# Is energy inversely proportional to mass?

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## Summary:

In some equations, it seems that energy is proportional to mass, and in others, it seems like energy is inversely proportional to mass. Why?

## Main Question or Discussion Point

Hello. I'm a graduate student in electrical engineering, and I'm taking a class in semiconductor physics. My professor has used this equation for the energy of an electron in a semiconductor:
$$E=\frac{\hbar^2 k^2}{2m^*}$$
This seems to imply that energy is inversely proportional to mass, i.e., a heavier particle has less energy. However, we all know the following equation for kinetic energy from high school physics:
$$E=\frac{1}{2}mv^2$$
That equation actually indicates the opposite: energy is proportional to mass, and a heavier particle has more energy. Moreover, it seems like these two equations are derived from each other, using ##p=mv## and ##p=\hbar k##.
$$E=\frac{1}{2}mv^2=\frac{mv^2}{2}=\frac{m^2v^2}{2m}=\frac{p^2}{2m}=\frac{\hbar^2 k^2}{2m}$$
This is somewhat confusing, that these two equations are derived from each other, and yet they seem to be saying opposite things. So which one is true? Do heavier particles have more energy, or less energy? Or are both statements true, in different situations?

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That depends, if two particles that have different mass move at the same velocity, the more massive will have more kinetic energy. But if instead of moving at the same velocity, both particles have the same momentum (i.e, the most massive moves slower) then is the lightest particle who will have more kinetical energy.

PeroK
BvU
Homework Helper
2019 Award
they seem to be saying opposite things
but they don't: with ##\ p\propto m\ ## as you already indicate, order is restored and ##\ E\propto m\ ##.

And even in a much broader sense: the famous EInstein ## \ E = mc^2 ## ! But that's for later ...

ZapperZ
Staff Emeritus
Summary: In some equations, it seems that energy is proportional to mass, and in others, it seems like energy is inversely proportional to mass. Why?

Hello. I'm a graduate student in electrical engineering, and I'm taking a class in semiconductor physics. My professor has used this equation for the energy of an electron in a semiconductor:
$$E=\frac{\hbar^2 k^2}{2m^*}$$
This seems to imply that energy is inversely proportional to mass, i.e., a heavier particle has less energy. However, we all know the following equation for kinetic energy from high school physics:
$$E=\frac{1}{2}mv^2$$
That equation actually indicates the opposite: energy is proportional to mass, and a heavier particle has more energy. Moreover, it seems like these two equations are derived from each other, using ##p=mv## and ##p=\hbar k##.
$$E=\frac{1}{2}mv^2=\frac{mv^2}{2}=\frac{m^2v^2}{2m}=\frac{p^2}{2m}=\frac{\hbar^2 k^2}{2m}$$
This is somewhat confusing, that these two equations are derived from each other, and yet they seem to be saying opposite things. So which one is true? Do heavier particles have more energy, or less energy? Or are both statements true, in different situations?
The problem here is that you are not focusing on what is being kept CONSTANT. And I'm surprised this is propping up now, because I went over something like this in my General Physics mechanics topic.

We know that

p = mv

and that

K = 1/2 mv2 = p2/2m.

If you have 2 masses, where m1 > m2, but you have it so that p is a constant for both masses (meaning that they also have different velocity), then obviously

p1 = p2

However, one can already tell that the kinetic energy for them is not identical, i.e.

K1 < K2.

The one with the greater mass, i.e. m1, may have the same momentum, but it has a smaller kinetic energy.

This is no different than the situation you brought up. It is just that you did not realize that in your question, you are implicitly making the momentum, i.e. p = ħk to be a constant.

Zz.

BvU and PeroK