Khushidebo
When the life time of a meson is given and we are asked to find the accuracy if its rest mass, we solve it with the following assumptions:-

1. Life time is taken as the accuracy in time. Why? Please explain in detail.

2. Accuracy in energy is taken as the accuracy in rest mass. Why? Please explain in detail.
rpenner
Perhaps you should state the whole problem and show the steps you are succeeding at.

Average lifetime is the "shortest time scale on which we will be able to notice changes" in state.
Mass, momentum and energy for a relativistic free particle are related by:
E² − ( m c² )² = ( p c )²
So if the uncertainty in p is relatively low, all the uncertainty in E must be equal to the uncertainty of m.

QUOTE (Matthew Donald+)
Let H be the (time-independent) Hamiltonian of some non-relativistic system. Let ψ be a wavefunction and let A be some other observable. ...
2 (ΔA) (ΔH) ≥ |<[H,A]>|  = | <ψ, [H,A] ψ> | = |  iℏ d<A>/dt |
(ΔT) (ΔE) = (ΔA / | d<A>/dt | ) (ΔH) ≥ ℏ/2
Khushidebo
The problem is the simple one: "A RECENTLY DISCOVERED MESON HAS A LIFE TIME OF 10^(-22) SECONDS. TO WHAT ACCURACY CAN ITS REST MASS BE KNOWN?"

I didn't have this idea about the average lifetime as the "shortest time scale on which we will be able to notice changes", that's why was confused about why we are taking it as one.

The relation between accuracy in energy and rest mass depends on how small is the accuracy in momentum.
CAN ANY SUCH INFORMATION BE GATHERED FROM THIS QUESTION?

I have another question: WHILE WRITING ABOUT OR ASKING QUESTIONS ABOUT UNCERTAINTY PRINCIPLE, BOOKS EITHER USE UNCERTAINTY OR ACCURACY. BUT THESE ARE TWO OPPOSITE TERMS. THOUGH STRICTLY WE TALK OF THE STANDARD DEVIATION WHICH IS BASICALLY UNCERTAINTY, WHY DO WE TALK ABOUT ACCURACY?
rpenner
The measure of accuracy can be listed in the same units as the quantity, in which case the measure accuracy is the smallness of the amount of uncertainty.

I promised a score of 730 plus or minus 1.5.

Or you can talk about relative uncertainty if the central figure is different from zero.
1.5/730 = 0.2%

Or you can talk about significant digits

- log (base 10) 1.5/730 = 1.7 so 730 has about 2 significant figures.

But its like a ruler that measures in two directions at once.
High uncertainty = Low accuracy (or precision)
High accuracy (and precision) = low uncertainty
Khushidebo
Yes, that is all right.
But then why do we use the uncertainty principle for both since they are two opposite things... is it because the uncertainty principle is based on multiplication?
Pleas explain in every detail.
rpenner
The uncertainty principle is a lower bound based on a multiplication of variances (uncertainties) in observables. As in my first post, detailed calculation shows that these lower bounds always apply due to the relationship between the quantum operators.
Khushidebo
So, since this is a multiplicative formula, we will have the same result for both accuracy and uncertainty.
This is it. Isn't it?
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