# How can $F_{\mu\nu}F^{\mu\nu} = 2(B^2-E^2)$ be proved? [closed]

How does $$F_{\mu\nu}F^{\mu\nu} = 2(B^2-E^2)$$?

$$F_{\mu\nu}=\pmatrix{ 0&E_x&E_y&E_z\\ -E_x&0&-B_z&B_y\\ -E_y&B_z&0&-B_x\\ -E_z&-B_y&B_x&0 }$$

$$F^{\mu\nu}=\pmatrix{ 0&-E_x&-E_y&-E_z\\ E_x&0&-B_z&B_y\\ E_y&B_z&0&-B_x\\ E_z&-B_y&B_x&0 }$$

The matrix product:

$$F_{\mu\nu}F^{\mu\nu}= \left( \begin{array}{cccc} \text{Ex}^2+\text{Ey}^2+\text{Ez}^2 & \text{Bz} \text{Ey}-\text{By} \text{Ez} & \text{Bx} \text{Ez}-\text{Bz} \text{Ex} & \text{By} \text{Ex}-\text{Bx} \text{Ey} \\ \text{By} \text{Ez}-\text{Bz} \text{Ey} & -\text{By}^2-\text{Bz}^2+\text{Ex}^2 & \text{Bx} \text{By}+\text{Ex} \text{Ey} & \text{Bx} \text{Bz}+\text{Ex} \text{Ez} \\ \text{Bz} \text{Ex}-\text{Bx} \text{Ez} & \text{Bx} \text{By}+\text{Ex} \text{Ey} & -\text{Bx}^2-\text{Bz}^2+\text{Ey}^2 & \text{By} \text{Bz}+\text{Ey} \text{Ez} \\ \text{Bx} \text{Ey}-\text{By} \text{Ex} & \text{Bx} \text{Bz}+\text{Ex} \text{Ez} & \text{By} \text{Bz}+\text{Ey} \text{Ez} & -\text{Bx}^2-\text{By}^2+\text{Ez}^2 \\ \end{array} \right)$$

So is $$F_{\mu\nu}F^{\mu\nu}$$ not a matrix product? How do I obtain $$2(B^2-E^2$$)?

• Isn't there a factor of 2 missing on RHS? Jun 30, 2020 at 1:56
• Matrix multiplication looks like this, ignoring up/down index placement: $C_{ik}=A_{ij}B_{jk}$. The contracted indices for the matrix multiplication are adjacent, representing forming scalar products of rows in the first with columns in the second. Jun 30, 2020 at 3:04
• Why did you put \text around each matrix element in the product? The product doesn’t look like MathJax, for obvious reasons. Jun 30, 2020 at 3:25
• $F_{\mu\nu}F^{\mu\nu}$ is fully contracted and thus a scalar. It thus cannot be a matrix. A matrix has two free indices. Jun 30, 2020 at 3:33
• @G.Smith Well I used Mathematica to compute the matrix multiplication and thats how the copy-paste in latex worked. Didn't feel line going item by item to remove the text notation. Jun 30, 2020 at 11:27

As pointed out in the other answer,you have to sum over the indices. This is equivalent to taking the trace over the matrix product(in this case). Note an overall negative sign in missing in your matrix product because because $$F_{\mu\nu}F^{\mu\nu}$$ is not the trace of the matrix product of the $$F_{\mu\nu}$$ matrix with the $$F^{\mu\nu}$$ matrix.It is the trace of the matrix product of one with the transpose of the other.After fixing the sign, You just need to take trace of the correct matrix product,and the result will be $$2(B^2-E^2)$$.

• The trace of the OP’s matrix product doesn’t have the correct sign, because $F_{\mu\nu}F^{\mu\nu}$ is not the trace of the matrix product of the $F_{\mu\nu}$ matrix with the $F^{\mu\nu}$ matrix. It is the trace of the matrix product of one with the transpose of the other. Jun 30, 2020 at 3:15
• You wrongly implied that what the OP did was fine. And even now you have not explained what the OP did wrong. Jun 30, 2020 at 3:20
• An the linear algebra matrix product as written is $F_{\mu\nu}F^{\nu\mu}$, i.e., one is transposed, hence the sign error.
– JEB
Jun 30, 2020 at 3:20
• No, I don’t think so. Your answer is confusing, but you can improve it. Jun 30, 2020 at 3:26
• It’s better now. Jun 30, 2020 at 3:41

In Einstein tensor notation, repeated incides in a product are implicitly summed over. So, explicitly written out, this reads:

$$\sum_{\mu,\nu}F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2)$$

• @G.Smith That's correct. I am stating what the statement currently reads, not what it should read. Jun 30, 2020 at 3:06
• @probably_someone I've edited it to the correct value, so feel free to edit your answer accordingly :) Jun 30, 2020 at 3:21

Since the sum over repeated indices is implied as other answers have pointed out, $$F_{\mu\nu}F^{\mu\nu}$$ is nothing but the sum of the elements of the element-wise product of the matrices $$F_{\mu\nu}$$ and $$F^{\mu\nu}$$. From this, the equality you are after follows immediately without even (matrix-)multiplying them, since the element-wise product is trivial to calculate.