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I What is Einstein's equation used for?

  1. Mar 18, 2017 #1
    I've been doing some research into Einstein's equation E2=(mc2)2 +(pc)2 but apart from in nucleur reactions, where you can use the simpler E=mc2 as momentum=0, I have been unable to find any applications.

    Thank you in advance
     
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  3. Mar 18, 2017 #2

    PeroK

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    It depends what you mean by an application. Beyond the atomic level rest mass is generally conserved. That said, the total of all the nuclear fusion in the Sun amounts to a huge loss of rest mass per second.
     
  4. Mar 18, 2017 #3
    Thank you for your response.
    Obviously it can be used to calculate the energy for a change of mass taking into account momentum, but I was hoping to find applications like in nuclear fusion where the equation can be demonstrated if that makes sense.
     
  5. Mar 18, 2017 #4

    PeroK

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    The particle experiments at CERN, for example, are continual demonstrations of special relativity, including the mass/energy relationship and the definition of relativistic KE and momentum.
     
  6. Mar 18, 2017 #5
    It can be used to calculate binding energies in atoms. There was a recent thread on this...

    https://www.physicsforums.com/threads/calculating-the-weight-of-1-atomic-mass-unit.905925/
     
  7. Mar 18, 2017 #6

    robphy

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    https://www.google.com/search?q=pet+scan+photon+energy

    https://en.wikipedia.org/wiki/Positron_emission_tomography
     
  8. Mar 18, 2017 #7
    Of course it's hard to find practical aplication because the formula makes sense under relativistic conditions only. But there are a lot of theoretical applications, e.g. for the thrust of a relativistic rocket engine.

    Just to be more precise: It’s the emission of particles and radiation that results in the loss of mass. The fusion itself has no effect on the mass of the Sun.
     
  9. Mar 18, 2017 #8

    jtbell

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    Particle physicists use that equation when they need to compute the mass, energy, or momentum of a particle from the other two quantitles.

    Example: accelerate an electron (mc2 = 511 keV) from rest through a potential difference of 1 million volts (1000 kV). What is its momentum? (which determines e.g. the radius of curvature of its path when it travels through a given magnetic field)

    Solution: From ##\Delta K = q \Delta V##, the kinetic energy is 1000 keV. The total energy (kinetic plus rest energy) is E = 1000 keV + 511 keV = 1511 keV. Therefore, the momentum (in energy units) is ##pc = \sqrt {E^2 - (mc^2)^2} = \sqrt {(1511~\rm{keV})^2 - (511~\rm{keV})^2} = 1422~\rm{keV}##.
     
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