# Klein-Gordon for a massless particle

1. May 20, 2012

### jabers

So I'm trying to find a solution of the Klein-Gordon equation for a massless particle. I reached the Klein-Gordon from the total energy-momentum equation. Then for a massless particle i get to this equation:
$${ \partial^2 \psi \over \partial t^2 } = c^2 \nabla^2 \psi$$How do I solve for psi? I was thinking about trying the Frobenius method, but I'm not sure how to do that. Any help would be appreciated.

Also how do I make my typed latex display in the latex format on this forum?

Last edited by a moderator: May 20, 2012
2. May 20, 2012

### vela

Staff Emeritus
3. May 20, 2012

### jabers

so for a one dimensional wave equation the equation is:

$${ \partial^2 \psi \over \partial t^2 } = { c^2 \nabla^2 \psi} = c^2 {\partial^2 \psi \over \partial x^2}$$

Right?

I don't see how I could use separation of variables. If it were something like this:

$${ \partial x \over \partial t } = { x }$$

Then I could say:

$${\frac{1}{x} \partial x } = {\partial t}$$

And then integrate but I dont see what you mean. What do you mean?

The only thing i could see doing is this:

$${\partial^2 \psi \partial^2 x} = c^2 {\partial^2 \psi \partial^2 t}$$

And integrating twice? But that doesnt really make sense to me.
I didnt think you could separate variables to solve a differential equation when there were double derivatives in the equation.

4. May 20, 2012

### vela

Staff Emeritus
5. May 20, 2012

### jabers

Cool, thank you.

6. May 22, 2012

### Morgoth

this equation looks pretty much like the wave-equation, for a wave moving with a speed v=c.
So I guess you can think for solutions like cos or sin, or better exponential.

If for example you say:

Ψ(r,t)= ei(-kr+ωt) which is a spherical wave, you can see how this thing behaves in your equation.