What is the relationship between scalar fields and tensor fields?

tionis
Gold Member
Messages
459
Reaction score
67
I've PMd some of you with this question, but I got some conflicting replies or no replies at all lol, so I'm posting it here. I also did a Google search and found this which I'm almost sure answers my question, but I just want to confirm with you guys:

''In general, scalar fields are referred to as tensor fields of rank or order zero whereas vector fields are called tensor fields of rank or order one.''

Since the Higgs is a scalar field, I guess it's a tensor field of rank 0, too. Is that correct?
 
Physics news on Phys.org
Maybe in some technical sense, but nobody calls it that.

And please ask questions on the forum and not by PMing random people.
 
Vanadium 50 said:
Maybe in some technical sense, but nobody calls it that.

Vanadium 50, could you please elaborate? This is exactly the kind of replies I was getting via PMs lol.
 
Your question is about language, not about physics/mathematics.
The mathematical concept of tensors has a parameter "nth order", which is used for tensors of 2., 3., ... order. You can consider "tensors of 1st order" and "tensors of 0th order", but everyone calls them vectors and scalars, respectively.
 
tionis said:
''In general, scalar fields are referred to as tensor fields of rank or order zero whereas vector fields are called tensor fields of rank or order one.''

Since the Higgs is a scalar field, I guess it's a tensor field of rank 0, too. Is that correct?

Yes. But people will look at you funny if you call it that. If you want more detail you might read about tensors on Wikipedia: http://en.wikipedia.org/wiki/Tensor

In the usual parlance, the field of a spin-0 particle like the Higgs is called a scalar field, the field of a spin-1 particle is called a vector field, and the field of a spin-2 particle is called a tensor field. Just saying "tensor" without specifying the rank is understood to mean a rank-2 tensor.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top