# Why E=mc^2 is different from E=1/2 mv^2?

• I
• sush
Einstein's equation for total energy ##E = \gamma mc^2## takes into account the energy due to motion as well as the rest energy of an object. The difference between the two equations is that ##\frac{1}{2}mv^2## only considers the energy due to motion, while Einstein's equation accounts for both motion and rest energy. This is because, according to Einstein, mass is equivalent to energy, and even at rest, an object's mass can be converted to energy if it comes into contact with antimatter. The term ##\gamma## in Einstein's equation represents the relativistic factor, which takes into account the effects of high speeds on an object's energy. But for speeds that are much smaller than the

#### sush

Why kinetic energy is ½ m v2? Why it is different from Einstein’s equation for Energy E= m c2?

First of all, Einstein's equation is for the total energy, while ##(1/2)mv^2## is the part of the energy just due to motion. Even when something is at rest, its mass is equivalent to energy, and in fact could be converted completely to energy if it came into contact with an equal amount of antimatter.

I prefer the notation and the language that "mass" refers to the invariant "rest mass". So when something is moving, its total energy is written as ##E = \gamma mc^2## where ##\gamma## is the relativistic factor ##1/\sqrt{1-(v/c)^2}## which is 1 when ##v=0## and approaches infinity as v approaches the speed of light.

That means that according to Einstein, the kinetic energy is the difference between total energy when moving and total energy when not moving.
$$KE = \gamma mc^2 - mc^2 = (\gamma - 1)mc^2$$.
It turns out that when v is small, then ##(1/2)mv^2## is a very good approximation to that. I'll post a followup with that proof.

• sush
Here's the proof.

##\gamma = 1/\sqrt{1-(v/c)^2} = [1-(v/c)^2]^{-(1/2)}##.

It can be shown with calculus that when ##\epsilon## is a small number, which we write as ##\epsilon \ll 1## or "##\epsilon## is much less than 1" then a good approximation to ##(1 - \epsilon)^n## is ##1 - n \epsilon##.

We have ##\epsilon = (v/c)## and ##n = -0.5##. So when ##(v/c) \ll 1##, ##\gamma## is approximately ##1 - (-0.5)(v/c)^2## or ##1 + 0.5(v/c)^2##, to a very good approximation.

How good? Let's say you're moving at 1/10 of the speed of light, ##(v/c)^2 = 0.01##. Then the exact value of ##\gamma## is 1.0050387... while the approximation gives ##1 + 0.5*0.01 = 1.005##. And the smaller v is, the better the approximation.

So ##(\gamma - 1)## is approximately ##1 + 0.5(v/c)^2 - 1 = 0.5v^2/c^2## and

##KE = (\gamma - 1)mc^2 = (0.5v^2/c^2)* mc^2 = 0.5mv^2##. To "very good approximation", for low velocities.

• sush
sush said:
Why kinetic energy is ½ m v2?

It's not! ##\frac{1}{2}mv^2## is approximately equal to the kinetic energy, but that approximation is valid only for speeds that are a small fraction of the speed of light.

## Why is E=mc^2 different from E=1/2 mv^2?

There are a few key differences between these two equations. First, E=mc^2 is known as the mass-energy equivalence equation, while E=1/2 mv^2 is the kinetic energy equation. This means that E=mc^2 relates to the relationship between mass and energy, while E=1/2 mv^2 relates to the relationship between an object's mass, velocity, and kinetic energy.

## What does the "c" represent in E=mc^2?

The "c" in E=mc^2 represents the speed of light, which is approximately 3 x 10^8 meters per second. This is a constant value that is crucial in understanding the relationship between mass and energy.

## Why is E=mc^2 considered one of the most famous equations in science?

E=mc^2 is considered one of the most famous equations in science because of its revolutionary implications. It showed that mass and energy are interchangeable and that even a tiny amount of mass could produce a vast amount of energy. This discovery led to advancements in nuclear energy and our understanding of the universe.

## Can E=1/2 mv^2 be used to calculate the total energy of an object?

No, E=1/2 mv^2 only calculates the kinetic energy of an object. To calculate the total energy of an object, you would need to also consider its potential energy, which is determined by its position and is not included in this equation.

## How does E=1/2 mv^2 differ from Einstein's Theory of Relativity?

Einstein's Theory of Relativity goes beyond the relationship between mass and energy and explains how the laws of physics are the same for all observers in uniform motion. E=1/2 mv^2 is a simpler equation that only applies to objects in motion and does not take into account the effects of relativity.