The dimension of fields

  • Thread starter jfy4
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Hi,

I was not entirely sure where to post this, but I think this will work.

With the gravitational field we have that
[tex]g^{\alpha\beta}g_{\alpha\beta}=4[/tex]
which is the dimension of the manifold I believe. I have normally heard of [itex]g_{\alpha\beta}[/itex] being interpreted as the gravitational field quantity (or the tetrad). For the other fields in physics (like [itex]A_{\mu}[/itex]), how does one compute the dimension, or does such a quantity not exist for anything other than the gravitational field?

Thanks in advance,
 

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  • #2
atyy
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The dimension is a property of the underlying manifold (ie. how many coordinates do you need to specify a point). All fields inherit their dimensionality from the manifold. In string theory, apparently you can get emergent space - where energy becomes a dimension - but I have no understanding of this!
 
  • #3
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Hi,

I was not entirely sure where to post this, but I think this will work.

With the gravitational field we have that
[tex]g^{\alpha\beta}g_{\alpha\beta}=4[/tex]
which is the dimension of the manifold I believe. I have normally heard of [itex]g_{\alpha\beta}[/itex] being interpreted as the gravitational field quantity (or the tetrad). For the other fields in physics (like [itex]A_{\mu}[/itex]), how does one compute the dimension, or does such a quantity not exist for anything other than the gravitational field?

Thanks in advance,
The contraction of the metric with its inverse indeed gives the dimension of the spacetime you are looking at. You can see this by taking a look at how the metric tensor is contracted twice with some vectorial quantity [itex]A_{\nu}[/itex] in a d-dimensional space:

[itex]g_{\mu\rho}g^{\mu\nu}A_{\nu}=A_{\rho}[/itex].

Therefore we can make the identification with a Kronecker delta of dimension d:

[itex]g_{\mu\rho}g^{\mu\nu}={\delta_{\rho}}^{\nu}[/itex].

The trace of the Kronecker delta gives just the dimension of the space, therefore

[itex]g_{\mu\nu}g^{\mu\nu}=d[/itex].
 
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