# E=mc^2, mass deficit

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#1
To introduce the topic of mass deficit in physics, my teacher said that when he jumps he gains potential energy. So that energy, E, divided by c^2 gives him his increase in mass.

I am confused by this because surely the energy he ‘gains’ in gravitational potential comes from stored chemical energy in our body, so the overall energy of him stays the same. If anything work is done against the ground when we jump so we lose energy. Is my teacher right that we gain mass when we jump?
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5 months ago
#2
(Original post by Maximus 190)
I am confused by this
Me too. Possibly, your teacher is also confused, but maybe someone with a better understanding can enlighten us.
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#3
(Original post by Hallouminatus)
Me too. Possibly, your teacher is also confused, but maybe someone with a better understanding can enlighten us.
Perhaps he was thinking that if we magically teleported upwards instead of jumping, then we would gain gravitational potential energy so our mass would increase?
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5 months ago
#4
(Original post by Maximus 190)
To introduce the topic of mass deficit in physics, my teacher said that when he jumps he gains potential energy. So that energy, E, divided by c^2 gives him his increase in mass.

I am confused by this because surely the energy he ‘gains’ in gravitational potential comes from stored chemical energy in our body, so the overall energy of him stays the same. If anything work is done against the ground when we jump so we lose energy. Is my teacher right that we gain mass when we jump?
The only way I can imagine this being true is the fact that when you jump your kinetic increases so your mass will momentarily be higher because of that but then you are right. If you take into account conservation laws then there hasn't really been an overall increase, if anything probably a decrease by jumping. I think your teacher is pretty confused. It do be like that some times.
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#5
(Original post by blackugo)
The only way I can imagine this being true is the fact that when you jump your kinetic increases so your mass will momentarily be higher because of that but then you are right. If you take into account conservation laws then there hasn't really been an overall increase, if anything probably a decrease by jumping. I think your teacher is pretty confused. It do be like that some times.
Okay, at least I wasn’t being dumb. My understanding of the meaning of E = mc^2 is still unclear though. Does it mean mass can turn into energy and vice versa. Or does it mean that if I have a bit of energy, it also has a mass. And if I have a bit of mass then it must ‘hold’ some sort of energy (potential, kinetic etc.)?
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5 months ago
#6
(Original post by Maximus 190)
Okay, at least I wasn’t being dumb. My understanding of the meaning of E = mc^2 is still unclear though. Does it mean mass can turn into energy and vice versa. Or does it mean that if I have a bit of energy, it also has a mass. And if I have a bit of mass then it must ‘hold’ some sort of energy (potential, kinetic etc.)?
Mass can turn into energy but technically mass is energy. The equation shows that mass and energy are in some sense equivalent. There are different interpretations but I just think of mass as a store of energy. Mass can turn into other forms of energy like kinetic. If you have a stationary particle, not in any field you would describe it as having no potential or kinetic energy but it still having mass.
Last edited by blackugo; 5 months ago
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5 months ago
#7
It’s been many years since I have studied A level physics & i do not have a physics degree but:

E = mc^2 refers to sub-atomic particles, its quantum mechanics which does not translate to Newtonian/classical mechanics like a human jumping in the air.

Additionally E = mc^2 is an approximation of E^2 = (mc^2)^2 + (pc)^2 ; as you also have to account for the particles momentum.
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5 months ago
#8
Yh just to add to that the rest mass is independent of momentum. So in a rest frame, you don't need to use E² = (m0c²)² + (pc)²

m0 is rest mass
When velocity is low then momentum or velocity will be close to 0 so that term is small. As p = m0v / sqrt√( 1 - (v²/c²) ). For low values of v momentum is low and when momentum is 0 it reduces to E=mc²

(Original post by mnot)
It’s been many years since I have studied A level physics & i do not have a physics degree but:

E = mc^2 refers to sub-atomic particles, its quantum mechanics which does not translate to Newtonian/classical mechanics like a human jumping in the air.

Additionally E = mc^2 is an approximation of E^2 = (mc^2)^2 + (pc)^2 ; as you also have to account for the particles momentum.
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