# Tag Info

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After giving myself a crash course on the deformation gradient tensor (I hadn't heard of it before) from the "Finite Strain Theory" wikipedia page, it seems evident that these two notions are not related at all. In summary, the Deformation Gradient Tensor is to do with the best fit linear approximation of a vector field around a point. In contrast, the ...

0

The covariant derivative of a field have values in the Lorentz algebra in the representation of the field, spinorial representation in this case, I.e., $$\mathcal{J}^{ab} = - \frac{\imath}{4}[\gamma^a,\gamma^b].$$ Thus the covariant derivative for spinors is $$\nabla_\mu \epsilon = \partial_\mu \epsilon + \frac{\imath}{2} \omega_\mu{}^{ab}\mathcal{J}_{ab},$$ ...

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If you've studied electric and magnetic fields before then you know they are related to forces on charged particles. However, if you look at how the electromagnetic field $F$ interacts with charges to produce forces you see the metric get involved. So the metric is essential to the field as a thing that exert forces, which is the actual definition. There is ...

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This is not that difficult: A scalar is something that is just a number and a Lorentz scalar is a scalar that is invariant under Lorentz transformations. For example distance is a scalar but not a Lorentz scalar and a proper time interval is a scalar and a Lorentz scalar. To find out if something is a Lorentz scalar we simple check how it transforms under a ...

3

Tensors in physics One of my professors at Cornell told me, possibly influenced by Anthony Zee, that the definition of a tensor in physics is A tensor is anything which transforms like a tensor. Our whole class laughed, which irked him, because, as he went on to point out: it's not quite circular. Once you know how one vector rotates under a ...

1

This might not be the final answer, but it might help you understand the issue. I will rely on other users to help me get the whole picture. There is a certain rule of thumb as to know when a quantity is a scalar or not: if the indices go through all space-time dimensions, i.e., $\alpha=0,1,2,3$, then the expression is probably a scalar. Your formula is not ...

0

Maybe it helps if we can place this in a context. You can have spaces like $A$ and $B$ and then you can make product spaces like $A \otimes B$. You can make linear operators like $S:A\rightarrow C$ and $T:B\rightarrow D$ and then since an arbitrary thing in $A \otimes B$ is spanned by things like $a \otimes b$ (with $a\in A$ and $b\in B$) you can clearly ...

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