Geodesics in Schwarzschild metric

In summary, PeterDonis explains that the energy parameter ##E## in equations for the circular orbits in the Schwarzschild metric is energy at infinity, not energy measured by a hovering observer at rest at the orbital radius.
  • #1
Haorong Wu
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TL;DR Summary
How to understand the circular orbits in Schwarzschild metric for photons?
Hello, there. I am learning the chapter, The Schwarzschild Solution, in Spacetime and geometry by Caroll. I could not grasp the idea of circular orbits.

It starts from the equations for ##r##, $$\frac 1 2 (\frac {dr}{d \lambda})^2 +V(r) =\mathcal E$$ where $$V(r)=\frac {L^2}{2r^2}-\frac {GML^2}{r^3}$$ and $$ \mathcal E =\frac 1 2 E^2.$$ ##L## is the angular momentum and ##E## is the conserved energy.

By ##dV/dr=0##, it can be solved to obtain the circular orbits ##r_c=3GM##.

All above are from the textbook. Now I am trying to solve the variables for the circular orbits for photons.

First, in the circular orbits, ##dr/d\lambda## should be zero, so ##V(r)=\mathcal E##. After substituting ##r_c## into ##r##, I have $$\frac {L^2}{27 G^2 M^2}=E^2.$$ Next, for photons, ##L=r_c \cdot p## and ##E=pc##. But now, the equation becomes $$1=\frac {1}{3c^2}.$$ I am not sure what goes wrong.

My guess is that ##E## is not the energy for photons. I mean not the energy just related to the frequency of the photons. From its definition, $$E=(1-\frac {2GM}{r})\frac {dt}{d\lambda},$$ maybe ##\frac {dt}{d\lambda}## is the part from the frequency, while ##-\frac {2GM}{r}\frac {dt}{d\lambda}## is the gravitational potential energy. Should I subtract the potential energy part from the frequency part?

Thanks!
 
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  • #2
E is not the photon energy and the relation between energy and momentum components (really, components of the 4-momentum) is not the one you have quoted but one that depends on the Schwarzschild metric.
 
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  • #3
##E=pc## comes from the fact that ##\eta_{\mu\nu}p^\mu p^\nu=0## and applies in flat spacetime. The relation in general is ##g_{\mu\nu}p^\mu p^\nu=0##, but you don't necessarily interpret ##p^0## as the energy. In general the energy is ##g_{\mu\nu}p^\mu U^\nu## where ##U## is the four velocity of the measurement apparatus - which, in flat spacetime for an apparatus at rest, does lead to ##p^0=E##.
 
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  • #4
Hi, @Orodruin, @Ibix. I am still confused.

For a circular orbit at the equatorial plane, from ##g_{\mu\nu}p^\mu p^\nu=0##, I have $$-(1-\frac {2GM}{r}) (p^t)^2+r^2 (p^\phi)^2=0,$$so $$p^t=(\frac {r^2}{1-2GM/r})^{1/2} p^\phi.$$

A rest observer at radius r will measure the energy to be $$E=-g_{\mu\nu}p^\mu U^\nu=(1-\frac {2GM}{r})p^t=\sqrt{1-\frac {2GM}{r}} r p^\phi.$$

The radius of the circular orbit is ##r_c=3GM##, so ##E=L/\sqrt 3## where ##L=r_c p^\phi##. Therefore, I still could not get the equation $$\frac {L^2}{27 G^2 M^2}=E^2.$$
 
  • #5
Haorong Wu said:
Summary:: How to understand the circular orbits in Schwarzschild metric for photons?
I don't understand your approach. You are looking for the lightlike geodesics, and these can be calculated directly from the metric.
 
  • #6
Hi, @Nugatory. I am learning GR now. I will take a GR course next month. GR is so fascinating and yet difficult. I am learning it for the second time, and I still find that there are too many things I missed in my first time. Also, I find that it would be more illustrative if I do some calculations rather than just read the book. So I am trying to solve the ODEs given above to obtain ##r## and ##\phi## in Matlab to draw geodesics for light and see how they look like.

