Lorentz Transformations and Angular momentum | Tong's QFT notes

In summary: J^{\mu \nu}## can be contracted with ##\eta^{\mu \nu}## resulting in (1.56).In summary,In summary, the conversation discusses an example in which the speaker is trying to understand the analogy in field theory to the conservation of angular momentum in classical mechanics. The example involves infinitesimal Lorentz transformations and the application of the general Lorentz transformation condition. The speaker also asks for clarification on various equations and concepts, including the number of 4x4 antisymmetric matrices and the transformation of a scalar field. The expert then provides a detailed explanation and justification for each of the questions raised.
  • #1
JD_PM
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TL;DR Summary
I was reading example 1.3.3 of Tong's lecture notes on QFT and I did not understand several points of it.

I have attached the PDF.
I am reading Tong's lecture notes and I found an example in which there are several aspects I do not understand.

This example is aimed at:

- Understanding what is the analogy in field theory to the fact that, in classical mechanics, rotational invariance gives rise to conservation of angular momentum.

Let's start then.

The infinitesimal form of Lorentz transformations is given as

$$\Lambda^{\mu}_{ \ \nu} = \delta^{\mu}_{ \ \nu} + \omega^{\mu}_{ \ \nu}$$

Where ##\omega^{\mu}_{ \ \nu}## is infinitesimal.

We know that the general Lorentz transformation condition is as follows (let's forget about priming indices in this post):

$$\eta^{\mu \nu} = \Lambda^{\mu}_{ \ \sigma} \Lambda^{\nu}_{ \ \tau} \eta^{\sigma \tau}$$

Thus, applying it to the infinitesimal case we get

$$(\delta^{\mu}_{ \ \nu} + \omega^{\mu}_{ \ \nu})(\delta^{\nu}_{ \ \nu} + \omega^{\nu}_{ \ \nu}) \eta^{\sigma \tau} = \eta^{\mu \nu}$$

Let me skip the Algebra; just note that I have dropped the term ##\omega^{\mu}_{ \ \sigma} \omega^{\nu \sigma}##. Thus we get

$$\omega^{\mu \nu} = - \omega^{\nu \mu}$$

Thus ##\omega^{\mu \nu}## is antisymmetric. I understand everything at this point.

Then Tong says we can make a check to see that the number of 4x4 antisymmetric matrices is equal to the total number of LTs: 4x3/2=6. This is a basic Linear algebra issue but I do not see it.

Now he shows that the transformation on a scalar field is given by

$$\phi(x) \rightarrow \phi'(x) = \phi(\Lambda^{-1}x) = \phi(x^{\mu} - \omega^{\mu}_{ \ \nu} x^{\nu}) = \phi(x^{\mu}) - \omega^{\mu}_{ \ \nu} x^{\nu} \partial_{\mu} \phi (x)$$

What I do not get is why

$$\phi(\Lambda^{-1}x) = \phi(x^{\mu} - \omega^{\mu}_{ \ \nu} x^{\nu})$$

We know that

$$\Lambda^{-1} = \Lambda^{\nu}_{ \ \mu} = \delta_{ \ \mu}^{\nu} + \omega_{ \ \mu}^{\nu}$$

Thus I get the same but with a positive sign

$$\phi(\Lambda^{-1}x) = \phi(x^{\mu} + \omega^{\mu}_{ \ \nu} x^{\nu})$$

He justifies the sign to be positive when we apply active transformations. Then I think he made a typo and it should be positive (I hqve to say I do not understand why we define two kinds of transformations though: active and positive).

Please let me attach an image of the rest of the example I want to comment:

Captura de pantalla (992).png


From here I basically do not get the following:

- I do not get the last equality on 1.53 (i.e. why the last equality follows because ##\omega^{\mu}_{\mu}=0##; I do not see the antysimmetry argument).

- I do not see neither how to get 1.55 nor ##\partial_{\mu}(J^{\mu})^{\rho \sigma} = 0##

I know there are many questions here, so let's summarize:


1) Why The number of 4x4 antisymmetric matrices is equal to the total number of LTs: 4x3/2=6.

2) Why ##\phi(\Lambda^{-1}x) = \phi(x^{\mu} - \omega^{\mu}_{ \ \nu} x^{\nu})##

3) How to get equation 1.53

4) How to get equation 1.55

Thank you very much.
 

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  • #2
JD_PM said:
I have attached the PDF.

