Mass, Energy and a compressed spring

In summary: 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?It's true that you have more energy available in a compressed spring. Try using a compressed spring to launch a toy...you'll see an increase in energy.
  • #1
RobbyQ
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TL;DR Summary
Mass Energy and a compressed spring
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 ?
 
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  • #2
If the energy to compress that spring comes from outside the box, then yes. The difference is too small to measure even with the best scales.
 
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  • #3
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 ?
You can calculate the additional mass yourself and then consider whether it may be measurable.

You might also ask whether other processes might exceed the difference in mass through the springs PE. E.g. damage to the spring during compression ("wear and tear").

My guess is that for a macroscopic spring and clamp, the change is immeasurably small.
 
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PS even tiny variations in the surface gravity over time may be more significant.
 
  • #5
PeroK said:
PS even tiny variations in the surface gravity over time may be more significant.
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?
 
  • #6
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?
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.
 
  • #7
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.
And the mass deficit is the binding energy? So I cannot think of the spring system in this way?
 
  • #8
RobbyQ said:
And the mass deficit is the binding energy?
The formation of a hydrogen atom involves loss of PE. Hence, a reduction in rest mass.
RobbyQ said:
So I cannot think of the spring system in this way?
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.
 
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  • #9
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.
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)?
 
  • #10
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)?
Google is your friend, as they say.

Look up "measure the mass of a hydrogen atom".
 
  • #12
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?
It's true that you have more energy available in a compressed spring. Try using a compressed spring to launch a toy car.
 

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