You can calculate the additional mass yourself and then consider whether it may be measurable.RobbyQ said:TL;DR Summary: Mass Energy and a compressed spring
Reference: https://www.physicsforums.com/forums/high-energy-nuclear-particle-physics.65/post-thread
If I take a spring with clamps and I weight that system accurately. Then I compress the spring and clamp it thus giving it potential energy. If I now weigh the clamped spring I should see an increase in mass because of the added energy. Is this the case and something that could be proved in the lab ?
Yes understandable PeroK. Hence why I wondered if this could be measured even. But from a mass/energy equivalence is it conceptually correct i.e if I were to convert both the uncompressed and compressed systems into pure energy would I observe the added PE difference?PeroK said:PS even tiny variations in the surface gravity over time may be more significant.
There's no such thing as "pure energy". Energy is a property of a system.RobbyQ said:Yes understandable PeroK. Hence why I wondered if this could be measured even. But from a mass/energy equivalence is it conceptually correct i.e if I were to convert both the uncompressed and compressed systems into pure energy would I observe the added PE difference?
And the mass deficit is the binding energy? So I cannot think of the spring system in this way?PeroK said:There's no such thing as "pure energy". Energy is a property of a system.
The proof of "mass-energy equivalence" is that the rest mass of the Hydrogen atom is less than the rest mass of the proton plus the rest mass of the electron. This has been measured to a very high degree.
The formation of a hydrogen atom involves loss of PE. Hence, a reduction in rest mass.RobbyQ said:And the mass deficit is the binding energy?
Yes, but it's too small to measure relative to the overall mass of the system in a macroscopic environment.RobbyQ said:So I cannot think of the spring system in this way?
What experimental setup do we use to measure PE loss in Hydrogen atom formation. Or, in fact, the PE needed to separate the electron and proton until they reach their rest mass.PeroK said:The formation of a hydrogen atom involves loss of PE. Hence, a reduction in rest mass.
Yes, but it's too small to measure relative to the overall mass of the system in a macroscopic environment.
If you jump off the ground, then technically the centre of mass of the Earth moves a tiny amount in the opposite direction. But, that's too small to measure, especially given that lots of other things are happening all the time on the Earth.
Google is your friend, as they say.RobbyQ said:What experimental setup do we use to measure PE loss in Hydrogen atom formation. Or, in fact, the PE needed to separate the electron and proton until they reach their rest mass.
Is it an implementation of the Einstein's Box (which I understand was a thought experiment)?
It's true that you have more energy available in a compressed spring. Try using a compressed spring to launch a toy car.RobbyQ said:But from a mass/energy equivalence is it conceptually correct i.e if I were to convert both the uncompressed and compressed systems into pure energy would I observe the added PE difference?
According to Einstein's theory of relativity, mass and energy are equivalent as expressed by the equation \(E=mc^2\). When a spring is compressed, the potential energy stored in the spring increases. This increase in energy results in a slight increase in the mass of the spring, although this change is typically too small to be measured with ordinary instruments.
Energy is stored in a compressed spring as elastic potential energy. When a spring is compressed or stretched from its equilibrium position, work is done on the spring, and this work is stored as potential energy. The amount of stored energy can be calculated using Hooke's Law, which states that the energy \(E\) is proportional to the square of the displacement \(x\) from the equilibrium position: \(E = \frac{1}{2} k x^2\), where \(k\) is the spring constant.
Yes, compressing a spring does increase its mass, but the increase is extremely small. The additional mass comes from the energy stored in the spring as elastic potential energy. According to the mass-energy equivalence principle, this energy contributes to the total mass of the system. However, this change is so minuscule that it is practically negligible for everyday purposes.
Yes, the energy stored in a compressed spring can be converted to other forms of energy. When the spring is released, the stored elastic potential energy can be transformed into kinetic energy as the spring returns to its equilibrium position. This kinetic energy can then be transferred to other objects, causing motion, or converted into other forms of energy such as thermal energy due to friction.
The spring constant \(k\) is a measure of the stiffness of the spring. A higher spring constant means the spring is stiffer and requires more force to compress or stretch it. The energy stored in a compressed spring is directly proportional to the spring constant. Specifically, the elastic potential energy \(E\) stored in the spring is given by \(E = \frac{1}{2} k x^2\). Therefore, for a given displacement \(x\), a stiffer spring (higher \(k\)) will store more energy than a less stiff spring (lower \(k\)).