Lorentz transformation for derivatives

[Haorong Wu]

TL;DR Summary

should $\frac {\partial} {\partial x ^{\mu}}$ transform to $\frac {\partial} {\partial y ^{\mu}}$ or remain unchanged?

Hello again. I am sorry I got another problem when learning QFT regarding the Lorentz transformation of derivatives.

In David Tong's notes, he says

Consider a real scalar field transformed as $\phi \left ( x \right ) \rightarrow \phi ^{'} \left ( x \right ) =\phi \left ( \Lambda ^{-1 }x \right )$.

Then the derivative of the scalar field transforms as a vector, meaning
$\left ( \partial _{\mu} \phi \right ) \left ( x \right ) \rightarrow \left ( \Lambda ^{-1 }\right ) ^{\nu} {}_{\mu} \left ( \partial _{\nu} \phi \right ) \left ( y \right )$ where $y=\Lambda ^{-1 }x$

I am not sure how to transform the partial derivatives. Explicitly, should $\frac {\partial} {\partial x ^{\mu}}$ transform to $\frac {\partial} {\partial y ^{\mu}}$ or remain unchanged? And why?

At first, I think since $x$ is transformed to $y$, then $\partial x^\mu$ should transform to $\partial y^\mu$ as well. However, after I tried the calculation, I found that, the derivative does not transform and it is still $\partial x^\mu$. I could not find a proper explanation to convince myself.

 

[PeroK]

Let's take an example:
$$t = \gamma(t' + vx'), \ \ \ x = \gamma(x' + vt')$$
We need the value of the field at the transformed coordinates to be the same physical value as the original field at the original coordinates. Hence:
$$\phi'(t', x') = \phi(t, x) = \phi(\gamma(t' + vx'), \gamma(x' + vt'))$$
Now the derivatives of the field are:
$$\frac{\partial\phi'(t', x')}{\partial t'} = \frac{\partial\phi(\gamma(t' + vx'), \gamma(x' + vt'))}{\partial t'} = \frac{\partial\phi(t, x)}{\partial t}\gamma + \frac{\partial\phi(t, x)}{\partial x}\gamma v$$
Hence:
$$\frac{\partial\phi'(t', x')}{\partial t'} = \gamma \frac{\partial\phi(t, x)}{\partial t} + \gamma v\frac{\partial\phi(t, x)}{\partial x}$$
Now, you just have to generalise that using $\Lambda$ and upper and lower indices on the spacetime coordinates.

 

[Haorong Wu]

Thank you very much, @PeroK !I never think it in this way,especially your equation $\phi'(t', x') = \phi(t, x)$ guide me through the problem. I also learn from them that $\partial _ \mu \phi^{'}$ means $\frac {\partial \phi \left ( y \right )} {\partial y^\mu}$.

I think the equations in David Tong's note are somehow confusing. After I rewrite it , everything works out.

Let $\phi \left ( x \right ) \rightarrow \phi ^{'} \left ( x^{'} \right ) =\phi \left ( x \right )$ where $x^{'}=\Lambda x$. Then
$\left ( \partial _{\mu} \phi \right ) \left ( x \right ) = \frac {\partial \phi \left ( x \right )} {\partial x^\mu} \rightarrow \frac {\partial \phi^{'} \left ( x^{'} \right )} { \partial x^{' \mu}}=\frac {\partial \phi \left ( x \right )} { \partial x^{' \mu}}=\frac {\partial \phi \left ( x \right )} { \partial x^{ \sigma}} \frac {\partial x^{ \sigma}} {\partial x^{' \mu}}=\left ( \Lambda ^{-1} \right ) ^\sigma {} _\mu \partial _\sigma \phi$

Cheers!

 

[Haorong Wu]

Hello, @PeroK. I would like to ask some advice about the derivatives in tensor forms.

Given $\partial _ \mu \phi ^ \nu$, how to calculate $\partial ^ \mu \phi ^ \nu$, $\partial _ \mu \phi _ \nu$, and $\partial ^ \mu \phi _ \nu$?

