Suppose a unknown quantity (whose dimension we know) depends on known three quantities like (Length of object L) , (Energy of Object E), (Density of Object D), when we try to get the relation among these four by dimensional analysis , we do like this : letting Unknown = $L^{alpha} * E^{beta} * D ^{gamma}$ we equate the dimensions to solve for alpha ,beta and gamma , but my question is why cant we consider situation where it might be like this : $Unknown(E,D,L)$=$ (E+aE^2 +bE^3...)^{beta} ....$ . Where a , b etc. are such that aE^2 has dimension E , similarily others . Is it not possible to have forms like this ? As such one can observe such type can be seen in mechanics problems where $v = at +bt^2$ etc type problems .
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I hope I'm understanding your question correctly.
If you add terms, you do not change the dimensionality of the result. For example, if I add 2 meters to 6 meters ($2 m + 6 m$) the result is still expressed in m (a length). However, if I multiply them, things change. For example, 2 meters by 6 meters ($2m \times6m$) is an area, not a length.
Now consider your example of $ (E+aE^2 +bE^3...)^{beta} ....$
If $E$ had the dimension of $m$, $E^2$ would be an area, $E^3$ would be a volume, etc. It is obvious those things cannot be summed.
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$\begingroup$ a and b has dimensions of E^-1 , E^-2 respectively , i meant hdhondt $\endgroup$ Commented Mar 14, 2022 at 11:52
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$\begingroup$ That still doesn't alter the fact that a factored dimension is a different thing. $1/m$ is not a length and $1/m^2$ is not an area. $\endgroup$– hdhondtCommented Mar 14, 2022 at 22:29
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$\begingroup$ I mean to say cant we have formulas like v= aT+bT^2 +cT^3 ? dimensional analysis will just tells its dimensional formula that is of form v=a'T but how exact like the above? $\endgroup$ Commented Mar 14, 2022 at 22:35
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$\begingroup$ a,b,c have dimensions such that overall dimension of each term matches v $\endgroup$ Commented Mar 14, 2022 at 22:36
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$\begingroup$ Not sure I understand... However, if "overall dimension of each term matches v", then you have something like $v=aT^2+bT^2+cT^2$, which reduces to $v=pT^2$ $\endgroup$– hdhondtCommented Mar 14, 2022 at 22:43