Understanding Lorentz Transformation on Scalar Fields

In summary: And the argument of ##\phi## is always seen from S. And ##x'## is always seen from S. So ##x## is always seen from S. Sorry for the confusion. You can (and should) fix a frame from which you work, and that is fixed. The other frame is the moving one (and can be rotated, boosted...).But the argument of ##\phi'## is always seen from S', right?Yes, sure. And the argument of ##\phi## is always seen from S. And ##x'## is always seen from S. So ##x## is always seen from S. Sorry for the confusion. You can (and should) fix a frame from which you work,
  • #1
Silviu
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Hello! Can someone explain to me how does a scalar field changes under a Lorentz transformation? I found different notations in different places and I am a bit confused. Thank you!
 
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  • #2
If ##x'=\Lambda x##, where ##\Lambda## is a Lorentz-transformation matrix, then a scalar field obeys by definition the transformation law
$$\phi'(x')=\phi(x)=\phi(\Lambda^{-1} x').$$
 
  • #3
vanhees71 said:
If ##x'=\Lambda x##, where ##\Lambda## is a Lorentz-transformation matrix, then a scalar field obeys by definition the transformation law
$$\phi'(x')=\phi(x)=\phi(\Lambda^{-1} x').$$
Thank you for your reply. This makes sense. However I found in Peskin's book on QFT a definition that is different from yours by a prime ( ' ) - I attached a screenshot of it. That is what got me confused. Do you know what does he mean by his notation?
 

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  • #4
Silviu said:
Thank you for your reply. This makes sense. However I found in Peskin's book on QFT a definition that is different from yours by a prime ( ' ) - I attached a screenshot of it. That is what got me confused. Do you know what does he mean by his notation?

The expression that vanhees71 wrote is equivalent to the expression in Peskin and Schroeder, and both expressions are equivalent to

$$ \phi'(Fred)=\phi(\Lambda^{-1} Fred).$$

Why?
 
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  • #5
George Jones said:
The expression that vanhees71 wrote is equivalent to the expression in Peskin and Schroeder, and both expressions are equivalent to

$$ \phi'(Fred)=\phi(\Lambda^{-1} Fred).$$

Why?
Sorry I am a bit confused. What is primed and what is unprimed?
 
  • #6
Just read the equation as a whole. My final equation was
$$\phi'(x')=\phi(\Lambda^{-1} x').$$
Now rename ##x'## back to ##x##, and you get Peskin Schroeder's formula. You can name it "Fred" as suggested in #4 (although that's a bit unusual ;-)).
 
  • #7
vanhees71 said:
Just read the equation as a whole. My final equation was
$$\phi'(x')=\phi(\Lambda^{-1} x').$$
Now rename ##x'## back to ##x##, and you get Peskin Schroeder's formula. You can name it "Fred" as suggested in #4 (although that's a bit unusual ;-)).
What do you mean by rename? If we have a frame S stationary with respect to the field and S' moving with respect to the field (so ##\phi'## and x' are measured in S') then we have by definition ##\phi'(x')=\phi(x)## and by the Lorentz transformation we also have ##\phi(x)=\phi(\Lambda^{-1} x')##. This make sense. But in Peskin he has a mix of both ##\phi'(x)## and this is what confuses me. How does he get to a mix of primed and unprimed indices without a ##\Lambda## factor somwhere? And I am not sure how can you rename x' to x, once you decided which moves and which is fixed. (so to be clear, I understand that choosing prime and unprime as moving or not moving is arbitrary, but Pesking seems to mix them, which confuses me). Thank you for help!
 
  • #8
The formula of itself is unique. You can name the argument as you like. You can as well write the law as
$$\phi'(y)=\phi(\Lambda^{-1} y).$$
Of course, the prime at the field symbol on the left-hand side is crucial!
 
  • #9
vanhees71 said:
The formula of itself is unique. You can name the argument as you like. You can as well write the law as
$$\phi'(y)=\phi(\Lambda^{-1} y).$$
Of course, the prime at the field symbol on the left-hand side is crucial!
But the argument of ##\phi'## is always seen from S', right?
 
  • #10
Yes, sure.
 

1. What is the Lorentz Transformation on Scalar Fields?

The Lorentz Transformation on Scalar Fields is a mathematical formula that describes how quantities such as time, distance, and energy change when viewed from different reference frames in special relativity.

2. Why is understanding Lorentz Transformation on Scalar Fields important?

Understanding Lorentz Transformation on Scalar Fields is important because it helps us make accurate predictions and measurements in the realm of special relativity. It also serves as the basis for many other important concepts, such as time dilation and length contraction.

3. How does the Lorentz Transformation affect scalar fields?

The Lorentz Transformation affects scalar fields by changing their values in different reference frames. This is due to the fact that scalar fields are quantities that do not have direction, and thus do not transform in the same way as vector quantities under the Lorentz Transformation.

4. Can you give an example of how the Lorentz Transformation affects a scalar field?

One example of the Lorentz Transformation affecting a scalar field is when we measure the length of an object in two different reference frames. In the reference frame where the object is moving, the length will appear shorter due to length contraction, while in the stationary frame, the length will remain unchanged. This is because the scalar field of length is being transformed by the Lorentz Transformation.

5. How do scalar fields behave under Lorentz Transformation when compared to vector fields?

Unlike vector fields, which transform by multiplication with a matrix, scalar fields transform by a simple scaling factor known as the Lorentz factor. This means that the values of scalar fields change in a more straightforward manner when viewed from different reference frames compared to vector fields.

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