B Work/Kinetic Energy/E=mc^2

1. Dec 11, 2017

learning_physica

Hey Everyone!

I hope all of y’all are having a great day so far! I take AP Physics 1 and I was just wondering (not homework related at all) why isn’t work also equal to Einstein’s famous equation? I know that that W=change in kinetic energy, but why can’t that also equal mc^2? I would really appreciate a response.... this is my first time being here!

From,

Learning_physica

2. Dec 11, 2017

Staff: Mentor

If you want to find the mass of a system, you have to take its total energy at rest (and vice versa). All types of energy inside the system contribute.

3. Dec 11, 2017

learning_physica

Could you please explain that a little further if you don’t mind?

4. Dec 11, 2017

Ibix

If you do work on a system and increase its energy (e.g. heat it up) its mass does increase because E has increased so m must also. The effect is so tiny (1J gives around 10-17kg) that you can usually ignore it.

5. Dec 11, 2017

Orodruin

Staff Emeritus
Because of the title, I think it is worth noting that purely boosting the entire system does not change its mass. However, giving different oarts of the system kinetic energy while maintaining the overall momentum constant (in other words, heating - as stated in #4) will increase the mass.

6. Dec 11, 2017

Ibix

@learning_physica - you will find sources (mostly older ones, and pop sci sources that focus on "cool" over "helpful") disagreeing with this, saying that "relativistic mass" increases with velocity. Relativistic mass is a concept that has largely been dropped because it's enormously confusing. Both Orodruin and I are using mass in the modern sense, meaning what is sometimes called "rest mass" or "invariant mass".

Last edited: Dec 11, 2017
7. Dec 11, 2017

jbriggs444

In relativity, total energy (rest energy plus kinetic energy) for a moving object is given by $$E=mc^2\ \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
[Thank you, @Orodruin for the correction]

Where m is the object's mass. In modern treatments the term "mass" is used to refer to an objects's invariant mass, also known as its rest mass.

If you turn this formula into an Taylor series in v2, the first two terms are: $mc^2$ and $\frac{1}{2}mv^2$

The object's rest energy is $mc^2$.
The object's kinetic energy is $\frac{1}{2}mv^2$ plus other trailing terms that are negligible as long as v is small compared to the speed of light.

Last edited: Dec 11, 2017
8. Dec 11, 2017

Orodruin

Staff Emeritus
Correction
$$E=mc^2\ \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

9. Dec 11, 2017

Mister T

Because $mc^2$ is not equal to the change in kinetic energy.

Do you want $mc^2$ to equal the rest energy, which is the true expression of the Einstein mass-energy equivalence, or do you want it to equal the total energy? The former is the more acceptable choice, but the latter is also prevalent in some older books and in some popular science books. Either way, though, $mc^2$ is not equal to the change in kinetic energy.