Lorentz Transformations

  • #1
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a)So I'm reading over my notes and they say that under the Lorentz transformation L, [itex]\phi \rightarrow \phi'[/itex] where [itex]\phi'(x)=\phi(x')[/itex] where [itex]x'^\mu = (L^{-1})^\mu{}_\nu x^\nu[/itex]

I don't really understand why this is true.

Why is it not just [itex]\phi'(x)= L \phi(x)[/itex]
Clearly this fails because the LHS is a scalar and the RHS should have indices on the L and so it won't be a balanced tensor equation but to my mind this is the right "form" that the equation should have.

Can anyone explain this to me?


b)However, earlier in my notes when Lorentz transformations were introduced it says
A Lorentz transformation is a linear transformation on space and time

[itex]x^\mu \rightarrow x'^\mu = L^\mu{}_\nu[/itex]
which preserves the relativistic line element [itex]ds^2[/itex]
and so
[itex]L^\mu{}_\sigma L^\nu{}_\tau g^{\sigma \tau} = g^{\mu \nu}[/itex] (*)

Why does it define the transformation as [itex]x^\mu \rightarrow x'^\mu = L^\mu{}_\nu[/itex] here and as [itex]x'^\mu = (L^{-1})^\mu{}_\nu x^\nu[/itex] in the other part of my notes (as I mentioned in part a) of my post) - are these equivalent? If so, how?

And how do we derive the equation (*)


Thanks a lot.
 
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Answers and Replies

  • #2
G01
Homework Helper
Gold Member
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a)So I'm reading over my notes and they say that under the Lorentz transformation L, [itex]\phi \rightarrow \phi'[/itex] where [itex]\phi'(x)=\phi(x')[/itex] where [itex]x'^\mu = (L^{-1})^\mu{}_\nu x^\nu[/itex]

I don't really understand why this is true.

Why is it not just [itex]\phi'(x)= L \phi(x)[/itex]
Clearly this fails because the LHS is a scalar and the RHS should have indices on the L and so it won't be a balanced tensor equation but to my mind this is the right "form" that the equation should have.

Can anyone explain this to me?

I think your just forgetting that [itex]\phi[/itex] is a scalar field. Scalars don't transform. That is the definition of a scalar. We don't multiply by a lambda because, by definition, our scalar quantity is invariant!

Think of the definition of tensors (Don't think of these tensors as fields yet.) by their transformation properties:

Rank 0 (Scalar): [itex]\phi\rightarrow\phi[/itex]

Rank1(Vector):[itex]x^{\mu}\rightarrow\Lambda^{\nu}_{\mu}x^{\mu}[/itex]

Rank2:[itex]A_{\mu\nu}\rightarrow\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}A_{\mu\nu}[/itex]

Again, we don't have a Lambda matrix multiplying [itex]\phi[/itex] because it is defined as a scalar. This is just what a scalar is!


However,in our case, our scalar is a field, defined at all points in space. This means it depends on spatial coordinates, which transform like vectors. Thus, we need to multiply the coordinates by lambda^-1 to transform them correctly.

I hope this helps explain why there is a lambda "inside" but not "outside."


b)However, earlier in my notes when Lorentz transformations were introduced it says
A Lorentz transformation is a linear transformation on space and time

[itex]x^\mu \rightarrow x'^\mu = L^\mu{}_\nu[/itex]
which preserves the relativistic line element [itex]ds^2[/itex]
and so
[itex]L^\mu{}_\sigma L^\nu{}_\tau g^{\sigma \tau} = g^{\mu \nu}[/itex] (*)

Why does it define the transformation as [itex]x^\mu \rightarrow x'^\mu = L^\mu{}_\nu[/itex] here and as [itex]x'^\mu = (L^{-1})^\mu{}_\nu x^\nu[/itex] in the other part of my notes (as I mentioned in part a) of my post) - are these equivalent? If so, how?

And how do we derive the equation (*)


Thanks a lot.

I'm new to field theory myself, so I'll think about this and hopefully get back to you. If not, I'll inform another homework helper about your question.
 
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  • #3
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I think your just forgetting that [itex]\phi[/itex] is a scalar field. Scalars don't transform. That is the definition of a scalar. We don't multiply by a lambda because, by definition, our scalar quantity is invariant!

Think of the definition of tensors (Don't think of these tensors as fields yet.) by their transformation properties:

Rank 0 (Scalar): [itex]\phi\rightarrow\phi[/itex]

Rank1(Vector):[itex]x^{\mu}\rightarrow\Lambda^{\nu}_{\mu}x^{\mu}[/itex]

Rank2:[itex]A_{\mu\nu}\rightarrow\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}A_{\mu\nu}[/itex]

Again, we won't have a Lambda matrix multiplying [itex]\phi[/itex] because it is defined as a scalar. This is just what a scalar is!


However,in our case, our scalar is a field, defined at all points in space. This means it depends on spatial coordinates, which transform like vectors. Thus, we need to multiply the coordinates by lambda^-1 to transform them correctly.