Don't attach it; link to it. Where online did you get it from?
 
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  • #3
It's here:

https://www.damtp.cam.ac.uk/user/tong/qft.html

Then Tong says we can make a check to see that the number of 4x4 antisymmetric matrices is equal to the total number of LTs: 4x3/2=6. This is a basic Linear algebra issue but I do not see it.
Think about ##\omega^{\mu \nu}## as a ##\mathbb{R}^{4 \times 4}## matrix. If it were unconstrained you'd have 16 independent real numbers in this matrix, but antisymmetry, ##\omega^{\mu \nu}=-\omega^{nu \mu}## are a lot of constraints. First of all the four diagonal elements of the matrix are 0 since antisymmetry means, e.g., ##\omega^{11}=-\omega^{11} \; \Rightarrow \; \omega^{11}=0##. This leaves you with the upper triangle of the matrix as the independent entries, i.e., the ##\sigma^{\mu \nu}## with ##\mu<\nu##. These are easy to count: ##\sigma^{01}##, ##\sigma^{02}##, ##\sigma^{03}## (generators of the boosts) and ##\sigma^{12}##, ##\sigma^{13}##, ##\sigma^{23}## (generators of rotations). The other matrix elements are determined from these 6 numbers due to the antisymmetry of the matrix.

What I do not get is why
$$\phi(\Lambda^{-1}x) = \phi(x^{\mu} - \omega^{\mu}_{ \ \nu} x^{\nu})$$
You only need the inverse of ##\Lambda## to linear order in ##\omega##. You know that
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}.$$
Contracting with ##\eta^{\rho \alpha}## gives (setting brackets for didactical clarification)
$$(\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} \eta^{\rho \alpha}) {\Lambda^{\nu}}_{\sigma} = \delta_{\sigma}^{\alpha}$$
or, more conveniently,
$${\Lambda_{\nu}}^{\alpha} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.$$
On the other hand
$${(\Lambda^{-1})^{\alpha}}_{\nu} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.$$
Since the inverse of a matrix is unique, this implies
$${(\Lambda^{-1})^{\alpha}}_{\nu}={\Lambda_{\nu}}^{\alpha}$$
and thus for an infinitesimal transformation
$${(\Lambda^{-1})^{\alpha}}_{\nu}=\eta_{\nu \mu} \eta^{\alpha \rho} {\Lambda^{\mu}}_{\rho} = \eta_{\nu \mu} \eta^{\alpha \rho}(\delta_{\rho}^{\mu} + {\omega^{\mu}}_{\rho}) = \delta_{\nu}^{\alpha} + {\omega_{\nu}}^{\alpha}=\delta_{\nu}^{\alpha} + \eta_{\nu \beta} \omega^{\beta \alpha} = \delta_{\nu}^{\alpha} -\eta_{\nu \beta} \omega^{\alpha \beta}=\delta_{\nu}^{\alpha} -{\omega^{\alpha}}_{\nu}.$$
This implies
$$(\Lambda^{-1} x)^{\mu} = x^{\mu} - {\omega^{\mu}}_{\nu} x^{\nu}.$$
From this (in linear order in ##\omega##) through Taylor expansion around ##x##
$$\phi'(x)=\phi(\Lambda^{-1} x)=\phi(x) - {\omega^{\mu}}_{\nu} x^{\nu} \partial_{\mu} \phi(x) + \mathcal{O}(\omega^2).$$
- I do not get the last equality on 1.53
$$\partial_{\mu} ({\omega^{\mu}}_{\nu} x^{\nu} \mathcal{L}) = {\omega^{\mu}_{\nu}}(\delta_{\mu}^{\nu} \mathcal{L} + x^{\nu} \partial_{\mu} \mathcal{L})={\omega^{\mu}_{\nu}}x^{\nu} \partial_{\mu} \mathcal{L},$$
because due to the antisymmetry of ##\omega^{\alpha \beta}## as well ass the symmetry of ##\eta_{\mu \nu}##
$${\omega^{\mu}_{\nu}}\delta_{\mu}^{\nu} =\omega^{\mu \nu} \eta_{\mu \nu} = \frac{1}{2} (\omega^{\mu \nu} - \omega^{\nu \mu}) \eta_{\mu \nu} = \eta_{\mu \nu} \omega^{\mu \nu} - \omega^{\nu \mu} \eta_{\nu \mu} = 0.$$
Noether's theorem tells you that (1.55) is conserved, but you want to get rid of the matrix ##\omega##. But the matrix ##\omega## can be arbitrarily chosen under the constraint that ##\omega_{\mu \nu}## is antisymmetric. So we rewrite the current a bit
$$j^{\mu} = -\omega_{\rho \nu} T^{\mu \rho} x^{\nu}=-\frac{1}{2} \omega_{\rho \nu} (T^{\mu \rho} x^{\nu} - T^{\mu \nu} x^{\rho}).$$
Now you can put for ##\omega_{\rho \nu}## all antisymmetric matrices, thus ##\partial_{\mu} j^{\mu}=0## implies that the ##(\mathcal{J}^{\mu})^{\rho \sigma}## are indeed 6 independent conserved currents, ##\partial_{\mu} (\mathcal{J}^{\mu})^{\rho \sigma}=0##.

