What Are the Values of \(a\) and \(b\) in the Schwarzschild 4-Velocity?

AI Thread Summary
The discussion focuses on determining the values of \(a\) and \(b\) in the 4-velocity \(U = a\partial_t + b\partial_\phi\) for a particle in a Schwarzschild spacetime. The participants explore the properties of the Schwarzschild metric and the conditions for a timelike geodesic, emphasizing the importance of the normalization condition \(U \cdot U = -1\). They discuss the role of Killing vectors in deriving constants of motion related to energy and angular momentum, leading to equations involving \(a\) and \(b\). Ultimately, the conversation highlights the need to combine these equations with the normalization condition to find the correct values for \(a\) and \(b\). The discussion concludes with a suggestion to apply the derived equations to solve for the 4-velocity components.
Omikron123
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Homework Statement
I have an object orbiting in free fall with constant radius ##r## in the plane ##\theta = \frac{\pi}{2}##.
I am supposed to prove that the 4-velocity ##U = a\partial _t + b\partial _\phi## and find the values of ##a## & ##b## for a free falling object in the plane ##\theta = \pi/2##
Relevant Equations
The Schwarzchild line element is given by $$ ds^2 = (1- \frac{R_s}{r})dt^2 - (1- \frac{R_s}{r})^-1dr^2 - r^2d\Omega ^2$$
So the line element is given by $$ ds^2 = (1- \frac{R_s}{r})dt^2 - (1- \frac{R_s}{r})^{-1}dr^2 - r^2d\Omega ^2$$
The object is orbiting at constant radius ##r## in the plane ## \theta = \frac{\pi}{2}##. I am supposed to find the values of ##a## and ##b## in the 4-velocity given by: $$U = a\partial _t + b\partial _\phi$$.
Im pretty new the general relativity and Schwarzschild geodesics but here is my attempted solution:

For a massive particle the squared 4-velocity ##U^2 = -1##, space-like which i can expand with the Schwarzschild metric, which is diagonal:
$$ U^2 = U^tU_t + U^\phi U_\phi = (g^{ta}U_a)U_t + (g^{\phi a}U_\phi) = -1$$ Since the metric is diagonal only ##g^{tt}, g^{\phi \phi}## are non-zero with values $$g^{\phi \phi} = \frac{1}{g_{\phi \phi}} = -\frac{1}{r^2}, g^{tt} = ... = \frac{1}{1-\frac{R_s}{r}}$$ At this point I am not sure how to continue, because I am not sure if ##U_\phi = \partial _\phi## etc. One idea was to compare the following:
$$U^2 = g^{tt}U_tU_t + g^{\phi \phi}U_\phi U_\phi = -\frac{1}{r^2}\partial _t^2 + \frac{1}{1-\frac{R_s}{r}}\partial _\phi ^2$$ and $$U^2 = (a\partial _t + b\partial _\phi)^2 = a^2\partial _t ^2 + b^2\partial _\phi^2$$ give ## a = \sqrt{\frac{1}{1-\frac{R_s}{r}}}, b = \sqrt{-\frac{1}{r^2}}## (since ##2ab\partial _t\partial _\phi = 0 ## due to diagonal metric??) As I said I don't really know what I am doing here so there might be some major errors in my thinking..
 
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Oh sorry of course, thank you I couldn´ t figure out how to use it!
So the line element is given by $$ ds^2 = (1- \frac{R_s}{r})dt^2 - (1- \frac{R_s}{r})^-1dr^2 - r^2d\Omega ^2$$
The object is orbiting at constant radius ##r## in the plane ## \theta = \frac{\pi}{2}##. I am supposed to find the values of ##a## and ##b## in the 4-velocity given by: $$U = a\partial _t + b\partial _\phi$$.
Im pretty new the general relativity and Schwarzschild geodesics but here is my attempted solution:

