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# 物理代写|相对论代写Theory of relativity代考|FYS2004 The Scalar Product of a Vector and Covector

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## 物理代写|相对论代写Theory of relativity代考|The Scalar Product of a Vector and Covector

The Scalar Product of a Vector and Covector. The scalar product of a vector with components $A^\mu$ and a covector with components $B_\mu$ is defined to be $A^\mu B_\mu$ (with the implied sum over $\mu$ ). This scalar product is interesting because its value is independent of coordinate system. The proof is direct:
$$A^{\prime \mu} B_\mu^{\prime}=\left[\frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha\right]\left[\frac{\partial x^\beta}{\partial x^{\prime \mu}} B_\beta\right]=\frac{\partial x^\beta}{\partial x^{\prime \mu}} \frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha B_\beta=\delta^\beta{ }\alpha A^\alpha B\beta=A^\alpha B_\alpha$$
In fact, equation $6.5$ means that this is really the same as $A^\mu g_{\mu v} B^v$, which we have already defined as the scalar product of two vectors, and which we have already seen is frame independent for Lorentz transformations. We now see that $A^\mu B_\mu=A^\mu g_{\mu \nu} B^\nu$ has a coordinate-independent value not only in special relativity, but in any arbitrary coordinate system we use (as long as we use a coordinate basis)!

The Inverse Metric. We have already seen that the components of the metric tensor transform as follows (see equation 5.11):
$$g_{\mu \nu}^{\prime}=\frac{\partial x^\alpha}{\partial x^{\prime \mu}} \frac{\partial x^\beta}{\partial x^{\nu v}} g_{\alpha \beta}$$
Compare this to the transformation equation $6.2$ for a covector. Note that $g_{\alpha \beta}$, which has two subscript indices, has a transformation law that is essentially a doubled version of equation 6.2: there is a partial derivative factor associated with each lower index that has the same form as the single factor associated with the single lower index in equation $6.2$.

## 物理代写|相对论代写Theory of relativity代考|The Gradient of a Tensor Is Not a Tensor

The Gradient of a Tensor Is Not a Tensor. Note that while the gradient of a scalar invariant yields a covector, the gradient of any other tensor does not yield another tensor. For example, imagine that $A^\mu$ is a four-vector. The components of the gradient of $A^\mu$ transform as follows when we change coordinate systems:
\begin{aligned} \partial_v^{\prime} A^{\prime \mu} & \equiv \frac{\partial A^{\prime \mu}}{\partial x^{\prime \nu}}=\frac{\partial}{\partial x^{\prime \nu}}\left(\frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha\right)=\frac{\partial x^\beta}{\partial x^{\prime \nu}} \frac{\partial}{\partial x^\beta}\left(\frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha\right) \ & =\frac{\partial x^\beta}{\partial x^{\prime \nu}} \frac{\partial^2 x^{\prime \mu}}{\partial x^\beta \partial x^\alpha} A^\alpha+\frac{\partial x^\beta}{\partial x^{\prime \nu}} \frac{\partial x^{\prime \mu}}{\partial x^\alpha}\left(\partial_\beta A^\alpha\right) \end{aligned}
by the product rule. The second term appearing in the bottom line, if it were alone, would be the transformation rule for a tensor. However, if the coordinate transformation factors $\partial x^{\prime \mu} / \partial x^\nu$ are not constants, then the second derivative appearing in the first term will not be zero, meaning that the gradient of a vector does not transform as a tensor, and therefore is not a tensor. The same issue arises with derivatives with respect to a particle’s proper time $\tau$ (see problem P6.7). This is an issue we must address in the future, because most physical equations involve such derivatives.
(However, the components of the Lorentz transformation are constants, so if we limit ourselves to transformations between IRFs in cartesian coordinates, the first term in equation $6.14$ is zero and the gradient of a vector does transform like a tensor. This also applies to the gradients of higher-rank tensors.)

## 物理代写|相对论代写Theory of relativity代考|The Scalar Product of a Vector and Covector

$$A^{\prime \mu} B_\mu^{\prime}=\left[\frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha\right]\left[\frac{\partial x^\beta}{\partial x^{\mu \mu}} B_\beta\right]=\frac{\partial x^\beta}{\partial x^{\prime \mu}} \frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha B_\beta=\delta^\beta \alpha A^\alpha B \beta=A^\alpha B_\alpha$$

$$g_{\mu \nu}^{\prime}=\frac{\partial x^\alpha}{\partial x^{\prime \mu}} \frac{\partial x^\beta}{\partial x^{\nu v}} g_{\alpha \beta}$$

## 物理代写|相对论代写Theory of relativity代考|The Gradient of a Tensor Is Not a Tensor

$$\partial_v^{\prime} A^{\prime \mu} \equiv \frac{\partial A^{\prime \mu}}{\partial x^{\prime \nu}}=\frac{\partial}{\partial x^{\prime \nu}}\left(\frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha\right)=\frac{\partial x^\beta}{\partial x^\mu} \frac{\partial}{\partial x^\beta}\left(\frac{\partial x^{\prime \mu}}{\partial x^\alpha} A^\alpha\right) \quad=\frac{\partial x^\beta}{\partial x^{\prime \nu}} \frac{\partial^2 x^{\prime \mu}}{\partial x^\beta \partial x^\alpha} A^\alpha+\frac{\partial x^\beta}{\partial x^{\prime}} \frac{\partial x^{\prime \mu}}{\partial x^\alpha}\left(\partial_\beta A^\alpha\right)$$

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