# What exactly is invariance?

1. Nov 11, 2015

### B3NR4Y

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 $A^{i'} = \lambda _{i} ^{i'} A^{i}$

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
$$S^{i' \, j'} = \lambda_{i}^{i'} \lambda_{j}^{j'} S^{i \, j}$$

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 $D^{\mu \, \mu}$ and $D_{\mu \, \mu}$ are not invariant under Lorentz transformations but $D_{\mu}^{\mu}$ is.

The first thing I noticed is that $D_{\mu}^{\mu}$ can be contracted to a scalar by summing over μ, so it's obviously invariant. But I also saw that following their transformation law, $D_{\mu '}^{\mu '} = \lambda_{\mu}^{\mu '} \lambda_{\mu'}^{\mu} D_{\mu}^{\mu}$ 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.

2. Nov 11, 2015

### PWiz

A vector in any coordinates consists of components and basis vectors in those coordinates (which may not necessarily be orthogonal/normalised). A separate transformation must be applied to both the components and basis vectors of the new coordinate system when changing coordinates, and the component - basis transformations are inverses of each other, so the "length" of the vector stays preserved.

Let's take the four velocity for example:
Upon a coordinate transformation $x^μ \rightarrow x^{μ'}$, the new four velocity is $\frac{∂ x^{μ'}}{∂x^μ} U^μ$, but the square of the length of the vector $g_{μ'ν'} U^{μ'} U^{ν'} = \frac{∂x^μ}{∂x^{μ'}} \frac{∂x^ν}{∂x^{ν'}} g_{μν} \frac{∂ x^{μ'}}{∂x^μ} U^μ \frac{∂ x^{v'}}{∂x^v} U^v = g_{μν} U^μ U^ν$ stays the same. (-1) [Note how the metric, which is composed of the basis vectors, has a transformation which cancels the four-velocity component transformation]

Last edited: Nov 11, 2015
3. Nov 11, 2015

### pervect

Staff Emeritus
What's invariant isn't the Lorentz transform. What's invariant is the Lorentz interval (which is a scalar) not the Lorentz transform. Consider the following analogy with 3-vectors X^i, and linear transforms R which represent rotations. We say that lengths are invariant under rotational transformations. If you have a vector X, the rotation of that vector can be represented by some rotational matrix R, $R^i{}_j$ in tensor notation, so that if we start with some vector $X^j$, and apply the rotational transform $R^i{}_j X^j$ to get $Y^i$, the length of Y, |Y| is the same as the length of X, |X|. Thus we say the length of a vector is invariant under rotation. There is a similar quantity that's invariant under the Lorentz transform, this quantity is the Lorentz interval. In cartesian coordinates (t,x,y,z), it has the familiar form $dx^2 + dy^2 + dz^2 - dt^2$ (modulo sign conventions), in general coordinates we write instead $g_{ij} dx^i dx^j$.

You may be confusing invariant with covariant, which is a related concept. In fact, Wiki mentions that general covariance is sometimes (confusingly) called general invariance, thus the confusion is understandable.

As an example we can say that Maxwell's equations are covariant under the Lorentz transform, they have the same form if you apply a Lorentz transform. Maxwell's equations are not covariant under the Galilean transform.

Tensor notation can be regarded as a tool that guarantees general covariance. The components of a tensor transform in a well defined manner, a manner defined by the tensor transformation laws. Hopefully your textbook is specific enough about how tensors transform that you can tell when something is a tensor, and when it isn't.

You may see tensor transformation laws written in terms of partial derivatives, or in terms of how basis vectors transform. The last confusing thing (IMO) is how partial derivatives are related to basis vectors, but getting into that would be a lot of work and also take us too far off-topic.

