Question about inner products of tensors and Einstein summation convention

So I am studying Special Relativity and basic tensor calculus and got stuck at an exercise. $$F^{\mu \nu}: = \left[ \begin {array}{cccc} 0&-{\it E_x}&-{\it E_y}&-{\it E_z} \\ {\it E_x}&0&-c{\it B_z}&c{\it B_y} \\ {\it E_y}&c{\it B_z}&0&-{\it cB_x} \\ {\it E_z}&-c{\it B_y}&c{\it B_x}&0\end {array} \right]$$

Now the question asks to find an explicit expression for $$F^{\mu}_{\,\,\nu}$$. My attempt is the following: $$F^{\mu}_{\,\,\nu} = g_{\nu\rho}F^{\mu\rho} = F^{\mu\rho}g_{\rho\nu}$$

Now from here I recognize this to be a dot product between $$F$$ and $$g$$ (the Minkowski metric tensor) so the result should be: $$\sum_{\rho = 0}^3(F^{\mu\rho}g_{\rho\nu})$$ Now this should be a scalar product right (not a matrix???) ? However, my professor's solution to the answer is the following: $$F^{\mu}_{\,\,\nu} = (F\cdot g)^{\mu}_{\,\,\nu}= \left[ \begin {array}{cccc} 0&{\it E_x}&{\it E_y}&{\it E_z} \\ {\it E_x}&0&c{\it B_z}&-c{\it B_y} \\ {\it E_y}&-c{\it B_z}&0&c{\it B_x} \\ {\it E_z}&c{\it B_y}&-{\it cB_x}&0\end {array} \right]$$

However, I am struggling to understand how a "dot product" between $$F$$ and $$g$$ result in that matrix.

• It might help to notice that with a scalar product you are never left with free indices. Is this the case? – jinawee Oct 30 '18 at 19:38
• @jinawee yes that makes sense. The thing I am confused is how the resulting sum converts to the 4x4 matrix above, how is that possible? – daljit97 Oct 30 '18 at 19:42
• $F^{\mu \rho}g_{\rho \nu}$ is basically matrix multiplication. Multiplying two matrices gives a matrix. – DanielSank Oct 30 '18 at 19:43
• @DanielSank but how can a sum give rise to matrix? That is the main confusion in my mind. – daljit97 Oct 30 '18 at 19:46
• I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. – Ben Crowell Oct 31 '18 at 2:21

You are summing matrices in your expression. If $$A$$ and $$B$$ are matrices, so is $$A+B$$. Each element $$F^{\mu\rho}g_{\rho\nu}$$ is a matix itself. For any given $$\rho$$, lets say $$\rho=0$$ you would have $$F^{\mu0}$$, which is a column vector, and $$g_{0\nu}$$ which is a row vector. You can see by basic matrix multiplication that $$\begin{pmatrix}* \\ * \\*\end{pmatrix} \begin{pmatrix}* & * &*\end{pmatrix}$$ is a matrix, this is the outter product and is what you have in each term of your sum, you are confusing it with $$\begin{pmatrix}* & * &*\end{pmatrix}\begin{pmatrix}* \\ * \\*\end{pmatrix}$$, which is the inner product, and would return you a number.

• That makes sense. So $F^{\mu}_{\,\,\nu} = F^{\mu 0}g_{0 \nu} + F^{\mu 1}g_{1 \nu} + F^{\mu 2}g_{2 \nu} + F^{\mu 3}g_{3 \nu}$. I understand that I can get the result this way, but how can I get it by doing matrix multiplication (I have noticed that $F^T g$ gives me the correct answer). – daljit97 Oct 31 '18 at 22:56

Now from here I recognize this to be a dot product between $$F$$ and $$g$$.

It is very difficult to write an answer without knowing your mathematical background. In my opinion those who answered before me approached the difficulty by doing some guesses, one different from another.

I was impressed by your speaking of a "dot product". Apparently you have never seen row-column multiplication of matrices. If you didn't have a course in linear algebra, I can't understand how you can follow tensor calculus. But I want to be positive,so I'll give you some hints, without oversimplifying the matter, which wouldn't help you.

@DanielSank rightfully said that $$F^{\mu\rho}\,g_{\rho\nu}$$ is (basically) a matrix product. Your answer showed this was novel to you. Wasn't it?

Well, matrices may be multiplied (row by columns) if only number of columns of the first equates number of rows of the second. In your case it's OK, since all these numbers are 4. And definition of matrix multiplication is exactly what is written in the expression $$F^{\mu\rho}\,g_{\rho\nu}$$, which with Einstein convention is a shorthand for $$\sum_{\rho=0,3} F^{\mu\rho}\,g_{\rho\nu}.$$ Result is a $$4\times4$$ matrix, with indices $$\mu$$ high and $$\nu$$ low.

In the actual case calculation is made simpler as $$g$$ is diagonal, so that for given $$\nu$$ there is only one term in the sum: the one with $$\rho=\nu$$. You have only to remember that the four diagonal components of $$g_{\rho\nu}$$ are not all $$=1$$. I guess that you have been taught $$g_{00}=-1$$.

I remember getting hung up on my intuitive notions of tensor manipulation as an undergrad. Geometrical pictures (e.g. matrix multiplication, dot products, etc) are useful, but sometimes you just need to sit down and calculate to see how things work out.

You are correct that $$F^\mu_{\ \ \nu}=g_{\nu \rho} F^{\mu \rho}$$. Clearly, this object has two spacetime indices (one upstairs, one downstairs) which means that it is an object with 16 components.

$$\mu=0,\nu=0 \implies F^0_{\ \ 0} = \sum_{\rho = 0}^3 g_{0\rho}F^{0\rho} = g_{00}F^{00}+g_{01}F^{01}+g_{02}F^{02}+g_{03}F^{03} = 0$$

$$\mu=0,\nu=1 \implies F^0_{\ \ 1} = \sum_{\rho = 0}^3 g_{1\rho}F^{0\rho} = g_{10}F^{00}+g_{11}F^{01}+g_{12}F^{02}+g_{13}F^{03} = (-1)(-E_x) = E_x$$

so on and so forth. At this point you've probably already noticed that most of the terms are going to vanish because $$g_{\nu\rho}$$ is zero if $$\nu\neq \rho$$ (in Minkowski spacetime, that is - for more general metrics, this is not true).

Once you see the patterns and identify the symmetries, this work becomes faster - but I find that it's most helpful to go through the work step by step first, and let the speed and shortcuts come with practice.

• Is there a way to get the result just doing "normal" matrix multiplication. I have noticed the $F^T g$ gives me the correct answer, but I don't understand why is this the case. – daljit97 Oct 31 '18 at 22:59
• Yes. Normal matrix multiplication follows the rule $(A\cdot B)^i_{\ \ j} = \sum_k A^i_{\ \ k} B^k_{\ \ j}$ so if you arrange two second rank tensors so that you're contracting the second index of the first with the first index of the second, the result can be thought of as normal matrix multiplication. However, be aware that this (obviously) doesn't work for tensors with rank higher than 2, and it doesn't work for contractions within a single tensor (like $R^\mu_{\ \ nu \mu \sigma}$, which appears in GR). It also doesn't follow the familiar transformation rules under changes of basis [...] – J. Murray Oct 31 '18 at 23:07
• [...] unless each of your tensors has one upstairs index and one downstairs index. – J. Murray Oct 31 '18 at 23:08