# I Scalar field

1. May 16, 2017

### Silviu

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!

2. May 16, 2017

### vanhees71

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. May 16, 2017

### Silviu

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. May 16, 2017

### George Jones

Staff Emeritus
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?

Last edited: May 16, 2017
5. May 17, 2017

### Silviu

Sorry I am a bit confused. What is primed and what is unprimed?

6. May 17, 2017

### vanhees71

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. May 17, 2017

### Silviu

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. May 17, 2017

### vanhees71

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. May 17, 2017

### Silviu

But the argument of $\phi'$ is always seen from S', right?

10. May 17, 2017

Yes, sure.