# Acceleration Potential and momentum of photon.

1. Aug 22, 2010

### Smarty7

We had a test a week ago, it asked for Acceleration Potential of an photon to achieve wavelenth of 0.005 picometer.
Not so surprisingly both of our teachers don't have a clue about what it means. While according to their solution we had to find the kinetic energy.

One more query, you have for photon
p=h/$$\lambda$$
i know how you get here with Einstein's equation E^2=(pc)^2+m^2c^4
but again when you put p=mv in momentum of photon it gives 0=h/$$\lambda$$ (since m=0; mv=0) So i would like to understand what would be momentum, because for a photon again to the eistein's equation both terms are indeed (pc)^2 (since p=mc).

2. Aug 22, 2010

### vanhees71

A photon is defined as a one-particle state in the Fock space of the electromagnetic field. A possible basis is the momentum-helicity basis $$|\vec{p},\lambda \rangle$$, where $$\vec{p} \in \mathbb{R}^3$$ and $$\lambda \in \pm 1$$.

This state is an energy-eigenstate of the electromagnetic field with eigenvalue $$E=|\vec{p}| c$$.

The flaw in your thinking about energy and momentum comes from a misconception of momentum. For classical point particles (a photon is very far from being a classical point particle by the way), energy and momentum are related to the three-velocity with respect to an inertial reference frame by

$$E=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}},$$
$$\vec{p} = \frac{m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}.$$

From this it follows the relation between energy and momentum,

$$E=c \sqrt{(m c)^2+\vec{p}^2},$$

and the three-velocity in terms of energy and momentum is obviosly given by

$$\vec{v}=c^2 \frac{\vec{p}}{E}.$$

Here, $$m$$ is the invariant mass, and the equations are written for a massive particle. The limit $$m \rightarrow 0$$ is most easily understood starting from energy and momentum. The relation between energy and momentum simply becomes as written above for the photon-energy-helicity-Fock eigenstates

$$E=p c$$

and the three-velocity of a massless particle is given by

$$\vec{v}=c^2 \vec{p}{E} \; \Rightarrow \; |\vec{v}|=c,$$

i.e., independent of momentum a massless particle always runs with the speed of light, and thus the limit $$m \rightarrow 0$$ from the formulae connecting energy and momentum with the three-velocity have to be taken by fixing the energy since the Lorentz factor, $$\gamma=1/\sqrt{1-\vec{v}^2/c^2}$$ diverges with $$m \rightarrow 0$$.

3. Sep 1, 2010

### Smarty7

I am really sorry for the late reply, but i was only trying to understand what you had already written. Yet, i don't think i understand much of the key concepts you mentioned in here. It will take time for me to study the modern physics and then start Quantum Mechanics

4. Sep 1, 2010

### Staff: Mentor

Are you sure the question was about a photon and not a proton?

You can't "accelerate" photons in an electric field because (a) they don't carry charge so they're not affected by an external electric field; (b) they always travel at speed c anyway.

On the other hand, protons have charge so an electric field can accelerate them, and their speed can be anything between 0 and c.

5. Sep 1, 2010

### Smarty7

We had the word Photon in the question in our test paper, perhaps it is wrong because a similar problem one of the helpbooks had was about calculation Acceleration Potential of a proton. Thus, it can be proton.
Also, I am getting the idea of what Acceleration Potential really is, so it is the measure of electric field that can give a proton of acceleration so as to have the specified wavelength. Right? How would i arrive on the units newtons per Coulomb then.

Is Acceleration Potential called by any other name, when i first searched for it on google. Turned out to give no meaningful results. only some unclear Wikipedia mentions.

Last edited: Sep 1, 2010