# Tag Info

1

You have a couple of mistakes. First, if you say that $G = 6\times 10^{-11} k$, then $k$ should be $\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}$. If we instead define $k$ as you did, then it is a dimensionless number: the conversion factor between the two sets of units. You should rather have said that $6\times 10^{-11} / k$ is the value of $G$ in the ...

3

The short answer is that it can, if $M = 1 = M^{-1}$. In this way of looking at it, all quantities in Planck units are pure numbers. The longer answer is that there are two different ways of thinking about natural unit systems. Natural unit systems in terms of standard units One of them, and perhaps the easier one to understand, is that you're still ...

0

Perhaps you are confused between dimension and unit. Note that $cm$ and $m$ are different units but have same dimension of length. See? It's simple. They have only different magnitudes. You have to understand that you cannot subtract or add 1 kg from 1 metre. Makes no sense, right? Suppose you want to know about speed. You know that it is $\frac ... 0 I'm not sure exactly what you are asking, so I will simply write some relevant facts that might answer your question. Dimensional analysis is a powerful tool for solving problems in physics. If we want the formula for a quantity$Q$, we can guess the formula for$Q$by first writing a product of all relevant dimensional quantities raised to unknown powers ... 3 Maxwell is being misunderstood. First, Maxwell makes very clear that length, time and mass are the fundamental types of units. Then he discusses a totally different convention that isn't used today, saying "in the astronomical system, the unit of mass is defined with respect to its attractive power". In other words, Maxwell is talking about a concept of ... 3 The explanation is Maxwell's text: If, as in the astronomical system, the unit of mass is defined with respect to its attractive power, the dimensions of$[M]$are$[L^3T^{-2}]$. To motivate this, it is perhaps useful to be aware of some of the different systems of units for electromagnetism used historically. One of the units of charge commonly in use ... 0 Consider two bodies,$m_{1}$and$m_{2}.$From Newton laws, we have: $$F=G\frac{m_{1}m_{2}}{s^{2}}\tag1$$ but $$F =m_{1}a,\tag2$$ and combining$(1)$and$(2)$two we obtain $$m_{1}a=G\frac{m_{1}m_{2}}{s^{2}}.\tag3$$ We also know that $$a=\frac{s}{t^{2}}.$$ Then,$(3)$becomes $$m_{1}\frac{s}{t^{2}}=G\frac{m_{1}m_{2}}{s^{2}},$$ which, after a little bit ... 2 I assume you're thinking about Minkowski space, i.e. the metric$\eta_{\mu\nu}=\text{diag}(c^2,-1,-1,-1)$. You should be aware that the dot notation is purely a notational shorthand, and has no other information contained in it. In particular, by definition we have $$\dot{A}\equiv\partial_0A=\frac{1}{c}\frac{\partial A}{\partial t}$$ Thus, there is no ... 1 Radius is usually measured in [m], but for rotational movement it's unit is different to length namely [m/rad]. Hence the unit for torque is [Nm/rad]. Torque times angle will come out as energy. I do not know why radians are omitted, causing confusion for the understanding population. 2 The diffusion equation takes the form $$\frac{\partial f}{\partial t}=D\frac{\partial^2 f}{\partial x^2}$$ where$f$is some function. For simplicity, let's let$f=\rho$the mass-density. In this case, then we have, units-wise, $$\frac{{\rm kg/m^3}}{\rm s} = \left[D\right]\frac{\rm kg/m^3}{\rm m^2}$$ Thus, in order to have the correct units on both ... 5 We have observed that the underlying level of nature is quantum mechanical. Quantum means "a definite quantity" of something so definite quantities can be counted and so integral numbers play a role : a) in the number of particles , in the number of energy levels characterized by quantum numbers ( i.e. integer numbers).b) There are the fields which are ... 0 Generally speaking, particles are quantized: you have$0,1,2,3,4\dots $of them. Even in quantum mechanics, you may not know how many you have, but if you measure it you will get a natural number. Most other quantities are real numbers and can take any real value. It is very hard to say why-some would say that is the definition of a particle. 1 When doing surface integrals like you say, you would always normalise by$R^2$. So, if you give a ray's direction by the spherical co-ordinates$\theta,\,\phi$, and you want the solid angle subtended by a bundle of such rays (e.g. a light field) converging on some point, then it is by definition the area of the part of the sphere "pierced" by the bundle. ... 0 John Rennie's answer seems fine to me (+1). I'll only add the relevant pieces of the BIPM brochure (PDF, p. 118). BIPM rules. 6 The solid angle is defined as the area on the unit sphere subtended by the angle divided by one unit area. It's a ratio so it's a single dimensionless number. I see why you think it should be a 2D quantity, because the surface of a sphere, and any patch on it, is a 2D manifold and you need two quantities (traditionally$\theta$and$\phi\$) to map it. When ...

Top 50 recent answers are included