I hope this helps a bit further.
 
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  • #4
vanhees71 said:
It's here:

https://www.damtp.cam.ac.uk/user/tong/qft.htmlThink about ##\omega^{\mu \nu}## as a ##\mathbb{R}^{4 \times 4}## matrix. If it were unconstrained you'd have 16 independent real numbers in this matrix, but antisymmetry, ##\omega^{\mu \nu}=-\omega^{nu \mu}## are a lot of constraints. First of all the four diagonal elements of the matrix are 0 since antisymmetry means, e.g., ##\omega^{11}=-\omega^{11} \; \Rightarrow \; \omega^{11}=0##. This leaves you with the upper triangle of the matrix as the independent entries, i.e., the ##\sigma^{\mu \nu}## with ##\mu<\nu##. These are easy to count: ##\sigma^{01}##, ##\sigma^{02}##, ##\sigma^{03}## (generators of the boosts) and ##\sigma^{12}##, ##\sigma^{13}##, ##\sigma^{23}## (generators of rotations). The other matrix elements are determined from these 6 numbers due to the antisymmetry of the matrix.You only need the inverse of ##\Lambda## to linear order in ##\omega##. You know that
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} = \eta_{\rho \sigma}.$$
Contracting with ##\eta^{\rho \alpha}## gives (setting brackets for didactical clarification)
$$(\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} \eta^{\rho \alpha}) {\Lambda^{\nu}}_{\sigma} = \delta_{\sigma}^{\alpha}$$
or, more conveniently,
$${\Lambda_{\nu}}^{\alpha} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.$$
On the other hand
$${(\Lambda^{-1})^{\alpha}}_{\nu} {\Lambda^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.$$
Since the inverse of a matrix is unique, this implies
$${(\Lambda^{-1})^{\alpha}}_{\nu}={\Lambda_{\nu}}^{\alpha}$$
and thus for an infinitesimal transformation
$${(\Lambda^{-1})^{\alpha}}_{\nu}=\eta_{\nu \mu} \eta^{\alpha \rho} {\Lambda^{\mu}}_{\rho} = \eta_{\nu \mu} \eta^{\alpha \rho}(\delta_{\rho}^{\mu} + {\omega^{\mu}}_{\rho}) = \delta_{\nu}^{\alpha} + {\omega_{\nu}}^{\alpha}=\delta_{\nu}^{\alpha} + \eta_{\nu \beta} \omega^{\beta \alpha} = \delta_{\nu}^{\alpha} -\eta_{\nu \beta} \omega^{\alpha \beta}=\delta_{\nu}^{\alpha} -{\omega^{\alpha}}_{\nu}.$$
This implies
$$(\Lambda^{-1} x)^{\mu} = x^{\mu} - {\omega^{\mu}}_{\nu} x^{\nu}.$$
From this (in linear order in ##\omega##) through Taylor expansion around ##x##
$$\phi'(x)=\phi(\Lambda^{-1} x)=\phi(x) - {\omega^{\mu}}_{\nu} x^{\nu} \partial_{\mu} \phi(x) + \mathcal{O}(\omega^2).$$

