General Relativity Geodesic Problem

AI Thread Summary
The discussion focuses on proving that the curve x1 = a sec x2 is a geodesic in a 2D Euclidean metric expressed in polar coordinates. The Euclidean metric is defined as ds^2 = dr^2 + r^2 dθ^2, and the geodesic equations involve calculating derivatives and Christoffel symbols. The calculations show that the second derivatives lead to equations that can be solved for the derivatives of r and θ, ultimately demonstrating that r = a and θ = ln(λ) + b satisfy the original equation. Thus, the curve is confirmed to be a geodesic. The conclusion emphasizes the simplicity of the result once the correct approach is taken.
regretfuljones
Messages
4
Reaction score
0
Show that x1=asecx2 is a geodesic for the Euclidean metric in polar coordinates.

So I tried taking all the derivatives and plugging into polar geodesic equations. Obviously, bad idea.

Now I'm thinking I need to use Dgab/du=gab;cx'c and prove that the lengths of some vectors and their dot product are invariants under parallel transport, but I don't know how to go about doing that. Any advice on how to relate these concepts would be appreciated.
 
Physics news on Phys.org
this turned out to be very simple.
since it is in 2D space, it is a straight line therefore a geodesic.
 


First, let's define the Euclidean metric in polar coordinates as:

ds^2 = dr^2 + r^2dθ^2

Now, we want to show that x1 = asecx2 is a geodesic for this metric. To do this, we need to use the geodesic equations:

d^2x^a/dλ^2 + Γ^a_bc(dx^b/dλ)(dx^c/dλ) = 0

where λ is an affine parameter, Γ^a_bc are the Christoffel symbols, and x^a are the coordinates. In our case, we have x^1 = r and x^2 = θ.

We can start by calculating the first derivatives of x^1 and x^2 with respect to λ:

dx^1/dλ = dr/dλ = r'
dx^2/dλ = dθ/dλ = θ'

where ' denotes differentiation with respect to λ.

Now, let's calculate the second derivatives:

d^2x^1/dλ^2 = d(r')/dλ = r''
d^2x^2/dλ^2 = d(θ')/dλ = θ''

Next, we need to calculate the Christoffel symbols for our metric. For the Euclidean metric in polar coordinates, we have:

Γ^1_11 = 0
Γ^1_22 = -r
Γ^2_12 = Γ^2_21 = 1/r

Now, let's plug everything into the geodesic equations:

r'' + (-r)(r')^2 = 0
θ'' + (1/r)(r')^2 = 0

Since we have two equations and two unknowns (r and θ), we can solve for r' and θ':

r' = 0
θ' = 1/r

Integrating with respect to λ, we get:

r = a
θ = ln(λ) + b

where a and b are constants of integration.

Now, let's substitute these values back into our original equation x^1 = asecx^2:

r = a
θ = ln(λ) + b
x^1 = a
x^2 = ln(λ) + b

We can see that this satisfies our original equation, so x^1 = a
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top