Timeline for Representing dimensions in Dirac delta function results
Current License: CC BY-SA 3.0
17 events
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| Jun 14, 2017 at 20:43 | comment | added | user8153 | @Nat Here is what I had in mind: Let's say you have $\delta(x)$, where $x$ is a length. Now I could choose a length unit A (say, one meter), and write $\delta(x) = (1/A) \delta(x/A)$. I could also choose another length unit B (say, one lightyear) and get $\delta(x) = (1/B) \delta(x/B)$. To convert between these two expressions I need a constant scale factor: $\delta(x/B) = (B/A) \delta(x/a)$. | |
| Jun 14, 2017 at 18:58 | comment | added | Nat | @user8153 I'm having trouble seeing how shifting units in that equation Ariana provided would require the unit proportionality factor to maintain correctness. Could you write Ariana's equation, but with values in place of the variables, to help me see the issue? | |
| Jun 14, 2017 at 18:09 | comment | added | user8153 | @Nat Ariana's comment contradicts the statement in your response that it would be a mistake to insert a unitless conversion factor if one changes the scale of the variable, for example by going from meters to lightyears. In fact, that is precisely what one should do. | |
| Jun 14, 2017 at 17:32 | comment | added | Nat | @Noiralef But it doesn't have units of $x^{-1}$; you can treat it as a function without resorting to that mistake. If it helps, you can say that the integration results in the inverse units rather than simply $1$. | |
| Jun 14, 2017 at 17:28 | comment | added | Noiralef | @Nat What you are basically saying is that $\delta(x)$ does not have a meaning on its own, only $\int \bullet \delta(x) dx$ which is unitless. But it is often helpful to treat $\delta(x)$ as if it were a function, and then it has units of $x^{-1}$. Also, there's no reason why the identity Ariana mentioned should only hold for scalar $\alpha$. | |
| Jun 14, 2017 at 17:27 | comment | added | Nat | @ArianaGrande If you read the fine print on that identity, it only applies when $\alpha$'s a non-zero scalar, i.e. unitless. | |
| Jun 14, 2017 at 17:24 | history | edited | Nat | CC BY-SA 3.0 |
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| Jun 14, 2017 at 17:23 | comment | added | Ariana | @Nat Saying the dirac delta function does not have units fail to the identity $\delta(ax)=\frac{1}{|a|}\delta(x)$ for $a\neq 0$ Furthermore, another defination of the the dirac delta function is the fourier transform of 1(units). Which also imples the delta function has the inverse of the units of its input | |
| Jun 14, 2017 at 17:16 | history | edited | Nat | CC BY-SA 3.0 |
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| Jun 14, 2017 at 17:09 | comment | added | Nat | @BySymmetry Thanks for pointing out the issue. Does the update fix it? | |
| Jun 14, 2017 at 17:03 | history | edited | Nat | CC BY-SA 3.0 |
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| Jun 14, 2017 at 16:54 | comment | added | Nat | @BySymmetry Ah, I get the confusion - I'll update the answer. | |
| Jun 14, 2017 at 16:54 | history | undeleted | Nat | ||
| Jun 14, 2017 at 16:47 | history | deleted | Nat | via Vote | |
| Jun 14, 2017 at 16:46 | comment | added | By Symmetry | The Dirac delta does indeed have units of $x^{-1}$. This can clearly be seen from the defining relation of the delta function $\int dx\; \delta(x) f(x) = f(0)$. Since $dx$ has units of $x$, $\delta(x)$ must have units of $x^{-1}$ | |
| Jun 14, 2017 at 16:45 | comment | added | Nat | Could downvoters explain? | |
| Jun 14, 2017 at 16:42 | history | answered | Nat | CC BY-SA 3.0 |