I start from the easy part, the circular orbit. But I cannot have the right answer with ##r## unaltered. So there must be some concepts I misunderstood.
 
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  • #7
Haorong Wu said:
A rest observer at radius r will measure the energy to be
Something that is not the same as the "energy" parameter ##E## of the orbit. The ##E## in your equations for the orbit is energy at infinity, not energy measured by a hovering observer at rest at the orbital radius.
 
  • #8
Thanks, @PeterDonis. But in the textbook, ##E## is defined by $$E=-K_\mu \frac {dx^\mu}{d\lambda}=(1-\frac {2GM} r) \frac {dr}{d\lambda}$$ where ##K_\mu=(-(1-2GM/r),0,0,0)## is the Killing vector corresponding to energy. It seems that ##E## is related to ##r##. I am not sure how to interpret it as energy at infinity.
 
  • #9
Haorong Wu said:
in the textbook, ##E## is defined by $$E=-K_\mu \frac {dx^\mu}{d\lambda}=(1-\frac {2GM} r) \frac {dr}{d\lambda}$$ where ##K_\mu=(-(1-2GM/r),0,0,0)## is the Killing vector corresponding to energy.
Yes, that's what "energy at infinity" means. See further comments below.

Haorong Wu said:
It seems that ##E## is related to ##r##.
Yes, because ##r## tells you how deep in the gravity well the orbit is, which is what determines the orbit's energy at infinity. The deeper in the gravity well the orbit is, the lower its energy at infinity is.

Haorong Wu said:
I am not sure how to interpret it as energy at infinity.
"Energy at infinity" is just a technical term for the conserved quantity corresponding to the timelike Killing vector field in a stationary spacetime. The physical interpretation is clearer for a massive particle than for a photon, since you can imagine bringing a massive particle to rest at infinity and seeing how much energy you either had to add to the particle or extract from it to do that. A photon can't be brought to rest anywhere, so that direct physical interpretation doesn't work for photons, but the basic underlying idea is the same: it's the energy relative to some "reference" state that is "at infinity" and therefore unaffected by the gravitational field of the massive body.
 
  • #10
Oh, thanks! @PeterDonis. I got the idea now.

Then ##E## could be measure by a rest observer at infinity, giving $$E=\lim_{r \rightarrow \infty} g_{\mu\nu} p^\mu U^\nu=p^t$$ with ##p^\mu=((\frac {r_c^2}{1-2GM/r_c})^{1/2} p^\phi,0,0,p^\phi)=(\sqrt 3 r_c p^\phi,0,0,p^\phi)##. If ##r_c p^\phi=L##, then ##V(r_c)=\mathcal E## is still not satisfied.

I guess I should not obtain ##E## in this method.
 
  • #11
Haorong Wu said:
Then ##E## could be measure by a rest observer at infinity
Not directly, because the photon itself is not at infinity. That's why your attempt here doesn't work.

Haorong Wu said:
I guess I should not obtain ##E## in this method.
That's right. The proper way to obtain ##E## is using the timelike Killing vector field, the way you quoted from your textbook before.
 
  • #12
To study orbits in Newtonian gravity you would use that KE+PE is a constant for a freefall orbit. That constant is the Newtonian approximation to ##E##, defined formally in GR in terms of the relationship of the orbit and the timelike KVF. That quantity is not the same as how much it hurts when the free falling object crashes into you, which is all about KE and in GR is formally the quantity ##g_{\mu\nu}p^\mu U^\nu## (although note that includes rest mass energy where that is applicable). Confusingly, both of these things are called "energy", but are only equal when the object is at infinity - which is why ##E## is often called the energy at infinity (even if that doesn't really make sense for a closed orbit that is never at infinity).
 
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  • #13
Thanks, @PeterDonis and @Ibix ! I get it now!