I have done the calculation. I have
$\partial ^ \mu \phi ^ \nu = \eta_{\mu \lambda} \partial _ {\lambda} \phi ^ \nu$,
$\partial _ \mu \phi _ \nu = \eta_{\nu \lambda} \partial _ {\mu} \phi ^ \lambda$, and
$\partial ^ \mu \phi _ \nu = \eta_{\mu \sigma} \eta_{\nu \lambda} \partial _ {\sigma} \phi ^ \lambda$.

I find these derivatives quite useful, but I do not know whether I have the right derivatives, and I do not know where to go to check them. Do you have any information about them?

Thanks!

 

[anuttarasammyak]

$$\phi^{\nu,\mu}=\eta^{\mu \lambda} \phi^{\nu}_{\ \ ,\lambda}$$
$$\phi_{\nu\ ,\mu}=\eta_{\nu\lambda} \phi^{\lambda}_{\ \ ,\mu}$$
$$\phi^{\ \ ,\mu}_{\nu}=\eta^{\sigma \mu} \eta^\nu_\lambda\phi^\lambda_{\ \ ,\sigma}$$
where
$$X_{,\mu}=\frac{\partial X}{\partial x^\mu}$$
$$X^{,\mu}=\frac{\partial X}{\partial x_\mu}$$

EDIT correction of the third equation
$$\phi^{\ \ ,\mu}_{\nu}=\eta^{\sigma \mu} \eta_{\nu\lambda}\phi^\lambda_{\ \ ,\sigma}$$
as pointed out.

 

[Haorong Wu]

$$\phi^{\nu,\mu}=\eta^{\mu \lambda} \phi^{\nu}_{\ \ ,\lambda}$$
$$\phi_{\nu\ ,\mu}=\eta_{\nu\lambda} \phi^{\lambda}_{\ \ ,\mu}$$
$$\phi^{\ \ ,\mu}_{\nu}=\eta^{\sigma \mu} \eta^\nu_\lambda\phi^\lambda_{\ \ ,\sigma}$$
where
$$X_{,\mu}=\frac{\partial X}{\partial x^\mu}$$
$$X^{,\mu}=\frac{\partial X}{\partial x_\mu}$$

Thanks, @anuttarasammyak .

If I use the metric tensor $\eta ^{\mu \nu}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\\0 & 0 &-1 & 0\\0 & 0 & 0 & -1 \end{pmatrix}$, then whether the indices are upper or lower would be irrelavent. After checking with your equations, I think I got them right.

Thanks!

 

[PeroK]

Hello, @PeroK. I would like to ask some advice about the derivatives in tensor forms.

Given $\partial _ \mu \phi ^ \nu$, how to calculate $\partial ^ \mu \phi ^ \nu$, $\partial _ \mu \phi _ \nu$, and $\partial ^ \mu \phi _ \nu$?

I have done the calculation. I have
$\partial ^ \mu \phi ^ \nu = \eta_{\mu \lambda} \partial _ {\lambda} \phi ^ \nu$,
$\partial _ \mu \phi _ \nu = \eta_{\nu \lambda} \partial _ {\mu} \phi ^ \lambda$, and
$\partial ^ \mu \phi _ \nu = \eta_{\mu \sigma} \eta_{\nu \lambda} \partial _ {\sigma} \phi ^ \lambda$.

I find these derivatives quite useful, but I do not know whether I have the right derivatives, and I do not know where to go to check them. Do you have any information about them?

Thanks!

You need to find a reference on tensor notation. The free indices must match on each side of the equation, and you sum over pairs of one upper and one lower index. Your formulas will work out because the metric tensor is the same in upper and lower format.

