Transformation to a local inertial Frame

In summary, the author has been working on a problem involving the path of a particle and the path of an observer. The author has found that they need to use coordinate transformations to solve the problem.
  • #1
JabberWalkie
16
0
So I've been working on this problem. I am given the metric in Kruskal coordinates, so

ds^2=32M^2exp(-r/2M)/r(-dT^2+dX^2)+r^2(dθ^2+sin^2(θ)dΦ^2)

And the path of a particle is

X=0 T=λ θ=π/2 Φ=0

And the path of the observer is

X=-1/2*T+1/2 θ=π/2 Φ=0

And I am asked to find the 3 velocity of the particle as seen by the observer when the two intersect. So far this is what i have

The 4-velocity of the observer in Kruskal Coordinates is

u=√[4/3*r/(32M^2exp(-r/2M)](1,-1/2,0,0)

So, I think i need to find a local intertial frame, but I am unsure how to do that and I am not sure what to do with it once I've found it! Thanks in advance!
 
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  • #2
Well, the first thing I would do is to write the Kruskal metric as an orthonormal basis of one forms.

This is actually easy to do once you know how. You basically have a diagonal metric

ds^2 = -A dT^2 + B dx^2 + C d theta^2 + D d phi^2

so w0=sqrt(A) dT, w1= sqrt(B) dX, w2=sqrt(C) d theta, and w3=sqrt(D) d phi is such an orthonormal basis.

Do you know about one-forms (also called cotangent vectors?), and do you know how to take the product of a one-form and a vector to get a scalar? Do you see why dT and dX are one-forms, and why w0, w1, w2, and w3 are othornormal?

(hint: the length of a vector is g_ij x^i x^j, the length of a one-form is g^ij x_i x_j). w0 has components (w0)_i, i=0,1,2,3

Do you see why the duals to w0, w1, w2, and w3 (which we can call e0, e1, e2, and e3) are unit vectors i.e. e0 = [itex]\hat{T}[/tex], e1 = [itex]\hat{X}[/itex], e2 = [itex]\hat{\theta}[/itex], e3 = [itex]\hat{\phi}[/itex]

Finally, do you see how
[tex]
v = (w0 v) \hat{T} + (w1 v) \hat{X} + (w2 v) \hat{\theta} + (w3 v) \hat{phi}
[/tex]

Lastly, do you see how this solves your problem? :-)
 
  • #3
So, i really havn't done anything with one forms...ive looked them up, but they don't really make any sence. I am not sure how this helps :cry:Is there a way to do this without 1 forms?...we havn't been taught those in class...so there should be a way to do the problem without that...
 
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  • #4
Hmmm. Well, while you should really learn about one-forms (or cotangent vectors or dual vectors) there is a work-around that gets the same answer, but you'll have to do coordinate transformations.

How do you deal with tensors at all without knowing about duals, though?

What you can do instead is to introduce new scaled coordinates, say x1, t1, theta1, phi1 in such a way that you have a locally Minkowskian metric (i.e. in this case so that you normalize the metric coefficients to unity) in terms of your new coordinates. Here x1= alpha x, t1 = beta t, etc.

Then dx1/dt1 = (dx1/dtau) / (dt1 / dtau) will be your desired velocity. See for instance https://www.physicsforums.com/showpost.php?p=602558&postcount=29 where I get the right answer after a few false starts for the velocity of an object falling into a black hole relative to a "hovering" observer. You'll probably need to read the entire thread to get the background of the problem, however, earlier attempts to solve the problem got the wrong answer. At the end of the theread, George Jones also works the problem via a slightly different means to get the same answer.

If you use something like GrTensorII, it's really easy to specify a metric by specifiying the ONB of one forms rather than the metric itself. You can then have a choice of working with either coordinate basis vectors, or a local orthonormal tetrad of unit vectors.

You can also try reading http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity which uses slightly different language to take the same approach I did.
 
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  • #5
Thanks for the help, i was able to find a transformation that made the metric locally minkowski, and everything just fell out from there! Thanks!
 

Related to Transformation to a local inertial Frame

1. What is a local inertial frame?

A local inertial frame is a reference frame in which the laws of physics are valid and hold true. This means that objects in this frame will obey the laws of motion without any external forces acting on them.

2. Why is transformation to a local inertial frame important in science?

Transformation to a local inertial frame is important because it helps us understand and analyze the behavior of objects in motion. It allows us to simplify complex systems and make accurate predictions based on the laws of physics.

3. How is a local inertial frame different from a global inertial frame?

A global inertial frame is a reference frame that is fixed in space and does not accelerate, while a local inertial frame is a reference frame that is non-inertial and may experience acceleration. In a local inertial frame, the laws of physics may not hold true for objects that are far from the frame's origin.

4. What is the process of transforming to a local inertial frame?

The process of transforming to a local inertial frame involves using mathematical equations to convert measurements and observations from a non-inertial frame to an inertial frame. This allows us to analyze the motion of objects in a simplified and accurate manner.

5. Can a local inertial frame exist in the real world?

Yes, local inertial frames exist in the real world. For example, a car moving at a constant speed on a straight road can be considered a local inertial frame. However, as the car accelerates or turns, it is no longer a local inertial frame and the laws of physics may not hold true for objects within the car.

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