From the definitions, $$E=(1-\frac {2GM}{r}) \frac {dt}{d\lambda}$$ $$L=r^2 \frac {d\phi}{d\lambda}$$ and $$p^0=\frac {dt}{d\lambda}$$ $$p^\phi=\frac {d\phi}{d\lambda}$$ with possible factors being omitted, along with the normalization of 4-momentum, $$0=g_{\mu\nu} p^\mu p^\nu,$$ I have $$E=\sqrt{1-\frac {2GM} r} \frac L r$$ which give the correct answer for ##E## when ##r=r_c##. Cheers!

Wait...No...I got another problem now. Both ##E## and ##L## should be conserved quantities, but from this equation, they could not both be conserved. I must make another mistake. I got to think harder.
 
  • #14
Haorong Wu said:
I must make another mistake.
Yes, the normalization of 4-momentum does not give you the equation you wrote down. It just gives you back the original effective potential equation you wrote in the OP. In other words, the effective potential equation contains the same information as the normalization of 4-momentum; there is no point in trying to use the second when you have already used the first.

What exactly are you trying to accomplish? You already derived the relationship between ##L## and ##E##, which are the two conserved quantities in the problem, from the effective potential equation in the OP. What else do you need?
 
  • #15
Morning, @PeterDonis. Since I am learning geodesics of Schwarzschild, I think it would be instructive and interesting to trace photons transmitting in a geodesic. Hence I am using Matlab to solve the ODEs to have variables ##r## and ##\phi##, and then I will plot them to see how photons travel, so I can grasp those conserve quantities, equations, relations, etc., more firmly.

Also, I find my mistake. From ##0=g_{\mu\nu} p^\mu p^\nu##, I have mistakenly assumed ##dr/d\lambda=0##. As you have pointed out, this normalization relation will give the equation in OP. Only when ##r=r_c##, ##E## will be equal to ##\sqrt{1-\frac {2GM} r} \frac L r##.

Thanks for your patient help. I have learned a lot.
 
  • #16
Haorong Wu said:
From ##0=g_{\mu\nu} p^\mu p^\nu##, I have mistakenly assumed ##dr/d\lambda=0##.
You are assuming ##dr / d\lambda = 0## because you are trying to find a circular orbit, not because ##0 = g_{\mu \nu} p^\mu p^\nu##. You will have ##dr / d\lambda = 0## for any circular orbit, not just the circular orbit followed by a light ray in Schwarzschild spacetime.

Haorong Wu said:
Only when ##r=r_c##, ##E## will be equal to ##\sqrt{1-\frac {2GM} r} \frac L r##.
Yes. But this is perfectly consistent with ##dr / d\lambda = 0##.
 
  • #17
@PeterDonis

Sorry I did not state it correctly. After expanding ##0=g_{\mu\nu} p^\mu p^\nu##, I have a term containing ##dr/d\lambda##, and I remove it since I wrongly assume ##dr/d\lambda=0##.

The lovely circle is now plotted. And next, I will alter ##E## and ##L## to see other orbits.

Thanks, again.
 
  • #18
Haorong Wu said:
@PeterDonis

After expanding ##0=g_{\mu\nu} p^\mu p^\nu##, I have a term containing ##dr/d\lambda##, and I remove it since I wrongly assume ##dr/d\lambda=0##.
I don't understand why you say you "wrongly" assume ##dr / d\lambda = 0##. You have to assume that if you are talking about a circular orbit. It's not wrong.
 
  • #19
PeterDonis said:
I don't understand why you say you "wrongly" assume ##dr / d\lambda = 0##. You have to assume that if you are talking about a circular orbit. It's not wrong.
Because my final goal and my program is for arbitrary orbits. The circular orbit is just the first step since it is easier and special.
 
  • #20
Haorong Wu said:
my final goal and my program is for arbitrary orbits.
Then you should be asking about arbitrary orbits, not circular orbits.

Haorong Wu said:
The circular orbit is just the first step since it is easier and special.
And if you are just considering that special case, then my statement stands. If you are not, then you are not, and pretty much all of the discussion in this thread is irrelevant since it only applies to circular orbits. Solving a circular orbit is very little help at all in learning how to solve an arbitrary orbit.
 
