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Seeking derivation of real scalar field Lagrangian

by snoopies622
Tags: derivation, field, lagrangian, real, scalar, seeking
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snoopies622
#1
Jan20-13, 10:07 AM
P: 611
Here and there I come across the following formula for the Lagrangian density of a real scalar field, but not a deriviation.

[tex]
\mathcal{L} = \frac {1}{2} [ \dot \phi ^2 - ( \nabla \phi ) ^2 - (m \phi )^2 ]

[/tex]

Could someone show me where this comes from? The m squared term in particular is a mystery.
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snoopies622
#2
Jan21-13, 10:18 AM
P: 611
The first two terms look like kinetic and potential energy, but the third one . . all that comes to mind is the mass-energy relationship of special relativity, but I thought this formula was from classical mechanics.
The_Duck
#3
Jan21-13, 12:19 PM
P: 844
In what context are you studying this? Most often this comes up at the beginning of a quantum field theory course.

The Lagrangian is more a definition than anything; when we write down the Lagrangian we are saying, "Let's imagine a field whose dynamics are described by this Lagrangian. How does it behave?" And we can write down any Lagrangian we want.

However, there are some constraints we tend to impose on what Lagrangians we write down and study. One is Lorentz invariance: the Lagrangian should describe a field whose dynamics are consistent with special relativity. This limits what terms can appear in the Lagrangian, but doesn't determine it fully. If we ask for the simplest Lorentz invariant Lagrangian for a scalar field, we get the one you wrote down above. This is why this Lagrangian comes up often.

m is a free parameter. In quantum field theory, a scalar field with this Lagrangian has excitations which turn out to be particles of mass m.

snoopies622
#4
Jan21-13, 05:44 PM
P: 611
Seeking derivation of real scalar field Lagrangian

Yes, I bumped into it studying quantum field theory. It's supposed to have the same form as the Klein-Gordon Lagrangian, yet come about without any quantum assumptions.

I assumed that there was some physical situation that corresponds to it, as

[tex]

L = \frac {1}{2}m \dot {x}^2 - \frac {1}{2}kx^2

[/tex]

corresponds to a mass on a spring obeying Hooke's law.
andrien
#5
Jan22-13, 07:00 AM
P: 1,020
why not.
the first two terms form a kinetic energy part
1/2(∂ψ/∂t)2-1/2(∇ψ)2 can be written as 1/2(∂μψ)2 which corresponds to kinetic energy and rest is of course some potential energy of spring type thing 1/2(mψ)2.
snoopies622
#6
Jan22-13, 09:53 AM
P: 611
Oh I see — the first two terms are analogous to the Minkowski displacement vector with magnitude [itex]ds^2 = (c dt) ^ 2 - dx ^2 [/itex] and the third term is the potential.

Yes, that makes some sense. Thanks, Andrien! :)
snoopies622
#7
Jan22-13, 11:03 AM
P: 611
That's funny, I just noticed that it doesn't have the same form as the Klein-Gordon Lagrangian at all. One has second derivatives and the other second powers.

Now I'm wondering where I read that . .

Edit: Oops, one's a Lagrangian and one isn't. Moderators, feel free to delete this entry #7.


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