Why is E = p²/2m but not E = pv/2?

[TheCelt]

Homework Statement

Relate energy equation to the momentum equation for particles

Relevant Equations

$$\lambda = h/p$$
$$p = mv$$
$$E=1/2mv^2$$

I am trying to relate these equations to get the energy with respect to momentum based on particle wavelengths.

I did the following:

$$\lambda = h/p$$ so $$p= h/ \lambda $$

Then

$$p=mv$$ and $$E=1/2mv^2$$

So

$$E = pv/2 $$

But the answer was:

$$E = p^2/2m $$

I don't understand how they got this answer or why mine is wrong ?

 

[jbriggs444]

$$E=1/2mv^2$$

That equation is for the kinetic energy of a massive object at non-relativistic speeds.

It seems that you are trying to use the momentum of a massless particle (e.g. a burst of light) to compute the energy of that burst.

Let us look at the correct formulation for the relativistic kinetic energy first and see where the $E=\frac{1}{2}mv^2$ comes from...

You are, of course, familiar with $E=mc^2$. That is the formula for the energy of a massive particle at rest.

When the particle is moving, the total energy (rest energy plus kinetic energy) is given by $\gamma mc^2$ where $\gamma$ is the factor: $\frac{1}{\sqrt{1-v^2/c^2}}$

If you subtract out the rest energy, what is left is the kinetic energy. This is given by $(\gamma - 1)mc^2$. If you work through taking the derivatives of this with respect to v and turn it into a Taylor series, the first terms in the series will cancel out to zero and the first non-zero term will turn out to be $\frac{1}{2}mv^2$.

But as I started out saying, we are not dealing with massive particles. We are dealing with massless particles moving at the speed of light. The relativistic $\gamma$ goes infinite. We cannot use the low speed approximation ($E=\frac{1}{2}mv^2$).

A proper way to proceed is with the energy-momentum equation: $$E^2 = m^2c^4 + p^2c^2$$Or:$$m^2c^4 = E^2 - p^2c^2$$In the latter form, this is a statement that "Invariant mass is the magnitude of the Energy-Momentum 4-vector".

 

[TheCelt]

The question involved non relativistic particles. Your answer is very different to the answer they give.

 

[hutchphd]

Are you requested to use the wavelength to do the problem? If you do it using "waves" you will need to use the group velocity.

 

[TheCelt]

I am unfamiliar with what group velocity is ?

 

[George Jones]

Then

$$p=mv$$ and $$E=1/2mv^2$$

Solve the first equation for $v$, and substitute the result into the second equation.

 

[hutchphd]

I am unfamiliar with what group velocity is ?

Then you probably shouldn't use quantum mechanical methods to prove the question. You should manipulate the classicle definitions as mentioned by @George Jones. Given mass momentum and KE you should be able to express each as a combination of the other two with facility.