# Lorentz transformation matrix applied to EM field tensor

1. May 20, 2013

### qtm912

In a recent course on special relativity the lecturer derives the Lorentz transformation matrix for the four vector of position and time. Then, apparently without proof, the same matrix is used to transform the EM field tensor to the tensor for the new inertial frame. I am unclear whether it should be obvious (if so why?) that the two are the same or whether the proof is non trivial (sketch of proof would help) and was just omitted.

Thank you.

2. May 20, 2013

### WannabeNewton

I'm not sure what you are asking. What do you want a proof of and what do you think should or should not be the same under the Lorentz transformation? If $F_{\mu\nu}$ are the components of the EM field tensor and $\Lambda_{\mu\nu}$ are the components of the Lorentz transformation then the components of the EM field tensor transform as $F_{\mu'\nu'} = \Lambda^{\alpha}_{\mu'}\Lambda^{\beta}_{\nu'}F_{\alpha\beta}$. The components $F_{\mu'\nu'}$ are not in general the same as the components $F_{\mu\nu}$ if that is what you are asking.

3. May 20, 2013

### qtm912

Thanks for the reply. I understand that the components of the transformed field tensor will be different. What I was unclear about is why the lambda matrix is the same for the EM field as it is for the position four vector.

4. May 20, 2013

### Mentz114

I was puzzled by the mention of a position vector.

It follows from $F_{\mu'\nu'} = \Lambda^{\alpha}_{\mu'}\Lambda^{\beta}_{\nu'}F_{\alpha\beta}$ that $F^{\mu'\nu'}F_{\mu'\nu'}=F^{\mu\nu}F_{\mu\nu}$

5. May 20, 2013

### WannabeNewton

If we have a coordinate transformation $x^{\mu}\rightarrow x^{\mu'}$ then the components of all tensors $T^{a_1...a_n}_{b_1...b_m}$ will transform as $$T^{\mu_1'...\mu_n'}_{\nu_1'...\nu_m'} = \frac{\partial x^{\mu_{1}'}}{\partial x^{\mu_{1}}}...\frac{\partial x^{\mu_{n}'}}{\partial x^{\mu_{n}}}\frac{\partial x^{\nu_{1}}}{\partial x^{\nu_{1}'}}...\frac{\partial x^{\nu_{n}}}{\partial x^{\nu_{m}'}}T^{\mu_1...\mu_n}_{\nu_1...\nu_m}$$

So if we map $x^{\mu}\rightarrow x^{\mu'} = \Lambda^{\mu'}_{\nu}x^{\nu}$, where $\Lambda^{\mu'}_{\nu}$ are the components of the Lorentz transformation, then $$\frac{\partial x^{\alpha}}{\partial x^{\mu'}} = \frac{\partial }{\partial x^{\mu'}}(\Lambda^{\alpha}_{\nu'}x^{\nu'}) = \Lambda^{\alpha}_{\nu'}\delta^{\nu'}_{\mu'} = \Lambda^{\alpha}_{\mu'}$$ thus $$F_{\mu'\nu'} = \frac{\partial x^{\alpha}}{\partial x^{\mu'}}\frac{\partial x^{\beta}}{\partial x^{\nu'}}F_{\alpha\beta} = \Lambda^{\alpha}_{\mu'}\Lambda^{\beta}_{\nu'}F_{\alpha\beta}$$

6. May 20, 2013

### qtm912

Thanks, this is what I was looking for. Thank you for clarifying and sorry if the initial question was unclear.

7. May 20, 2013

### WannabeNewton

No problem mate! Feel free to ask any further questions you may have.

8. May 20, 2013

### qtm912

Dear Mentz
Thanks for taking the trouble to reply. I meant coordinate transformation and I should not have mentioned the position vector. Anyway you have both addressed my question, thanks for that.

9. May 22, 2013

### HomogenousCow

That's quite an elegant solution.