- #1
B3NR4Y
Gold Member
- 170
- 8
In my GR book they discuss things that are invariant, and I know from my math classes that invariant things are very useful. However, my intuition with invariance is that when a coordinate transformation is applied, the object is the same. Scalars are the same scalar in one frame as another. Vectors may have different components, but the length is the same. I am okay with this, but when they discuss transformations with the Lorentz transformation I get a bit confused. I've seen the standard undergrad Lorentz transformations with gamma and stuff, but they give the definition of transformations, of A, as [itex]A^{i'} = \lambda _{i} ^{i'} A^{i} [/itex]
The λ is the Lorentz transformation matrix/tensor. The summation convention doesn't confuse me, what confuses me is how does this show invariance? Is that just how it's defined? Or is there some fundamental thing I'm missing. I considered writing out each component and finding the length and showing they're equivalent, but the book explicitly says that vectors are invariant. But that's not obvious to me looking at their transformation law. They then have a tensor transforming as
[tex] S^{i' \, j'} = \lambda_{i}^{i'} \lambda_{j}^{j'} S^{i \, j} [/tex]
Which I'm okay with, I've seen before but it's not obvious that the tensor is invariant. I know that tensors describe properties of something that don't usually depend on coordinates, like the inertia tensor, but it's not easy for me to know what's invariant about this tensor.
And a problem that's given is
Show [itex] D^{\mu \, \mu}[/itex] and [itex] D_{\mu \, \mu}[/itex] are not invariant under Lorentz transformations but [itex] D_{\mu}^{\mu}[/itex] is.
The first thing I noticed is that [itex] D_{\mu}^{\mu}[/itex] can be contracted to a scalar by summing over μ, so it's obviously invariant. But I also saw that following their transformation law, [itex] D_{\mu '}^{\mu '} = \lambda_{\mu}^{\mu '} \lambda_{\mu'}^{\mu} D_{\mu}^{\mu} [/itex] the lambdas are orthogonal so they equal the kronecker delta when multiplied by each other, which is one because the μs are equivalent, but I don't see how the other two aren't invariant. The tensor transformation law they gave used gave us a tensor with two upper indices. The two not invariant tensors they gave are just the traces of their respective matrix representations, but they still follow the transformation law. The diagonal of the Lorentz transformation is not zero. So I am officially confused.
The λ is the Lorentz transformation matrix/tensor. The summation convention doesn't confuse me, what confuses me is how does this show invariance? Is that just how it's defined? Or is there some fundamental thing I'm missing. I considered writing out each component and finding the length and showing they're equivalent, but the book explicitly says that vectors are invariant. But that's not obvious to me looking at their transformation law. They then have a tensor transforming as
[tex] S^{i' \, j'} = \lambda_{i}^{i'} \lambda_{j}^{j'} S^{i \, j} [/tex]
Which I'm okay with, I've seen before but it's not obvious that the tensor is invariant. I know that tensors describe properties of something that don't usually depend on coordinates, like the inertia tensor, but it's not easy for me to know what's invariant about this tensor.
And a problem that's given is
Show [itex] D^{\mu \, \mu}[/itex] and [itex] D_{\mu \, \mu}[/itex] are not invariant under Lorentz transformations but [itex] D_{\mu}^{\mu}[/itex] is.
The first thing I noticed is that [itex] D_{\mu}^{\mu}[/itex] can be contracted to a scalar by summing over μ, so it's obviously invariant. But I also saw that following their transformation law, [itex] D_{\mu '}^{\mu '} = \lambda_{\mu}^{\mu '} \lambda_{\mu'}^{\mu} D_{\mu}^{\mu} [/itex] the lambdas are orthogonal so they equal the kronecker delta when multiplied by each other, which is one because the μs are equivalent, but I don't see how the other two aren't invariant. The tensor transformation law they gave used gave us a tensor with two upper indices. The two not invariant tensors they gave are just the traces of their respective matrix representations, but they still follow the transformation law. The diagonal of the Lorentz transformation is not zero. So I am officially confused.