  • #21
PeterDonis said:
If you are not, then you are not, and pretty much all of the discussion in this thread is irrelevant since it only applies to circular orbits. Solving a circular orbit is very little help at all in learning how to solve an arbitrary orbit.
Well, personally, I could not agree with that. The discussion does help me strengthen my understanding of those ODEs, and particularly of the concepts of energy and how I can calculate the conserved energy and angular momentum. I am quite grateful for that.
 
  • #23
A topic related to Schwarzschild metric. Starting from this insight series and the following derivation I've not a clear understanding about the meaning of the coordinates used in (global) Schwarzschild chart.

From a general point of view we know that in GR coordinates do not need to have a physical significance. We can possibly 'attach' a physical significance to them only when we know the solution (namely the metric tensor components in the given coordinate chart) as in the case of Schwarzschild spacetime in Schwarzschild chart.

Nevertheless it seems to me ##\phi## and ##\theta## coordinates in Schwarzschild chart do actually have 'in advance' a physical significance.

$$ds^2= - \left (1 - \frac {2M} {r} \right )dt^2 + \frac {1} {1 - 2M/r} dr^2 + r^2(d \theta^2 + sin^2\theta d \phi^2)$$
My doubt is: do we assume 'in advance' the standard interpretation for ##\phi## and ##\theta## coordinates (namely the same physical significance as in 3D Euclidean space) just because the spherically symmetric spacetime property ?
 
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  • #24
The angular coordinates are chosen as the standard spherical coordinates by default.
 
  • #25
Orodruin said:
The angular coordinates are chosen as the standard spherical coordinates by default.
ok, but the point is: do not standard spherical coordinates somehow 'assume' a 3D Euclidean space ?
Fixing a t-coordinate the spacelike 3D space 'slice' we get from the Schwarzschild metric is not actually Euclidean, I mean.
 
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  • #26
cianfa72 said:
do not standard spherical coordinates somehow 'assume' a 3D Euclidean space ?
No. The standard spherical coordinates just relate to the topology of the sphere and the metric being on the form ##ds^2 = d\theta^2 + \sin^2(\theta) d\varphi^2##. That this metric can be induced by a particular embedding in Euclidean space is secondary.
 
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  • #27
Orodruin said:
No. The standard spherical coordinates just relate to the topology of the sphere and the metric being on the form ##ds^2 = d\theta^2 + \sin^2(\theta) d\varphi^2##. That this metric can be induced by a particular embedding in Euclidean space is secondary.
ok, so just the very 'spherically symmetric spacetime property' actually sufficies to assume the metric form $$ds^2 = d\theta^2 + \sin^2(\theta) d\varphi^2$$ for fixed values of t-coordinate and r-coordinate. In other words we can assume that metric for the nested 2-sphere foliation of the 3D space slice we get as level spacelike hypersurfaces of constant t-coordinate, I believe.

Is that right ? Thanks.
 
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  • #28
cianfa72 said:
ok, so just the very 'spherically symmetric spacetime property' actually sufficies to assume the metric form $$ds^2 = d\theta^2 + \sin^2(\theta) d\varphi^2$$ for fixed values of t-coordinate and r-coordinate. In other words we can assume that metric for the nested 2-sphere foliation of the 3D space slice we get as level spacelike hypersurfaces of constant t-coordinate, I believe.

Is that right ? Thanks.
Well, if you want to be formally complete you would also need other charts to cover the full sphere. But the general idea is that, you have a 2D manifold with the standard two-dimensional closed surface with maximal symmetry and impose some standardised coordinates on that.
 
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  • #29
Orodruin said:
Well, if you want to be formally complete you would also need other charts to cover the full sphere. But the general idea is that, you have a 2D manifold with the standard two-dimensional closed surface with maximal symmetry and impose some standardised coordinates on that.
You mean the 2D manifold we get fixing t-coordinate and r-coordinate.

What about the r-coordinate ? From my understanding it basically 'labels' the family of nested 2-sphere foliating the 3D spacelike slice for a given t-coordinate value.

