Physics: I am asking for help to derive a general expression for the total amount of energy lost as a function of time from a radiating object.

I'll simplify my problem like this: Say for example that an object's energy $E$ and velocity $v$ are related by:

$$ E = m v, $$

where $m$ is a constant. The object loses energy at a rate that is proportional to its velocity:

$$ \frac{dE}{dt} = - k v, $$

where $k$ is another constant. Physically, this means that the object's energy will decrease, but will do so at an ever decreasing rate as the velocity decreases along with the falling energy.

If the original ($t=0$) velocity was $v_0$ (or equally, original energy was $E_0$), then can one express the total amount of energy $\Delta E$ radiated (lost) up to an arbitrary time, $t$?

Thanks for any help... this question is similar to that asked here, but I struggled to understand it in terms of my particular problem.

  • $\begingroup$ If $E(t)\propto -E'(t)\propto v$, then $\dfrac{E'}{E}=\dfrac{d}{dt}(\ln E)$ is some (negative) constant. That's readily integrated. $\endgroup$ – Semiclassical Oct 22 '14 at 14:56
  • $\begingroup$ I was hoping for a run through. Have you jumped a few intermediate steps (that may make the final answer more digestible)? $\endgroup$ – Ben Oct 22 '14 at 16:06
  • $\begingroup$ There's really not much to add: write out $E'/E=\frac{d}{dt}\ln E$, show that this ratio is just a constant, and then integrate both sides. From that, you can find $E(t)$ since you know $E(0)=E_0=mv_0$. Then it's just a matter of writing out $\Delta E=E_0-E(t)$ explicitly. $\endgroup$ – Semiclassical Oct 22 '14 at 16:09
  • $\begingroup$ I end up with the following: $[\ln{E(t)} - \ln{E(0)}] = -\frac{k}{m}[t-0]$. Is this ok? $\endgroup$ – Ben Oct 22 '14 at 16:17
  • $\begingroup$ Yep! Now you just need to exponentiate both sides to get $E(t)$. $\endgroup$ – Semiclassical Oct 22 '14 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.