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I have studied that a dimensionally correct formula/equation may or may not be correct. But in order for a formula or an equation to be correct, it must be dimensionally correct, according to the principle of homogeneity. What about so many $x$-$t$, $v$-$t$, $a$-$t$, and $a$-$v$ relations then? For example, displacement as a function of time is given by x = 2t, or x = t² - t ?? Likewise, v = 3t, v = t², a = 1/x² ...... etc. Just a few examples.

In all of the above examples, from what I see, they are dimensionally incorrect. It's pretty obvious to figure it out, isn't it? My question is, why are they being used in my books then? They show up often in illustrations, or when I solve numerical problems.

EDIT : I see my question has been marked as a duplicate. I'm totally surprised. Because my question is different, from the one asked in the other thread. I know that the equation for working out the displacement in nth second is dimensionally correct, there's a hidden (1s) in three places on the right hand side. And no, we're not choosing this (1s) arbitrarily. It can be proved using the 2nd equation of motion, that there's a hidden (1s) in three places on the right hand side. I already know, that equation is dimensionally correct. What I've asked is completely different. I mean, the answers I got, were something like, there's a hidden 1 m/s², or 1 m/s. How do I know for sure that the hidden constant has the same dimensions as m/s² or m/s, or something else? Is there a proof? Because as I said, the equation for working out the displacement in nth second is dimensionally correct and can be proved using the 2nd equation of motion, that there's a hidden (1s) in three places on the right hand side. We're not just choosing this (1s) ''arbitrarily'' just to make the equation dimensionally correct.

4d_
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  • I'm sure my question is different, not a duplicate. I know that the equation for working out the displacement in nth second is dimensionally correct, there's a hidden (1s) in three places on the right hand side. And no, we're not choosing this (1s) arbitrarily. It can be proved using the 2nd equation of motion, that there's a hidden (1s) in three places on the right hand side. I already know, that equation is dimensionally correct. What I've asked is completely different. – 4d_ Oct 05 '18 at 17:32
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    PROPERLY done, physics doesn't contain "hidden" units. x does NOT equal 2t, and v does NOT equal 3t or $t^2$! These equations are dimensionally inconsistent, so they are definitely incorrect. And note - physics is NOT math. In a math class, you can multiply anything together with total disregard for units. In a physics class, your equations are modeling a real physical phenomenon, so the answer HAS to have a physical reality to it. – David White Oct 05 '18 at 18:07
  • I think the idea of dimensional analysis is applicable to physical equation and not to numerical equations. The function y=3t simply gives how numerical value of y coordinate change with time. In general it is not any physical equation describing the physical world . Now saying 1 with unit of m/s is hidden inside seems something odd. In physical equations nothing is hidden. Similarly the numerical equation which is used to find the numerical value of displacement traveled in Nth second is written as 4th equation and it is said 1 with dimensionof time is hidden at right places in the equation – Shinnaaan Jan 07 '22 at 07:15

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All of those equations seem to have incorrect dimension because the dimensions of the constants in them are not explicitly specified. For example, for $x(t)=2*t$, by taking the derivative you can check that $\frac d{dt}x(t)=v(t)=2$. So the 2 in the equation actually has dimension of velocity, and then you have that the dimension of x is:$$dim(x)=\frac {distance}{time}*time=distance$$ So the equation has the right dimensionality, the apparent problem comes from omitting the dimensions of the constant "2". And the same argument can be used for all the equations that you presented.

ADDED: I think you need to go back and try to understand what you mean by the equation $x(t)=2t$ or $x(t)=t^2$ in the first place. You can look at it as a simple function of one argument, with no physical meaning, and then that 2 and the 1 are just numbers. Or you can be trying to model some physical situation with the equation. If what you are trying to model with $x(t)=2t$ is the position of a particle as a function of time, then x has to have dimension of length, just as t must have dimension of time, and because of that 2 has to have dimension of velocity, because if not, the equation would be meaningless. The 2 in the equation does not show dimensions explicitly because it would just clutter notation, and it would be useless to cary it around when doing calculations.

Dimensional analysis is important, but is trivial in these simple self-explanatory cases, no information can be taken from it. The real importance of dimensional analysis comes when using fundamental constants, like $c$ or $h$, because then there might be unic, or at least a few, ways of combining these to get the units of some quantity you are trying to find.

Hugo V
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  • Thanks. I understand it slightly better now, but it still doesn't make complete sense to me. How do we know that the '2' in the equation v(t) = 2 has the dimensions of velocity? How do we know that for sure that it really does have the dimensions of velocity only? Can you please elaborate it a bit more? – 4d_ Oct 05 '18 at 16:57
  • For example, let's say x(t) = t², on differentiating it once, wrt time, we get v(t) = 2t. Apparently, here 2 has the dimensions of acceleration (as per what you said), because on differentiating it again wrt time, we get a(t) = 2. So 2 has the dimensions of acceleration. What about x(t) = t² then? There's no '2' in this equation. How do I know if it's dimensionally correct? – 4d_ Oct 05 '18 at 17:18
  • You'd say that there is a hidden "1" there, which would actually be " 1 m/s² ". This is done to simplify notation, and it's acceptable as long as you can always be sure that "x=2" means "x=2m". – FGSUZ Oct 05 '18 at 17:23
  • You can also see it as working with non-dimensional variables. Let $x,\ t,\ v,\ a$ be non-dimensional. That's the same as saying $x=position/(1m)$, $t=time/(1s)$ and so on. – FGSUZ Oct 05 '18 at 17:24
  • You probably didn't get my question. Yes, these's a hidden "1" in that equation, but how do I know for sure that it is 1 m/s² ?? Aren't we choosing the dimensions ( m/s² ) arbitrarily, just to make the equation dimensionally correct? In other words, we are 'making' this equation look dimensionally correct. From what I understand. Is there a concrete explanation that the hidden "1" does have the same dimensions as m/s² ?? That's what I'm trying to understand – 4d_ Oct 05 '18 at 17:29
  • Oh sorry, now I get you. OF course, units must be specified somehow. Maybe short sentences like "all in SI units" at the beginning of the page/book/chapter.... but it must be anywhere, otherwise you couldn't tell. – FGSUZ Oct 05 '18 at 17:43
  • I added some more information to try to make it clearer – Hugo V Oct 05 '18 at 18:01