(Special relativity) Two masses connected by spring

AI Thread Summary
The discussion revolves around a problem from introductory special relativity involving two masses connected by a spring, where radiation emitted from one mass affects the system's oscillation. The participant attempts to analyze the recoil effects and derive the maximum extension of the spring using conservation of energy and momentum principles. Confusion arises regarding the treatment of mass changes and energy conservation, particularly at maximum extension. The conversation highlights the importance of simplifying assumptions, such as treating the recoil as simultaneous, and the realization that second-order effects may not need to be considered for this problem. Overall, the thread emphasizes the complexities of applying special relativity concepts to mechanical systems.
vielendank
Messages
3
Reaction score
1

Homework Statement


Hi all--humanist here. Now that the semester is over, I am taking the opportunity to (attempt to) self-study introductory SR. This is problem 12 in chapter 1 of Special Relativity by AP French.

(1-12) A body of mass m1+Δm is connected to a body of mass m2-Δm by a spring of spring constant k and negligible mass. The system is at rest on a frictionless table. A burst of radiation is emitted by the first body and absorbed by the second, changing the masses to m1 and m2 and setting the system into oscillations. If the time of transit of the radiation is negligibly small compared to the period of oscillation, show that the maximum extension of the spring is given by
$$x=c \Delta m ({\frac{m_1 + m_2}{m_1 m_2 k} })^{\frac{1}{2}}$$

Homework Equations


E=cp
E=mc2
And I guess he's assuming we'll use Hooke's law. (?) (Or else potential energy ##U= \frac{1}{2} kx^2##.)

The Attempt at a Solution


The problem seems to call for an analysis of the kind French used to treat "Einstein's box" earlier in the chapter. The radiation at the first mass causes a recoil pushing it to the left, and when the radiation is absorbed on the right, it imparts a recoil there as well.

I took the assumption about the time of transit to mean that, to a good approximation, we can assume that the recoil on the left and right happen "at the same time". (I may be wrong, though; it just seems like it would get heinously complicated without using this approximation.)

Given this, I tried to write equations for the positions x1 and x2 as functions of time, after the recoil, using the above equation to express the momentum imparted, as well as Hooke's law for the acceleration.
$$x_1(t)= \frac {-\frac{E}{c}} {m_1} t + \frac {k(x_2 - x_1 - X)} {2 m_1} t^2$$
$$x_2(t)= \frac {\frac{E}{c}} {m_2} t - \frac {k(x_2 - x_1 - X)} {2 m_2} t^2$$
where ##X## is the equilibrium position of the spring.

Then I took ##x_2 - x_1 \equiv x_{21}##, which led to an equation in the reduced mass system $$x_{21} (t)= \frac{Et}{c \mu} - \frac {k t^2}{2 \mu} (x_{21} (t) - X)$$

I did some other things, including changing variables to get rid of ##X##. Oh, and I took the derivative $$\dot x_{21} = \frac{E}{\mu c} - \frac{k}{2 \mu} (2 t x_{21} + t^2 \dot x_{21})$$
then used the fact that at maximum extension ##\dot x_{21}## is zero. This yielded $$x_{21} = \frac {E}{kct}$$

(My other thought was that at maximum extension, all the energy would be potential energy in the spring, which I could then equate with the energy "created" by ##\Delta m##, but this is not getting me the right result.)

And now I am realizing that if I take the period of oscillation ##T=2 \pi \sqrt{\frac{\mu}{k}}##, divide it by four (because it is presumably one-fourth of a period to get to maximum in SHM?), and substitute this value for ##t## in the above equation for ##x_{21}##, I get the right answer except a factor of ##\frac{2}{\pi}##! $$x_{21}=\frac{\Delta m c}{\sqrt{\mu k}} \frac {2}{\pi}$$

If you've managed to read this far and can put up with the confused thinking of a non-physicist I (congratulate you and) would be grateful for any guidance.
 
  • Like
Likes Merlin3189
Physics news on Phys.org
vielendank said:
The problem seems to call for an analysis of the kind French used to treat "Einstein's box" earlier in the chapter. The radiation at the first mass causes a recoil pushing it to the left, and when the radiation is absorbed on the right, it imparts a recoil there as well.

I took the assumption about the time of transit to mean that, to a good approximation, we can assume that the recoil on the left and right happen "at the same time".
Right.
vielendank said:
Given this, I tried to write equations for the positions x1 and x2 as functions of time, after the recoil, using the above equation to express the momentum imparted, as well as Hooke's law for the acceleration.
Where does that formula comes from? The t2 part would need a constant acceleration, which does not happen.

You can set up differential equations and solve them, but conservation of energy is so much easier.
 
  • Like
Likes vielendank
mfb said:
Where does that formula comes from? The t2 part would need a constant acceleration, which does not happen.
D: Yes, that was silly of me.

I think my difficulty is a more basic conceptual confusion, getting in the way of writing the conservation conditions. At the moment of maximum extension, there will be energy stored in the spring as well as the rest energy of the two masses, m1 and m2: ##\frac{1}{2} k x^2 + m_1 c^2 + m_2 c^2##. When they are at rest initially, they have rest energy ##m_1 + \Delta m## and ##m_2 - \Delta m##. But when you sum these, you will just get the rest energy of m1 and m2, because the ##\Delta m##s will cancel, and then when you equate these two expressions you get ##\frac{1}{2} k x^2 = 0##, which is wrong. Surely the (rest) masses must be less than ##m_1## and ##m_2## at maximum extension? Or perhaps my difficulty is whether ##m_1## and ##m_2## are supposed to be the masses including the initial kinetic energy imparted by the recoil? I just have little idea of what, physically, is supposed to be going on...

I might be beyond the pale, here. Thank you for your reply, though.
 
Well, to have exact energy conservation during the exchange, you have to consider the recoil of the masses - if one emits ##\Delta m c^2## of energy it will lose more as mass, and the other will gain less as mass as it recoils as well. Those are second-order effects, however, I don't think you have to take them into account. Consider the momentum exchange for objects at rest, then let the masses move according to their momentum and treat everything else nonrelativistically.
 
  • Like
Likes vielendank
mfb said:
Well, to have exact energy conservation during the exchange, you have to consider the recoil of the masses - if one emits ##\Delta m c^2## of energy it will lose more as mass, and the other will gain less as mass as it recoils as well. Those are second-order effects, however, I don't think you have to take them into account. Consider the momentum exchange for objects at rest, then let the masses move according to their momentum and treat everything else nonrelativistically.
Dear lord, I spent so much time trying to even conceive of how I would take into account those second order effects mathematically, and then the answer turned out to be (barely) two lines long... :eek:
Thank you for your help!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top