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Explicit x dependence in Lagrangian

  1. Jul 24, 2012 #1
    My friend and I have been getting all confused about the following problem with a Lagrangian. It comes from David Tong's online notes on QFT, but given it is about the Lagrangian, I figure it does well in this section.

    Ok, Tong is talking about Noether's theorem, and using the example of space-time translation
    $$x^\mu \rightarrow x^\mu - \epsilon^\mu\quad \Rightarrow \phi_a(x) \rightarrow \phi_a(x) + \epsilon^\nu \partial_\nu \phi_a$$
    He then says that the Lagrangian transforms the following way if it has no explicit [itex]x[/itex] dependence, but only depends on [itex]x[/itex] through the fields [itex]\phi_a(x)[/itex]

    $$\mathcal{L}(x) \rightarrow \mathcal{L}(x) + \epsilon^\nu \partial_\nu \mathcal{L}(x)$$

    My main question is simple: what is meant by "explicit [itex]x[/itex] dependence"? As far as I understand it [itex]\mathcal{L}[/itex] can be written in terms of [itex]\phi_a(x)[/itex] and [itex]\dot\phi_a(x)[/itex] alone (so no explicit [itex]x[/itex] dependence) or the formula for [itex]\phi_a(x)[/itex] and [itex]\dot\phi_a(x)[/itex] explicitly written out in terms of [itex]x[/itex], in which case, there is clearly explicit [itex]x[/itex] dependence. Given they're the same equation, my friend and I are somewhat confused. Our current thinking is that "explicit [itex]x[/itex] dependence" means that [itex]\mathfrak{L}[/itex] could not be written in terms of just [itex]\phi_a(x)[/itex] and [itex]\dot\phi_a(x)[/itex], but there'd be some additional [itex]x[/itex] terms floating around.

    Oh, and what confused us further is that if there is no explicit [itex]x[/itex] dependence, then doesn't this mean that [itex]\partial_\nu \mathcal{L}(x) = 0 [/itex] in the above equation?

    (On an aside note, in the above transformation, does [itex]\delta\phi_a =\epsilon^\nu \partial_\nu \phi_a[/itex] or [itex]\delta\phi_a =\partial_\nu \phi_a[/itex]? Tong has used both, and it's unclear. If you know about this, could anyone provide a mathematical definition of what [itex]\delta x [/itex] quantities are in the calculus of variations, or provide some rigourous resources for finding out what they are?)
     
  2. jcsd
  3. Jul 24, 2012 #2

    mfb

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    An explicit x-dependence would be an expression like [itex]x \phi_a(x)[/itex] - something which changes under the transformation given in your post. If everything consists of [itex]\phi_a(x)[/itex] and [itex]\dot\phi_a(x)[/itex] only, you do not have any explicit x-dependence.
    I would expect that [itex]\partial_\nu\mathcal{L}(x)=0[/itex] in this case, indeed.

    Concerning your last question, I think it is the former one, but the latter one might be possible to use, too.

    Classical physics, really? ;)
     
  4. Jul 24, 2012 #3
    OK, say [itex]\mathcal{L} = x \phi_a(x)[/itex]. But what if [itex]\phi_a(x)[/itex] is an invertible function of [itex]x[/itex], i.e. [itex]\phi_a(x) = 2x[/itex]? Then [itex]\mathcal{L} = \phi_a^2(x)/2[/itex]. So could you not argue that there is now no explicit [itex]x[/itex] dependence? Of course, this depends on [itex]\phi_a(x)[/itex], and I think it goes back to my first point: [itex]\mathcal{L}[/itex] cannot be written purely as a function of [itex]\phi_a(x)[/itex] and [itex]\dot\phi_a(x)[/itex], but has additional [itex]x[/itex] terms in it. Such and example would be [itex]\phi_a(x) = x^2 + x[/itex] with your given [itex]\mathcal{L}=x\phi_a(x)[/itex].

    However, this seems all too messy, and I'm sure there must be a better way of characterising this.

    I would also expect that [itex]\partial_\nu\mathcal{L}(x)=0[/itex]. However, if one does so, then the derivation of the stress-energy tensor is incorrect, as it includes the (presumably nonzero) terms derived from this variation of the Lagrangian.

    Oh, and as I said, this is classical fields and lagrangians, so I think it belongs here. If you've got a better place, I'm more than happy for it to be moved.
     
  5. Jul 24, 2012 #4
    I'm going through these notes at the moment and thinking about the same sort of stuff, so I'm not an expert or anything but I'll see if I can help. Explicit x dependence just means that an x appears somewhere in the expression for the Lagrangian.

    You seem to be confused by the fact that the Lagrangian always appears to depend on x via its dependence on the fields, as you ask why can't you just write the fields in terms of x and have the Lagrangian always explicitly dependent on x. But you can only do this when you know the specific specific given field configuration, like if you knew [itex]\phi(x) = x + 5[/itex] for example. Generally though, in the context you use the Lagrangian in, you don't have a specific field configuration like that. You instead put it in an integral to get the action, and you consider varying [itex]\phi[/itex] to find the specific configuration that extremises the action. I don't know the proper mathematical jargon for what the Lagrangian is, but it is like a kind of functional, its input is not (just) x, but a specific field configuration. The Lagrangian is a function over the space of all possible [itex]\phi[/itex].

    As for the question about whether [itex]\partial_\nu \mathcal{L}(x) = 0 [/itex], I disagree with the other poster. I don't think it is true in general, because that expression is used in the context where you do have a specific [itex]\phi[/itex], and you then consider applying a transformation to that specific [itex]\phi[/itex]. In this case the Lagrangian for this [itex]\phi[/itex] can be viewed as a function of x. It's like evaluating the function f(x) = x + 5, at a specific point like x=2. f isn't really a number, it is a function, but when you know x is 2 you can say f=7 and people know what you mean. The Lagrangian isn't really a function of just x, but when you are evaluating it for a given [itex]\phi[/itex] you can use the same letter, L, to refer to the function of x that this gives you and then talk about its gradient, which is non-zero in general.