I hope this helps explain why there is a lambda "inside" but not "outside

ok. but shouldn't it then be [itex]\phi'(x)=\phi(x)[/itex] as the transformation law???

or is the following correct:

consider [itex]\phi'(x)[/itex]
by defn this is equal to [itex]\phi(x')[/itex] (is this actually just a defn i should accept?)
and then we do the lorentz transformation of the spatial coords to get the desired result?

thanks again!
 
  • #4
G01
Homework Helper
Gold Member
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ok. but shouldn't it then be [itex]\phi'(x)=\phi(x)[/itex] as the transformation law???


Remember that you are still moving from one frame to another. Thus, we need to transform to the coordinates of the new frame. In the above statement, you are essentially say the following:

"Our field is a scalar. Therefore, The field value in frame 1 written in terms of frame 1's coordinates is equal to the field value in frame 2, also written in terms of frame 1's corrdinates."

This is not true! Obviously when we are working in a new frame, we want to work with those new coordinates. Otherwise we will be comparing field values for different spatial locations, which obviously don't have to be equal. So what we really want to say is the following:

"Our field is a scalar. Therefore, the field value in frame 1 written in terms of frame 1's coordinates is equal to the field value in frame 2, written in terms of frame 2's coordinates."

Or in math:

[itex]
\bar{\phi}(\bar{x})=\phi(\bar{x})=\phi(\Lambda^{-1}x)
[/itex]

Again, I'll get back to you on why we use lambda inverse.
 
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  • #5
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0
Remember that you are still moving from one frame to another. Thus, we need to transform to the coordinates of the new frame. In the above statement, you are essentially say the following:

"Our field is a scalar. Therefore, The field value in frame 1 written in terms of frame 1's coordinates is equal to the field value in frame 2, also written in terms of frame 1's corrdinates.'

This is not true! Obviously when we are working in a new frame, we want to work with those new coordinates. Otherwise we will be comparing field values for different spatial positions, which obviously don't have to be equal. So what we really want to say is the following:

"Our field is a scalar. Therefore, the field value in frame 1 written in terms of frame 1's coordinates is equal to the field in frame 2, written in terms of frame 2's corrdinates."

Or in math:

[itex]
\bar{\phi}(x)=\phi(\bar{x})=\phi(\Lambda^{-1}x)
[/itex]

surely you mean
[itex]\phi'(x)=\phi(x') = \phi( \Lambda^{-1} x)[/itex]

ok. but i don't get why we don't write

[itex]\phi'(x') = \phi(x)[/itex]

i guess i'm just confused by us having primes on both sides of the equation rather than having them all on one side - surely this woould make more sense if we took a prime to mean frame 1 then what i've written above would say field in frame 1 in frame 1 coords equals field in frame 2 in frame 2 coords but [itex]\phi'(x)=\phi(x')[/itex] means field in frame 1 in frmae 2 coords equals field in frame 2 in frame 1 coords....i'm a bit confused still.

sorry!
 
  • #6
G01
Homework Helper
Gold Member
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surely you mean
[itex]\phi'(x)=\phi(x') = \phi( \Lambda^{-1} x)[/itex]

Yes I had a typo, but what I meant was:

[itex]\phi'(x')=\phi(x') = \phi( \Lambda^{-1} x)[/itex]

Does this help? Sorry for the mistake. (I'm not much farther into this stuff than you. I'm afraid we may be an example of the blind leading the blind!)

Also, the appearance of the inverse lambda in the field argument has to do with the difference between active and passive transforms. Check out the "Lorentz invariance" section of these lecture notes from David Tong:

http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf

The whole set of notes is at:

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

I found these notes quite helpful, especially for the discussion of classical field theory and Lorentz invariance.
 
  • #7
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Yes I had a typo, but what I meant was:

[itex]\phi'(x')=\phi(x') = \phi( \Lambda^{-1} x)[/itex]

Does this help? Sorry for the mistake. (I'm not much farther into this stuff than you. I'm afraid we may be an example of the blind leading the blind!)

Also, the appearance of the inverse lambda in the field argument has to do with the difference between active and passive transforms. Check out the "Lorentz invariance" section of these lecture notes from David Tong:

http://www.damtp.cam.ac.uk/user/tong/qft/one.pdf

The whole set of notes is at:

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

I found these notes quite helpful, especially for the discussion of classical field theory and Lorentz invariance.

yeah. thanks for that. i now understand the inverse lambda term.

still not sure why [itex]\phi'(x)=\phi(x')[/itex]? is this just a definition?
 
  • #8
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0
ok. part b) actually just follows quite neatly as a consequence of special relativity. The relativistic line element ds^2 must be invariant under Lorentz transformations and if you write it out it all works out.

So I just need clarification of whether or not [itex]\phi'(x)=\phi(x')[/itex] is an identity that I should just accept or whether it is in fact derivable?
 

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