# Tag Info

84

The answers are no and no. Being dimensionless or having the same dimension is a necessary condition for quantities to be "compatible", it is not a sufficient one. What one is trying to avoid is called category error. There is analogous situation in computer programming: one wishes to avoid putting values of some data type into places reserved for a ...

83

It doesn't matter where the equation came from - a fit to experimental data or a deep string theoretic construction - or who made the equation - Albert Einstein or your next-door neighbour - if the dimensions don't agree on the left- and right-hand sides, it's nonsense. Consider e.g. my new theory that the mass of an electron equals the speed of light. It's ...

79

Physics is independent of our choice of units And for something like a length plus a time, there is no way to uniquely specify a result that does not depend on the units you choose for the length or for the time. Any measurable quantity belongs to some set $\mathcal{M}$. Often, this measurable quantity comes with some notion of "addition" or "concatenation"...

77

Whenever I think about this problem I go back to one of Joel Spolsky's articles, "Making Wrong Code Look Wrong", which talks about Hungarian notation. Not only the useless kind of Hungarian notation, where variables are named in a way that describes their types (f_pos for a float, d_pos for a double, etc.) - this is "Systems Hungarian" in the article - but ...

67

first of all, the question you are asking is very important and you may master it completely. Dimensionful constants are those that have units - like $c, \hbar, G$, or even $k_{\rm Boltzmann}$ or $\epsilon_0$ in SI. The units - such as meter; kilogram; second; Ampere; kelvin - have been chosen partially arbitrarily. They're results of random cultural ...

33

A standard argument to deny possibility of inserting dimensionful quantities into transcendental functions is the following expression for Taylor expansion of e.g. $\exp(\cdot)$: $$e^x = \sum_n \frac{x^{n}}{n!} = 1 + x +\frac{x^2}{2} + \dots\,.\tag1$$ Here we'd add quantities with different dimensions, which you have already accepted makes no sense. OTOH,...

33

It depends what you mean by "unit". If you mean something like "seconds", then no. Counterexample: 1 minute = 60 seconds has different units on both sides, but they're both representing a duration, so they can still be equal. If you mean something like "time", then yes. An equation means two things are equal, i.e. the same. For that to be true, they have ...

29

All right, I'm cashing in my comments to provide an answer: Let's start with an example that doesn't invoke dimensions, units, or physics at all. How do we evaulate the following expression? $$1 + \begin{pmatrix}5\\2\\-9\end{pmatrix} + \begin{pmatrix}a\ b\\c\ d\end{pmatrix}$$ The answer is, we don't. Not without defining some special convention, like ...

28


15

Your friend's question is perceptive but not at odds with your earlier answer. When you compare the length of something with a unit (1 meter), the ratio is indeed a unitless number. But then all numbers (1.5, $\pi$, 42) are unitless. When you want to determine speed you divide displacement by time - each of which has units. But what you enter into you ...

15

No. All equations have the same dimension on both sides. Dimensions are mass, distance, time, speed, acceleration, force, power, electric current, electric charge etc. As long as you work with symbolic relations, you only care about dimensions. The equation $$v = \frac{s}{t}$$ (velocity = distance / time) works with any units as long as they are units for ...

14

The infinitesimal length interval between two events in spacetime $ds$ is defined by $$ds^2=c^2 dt^2 - dx^2 - dy^2 - dz^2$$ The creature is dimensionally consistent, because time is multiplied with a speed. You can think of $(t,x,y,z)$ as the four coordinates of spacetime $(x^0,x^1,x^2,x^3)$ and $c$ appears naturally in the equations. However, the usual ...

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