What Causes the Negative Sign in the Schwarzschild Metric Lagrangian?

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SUMMARY

The discussion focuses on the derivation of the Lagrangian for a test particle in the Schwarzschild metric, specifically addressing the negative sign in the Lagrangian as presented in the book "Black Holes, White Dwarfs, and Neutron Stars" by Shapiro. The Schwarzschild metric is defined as ds² = - (1 - 2M/r) dt² + (1 - 2M/r)⁻¹ dr² + r² dθ² + r² sin²θ dφ². The correct Lagrangian is established as L = -m²/2, which is essential for deriving the equations of motion for the test particle, emphasizing the necessity of integrating over timelike curves rather than spacelike curves.

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I am trying to find the equations of motion for a test particle in the schwarzschild metric. However, I cannot find the correct first integral for the Lagrangian.

The Schwarzschild metric is:
ds^2 = -\left(1-\frac{2M}{r}\right)\,dt^2+\left(1-\frac{2M}{r}\right)^{-1}\,dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2
If we look at the action, and parameterise it using s=\frac{\tau}{m},
S=\int\,ds = \int_{\tau_i}^{\tau_f} \sqrt{g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}m^2}\left(\frac{1}{m}\,d\tau\right)
If we DEFINE our lagrangian to be L=\frac{1}{2}m^2g_{ab}\frac{dx^a}{d\lambda}\frac{dx^b}{d\lambda}, then
S = \int_{\tau_i}^{\tau_f} \frac{\sqrt{2L}}{m}\,d\tau
Due to reparameterisation invariance, we can set once again set \tau=s, and in that case we get \frac{\sqrt{2L}}{m}=1\Rightarrow L = \frac{m^2}{2}.

However, in the book I'm reading, "Black Holes, White Dwarfs, and Neutron stars (Shapiro)", they have L=-\frac{m^2}{2}. This is correct, since you need it to find the equation of motion for the test particle. However, I can't seem to get the minus sign. How does it come about?

Thanks
 
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With the signature you're using, gμνdxμ/dτ dxν/dτ is negative along a timelike curve, so you want to integrate ds = √-gμνdxμ/dτ dxν/dτ dτ
 
Thanks Bill_K, but I'm not exactly sure I understand.

By signature I'm guessing you mean
ds^2=-\left(1-\frac{2M}{r}\right)\,dt^2+\left(1-\frac{2M}{r}\right)^{-1}\,dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2
instead of
d\tau^2=\left(1-\frac{2M}{r}\right)\,dt^2-\left(1-\frac{2M}{r}\right)^{-1}\,dr^2-r^2\,d\theta^2-r^2\sin^2\theta\,d\phi^2

But when we take the action, why do we need to integrate over a timelike curve? What's wrong with integrating over a spacelike curve? After all, isn't the definition

S = \int\,ds
 
Particles don't move along space-like curves. They must move on time-like curves.
 
ah of course. cheers
 

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