For a massive particle the squared 4-velocity ##U^2 = -1##, space-like which i can expand with the Schwarzschild metric, which is diagonal:
$$ U^2 = U^tU_t + U^\phi U_\phi = (g^{ta}U_a)U_t + (g^{\phi a}U_\phi) = -1$$ Since the metric is diagonal only ##g^{tt}, g^{\phi \phi}## are non-zero with values $$g^{\phi \phi} = \frac{1}{g_{\phi \phi}} = -\frac{1}{r^2}, g^{tt} = ... = \frac{1}{1-\frac{R_s}{r}}$$ At this point I am not sure how to continue, because I am not sure if ##U_\phi = \partial _\phi## etc. One idea was to compare the following:
$$U^2 = g^{tt}U_tU_t + g^{\phi \phi}U_\phi U_\phi = -\frac{1}{r^2}\partial _t^2 + \frac{1}{1-\frac{R_s}{r}}\partial _\phi ^2$$ and $$U^2 = (a\partial _t + b\partial _\phi)^2 = a^2\partial _t ^2 + b^2\partial _\phi^2$$ give ## a = \sqrt{\frac{1}{1-\frac{R_s}{r}}}, b = \sqrt{-\frac{1}{r^2}}## (since ##2ab\partial _t\partial _\phi = 0 ## due to diagonal metric??) As I said I don't really know what I am doing here so there might be some major errors in my thinking..
 
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Likes PeroK
First, ##\partial_t## and ##\partial_{\phi}## are the Schwarzschild coordinate unit vectors. So, ##U = a \partial_t + b\partial_{\phi}## is an alternative way of writing: ##U = (\frac{dt}{d\tau}, 0, 0, \frac{d \phi}{d\tau})##. Where ##a = \frac{dt}{d\tau}## etc.

Your main problem is that you are only looking at timelike paths: ##U \cdot U = -1##. What you also need are the conditions for a timelike geodesic. Have you heard of Killing vectors?
 
Okay thank you. Yes that was another trail I tried to follow. From what I have learned both ## \partial _t ## and ##\partial _\phi## are killing fields due to the spherical symmetri in Schwarzschild spacetime? So the derivative with respect to proper time ##\tau## of a scalarproduct between one of the killing fields "##K##" should be zero:

$$\frac{d}{d\tau}(g(K,U)) = \nabla_Ug(K,U) = g(\nabla_UK,U) + g(K,\nabla _UU) = U^\mu U^\nu \nabla _\mu K_\nu + 0 = 0$$ since ##U^\mu U^\nu## is symmetric and ## \nabla _\mu K_\nu## antisymmetric, which mean that ##g(K,U)## is constant along the geodesic. However i don't know how I can decide the 4-velocity components from this.

If I remember correct, by skipping the derivative, ##g(\partial _t,U)## gives me the energy per unit mass, and ##g(\partial _\phi,U)## the angular momentum per unit mass, are these the velocity values ##a ## and ## b ##?
 
Omikron123 said:
Okay thank you. Yes that was another trail I tried to follow. From what I have learned both ## \partial _t ## and ##\partial _\phi## are killing fields due to the spherical symmetri in Schwarzschild spacetime? So the derivative with respect to proper time ##\tau## of a scalarproduct between one of the killing fields "##K##" should be zero:

$$\frac{d}{d\tau}(g(K,U)) = \nabla_Ug(K,U) = g(\nabla_UK,U) + g(K,\nabla _UU) = U^\mu U^\nu \nabla _\mu K_\nu + 0 = 0$$ since ##U^\mu U^\nu## is symmetric and ## \nabla _\mu K_\nu## antisymmetric, which mean that ##g(K,U)## is constant along the geodesic. However i don't know how I can decide the 4-velocity components from this.

If I remember correct, by skipping the derivative, ##g(\partial _t,U)## gives me the energy per unit mass, and ##g(\partial _\phi,U)## the angular momentum per unit mass, are these the velocity values ##a ## and ## b ##?
That's the theory of Killing vectors right enough. In this case you know that ##K_1 = (1, 0, 0, 0)## and ##K_2 = (0, 0, 0, 1)##.

And you also have a specific form for ##g_{\mu \nu}## .
 
