Talk:Relativistic Euler equations

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Shouldn't be defined as , without the factor multiplying the internal energy? Or is the internal energy per unit mass? If it is per unit mass, that should be stated. (I'm new at this and I'm not confident enough in my comment to just enough to just edit the page.)

Clay Spence 15:52, 3 January 2007 (UTC)

Hi Clay. I think you're right. I would recommend, in general, that you be bold and edit the page. But in this case I've done it for you.

Best wishes, and really, do be bold: that way, we all gain. Robinh 20:21, 3 January 2007 (UTC)

My tensor maths isn't that strong at the moment, but I have reason to believe that there are a number of errors in this article and its use of tensor notation. Firstly, the 4-derivative has one lower index, mu, but the function T has two upper indices, mu and nu. Since we've got a covariant-contravariant vector product we should end up with an invariant equantity (if they're to obey the Lorentz transform in special relativity) however you have a free index - this means the product is be non-zero (indeed, you'll end up with a vector with 1 free index). Furthermore, the definition of T-mu,nu itself seems wrong. The left-hand side of the equation you have upper indices, the right-hand side you have lower indices which you cannot have by the definitions of contra- and co-variant vectors.

From other areas of physics where I have seen the continuity equation applied to situations, you define a 4-vector containing a "density" like term (that you connect with the temporal component of 4-derivative) and a "current" like term (which you connect with the spatial components of the 4-derivative); this 40current only needs 1 free index. -- Lateralis (16/1/07, 22:10 GMT)

I think most of this page is wrong. especially the un-index-balanced tensor equation which has already been mentioned. I think perhaps someone should either ammend or remove this page

The contraction of T (which has 2 indices) with one index (from the covariant derivative) gives a vector - true; but this vector doesn't have to be zero. From physics, we stipulate that it be zero (or, more precisely, the zero four-vector). I've fixed the positions of the indices in the equations that needed this fixing. The page should not be removed, as it is an essential topic in relativity. I do agree that the page does needs to be improved significantly. MP (talk) 19:21, 9 February 2007 (UTC)

That is fine MP, but there's still a problem with indecies, and I appreciate I am probably being a pedant. However, the partial derivative with one index, acting on a matrix with two indecies leaves a qauntity with 1 free index. In general, this would be a non-zero vector. If you were to stipulate that this result is 0 - and hence demand that all elements of the vector are 0 - then it needs stating explicitly for the sake of completeness and clarity that the result is a vector with all elements identically 0. Otherwise it looks like an index has just been obliterated. -- Lateralis (15/2/07, 13:13 GMT)