4. Nov 11, 2015

### PWiz

I don't see how the defining equation is complicated though: $e_{\mu} = \frac {\partial }{\partial x^{\mu}}$

5. Nov 11, 2015

### B3NR4Y

Okay so when they say that the vector Xi is invariant they mean the lorentz interval is invariant under the lorentz transformation, which is given by $X^{i'} = \lambda_{i}^{i'} X^{i}$. I understand, I think. But I'm still confused about how I should go about showing $D_{\mu}^{\mu }$ is invariant but $D^{\mu \mu}$ is not, with a vector I know how to write the lorentz interval in tensor notation and then transform it, but I am not sure how to do that for tensors or if I need to do something else. This isn't homework, by the way, just an extra problem I saw that really highlighted what I don't understand.

Also, I was reading the next section and it mentions that Einstein's equation is invariant, it gives the Einstein equations in the form $R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8\pi G T_{\mu \nu}$. Does this mean applying transformation laws to either side will yield the same equation (which is actually covariance, I think? I've never been good at vocabulary)? Or, like vectors being invariant, there is some other property that is invariant, I assume the property that remains the same is how spacetime curves in the presence of matter, but I want to be sure I understand.

We also define $\lambda_{i}^{i'} = \frac{\partial x^{i'}}{\partial x^{i}}$. And use this to show
$$\partial _{\nu'} A_{\mu'} = \lambda_{\nu'}^{\nu} \lambda_{\mu'}^{\mu} \partial_{\nu} A_{\mu}$$
So does this mean that the derivative of a vector transforms like a tensor? So is the derivative itself a tensor?

6. Nov 12, 2015

### PWiz

Do you know the tensor transformation law? Under an arbitrary diffeomorphism (i.e., an arbitrary coordinate change) $x^{\mu} \rightarrow x^{\mu '}$ , the tensor $D^{\mu \mu}$ transforms like this: $\frac{\partial x^{\mu '}}{\partial x^{\mu}} \frac{\partial x^{\mu '}}{\partial x^{\mu}} D^{\mu \mu}$, which is not the same as the original tensor. However, $D^{\mu} _{\mu}$ transforms like this: $\frac{\partial x^{\mu '}}{\partial x^{\mu}} \frac{\partial x^{\mu}}{\partial x^{\mu'}} D^{\mu} _{\mu}= D^{\mu} _{\mu}$, so this tensor is said to be invariant under diffeomorphisms.
Yes. The EFE is a tensor equation, and is therefore invariant under diffeomorphisms.
No, the derivative of a vector does not transform like a tensor: $$\frac{\partial V^{\alpha}}{\partial x^{\mu}} \rightarrow \frac{\partial V^{\alpha '}}{\partial x^{\mu '}}= \left ( \frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial}{\partial x^{\mu}} \right ) \left ( \frac{\partial x^{\alpha '}}{\partial x^{\alpha}} V^{\alpha} \right ) = \frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial x^{\alpha '}}{\partial x^{\alpha}} \frac{\partial V^{\alpha} }{\partial x^{\mu}} + \frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial^2 x^{\alpha '}}{\partial x^{\alpha} \partial x^{\mu}} V^{\alpha}$$ , where I used the chain rule for the derivative transformation.

If we were to only use the tensor transformation law, we should not get the second term (it's zero only in flat space). To fix this problem, we introduce the covariant derivative:$$∇_{\mu} V^{\alpha} = \frac{\partial}{\partial x^{\mu}} V^{\alpha} + Γ^{\alpha} _{\mu \lambda} V^{\lambda}$$ , where $Γ^{\alpha} _{\mu \lambda}$ are a set of coefficient symbols (the Christoffel symbols) that do not transform like tensors, but like this: $$Γ^{\alpha '} _{\mu ' \lambda '} = \frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial x^{\lambda}}{\partial x^{\lambda '}} \frac{\partial x^{\alpha '}}{\partial x^{\alpha}} Γ^{\alpha} _{\mu \lambda} - \frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial x^{\lambda}}{\partial x^{\lambda '}} \frac{\partial ^2 x^{\alpha'}}{\partial x^{\mu} \partial x^{\lambda}}$$ . Then the covariant derivative transforms like a tensor.
Only in flat space (unless you're differentiating a scalar field).

Last edited: Nov 12, 2015