How Do You Calculate and Sketch the Light Cone in a 2D Space-Time Geometry?

[Imuck4u]

Homework Statement

We are given a 2 dimensional space time line element and we want to calculate the light cone at a point (x,y)

Homework Equations

ds^2=x(dy)^2-2(dy)(dx)

The Attempt at a Solution

For a light cone, ds^2=0 so x(dy)^2-2(dy)(dx)=0 now what?

 

[strangerep]

Apparently you have:
$$x dy^2 - 2dy dx ~=~ 0 ~.$$Since $dy\ne 0$ in general, you can divide throughout to get
$$x dy - 2 dx ~=~ 0$$Then you just need to rearrange the equation so you have only $y$ stuff on side, and only $x$ stuff on the other.
Finally, integrate both sides.

 

[Imuck4u]

So, dy=2dx/x, I integrate and get y=2 log(x)+c is that right?. I saw a solution to the problem given as dy/dx=0 and dy/dx=1/2x which I don't really get. Can someone explain?

 

[strangerep]

So, dy=2dx/x, I integrate and get y=2 log(x)+c is that right?

Looks reasonable to me, assuming you wrote the original problem correctly.

I saw a solution to the problem given as dy/dx=0 and dy/dx=1/2x which I don't really get. Can someone explain?

Regarding the 1st option, I was sloppy in my earlier advice. I should have told you to re-express the original equation as
$$dy(x dy - 2 dx) ~=~ 0 ~.$$From this, we must have either $dy = 0$, or what I said before. The case $dy = 0$, when integrated, is $y = const$. I.e., it's equivalent to $dy/dx=0$. I should not have blithely discarded the $dy=0$ possibility.

Regarding your $dy/dx=1/2x$, are you sure you've transcribed that correctly? If it was $dy/dx=2/x$, that would be equivalent to what I wrote earlier.

 

[Imuck4u]

I saw it in the solutions manual. It might just be a typo but it got me really confused. By the way, thanks a lot!

 

[strangerep]

I saw it in the solutions manual. [...]

What manual? Is it online? If so, can you post a link? (I'd like to see the full context.)

 

[Imuck4u]

http://web.physics.ucsb.edu/~phys131/HW4_soln.pdf

7.5

 

[strangerep]

http://web.physics.ucsb.edu/~phys131/HW4_soln.pdf
7.5

Hmm, not a very informative "solution". Can you ask a TA about how they get that factor of 2 in the denominator of dv/dx? (I'm curious now.)

[Edit: Ha! I was right. It's in the Errata List for the solutions manual. (It's always a good idea to check the author/publisher website to get most recent errata lists...)]

The "solution" to part(c) is also excessively cryptic. To see it better, one needs the full solution, i.e., $$v = 2\ln x + C$$and then remember the logarithm is undefined for real $x \le 0$. I don't know how you're supposed to see that just by looking at those infinitesimal lightcones from part(b).

 

[Imuck4u]

nice find! Glad it was wrong! By the way, how do you sketch those infinitesimal lightcones?

 

[strangerep]

[...] how do you sketch those infinitesimal lightcones?

Pick a value of v. Then think about the 2 possible slopes dy/dx of each lightline for various values of x. There's 2 lightlines through each point (v,x)...

To figure which parts are the "interior" of the light cone, take infinitesimal vectors along pairs of lightline directions anchored at the vertex where the the light rays intersect. Form their vector sum, and calculate its magnitude $ds^2$. Hartle seems to be using a (-,+,+,+) metric convention, so the lightcone interior is where $ds^2 < 0$.