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The notion of dimension is well-defined for a vector space. In Newtonian physics, you define time as a scalar, so it doesn't make sense to give it a dimension. In general, though, time should be seen as the component of a vector (think about a time interval). It is neither a scalar nor a vector. You can talk about the dimension of the vector itself, but not ...


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It's because the circumference of a circle with radius $r$ is $2\pi r$. It's easier to see with velocities, so I'll stick to that for now. If you want to know the tangential velocity of a particle rotating around some point, you have to divide the distance it travels by the time it takes to do that. At $1\,\mathrm{rpm}$, the distance traveled in $1\,\mathrm{...


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$t$ contains both a unit AND a number. You can't use a unit to cancel both a number and a unit. You don't know what unit you are going to put into the variable ahead of time (seconds vs hours, meters vs km, etc). And even if you did know, you can't cancel only the unit part of $t$ while leaving the number part of $t$ in the equation. That would leave you ...


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You need to think in terms of quantities. It looks like you are using some equation for distance without acceleration: $$x-x_0=vt$$ where $x$ and $x_0$ are quantities called "position" (specifically at time $t$ and $0$) and will have units of distance. $v$ is a quantity called "velocity" and has unit of distance/time, and $t$ is a ...


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You cannot cancel it because the s in 4 m/s is a unit (seconds) and t is a variable (time), which is measured in seconds. So the units cancel, but not the numeric value. Assume $t = 2s$, then you get $4 m/s \cdot 2 s = 8 m$.


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Essentially you are right in your last statement. Let's define $t_{1s} \equiv 1\ {\rm s}$. Then your question amounts to asking whether this equation is true for any $t$: \begin{equation} \frac{t}{t_{1s}} \stackrel{?}{=}1 \end{equation} The answer is, no, this equation is not always true. It is only true if $t=1{\rm s}$.


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The regularization issue that you mention is exactly where the solution of the problem lies. Path integrals are always formally divergent. In order to get a physically meaningful result, it is always necessary to take a ratio of two path integral expressions. For the case of a thermal partition function, what you normally want is the ratio of the partition ...


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This is a stark demonstration of the mess inflicted by failure to nondimensionalize. If your quantities are dimensional, write their units next to them, in your case: $$ M = \begin{pmatrix} 1 + xy & ym \\ x/m & 1 \end{pmatrix}, $$ which acts on vectors $( am, b)^T$, to produce like-dimensioned vectors, for numerical x,y,a,b. Having ensured ...


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I suppose you consider states $|k>$ in the continuous spectrum, otherwise you would not use the integration over $k$ but a summation. If so, the normalization of your $|k>$ probably looks like that $<k_i|k_j>=\delta(k_i-k_j)$ which means $|k>$ is not exactly dimensionless for the reasons you mentioned (Dirac delta is $[k]^{-1}$). Actually, you ...


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Note: the paper says dimension 6 operators, not order 6 operators. However, I should warn you that they are using a somewhat lazy definition of dimension here. Consider a term in the Lagrangian (in $D$ spacetime dimensions) of the form \begin{equation} \mathcal{L} \sim \frac{1}{M^{N-D}} \mathcal{O}_N \end{equation} where $\mathcal{O}_N$ is an expression with ...


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$\int \delta(x) dx=1$ shows that $\delta(x)$ has units of (length)$^{-1}$. As $V\delta(x)$ is a potential energy, $V$ must have units of (energy)$\times$ (length). This is consistent with the first-order energy shift as $|\psi|^2$ is probability (dimensionless) per unit length.


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I have learnt that the dimensionless quantities have no unit. Whoever you learnt this from is incorrect . Clearly, dimensionless quantities CAN have units, as you have figured out in the case of angles. Another example of dimensionless quantity with units is the relative abundance of particles which has units of ppm(parts per million) , ppb(parts per ...


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With our current definitions of meter and second, $$c = 299, 792 ,458\frac{m}{s}.$$ Let's say we were to use some other unit of second, let's call it the "zepond" and denote it with $z$ instead of $s$. Let's say one zepond is $2$ seconds, so $$ z = 2 s. $$ Then the speed of light, measured in meters per zepond (instead of meters per second) would ...


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Your constant can be written as: $\beta= \alpha J^{4/4} m^{5/4} $ If you want to use a different constant without fractional units, you can simply: $$\beta' = \beta^4 = \alpha^4 J^4 m^5$$ Taking the "1.25th" root would still give a fractional value for joules. Note that the equation is already in SI units, it just looks a bit ugly to use noninteger ...


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The integral of distance with respect to time is known as absement. It is one of the family of derivatives and integrals of position, and can be integrated further to get absity, abseleration and abserk. Absement appears when considering situations where a quantity depends on both how far something has moved or extended and how long the movement is ...


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As far as I know, distance $s$ multiplied by time $t$ or (as the differential analogue) the integral $$^{(*)}: \int_{t_0}^{t}s(\tilde{t})\text{d}\tilde{t}\ $$ are not used in the context of physics. Nonetheless, this does not mean that it cannot interpreted physically. If an object travels uniformly along a straight line, $^{(*)}$ will obviously be larger ...


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The binding energy can be calculated from the work needed to take shells of matter away from the star to infinity, until it's all gone (technically the negative of that). If $\rho$ is the density, the radius of the star that's left is $r$ then the mass of the star that is left is $\frac{4}{3}\pi r^3\rho$, and the work needed to remove a shell of mass $4\pi r^...


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Temperature is not a measure of energy. It seems what you are doing is you are interpreting temperature as a kind of "energy per unit mass", but that is not the case - otherwise we'd just use that: energy per unit mass (joules per kilogram, say), and we would not need a separate unit (kelvin) for temperature. It's far from unrelated to energy, and ...


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Energy is not temperature. The lack of kelvins in the definition of the joule is not weird when you consider that every material requires a different amount of joules to heat up by 1 kelvin, and during a phase change, there is no change in temperature (kelvins) at all even as you add or remove energy (joules). You can add a bunch of joules to a material, and ...


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The kinetic energy (in Joules), for each atom in a gas, is $\frac{3kT}{2}$, where $k$ is Boltzmann's constant and $T$ is the temperature in Kelvin. The number of atoms in each gram depends on the atomic mass; for example, 12 g of carbon has Avogadro's number ($6.02\times 10^{23}$) atoms of carbon. The total heat energy is then the kinetic energy times the ...


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The decimal system is conveniant in conversion of magnitudes. In a positional system it is easy to multiply and divide in factors of ten and ten itself is easy enough to conceptualise. This fact is used in the SI system where there is a system of prefixes that indicate these scale factors. For example, $k$ for kilo and which means multiplying by a factor of ...


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