The discussion centers on the treatment of time derivatives in variational calculus, specifically regarding the integral of energy density F[f(x), f'(x), df(x)/dt]. Participants clarify that if f is a function of a single variable x, then df(x)/dt equals zero. For non-zero derivatives with respect to both x and t, the notation f(x, t) must be used. The conversation emphasizes the importance of clearly stating the problem context to facilitate effective assistance.
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Physics problem. It's one of taking the variation of an energy of an object with energy density F in the OP varying in space. f(x) can be a displacement, x can be a coordinate. df(x)/dt can be a speed. Clear?jedishrfu said:Can you provide some more context about what problem you are working on or what class this came from?
These kinds of details help us decide on the level of response.
From your notation, ##f## appears to be a function of only one variable; namely, x. In that case ##\frac{d(f(x))}{dt} = 0##.feynman1 said:Taking the variation w.r.t f(x) of the integral over some x domain of F[f(x), f'(x), df(x)/dt], why doesn't df(x)/dt need to be taken a variational derivative and is treated as if it were constant?
right because I couldn't put a dot on top of f(x).Mark44 said:From your notation, ##f## appears to be a function of only one variable; namely, x. In that case ##\frac{d(f(x))}{dt} = 0##.
For ##f## to have nonzero (partial) derivatives wrt x and t, it should use this notation: ##f(x, t)##. Then ##\frac{\partial f}{\partial x} = f_x((x, t)## and ##\frac{\partial f}{\partial t} = f_t((x, t)## would make sense.
Well, not if ##x## itself depends on ##t##.Mark44 said:##f## appears to be a function of only one variable; namely, x. In that case ##\frac{d(f(x))}{dt} = 0##.
Sorry there's no direct reference as it's a work in progress. Physics problem. It's one of taking the variation of an energy of an object with energy density F in the OP varying in space. f(x,t) can be a displacement, x is a fixed coordinate not dependent on t. df(x,t)/dt can be a speed. Clear?ergospherical said:Well, not if ##x## itself depends on ##t##.
To @feynman1: could you point us to a suitable reference? I'm not yet sure what you are asking.
if that helpsergospherical said:Not clear at all, sorry. I can't help until there's a clear problem statement.
Which would show up in a clearly posed question. If someone writes "f(x)" without anything further, the reasonable assumption is that f depends only on x, with no relationship to t.ergospherical said:Well, not if x itself depends on t.
sorry unable to edit the OP any moreMark44 said:Which would show up in a clearly posed question. If someone writes "f(x)" without anything further, the reasonable assumption is that f depends only on x, with no relationship to t.
only difference from trivial cases: the integrand has time dependent terms but the integral isn't over time, which isn't covered by variational calculus. Clear?ergospherical said:I’m well aware of the calculus of variations! What I mean is that you have not adequately stated your problem.
No, not clear. You said "Physics Problem" more than once, but with no further detail, as if we're supposed to know clairvoyantly which physics scenario you're talking about. If the integral isn't over time (but assuming ##t## and ##x## are independent variables), then you treat ##g(x,t) := \dot f(x,t)## as simply another field in the integrand.feynman1 said:[...] because I couldn't put a dot on top of f(x). [...] only difference from trivial cases: the integrand has time dependent terms but the integral isn't over time, which isn't covered by variational calculus. Clear?
This...strangerep said:[Aside: You need to spell out your specific situation/problem explicitly. If you won't put more effort into composing your questions, why should other people put more effort into helping you? ]