What are the Conditions for a Massive and Massless Particle in Spacetime?

[Biest]

Homework Statement

The equations of motion are:
\frac{d^2 x^\mu}{d\tau ^2} = 0
Write down the tra jectory that’s the most general solution to them. What are the conditions
on your solution if the particle is:
(a) Massive?
(b) Massless?

The Attempt at a Solution

I am not too sure what is meant by write down, but judging from equation I can assume that the path is straight worldlinefor a massive particle in spacetime, since there is no four-force acting on it. What my main problem is the massless particle, as it is obvious that the acceleration is 0 from the fact that massless particles should always travel at c. Any idea what is meant by that. I am really sorry for asking, but the prof is the kind of person who states that everything is supposed to be from kindergarten.. string theorists don't you love them.

Cheers,

Biest

 

[StatusX]

The most general solution to the differential equation:

\frac{d^2 y}{dx^2} = 0

is:

y(x) = ax + b

So the solution to the equation you have is just something of the above form for every coordiante (ie, every value \mu can take). Yes, this is a straight line through spacetime.

This will give 2D arbitrary parameters (where D is the dimension of spacetime, which is probably 4, but you never know when a string theorist is teaching), specifying two vectors: an initial position and a velocity. The velocity must be timelike (ie, v^2 = - v_0^2 + v_1^2 + ... + v_D^2 < 0) for massive particles and lightlike ( v^2 = 0) for massless ones. This doesn't follow from the equation, but from the definition of mass:

m^2 = - p^2 \propto - v^2

where the proportionaliy constant is positive.

 

[Biest]

That is what i have figured. It would be something along the lines of

x^\mu = a^\mu \tau + b^\mu

Where a^\mu and b^\mu are the parameters, which we can not find cause of lack of information.

And D = 0 in our class

 

[Dick]

D=0? That's funny dimension for even a string theorist to be working in. In zero dimensions, no one can hear you scream. :)

 

[Biest]

D=0? That's funny dimension for even a string theorist to be working in. In zero dimensions, no one can hear you scream. :)

ups sry i really should wear my glasses, did not see what, well i thought it was the 0, 1, 2, 3 thing.

Even when i read his replies to my emails no one can hear me scream...

 

[Biest]

The most general solution to the differential equation:

\frac{d^2 y}{dx^2} = 0

is:

y(x) = ax + b

So the solution to the equation you have is just something of the above form for every coordiante (ie, every value \mu can take). Yes, this is a straight line through spacetime.

This will give 2D arbitrary parameters (where D is the dimension of spacetime, which is probably 4, but you never know when a string theorist is teaching), specifying two vectors: an initial position and a velocity. The velocity must be timelike (ie, v^2 = - v_0^2 + v_1^2 + ... + v_D^2 < 0) for massive particles and lightlike ( v^2 = 0) for massless ones. This doesn't follow from the equation, but from the definition of mass:

m^2 = - p^2 \propto - v^2

where the proportionaliy constant is positive.

I was just working on the problem of the massless particle again, if it is moving at c, shouldn't it be a null like dependence rather then a timelike?

 

[Dick]

I was just working on the problem of the massless particle again, if it is moving at c, shouldn't it be a null like dependence rather then a timelike?

Sure. StatusX just said 'lightlike' instead of null. It's the same thing.