When using propagators and potentials (like the Yukawa one), time is not explicit. Should we use different forms of the propagators and the potentials in a 4d spacetime with time being a 'regular' spacedimension (except for its irreversibility)? Does the incompleteness of particle physics theories come from the fact that time is always treated separately?
No Nathalie, we can't get rid of such a categorical concept (in Kant's sense) required in any description of reality. But I completely agree with the idea that a lack of an adecuate mathematical representation of time duality (absolute and relative) is the reason of theoretical difficulties in M4 physics.
No-there's a physical property of the phenomena described in particle physics called Lorentz invariance, that has been tested and is tested in any particle physics experiment and its description through quantum field theory in general and the Standard Model in particular. This is standard material in an introductory quantum field theory course. One should distinguish the mathematical description used, from the physical properties under study. One can use the mathematical tools one likes, as long as the physical properties are described properly.
Indeed, Stam, but as a matter of fact , time is treated as different from space dimension, even if Lorentz invariance holds.
The difference doesn't matter-what matters is invariance under Lorentz transformations and how the quantities that are physically relevant transform under them.
yes the difference matters, did you ever see time treated as a function (like for example x)? if it was, could we use Lagrangian? what about the Yukawa coupling? where is time in it? something is incomplete, I think.
The Lagrangian of a Lorentz invariant theory is a scalar under Lorentz transformations-and these include the Lorentz boosts. The Yukawa interaction term, in particular, is, also, a scalar under Lorentz transformations: it's the product of a scalar field and a fermion bilinear that's a scalar under Lorentz transformations. The scalar field and the fermion bilinear, of course, depend on the spacetime point, i.e. on time. The coupling constant is a constant, whose value can be computed from correlation functions of the theory. If a term in the Lagrangian depended on time alone, it wouldn't be Lorentz invariant-it can only depend on time through Lorentz invariant combinations. This is discussed in the introductory chapters of all courses on field theory.
Stam, I never said a term in a Lagrangian would depend on time alone...
well to summarize, I think that time is not only a variable. Of course, there is a problem with that in quantum field theory....
Once more: the fields in a field theory are functions of the spacetime point, thus functions of four variables, in four dimensions. These four variables transform as a 4-vector under Lorentz transformations. The fields themselves transform as scalars, spinors and 4-vectors (in the absence of gravity-the metric tensor transforms like a tensor). That's all. There isn't any problem in either classical or quantum field theory, where Lorentz invariance is a global symmetry in both cases.This implies, in particular, invariance under time translations, that is the statement that energy is conserved.
The statement ``time is not only a variable'' as such doesn't *mean* anything-just like ``space is (or not) only a variable'' would; time is the zeroth (or fourth...) component of a 4-vector that labels the points in spacetime (and space is specified by three numbers, the other three components of a 4-vector).
[Of course time is treated like a variable, just like space. And that doesn't affect the Lagrangian, since the Lagrangian is a scalar under Lorentz transformations.]
Lorentz transformations leave the Minkowski metric invariant-in field theory (absent gravity) Lorentz invariance is a global symmetry. In particular, the signature is preserved and time-like, null and space-like distances between points in spacetime have an invariant meaning, by definition of the signature. However, since Lorentz transformations mix the components of a 4-vector, the only meaningful statement is that a 4-vector has one time-like and three space-like components-but not which are which. In fact these properties *define* Minkowski spacetime and determine many of the properties of the dynamics that can take place in it.
Does it exist terms in a Lagrangian which only depend on space? if yes, why wouldn't it exist terms in a Lagrangian which only depend on time? you are going to reply that it is because of Lorentz invariance, maybe this last is incomplete (same difference between special relativity and general relativity, the last taking account of geometry)
The first staement doesn't specify *what* should exist, so it's impossible to say anything. A Lagrangian, whose fields depend only on spatial variables isn't a Lorentz scalar, because spatial variables in one frame are not spatial in all frames. So it cannot describe phenomena that are known to be described by Lorentz invariant Lagrangians. (One should *not* confuse this situation with Euclidian quantum field theory, when the time coordinate is rotated along the imaginary axis-this is a *mathematical* operation, that can be *proved* to be well defined, thanks to the analyticity properties of correlation functions. In this case Lorentz invariance of the original description maps to rotational invariance in the Euclidian description.)
In particle physics Lorentz invariance, once more, is a *global* symmetry (it's when gravitational effects are to be taken into account that it's a *local* symmetry-but gravitational effects are not *relevant* in the energy scales described by particle physics, so it's consistent to consider Lorentz invariance as a global symmetry and this statement can be given quantitative content). When Lorentz invariance is a *global* symmetry, there aren't any problems, either conceptual or calculational in dealing with the consequences of Lorentz invariance in quantum field theory.