PeroK said:
That's the theory of Killing vectors right enough. In this case you know that ##K_1 = (1, 0, 0, 0)## and ##K_2 = (0, 0, 0, 1)##.

And you also have a specific form for ##g_{\mu \nu}## .
Okay, so the only non-zero components of ##g_{\mu \nu}## are ## g_{tt} = 1- \frac{R_s}{r}## and ##g_{\phi \phi} = -r^2## which gives $$g(K_1,U) = g_{tt}\frac{dt}{d\tau} = (1-\frac{R_s}{r})\dot{t}$$ and $$g(K_4,U) = g_{\phi \phi}\frac{d\phi}{d\tau} = (-r^2)\dot{\phi}$$ which give ##a = 1-\frac{R_s}{r}## and ##b = -r^2##, $$ U = (1-\frac{R_s}{r})\partial _t -r^2\partial _\phi$$ Is this correct?
 
Omikron123 said:
Okay, so the only non-zero components of ##g_{\mu \nu}## are ## g_{tt} = 1- \frac{R_s}{r}## and ##g_{\phi \phi} = -r^2## which gives $$g(K_1,U) = g_{tt}\frac{dt}{d\tau} = (1-\frac{R_s}{r})\dot{t}$$ and $$g(K_4,U) = g_{\phi \phi}\frac{d\phi}{d\tau} = (-r^2)\dot{\phi}$$ which give ##a = 1-\frac{R_s}{r}## and ##b = -r^2##, $$ U = (1-\frac{R_s}{r})\partial _t -r^2\partial _\phi$$ Is this correct?
That's not right. You have calculated the constants of the motion: ##(1-\frac{R_s}{r})\dot{t}## and ##(r^2)\dot{\phi}##.

Note that you perhaps need to try to relate the mathematical formalism to the physics. For example, generally the first of these is analagous to conservation of energy along a geodesic and we can write: $$e = (1-\frac{R_s}{r})\dot{t}$$ And the second is conservation of angular momentum: $$l = (r^2)\dot{\phi}$$
That's what the geodescic equations give us (via the short-cut of using Killing vectors).

You need to combine that with the normalisation condition for a timelike path: ##U \cdot U = -1## to get the solution.

Note that your four velocity is not normalised.
 
PS You may also need to look at the condition for a stable circular orbit to give you another equation in ##l## and ##r##.

Have you seen the equations for Schwarzschild orbits generally?
 
  • #10
PeroK said:
PS You may also need to look at the condition for a stable circular orbit to give you another equation in ##l## and ##r##.

Have you seen the equations for Schwarzschild orbits generally?
In case you haven't here's the general approach to these problems:

1) You use the Killing vectors and timelike normalisation to get a formula for ##e, l, r##: $$e^2 = f(r)$$ where you have to calculate ##f(r)##. Note that this equation involves ##l^2## as well.

2) Now, for a stable circular orbit ##f(r)## must be at a minimum at the orbital radius. So, we need also ##f'(r) = 0##. This gives you an equation of the form $$l^2 = g(r)$$ where you have to calculate ##g(r)##.

3) If you plug ##g(r)## into your equation for ##e^2## you should get ##e## in terms of ##R_s## and ##r##. This gives you ##dt/d\tau##.

4) Then you can also solve for ##l##, which gives you ##d\phi/d\tau##.
 
  • #11
PeroK said:
In case you haven't here's the general approach to these problems:

1) You use the Killing vectors and timelike normalisation to get a formula for ##e, l, r##: $$e^2 = f(r)$$ where you have to calculate ##f(r)##. Note that this equation involves ##l^2## as well.

2) Now, for a stable circular orbit ##f(r)## must be at a minimum at the orbital radius. So, we need also ##f'(r) = 0##. This gives you an equation of the form $$l^2 = g(r)$$ where you have to calculate ##g(r)##.

3) If you plug ##g(r)## into your equation for ##e^2## you should get ##e## in terms of ##R_s## and ##r##. This gives you ##dt/d\tau##.

4) Then you can also solve for ##l##, which gives you ##d\phi/d\tau##.

Thank you very much for the help, I'll give it a try!
 
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