    In the notes this is then used to derive a conserved quantity, the stress energy tensor. If it were true that [itex]\partial_\nu \mathcal{L}(x) = 0 [/itex] then the stress energy tensor derived in the notes would look very different.
     
  6. Jul 24, 2012 #5
    Thanks, that's really helpful. So you are saying that while we may know the fields [itex]\phi_a[/itex] depend on [itex]x[/itex], we may not know their mathematical expression, in terms of [itex]x[/itex]? In this case then, I'd agree with you. And I guess when you have a variation such as

    $$\frac{\partial \mathcal{L}}{\partial \phi_a}\delta\phi_a$$

    then the variation due to [itex]x[/itex] is in the [itex]\delta\phi_a = (\partial\phi_a /\partial x) \delta x[/itex], but you can't evaluate this as you don't have a specific [itex]\phi[/itex].

    I am still unsure of this, though I agree that, in the notes [itex]\partial_\nu \mathcal{L}(x) \neq 0 [/itex]. Firstly, why is there a specific [itex]\phi[/itex] in this example? The transformation
    $$x^\mu \rightarrow x^\mu - \epsilon^\mu\quad \Rightarrow \phi_a(x) \rightarrow \phi_a(x) + \epsilon^\nu \partial_\nu \phi_a$$
    can be for any field [itex]\phi[/itex]. Furthermore, if in this case you are talking about a specific [itex]\phi[/itex] such that [itex]\mathcal{L}[/itex] is expressed explicitly in terms of [itex]x[/itex], then why can't you say the same for [itex]\mathcal{L}(x) \rightarrow \mathcal{L}(x) + \epsilon^\nu \partial_\nu \mathcal{L}(x)[/itex] (where [itex]\mathcal{L}[/itex] is said to have no explicit [itex]x[/itex] dependence)?

    I guess the answer is that "no explicit [itex]x[/itex] dependence" is meant for the general [itex]\mathcal{L}[/itex], where the specific field [itex]\phi[/itex] is unknown. (If there was explicit [itex]x[/itex] dependence in this, such as [itex]\mathcal{L}=x\phi[/itex], then there would be additional terms in the variation since [itex]x \rightarrow x - \epsilon[/itex].) When you have a specific [itex]\phi(x)[/itex], then you can evaluate it and [itex]\partial_\nu \mathcal{L}(x) [/itex] need not be zero.
     
  7. Jul 24, 2012 #6
    It can be for any field, but you apply it to a specific field... I think... I'm confusing myself with the language now.

    Here is how I would go about justifying this derivation, maybe this helps:

    Ok so you have some Lagrangian e.g. the Klein-Gordon Lagrangian, and say you are given a specific field configuration [itex]\phi(x)[/itex] and you apply an infinitesimal displacement to it, [itex]\delta x^{\mu}[/itex]. The Lagrangian in general depends on [itex]\phi[/itex] but once you have been given this specific [itex]\phi (x)[/itex] you can write the Lagrangian as a field depending just on x. You want to find what happens to this field under the transformation. There are two ways of writing the change in this Lagrangian, the first is simply:

    [itex]\delta L = \partial_{\mu}L \delta x^{\mu}[/itex] where you view L as a function just of x and treat it the same way you would any scalar field.

    The second is to remember the original form of the Lagrangian as a function of [itex]\phi, \partial_{\mu}\phi[/itex], and [itex]x^{\mu}[/itex] so the change in the Lagrangian becomes:

    [itex]\delta L = \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_{\mu} \phi)}\delta(\partial_{\mu}\phi) + \frac{\partial L}{\partial x^{\mu}}\delta x^{\mu}[/itex]

    Where [itex]\delta\phi = \partial_{\nu}\phi \delta x^{\nu}[/itex]

    Both of these expressions for [itex]\delta L[/itex] are equal. I think confusion might stem from the fact that the last term in the second expression looks a lot like the only term in the first expression, but in the second expression you differentiate with respect to x while holding [itex]\phi[/itex] and its derivatives constant whereas in the first expression you take account of the x dependence in [itex]\phi[/itex], so the two terms are not in fact really equal. If the Lagrangian doesn't depend explicitly on x, then this term in the second expression vanishes (not the one in the first) and you are left with:

    [itex]\delta L = \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_{\mu} \phi)}\delta(\partial_{\mu}\phi)[/itex]

    Now in the specific case where the [itex]\phi[/itex] you are given satisfies the Euler-Lagrange equations (meaning it extremises the action), then you can reduce this expression even further with some simple manipulation to get:

    [itex]\delta L = \partial_{\mu}(\frac{\partial L}{\partial(\partial_{\mu} \phi)}\delta\phi)[/itex]

    Equating this with the first expression for [itex]\delta L[/itex] gives you an equation saying that some quantity, the stress-energy tensor, is conserved in the case when [itex]\phi[/itex] obeys the Euler-Lagrange equations.
     
  8. Jul 24, 2012 #7
    Thanks TobyC, that was very clearly explained. I agree that the source of the confusion is as you state it, and you've cleared it up well. I know it's one of those little niggling things, and isn't a big detail, but it's most appreciated.

    I agree, but this requires [itex]\partial_\mu (\delta \phi) = \delta (\partial_\mu \phi)[/itex], doesn't it? I noticed this when the derivation was first done, and was a little puzzled. I assume it to be true, but am not sure why --- which is partly why I asked for more information regarding variations in the first post. Something to look into.
     
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