In flat spacetime we have a four vector in upper and lower index notation:
$$a^{\mu} = (a^0, a^1, a^2, a^3), \ \text{and} \ a_{\mu} = (a_0, a_1, a_2, a_3)$$
Where $a^0 = a_0 = a_t, a^1 = -a_1 = a_x , a^2 = -a_2 = a_y, a^3 = -a_3 = a_z$

This can be written using the metric tensor $\eta^{\mu\nu} = \eta_{\mu\nu} = diag(1, -1, -1, -1)$:
$$a^{\mu} = \eta^{\mu\nu} a_{\nu}$$
The derivatives are defined by:
$$\partial_{\mu} = \frac{\partial}{\partial x^{\mu}} \ \text{and} \ \partial^{\mu} = \frac{\partial}{\partial x_{\mu}}$$
And these are related:
$$\partial^{\mu} = \eta^{\mu\nu}\partial_{\mu}$$
The derivative of a scalar is a four-vector:
$$\partial^{\mu} \phi = \eta^{\mu\nu}\partial_{\mu}\phi$$
And the derivative of a four-vector has upper and lower components:
$$\partial^{\mu} \phi^{\lambda} = \eta^{\mu\nu} \eta^{\lambda\rho}\partial_{\mu}\phi_{\rho}$$

 

[Haorong Wu]

Thanks, @PeroK .

I intentially misuse the upper and lower indices since the metric tensor in QFT seems not to play an important role. (Am I right? I have not met the situations where they are important.)

However, I will use your expression since they would be much more easier to memorize.

Thanks!

From now on, I would just use a metric tensor to lower or raise a indice in the partial derivatives.

 

[anuttarasammyak]

If I use the metric tensor ημν=(10000−10000−10000−1), then whether the indices are upper or lower would be irrelavent.

And
$$\eta_{\mu}^\nu= \begin{pmatrix} 1 & 0 & 0& 0 \\ 0 & 1 & 0& 0 \\ 0 & 0 & 1& 0 \\ 0 & 0 & 0& 1 \\ \end{pmatrix}=\delta_{\mu}^\nu$$

 

[Haorong Wu]

And
$$\eta_{\mu}^\nu= \begin{pmatrix} 1 & 0 & 0& 0 \\ 0 & 1 & 0& 0 \\ 0 & 0 & 1& 0 \\ 0 & 0 & 0& 1 \\ \end{pmatrix}=\delta_{\mu}^\nu$$

But...I do not understand this equation
$\phi^{\ \ ,\mu}_{\nu}=\eta^{\sigma \mu} \eta^\nu_\lambda\phi^\lambda_{\ \ ,\sigma}$

I only get
$\partial ^ \mu \phi _\nu = \frac {\partial \phi _\nu} {\partial x_\mu}=\frac {\eta _{\nu \lambda} \partial \phi ^ \lambda} {\eta_{\mu \sigma} \partial x^ \sigma}=\eta _{\nu \lambda} \eta ^{\mu \sigma} \partial _\sigma \phi ^ \lambda$

 

[anuttarasammyak]

You are right. I wrote it with mistake.

 

[Haorong Wu]

You are right. I wrote it with mistake.

OK. Thanks, @anuttarasammyak .

 

[PeroK]

@Haorong Wu another important point is the defining property of a Lorentz Transformation:

$\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}$

 

[Haorong Wu]

@Haorong Wu another important point is the defining property of a Lorentz Transformation:

$\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}$

Oh, yes. David Tong mentioned it in his note. Also, I derived that $\eta_{\mu\nu} \left ( \Lambda ^{-1} \right )^{\mu}{}_{\rho}\left ( \Lambda ^{-1} \right )^{\nu}{}_{\sigma} = \eta_{\rho \sigma}$, hoping this would be correct.

 

[PeroK]

Oh, yes. David Tong mentioned it in his note. Also, I derived that $\eta_{\mu\nu} \left ( \Lambda ^{-1} \right )^{\mu}{}_{\rho}\left ( \Lambda ^{-1} \right )^{\nu}{}_{\sigma} = \eta_{\rho \sigma}$, hoping this would be correct.

Well, $\Lambda ^{-1}$ is also a Lorentz Transformation, so that equation must hold!