If you want to talk about time in particle physics you *are* dealing with Lorentz invariance, because that is the symmetry that allows time to be described in particle physics. If you do not want to talk about Lorentz invariance, any description of time in particle physics doesn't make sense and one is talking about something else. You may, of course, ask whether terms that *violate* Lorentz invariance, could have measurable effects in particle physics-and this has been studied. In such a case a separation into space and time coordinates would have measurable consequences. The absence of the former places constraints on the latter. Vice versa, a measurable way to enforce such a separation *implies* the appearence of Lorentz-violating terms, that can, then, be constrained, since they would, also, affect other processes.
The question of how time is treated (explicitly and implicitly) is fundamental. For example, each time you use an impulsion, space and time are not treated the same way even if this is implicit.
In a non-relativistic theory, like Newtonian mechanics, indeed-that's what non-relativistic *means*. In a relativistic theory, momentum is the spatial part of the energy-momentum 4-vector of a particle so doesn't have an invariant meaning-it's the 4-vector, whose invariant length is the rest mass (squared) that's meaningful. Once more, it's useful to express everything in terms of quantities that transform ``simply'' under Lorentz transformations: so, instead of (time, space)-> 4-vector of the spacetime point, instead of (energy,momentum)->4-vector of the energy-momentum of a particle. And, for infinite number of degrees of freedom, it's more useful to deal with the energy-momentum tensor.
The only invariant distinction between space and time is the signature: how many space-like dimensions, how many time-like dimensions. Lorentz transformations cannot change this. *Which* among them is which, isn't invariant.
In a Lorentz transformation, you don't treat time and space the same way.
Ok, I understand your confusion: when I say that in impulsion (momentum, velocity), time and space are not treated the same way, I was not talking about their (numerical) values but about their dimensions (e.g. p=ML/T)
Of course you treat them the same way, since, for example, the Lorentz transformations take you from (x,y,,z,t)->(x',y',z,'t')=(ax + b t, y,z,b x + a t), with a^2-b^2=1. It's useful to work in *units* where c=1. ``Treating them the same way'' *means* being able to give meaining to linear combinations-where there is, of course, a numerical factor-here the speed of light in vacuum.
What these equations mean is that x' has all the properties of a space-like variable and t' all the properties of a time like variable (formally t = c*time, t'=c*time').
No-the *units* change, the *dimensions* don't. One sets x'= a*x + b*c*t, c*t' = b*x + a*c*t. Since the speed of light has a non-zero value, one can *choose* units of length and time so that its numerical value is equal to 1-but this 1 has *dimensions* length/time, i.e. it *is* a velocity.
One isn't adding space and time; one is adding two lengths, or two times.But one isn't obliged to work in units where lengths are measured in meters or time lapses in seconds.
Indeed since the speed of light is the same in all frames, related by a Lorentz transformation, this conversion makes sense: (x,c*t)->(x', c*t'), with the same c. One can then write c*t=u, c*t'=u', with u and u' lengths.
and a velocity has a dimension of L/T so lengths and time are treated in a different way (I know that it implies many things). not only speaking of Lorentz transformations
The conversion factor is the speed of light-which means that x and c*t are treated in the same way, since they can then be *added* together and thus subject to linear transformations. Since, under Lorentz transformations, the speed of light doesn't change, (c*t)' = c*t'. Since x,y,z and c*t are components of a 4-vector, with the same dimensions, they may be mixed by linear transformations: the transformations are important and what they leave invariant. The individual components can't be given any invariant meaning.
(However these transformations wouldn't be valid were the velocity factor *not* the speed of light-because this factor would not remain invariant.)
One can get rid of time from particle physics if time is treated as a periodic phenomenon rather than linear time moving in one direction. This means the smallest unit of time is the period of the wave. T=1/f is the only fundamental relation in physics that connects the physical reality with imaginary time. This is the reason it is impossible to measure time without the help of some form of oscillator. This idea is introduced in the development of "Periodic quantum gravity and cosmology" theory where Hubble parameter is associated with the Planck frequency at planck time. Both have same units. This Hubble parameter rolls down to the present value at present epoch.
Article Periodic quantum gravity and cosmology
This is a very important question. Matter cannot exist outside of the time dimension. We can visit different points in space but never at the same time. Einstein is right in that space and time are inseparable but time is not the same as space. The space dimensions are isotropic but time has a direction defined by the entropy gradient.
Then there is the question of whether time is quantised The RG consensus is "not" but I think it must be, if energy is quantised.
There are at least two closely related questions in RG:
"Is time quantised?"
"Is the flow of time an illusion?"
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