagreeing about something very basic like this, why not just look it up in a reliable source like a textbook? The relation is v=\omega\times r. You can verify this using the right-hand rule. 1 This seems like a misapplication of the concept of pushforward and pullback. Carroll is speaking the language of physicists, but I think in the language of a modern differential geometer, tensors such as the metric do not transform under a change of coordinates. The tensor is invariant, but its components can be expressed in different bases. Assuming I'm ... 0 I.e. potential difference = change in potential / charge. \Delta V is the difference in potential (voltage) between points a and b. \Delta U is the change in potential energy, not the difference in potential. Regarding the second equation, the first part of the equation$$\Delta V=\frac{W}{Q} comes from the definition of potential difference, ...

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