# I What is Einstein's equation used for?

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1. Mar 18, 2017

### Ashley Warnes

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.

2. Mar 18, 2017

### PeroK

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.

3. Mar 18, 2017

### Ashley Warnes

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.

4. Mar 18, 2017

### PeroK

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.

5. Mar 18, 2017

### Comeback City

It can be used to calculate binding energies in atoms. There was a recent thread on this...

6. Mar 18, 2017

### robphy

https://en.wikipedia.org/wiki/Positron_emission_tomography

7. Mar 18, 2017

### DrStupid

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.

8. Mar 18, 2017

### Staff: Mentor

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 $pc = \sqrt {E^2 - (mc^2)^2} = \sqrt {(1511~\rm{keV})^2 - (511~\rm{keV})^2} = 1422~\rm{keV}$.