Define a new collection of functions, Dij by taking. Show that Dij, is a type 0, 2 tensor. Definition 6. We also call such a gizmo a symmetric bilinear form. A manifold endowed with a smooth inner product is called a Riemannian manifold.
Before we look at some examples, let us see how these things can be specified. We call this tensor the fundamental tensor or metric tensor of the Riemannian manifold. Question How does this effect the length of vectors? Answer We saw in Section 3 that, in En, we could think of tangent vectors in the usual way; as directed line segments starting at the origin. The role that the metric plays is that it tells you the length of a vector; in other words, it gives you a new distance formula:.
Geometrically, the set of all points in Euclidean 3-space at a distance r from the origin or any other point is a sphere of radius r. In Minkowski space, it is a hyperbolic surface. In Euclidean space, the set of all points a distance of 0 from the origin is just a single point; in M, it is a cone, called the light cone.
In particular, if M is any 3-dimensional manifold embedded in E4 with the metric shown above, then M inherits such a inner product. Following the line of reasoning in the previous example, we have. Thus, in particular, a Minkowski metric has signature 1, 1, 1, If we now define j new coordinates x— by. Note The non-degeneracy condition in Definition 6. Note that X 2 may be negative.
If X is not spacelike, then we can define. The expected inequality. Arc Length One of the things we can do with a metric is the following.
To show as we must that this is independent of the choice of chart x, all we need observe is that the quantity under the square root sign, being a contraction product of a type 0, 2 tensor with a type 2, 0 tensor, is a scalar. Proposition 6. In other words, t must be, up to a constant, arc length. But, by the Fundamental theorem of Calculus and the definition of L a, t ,. Exercise Set 6 1. Give an example of a Riemannian metric on E2 such that the corresponding metric tensor gij is not constant. Let aij be the components of any symmetric tensor of type 0, 2 such that det aij is never zero.
Show that this is a smooth inner product on M. We call such a field a null field.
Show that the resulting inverse of matrices gij is a type 2, 0 tensor. What is the inverse operation? What can be said about the columns of T in a given coordinate system x?
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In particular, we said that Minkowski space comes with a such a metric tensor having signature 1, 1, 1, In relativity, we take deal with 4-dimensional manifolds, and take the first three coordinates x1, x2, x3 to be spatial measuring distance , and the fourth one, x4, to be temporal measuring time. Let us postulate that we are living in some kind of 4-dimensional manifold M since we want to include time as a coordinate. By the way, we refer to a chart x at the point p as a frame of reference, or just frame.
Suppose now we have a particle—perhaps moving, perhaps not—in M. Assuming it persists for a period of time, 1 2 3 4 we can give it spatial coordinates x , x , x at every instant of time x. This path is called the world line of the particle.
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Note: t is not time; it's just a parameter. The velocity and speed of this particle as measured in our frame are given by. The problem is, we cannot expect v to be a vector—that is, satisfy the correct transformation laws. But we do have a contravariant 4-vector. T stands for tangent vector. Also, remember that t is not time. If the particle is moving at the speed of light c, then. T 2 of the vector T under the metric whose matrix is Now this looks like the norm-squared T. So, to check whether a particle is moving at light speed, just check whether T is null.
Answer Since physical units of time are usually not the same as physical units of space, we would like to convert the units of x4 the units of time to match the units of the other axes. Since relativity holds that the speed of light c is a universal constant, it seems logical to use c as this conversion factor. Now, if we happen to be living in a Riemannian 4-manifold whose metric diagonalizes to 2 something with signature 1, 1, 1, -c , then the physical property of traveling at the speed. Definition 7.
This motivates the following definition. It follows from the remark preceding the defintion that if x is any chart such that, at the point p, the metric has the nice form diag[1, 1, 1, -c2], then x is a Lorentz frame at the point p. Note that in general, the coordinates of T in the system x—i are given by matrix. Thus, very few frames are going to be Lorentz. Physical Interpretation of a Lorentz Frame What the definition means physically is that an observer in the x—-frame who measures a particle traveling at light speed in the x-frame will also reach the conclusion that its speed is c, because he makes the decision based on I , which is equivalent to II.
A Lorentz frame in locally Minkowskian space is any frame in which light appears to be traveling at light speed, and where we measure length in the usual way. Answer Yes, as we shall see below. Question OK. But if x and x— are two Lorentz frames at the point p, how are they related?
Answer Here is an answer. First, continue to denote a specific Lorentz frame at the point p by x. Theorem 7.
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We proceed by invoking condition a of Definition 7. These paths will correspond to sending out light rays in various directions. This gives. Let D be the change-of-basis i matrix to the other inertial frame x— ;. In other words, the first and fourth columns of D are orthogonal under the Minkowskian inner product.
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Similarly, by sending light beams in the other directions, we see that the other columns of D are orthogonal to the fourth column. Similarly by choosing other photons we can replace column 1 by either column 2 or column 3, showing that if we take. Therefore, we have. But, what is k? Let us invoke condition b of Defintion 7. To measure the length of a vector in the new frame, we need to transform the metric tensor using this coordinate T change. Now, since G is just a constant multiple of an elementary matrix, all it does is multiply the last row of P by c2.
So, when we take PT GP , we are really getting the funny dot product of the columns of P back again, which just gives a multiple of G. Now we invoke condition b in Definition 7. Recognize it? This proves b and also c , by the way.
We will call the transformation from one Lorentz frame to another a generalized Lorentz transformation. This is the first column of D. To keep things simple, let us take the next two columns to be the corresponding basis vectors e2, e3. Now we might be tempted to take the forth vector to be e4, but that would not be orthogonal to the above first vector.
By symmetry to get a zero inner product we are forced to take the last vector to be. In other words, it is moving in the x-direction with a velocity of. In other words, if a particle is moving at sub- light speed in any one Lorentz frame, then it is moving at sub-light speed in all Lorentz frames.
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Prove the claim in the proof of 7. If the x—i-system is moving with a velocity v in a certain direction with resepct to the xi- system, we call this a boost in the given direction.
Now do some reading to find the transformation for a pure boost in an arbitrary direction. Intuitively, by a parallel vector field, we mean a vector field with the property that the vectors at different points are parallel. Is there a notion of a parallel field on a manifold? For instance, in En, there is an obvious notion: just take a fixed vector v and translate it around. On the torus, there are good candidates for parallel fields see the figure but not on the 2- sphere. There are, however, parallel fields on the 3-sphere….
Let us restrict attention to parallel fields of constant length. Usually, we can recognize such a field by taking the derivatives of its coordinates, or by following a path, and taking the derivative of the vector field with respect to t: we should come up with zero.