$$\partial_{\mu} ({\omega^{\mu}}_{\nu} x^{\nu} \mathcal{L}) = {\omega^{\mu}_{\nu}}(\delta_{\mu}^{\nu} \mathcal{L} + x^{\nu} \partial_{\mu} \mathcal{L})={\omega^{\mu}_{\nu}}x^{\nu} \partial_{\mu} \mathcal{L},$$
because due to the antisymmetry of ##\omega^{\alpha \beta}## as well ass the symmetry of ##\eta_{\mu \nu}##
$${\omega^{\mu}_{\nu}}\delta_{\mu}^{\nu} =\omega^{\mu \nu} \eta_{\mu \nu} = \frac{1}{2} (\omega^{\mu \nu} - \omega^{\nu \mu}) \eta_{\mu \nu} = \eta_{\mu \nu} \omega^{\mu \nu} - \omega^{\nu \mu} \eta_{\nu \mu} = 0.$$
Noether's theorem tells you that (1.55) is conserved, but you want to get rid of the matrix ##\omega##. But the matrix ##\omega## can be arbitrarily chosen under the constraint that ##\omega_{\mu \nu}## is antisymmetric. So we rewrite the current a bit
$$j^{\mu} = -\omega_{\rho \nu} T^{\mu \rho} x^{\nu}=-\frac{1}{2} \omega_{\rho \nu} (T^{\mu \rho} x^{\nu} - T^{\mu \nu} x^{\rho}).$$
Now you can put for ##\omega_{\rho \nu}## all antisymmetric matrices, thus ##\partial_{\mu} j^{\mu}=0## implies that the ##(\mathcal{J}^{\mu})^{\rho \sigma}## are indeed 6 independent conserved currents, ##\partial_{\mu} (\mathcal{J}^{\mu})^{\rho \sigma}=0##.

I hope this helps a bit further.

I will read it carefully, thank you.
 
  • #5
Vanhees71 thank you very much, your explanation made me understand almost everything.

Here's my issue now: I do not see why the following holds

$$\partial_{\mu} ({\omega^{\mu}}_{\nu} x^{\nu} \mathcal{L}) = {\omega^{\mu}_{\nu}}(\delta_{\mu}^{\nu} \mathcal{L} + x^{\nu} \partial_{\mu} \mathcal{L})$$

I would expect instead to have the following

$$\partial_{\mu} ({\omega^{\mu}}_{\nu} x^{\nu} \mathcal{L}) = \omega^{\mu}_{\nu} x^{\nu} \partial_{\mu} \mathcal{L}$$

Thus, what I do not see is why the term ##\omega^{\mu}_{\nu}\delta_{\mu}^{\nu} \mathcal{L}## arises.
 
  • #6
It's from the product rule and
$$\partial_{\mu} x^{\nu} = \delta_{\mu}^{\nu}.$$
 
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  • #7
vanhees71 said:
It's from the product rule and
$$\partial_{\mu} x^{\nu} = \delta_{\mu}^{\nu}.$$

Ahh sorry Sr, I see it now! Thank you again.

I forgot that

$$\partial_{\mu} x^{\nu} = \frac{\partial x^{\nu}}{\partial x^{\mu}} = \delta_{\mu}^{\nu}.$$
 
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Related to Lorentz Transformations and Angular momentum | Tong's QFT notes

1. What are Lorentz Transformations?

Lorentz Transformations are mathematical equations that describe how the measurements of space and time change between two reference frames that are moving at a constant velocity relative to each other. They were developed by Hendrik Lorentz and are a fundamental concept in Einstein's theory of Special Relativity.

2. How do Lorentz Transformations affect the concept of time dilation?

Lorentz Transformations show that time is not absolute and can be experienced differently by observers in different reference frames. This is known as time dilation, where time appears to pass slower for objects moving at high speeds relative to an observer. Lorentz Transformations provide the mathematical framework for calculating this effect.

3. How are Lorentz Transformations related to the concept of length contraction?

Similar to time dilation, Lorentz Transformations also show that length is not absolute and can appear to contract for objects moving at high speeds. This is known as length contraction and is a direct consequence of the equations in Lorentz Transformations.

4. What is the significance of Lorentz Transformations in quantum field theory?

In quantum field theory, Lorentz Transformations play a crucial role in understanding the behavior of particles and fields at high speeds and in different reference frames. They are used to transform the equations of motion for particles and fields to different frames of reference, allowing for a deeper understanding of their behavior.

5. How are Lorentz Transformations related to angular momentum?

Lorentz Transformations are closely related to the concept of angular momentum, which is a measure of the rotational motion of a particle or system. This is because Lorentz Transformations include rotations in their equations, and thus, the transformations of angular momentum can be described using these equations.

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