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When adding or subtracting quantities I am told to add absolute uncertainties while for multiply and dividing quantities, we add percentage uncertainties. I am trying to gain intuition for why this is the case.

Consider 2 measurements. $3 \pm 0.1$ and $2 \pm 0.5$. ( Units ignored ). To me $3 \pm 0.1$ represents that the measurement can be anywhere in the range of $2.9$ to $3.1$. Similarly $2 \pm 0.5$ means the measurement can be anywhere from $1.5$ to $2.5$. When taking the upper bounds of these two measurements and lower bounds and adding them together, I found that this new quantity could be anywhere within the range $4.4$ to $5.6$. Hence adding absolute uncertainties to form $5 \pm0.6$ seems completely intuitive.

Now I tried to the same thing for multiplication. I found the new quantity to be in the range of $4.35$ to $7.75$. However, when calculating the percentage uncertainty of my initial measurements and adding them together, before converting back to absolute uncertainty, I got an answer of $6 \pm 1.7$.

It seems that adding percentage uncertainty only gives an approximate representation but a very close one. $7.75$ isn't in the range $6 \pm 1.7$. I struggle to understand the mathematics behind why adding percentage uncertainty gives such a close approximation. How did someone come up with adding percentage uncertainties when multiplying and dividing quantities?

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  • $\begingroup$ Even for addition or subtraction I would not necessarily say adding absolute uncertainties is correct. Depending on what you want the uncertainty to mean. If it means the 2sigma or some n*sigma spread in the measurement, you'd have to use the root sum square formula to get the corresponding 2sigma of the computed value $\endgroup$
    – RC_23
    Commented May 23, 2023 at 3:26

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This is more of an Algebraic problem, but since it relates to measurements I suppose it does have some bearing on Physics.

Let's generalize your problem a bit, let's call the two measured values $m_1$ and $m_2$, which correspond to real (exact) values $v_1$ and $v_2$. Respectively, denote the two measurement uncertainties to be $\Delta{x_1}$ and $\Delta{x_2}$, and we assume the uncertainties are already with the appropriate sign ($+$ or $-$).

We now express the sum of the real values via the measured values and their uncertainties:

\begin{align*} v_1+v_2 &= m_1+\Delta{x_1}+m_2+\Delta{x_2} \\ &= m_1+m_2+\Delta{x_1}+\Delta{x_2} \end{align*}

It is evident that, as you have correctly concluded, the sum has an uncertainty of $\left|\Delta{x_1}\right|+\left|\Delta{x_2}\right|$

Now with the above definitions in place, finding the uncertainty in the product of the measured values becomes a straightforward exercise, similar to what we did before, we express the product of the real values in terms of the measured values and their uncertainties:

\begin{align*} v_1v_2 &= (m_1 + \Delta{x_1})(m_2 + \Delta{x_2}) \\ &= m_1m_2 + m_1\Delta{x_2} + m_2\Delta{x_1} + \Delta{x_1}\Delta{x_2} \tag{1}\end{align*}

So it is clear from the above that in order to find the uncertainty, we need to find the lower and upper bounds of the expression $m_1\Delta{x_2} + m_2\Delta{x_1} + \Delta{x_1}\Delta{x_2}$. Clearly, the upper bound is given by:

$$|m_1\Delta{x_2}| + |m_2\Delta{x_1}| + |\Delta{x_1}\Delta{x_2}|$$

However, the lower bound is a bit more tricky to calculate, since when $\Delta{x_1}$ and $\Delta{x_2}$ differ in sign, two terms in $(1)$ are negative (the last one, and either the second or the third one). But on the other hand, when they are both negative, the last term is positive and again, this time the preceding two terms are negative. However, assuming that the measurement errors are small we can usually neglect the positive contribution that occurs when both errors are negative, that is, assume that:

$$ |\Delta{x_1}\Delta{x_2}| \lt\lt |m_i\Delta{x_j}|$$

So we conclude that a good estimate for the uncertainty in the product is given by:

$$ v_1v_2 = m_1m_2 \pm \left( |m_1\Delta{x_2}| + |m_2\Delta{x_1}| + |\Delta{x_1}\Delta{x_2}| \right)$$

Where again we note that the last term's $|\Delta{x_1}\Delta{x_2}|$ contribution is negligible and hence also the error in our estimate of the lower bound.

