Physics Friday 110

Today’s post is based on an important detail that came up in a physics problem I was discussing with a friend of mind, and which we both initially missed. Here, I will modify the situation from an earlier post to display this.

Given two masses, m1 and m2, connected to each other by a spring of spring constant k2, and with each mass connected to a fixed wall by a spring with spring constant k1, so that at (static) equilibrium position, the springs are all at their resting lengths.

So far, this differs from the problem here only by the fact that the masses now are not the same. With the horizontal displacements of the masses being, as before, x1 and x2, we have differential equations:

obtained via F=ma and Hooke’s Law.
Now, suppose that one of the masses, say m2, is decaying or evaporating, so that it obeys , so that it has mass m0 at t=0, and decays toward mf, with τ the time constant for this decay. So, to solve this situation, we just plug in this into the second differential equation above, to get a linear ordinary differential equation with variable (instead of constant) coefficients, right?


This is where the key point comes in. Our differential equations were made using , but this assumes that the mass is constant. We might be tempted to use the form, apply the product rule to take the derivative of the momentum to get , but this would also be incorrect. As Plastino and Muzzio note, this equation violates Galilean invariance.
Specifically, we have to consider the momentum of the mass leaving the system. If the center of mass of the body has velocity v, the escaping mass has velocity u relative to the center of mass, and F is the net external force, then we have
Note, that if the mass is lost isotropically (relative to the evaporating body), then the
total effect of the term(s) will be zero, and we do get
, and our original direct substitution would work. However, if the mass loss is not isotropic, then we will have another term in our differential equation.


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