Understanding Rates of Decay in Simple Terms

[paulo84]

In simple terms, rate of decay is kinda like:

dN/dt = -λN

Has anyone tried finding the rate of the rate, or the rate of the rate of the rate, etc, to make sure that the rate of the rate of decay (etc.) isn't changing??

 

[mjc123]

If dN/dt = -λN, then the rate of decay will decrease as the sample decays and N decreases. So the rate of the rate of decay is changing.
If you express the law as N = N₀e^-λt, I assume you are aware that if you differentiate an exponential you get an exponential, and this continues however many times you differentiate, so you never arrive at an nth derivative that is constant.

 

[lekh2003]

Finding the rate of decay is known as the derivative of the decay. The second derivative would be the rate of the rate. You can continue finding all of the rates of decay. Not sure if the rate of decay is changing...

Another example of this derivate business is in kinematics. Velocity is the first derivative, acceleration is the second, jerk is the third, which are followed by jounce, snap, crackle and pop. Quite interesting and equally as complicated.

Edit: @mjc123 beat me to it.

 

[paulo84]

Finding the rate of decay is known as the derivative of the decay. The second derivative would be the rate of the rate. You can continue finding all of the rates of decay. Not sure if the rate of decay is changing...

Another example of this derivate business is in kinematics. Velocity is the first derivative, acceleration is the second, jerk is the third, which are followed by jounce, snap, crackle and pop. Quite interesting and equally as complicated.

Edit: @mjc123 beat me to it.

Cheers, glad we've found a common interest! :)

 

[lekh2003]

Cheers, glad we've found a common interest! :)

Yup, derivatives and rates of rates seem so cheerful. "Jerk, jounce, snap, crackle, and pop! The higher derivatives of kinematics!" It is literally like an advertisement.

 

[Vanadium 50]

rate of decay is kinda like

"Kinda" is not a word.

What you wrote down is an equation for a constant rate of decay. One can write different equations for a non-constant rate of decay. But for the equation you wrote, the rate of decay is necessarily constant.

 

[DrStupid]

What you wrote down is an equation for a constant rate of decay.

The equation for a constant rate of decay would be dN/dt = -λ. That's not what he wrote.

 

[paulo84]

It occurred to me this may have relevance for carbon dating.

 

[fresh_42]

It occurred to me this may have relevance for carbon dating.

Yes. ${}^{14}C$ has a half life of 5730 years. Since living organisms eat, they have a regular intake of all kinds of carbon isotopes. This of course ends with their death. So if we count how many ${}^{14}C$ isotopes are left in them, compared to the average in the environment, we know how long it's been that they didn't have a normal intake of ${}^{14}C$.

https://en.wikipedia.org/wiki/Radiocarbon_dating#Principles

$dN/dt = -λN \quad \longleftrightarrow \quad N(t)= N_0 \cdot e^{- \lambda t}$

 

[mfb]

By measuring the decay rate (and decays are the only thing you can directly measure), you can also calculate all higher derivatives. It doesn’t matter, however. As long as the first derivative follows the expected shape you know all higher derivatives do so as well.

 

[DrStupid]

By measuring the decay rate (and decays are the only thing you can directly measure), you can also calculate all higher derivatives.

Until you get white noise.

 

[mfb]

Sure, no measurement is perfect and the discrete nature of the decays makes some things impossible to determine with reasonable sample sizes, but that is just a limit of our measurement devices.

 

[Vanadium 50]

The equation for a constant rate of decay would be dN/dt = -λ.

No, it's -λN. The number decaying is proportional to the number present.

 

[mfb]

It depends on what you call "rate of decay".
Constant decay rate of the material, or a constant number of decays per second.

 

[Vanadium 50]

Yes, but dN/dt has to be proportional to -N, no matter what you call it.

 

[mfb]

For radioactive decay: Yes.

For general situations where we could interpret "rate of decay" as decrease in N: not necessarily. Remove 1 atom per second out of a trap: dN/dt = -λ

 

[DrStupid]

No, it's -λN.

That's not constant.

 

[Vanadium 50]

You're going to keep riding this hobby horse, aren't you? The OP wrote down a perfectly correct differential equation for a decay occurring at a constant rate - i.e. a fixed probability per element per time. Forgive me for having the temerity to believe he meant what he wrote.

 

[DrStupid]

The OP wrote down a perfectly correct differential equation for a decay occurring at a constant rate

The integration of the differential rate equation

\dot N = - \lambda \cdot N

results in

N = N_0 \cdot \exp \left( { - \lambda \cdot t} \right)

and therefore in the rate

\dot N = - \lambda \cdot N_0 \cdot \exp \left( { - \lambda \cdot t} \right)

which is obviously not constant.

 

[mfb]

You two are using "constant" for two different things that can be constant. Of course the formulas depend on what you keep constant. Can you stop this side-discussion please or make your own thread for it?