Substituting the numbers in your question, we have $m_1=3$, $\Delta{x_1}=\pm 0.1$, $m_2=2$, $\Delta{x_2}= \pm 0.5$:

$$ v_1v_2 = 6 \pm \left( 1.5 + 0.2 + 0.05 \right) = 6 \pm 1.75 $$

So the real value of the product $v_1v_2$ is somewhere between $4.25$ and $7.75$.

Addressing your question about percentage uncertainty: percentage uncertainty is simply the relative uncertainty multiplied by $100$. The relative uncertainties in the above definitions will be: $\Delta{x_1}/m_1$ and $\Delta{x_2}/m_2$. From $(1)$ we can find a formula in terms of these quantities, that may shed light on why they're useful for estimating the product uncertainty:

\begin{align*} v_1v_2 &= (m_1 + \Delta{x_1})(m_2 + \Delta{x_2}) \\ &= m_1m_2 + m_1\Delta{x_2} + m_2\Delta{x_1} + \Delta{x_1}\Delta{x_2} \end{align*}

We subtract $m_1m_2$ from both sides and then divide by $m_1m_2$ to obtain:

\begin{align*} \frac{v_1v_2 - m_1m_2}{m_1m_2} &= \frac{\Delta{x_2}}{m_2}+\frac{\Delta{x_1}}{m_1}+\frac{\Delta{x_1}}{m_1}\frac{\Delta{x_2}}{m_2} \end{align*}

Now we denote $v_1v_2-m_1m_2$ as $\Delta{(m_1m_2)}$ because clearly this difference is nothing but the uncertainty in the product of the measured values. Hence, we have obtained the relation between the relative uncertainty of the product and the relative uncertainties of the individual values:

$$\frac{\Delta{(m_1m_2)}}{m_1m_2} = \frac{\Delta{x_1}}{m_1}+\frac{\Delta{x_2}}{m_2}+\frac{\Delta{x_1}}{m_1}\frac{\Delta{x_2}}{m_2} $$

From this, you can see that indeed, the relative uncertainty of a product is (approximately, if you neglect the last term) the sum of the relative uncertainties of each measured value, which I think is the main thing you were asking to clarify here :)

Indeed, if we now take percentages with the given values:

$100 \times 0.1/3 \approx 3.33 \%$ for the first measurement $m_1$, and $100 \times 0.5/2 = 25 \%$ for the second measurement $m_2$ their sum is $\approx 28.33\%$ which out of the product of measurements $6$ is indeed $\approx 1.7$ as you've mentioned in your question.

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\begin{align*} (x_1+\Delta x_1)\times (x_2+\Delta x_2) &= x_1x_2\left (1+\frac{\Delta x_1}{x_1}\right)\left (1+\frac{\Delta x_2}{x_2}\right ) \\ &= x_1x_2\left (1+\frac{\Delta x_1}{x_1}+\frac{\Delta x_2}{x_2}+\frac{\Delta x_1}{x_1}\frac{\Delta x_2}{x_2}\right ) \end{align*}

Thus, if you take the fractional error as being $\frac{\Delta x_1}{x_1}+\frac{\Delta x_2}{x_2}$ you are ignoring the second order term $\frac{\Delta x_1}{x_1}\frac{\Delta x_2}{x_2}$.

Note that you are probably overestimating the fraction error and a better estimate of the fractional error in $x_1x_2$ is $\sqrt{\left ( \frac{\Delta x_1}{x_1}\right)^2 + \left ( \frac{\Delta x_2}{x_2}\right)^2}$.

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