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? – Manas Dogra Jun 30 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. – G. Smith Jun 30 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. – G. Smith Jun 30 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. – G. Smith Jun 30 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. – Alexandre H. Tremblay Jun 30 at 11:27

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. – probably_someone Jun 30 at 3:06
• @probably_someone I've edited it to the correct value, so feel free to edit your answer accordingly :) – John Dumancic Jun 30 at 3:21

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. – G. Smith Jun 30 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. – G. Smith Jun 30 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 at 3:20
• No, I don’t think so. Your answer is confusing, but you can improve it. – G. Smith Jun 30 at 3:26
• It’s better now. – G. Smith Jun 30 at 3:41

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.