So, when at the beginning of the Schwarzschild solution derivation we write the metric for the 3D spacelike slice in the form: $$ ds^2 = A(r)dr^2 + r^2d\theta^2 + r^2sin^2\theta d\phi^2$$
we do not actually know -- let me say in advance -- the physical meaning of the r-coordinate; we just know that 'spherically symmetric property' plus 'static spacetime property' enforce us to choose the metric in the above form.
 
  • #30
cianfa72 said:
You mean the 2D manifold we get fixing t-coordinate and r-coordinate.

What about the r-coordinate ? From my understanding it basically 'labels' the family of nested 2-sphere foliating the 3D spacelike slice for a given t-coordinate value.

So, when at the beginning of the Schwarzschild solution derivation we write the metric for the 3D spacelike slice in the form: $$ ds^2 = A(r)dr^2 + r^2d\theta^2 + r^2sin^2\theta d\phi^2$$
we do not actually know -- let me say in advance -- the physical meaning of the r-coordinate; we just know that 'spherically symmetric property' plus 'static spacetime property' enforce us to choose the metric in the above form.
We do know the meaning of the ##r## coordinate - it is related to the area of the sphere labeled by ##r## as it appears as a multiplier of the metric of the unit sphere.
 
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  • #31
Let me also say that you can write down the general spherically symmetric vacuum solution and it will necessarily be the Schwarzschild metric that is static and asymptotically flat. This is Birkhoff's theorem.
 
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  • #32
Orodruin said:
We do know the meaning of the ##r## coordinate - it is related to the area of the sphere labeled by ##r## as it appears as a multiplier of the metric of the unit sphere.
You mean for ##r## constant, ##dr=0## hence ## ds^2 = r^2 (d\theta^2 + sin^2\theta d\phi^2)##.

So when at the beginning of the derivation we write down the spacetime metric in the form $$ds^2 = A(r)dr^2 + r^2 (d\theta^2 + sin^2\theta d\phi^2) + B(r)dt^2$$
we actually know the physical meaning of the coordinates ##r,\theta,\phi## including the time coordinate ##t## ?
 
  • #33
cianfa72 said:
You mean for ##r## constant, ##dr=0## hence ## ds^2 = r^2 (d\theta^2 + sin^2\theta d\phi^2)##.

So when at the beginning of the derivation we write down the spacetime metric in the form $$ds^2 = A(r)dr^2 + r^2 (d\theta^2 + sin^2\theta d\phi^2) + B(r)dt^2$$
we actually know the physical meaning of the coordinates ##r,\theta,\phi## including the time coordinate ##t## ?
The physical interpretation of ##r## as related to the area of the spheres is set, yes. For ##t## you would need to normalise properly at infinity to fix the interpretation on as the time for an observer at infinity.
 
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  • #34
cianfa72 said:
You mean the 2D manifold we get fixing t-coordinate and r-coordinate.
You're looking at it backwards. There are many spherically symmetric spacetimes other than Schwarzschild spacetime. If all we know is that a spacetime is spherically symmetric, then we can say that any point in the spacetime must lie on a 2-sphere on which we can choose standard spherical coordinates. That in itself is enough to show that we can use the standard ##\theta## and ##\phi## as two of our four coordinates on the spacetime. But without knowing more about the spacetime geometry, we don't know how to parameterize all the 2-spheres with the other two coordinates.
 
  • #35
PeterDonis said:
If all we know is that a spacetime is spherically symmetric, then we can say that any point in the spacetime must lie on a 2-sphere on which we can choose standard spherical coordinates.
Do you mean the spherically symmetric property basically amounts to the existence of a spacetime foliation with a 'two-fold' family of 2-sphere ?

PeterDonis said:
That in itself is enough to show that we can use the standard ##\theta## and ##\phi## as two of our four coordinates on the spacetime. But without knowing more about the spacetime geometry, we don't know how to parameterize all the 2-spheres with the other two coordinates.
but...standard ##\theta## and ##\phi## coordinates should not make sense just for 2-sphere in the 'space' slice (namely in the 3D spacelike hypersurfaces of constant coordinate time ##t##) ?
 

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