# Why is null geodesic in Eddington-Finkelstein coordinate system a $45^\circ$ straight line?

In the book Relativity, Gravitation and Cosmology by Ta-Pei Cheng, page 106-7, the Eddington-Finkelstein coordinate system is described as a coordinate system set up using a photon falling radially towards the gravitational source at the origin as the observer.

It was then said that the null geodesic $$ds^2=0$$ in the Eddington-Finkelstein (EF) coordinate system is a $$45^\circ$$ straight line. Why is that so? The author later on derived this result in a circular way. Is it possible to prove this statement by only using physical arguments?

Edit: The author "derived" this result in the following cicular way:

• Start with the fact that the null geodesic is a $$45^\circ$$ straight line in the EF coordinate system $$(\bar{t},r)$$. Thus $$c\bar{t}=-r+\text{constant}$$ is the worldline for an infalling photon.

• Compare this with the worldline of an infalling photon in Schwarzschild coordinates $$(t,r) :$$ $$ct+r^*\ln|r-r^*|=-r +\text{constant}$$. ($$r^*$$ is the Schwarzschild radius)

• Thus to get EF coordinates from Schwarzchild coordinates, one just make the coordinate transformation of $$ct\rightarrow c\bar{t}=ct+r^*\ln|r-r^*|$$.

• Use this coordinate transformation on the Schwarschild line element $$ds^2$$ to get the EF line element.

• Using the EF line element, the result $$c\bar{t}=-r +\text{constant}$$ is verified to be true for an infalling photon.

• Probably best not to link to pirated textbooks (if that is one) if that's where that link goes. Nov 19, 2020 at 15:16
• Could you please clarify on what you mean by "in a circular way"? Nov 20, 2020 at 2:32
• @NiharKarve I had edited my question. Nov 20, 2020 at 7:22

It is actually the converse. While trying to find null geodesics in Schwarzschild coordinate, you will end up getting

$$\pm ct=r+r^*ln|r^*-r|+C$$ $$\ \ \ \ \ \ \ \ \ \ \$$(+ for outgoing and - for ingoing)

So, In the Ingoing EF coordinates,

$$ct^{'}=ct+r^*ln|r^*-r|$$

and since the ingoing null geodesic is given by

$$ct=-r-r^*ln|r^*-r|+C$$

you apparently get straight 45$$^{\circ{}}$$ lines in the Ingoing EF coordinates. But the outgoing null geodesics form weird lines in the same coordinate.

In the Outgoing EF Coordinates, the opposite happens, Outgoing null geodesics form straight 45$$^{\circ{}}$$ lines and Ingoing null geodesics go crazy.