# Understanding the Relationship Between Momentum and Energy: A Classical Approach

• TheCelt
In summary, according to @George Jones, the equation for kinetic energy in relativistic mechanics is given by: $$E^2 = m^2c^4 + p^2c^2$$
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

$$E = pv/2$$

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

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

TheCelt said:
$$E=1/2mv^2$$
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".

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.

I am unfamiliar with what group velocity is ?

TheCelt said:
Then

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

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

TheCelt said:
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.

## 1. What is the relationship between momentum and energy?

The relationship between momentum and energy is that momentum is a measure of an object's motion, while energy is a measure of an object's ability to do work. In other words, momentum is the quantity that describes how much an object is moving, while energy is the quantity that describes how much work an object can do.

## 2. How is momentum related to kinetic energy?

Momentum and kinetic energy are related in that both are measures of an object's motion. Kinetic energy is a type of energy that an object possesses due to its motion, while momentum is a measure of the object's motion itself. The equation for kinetic energy (KE) is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity. The equation for momentum (p) is p = m * v, where m is the mass of the object and v is its velocity.

## 3. How does an object's mass affect its momentum and energy?

An object's mass has a direct relationship with its momentum and energy. The greater the mass of an object, the greater its momentum and energy will be. This is because both momentum and energy depend on an object's mass in their equations (p = m * v and KE = 1/2 * m * v^2).

## 4. Can momentum and energy be transferred between objects?

Yes, momentum and energy can be transferred between objects. When two objects collide, momentum can be transferred from one object to another. This is known as conservation of momentum. Similarly, energy can also be transferred between objects in a collision, but it may change forms (such as from kinetic energy to sound energy).

## 5. How is momentum conserved in a closed system?

In a closed system, the total momentum of all objects before and after a collision remains the same. This is known as the law of conservation of momentum. This means that if one object gains momentum, another object must lose the same amount of momentum in order to keep the total momentum of the system constant. This law applies to both elastic and inelastic collisions.

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