In which sense equations of motion are covariant?

Question

I read lots questions about what covariance is and I found out that, according to this topic Lorentz invariance of the Minkowski metric, we say an object is covariant if it doesn't take the same value on every frame of reference, but the different values are related in a well defined way: the components of a covariant object must satisfy the tensors transformation rule.

I understand these definitions but at the same time I heard many times about covariance of an equation. I tried to figure out what is a covariant equation and I noticed that if I have an equality where right and left terms are covariant objects than the equation remains true when I change the frame of reference because both sides transform equally. So i was tempted to say en equation is covariant if it's between covariant objects. On the other hand there are some equations that are said to be covariant but doesn't respect this definition. For example the equations of motion when the frame of reference is changed remains true but they are not made of objects that are covariant.

Answer 1

An example of the equations of motion that you describe as being covariant even though they use non-covariant objects is the geodesic equation: $$ \ddot x^μ +Γ^μ_{αβ} \dot x^α \dot x^β = 0 . $$ That equation can be written using covariant objects: $$ ∇_{\dot x^μ} \dot x^μ = 0 . $$

In general, all covariant equations can be written in terms of covariant objects. However, some covariant objects can be decomposed in a non-covariant way. When the equations use that non-covariant decomposition their covariance is not explicit. In the geodesic example, the derivative $∇_{\dot x^μ}$ includes $∂_τ$ --that is the part needed for $\ddot x^μ$-- but it also includes the addends with the Christoffel symbols. If the equation is to be covariant, the non-covariance of its objects has to cancel out. In fact, you can build covariant objects out of non-covariant ones as long as you cancel the contributions that would ruin the covariance during a coordinate transformation. For example, a single connection ${Γ^{(a)μ}}_{\!α\!β}$ is not covariant, but the difference of two connections ${Γ^{(a)μ}}_{\!α\!β} - {Γ^{(b)μ}}_{\!α\!β}$ is a perfectly good (1,2) tensor.

Answer 2

It will be helpful if you understand the difference between covariance and invariance. Covariance means form invariant. That is, the form of equations remains unchanged under some cordinate transformation. The form can include covariant and contravariant components as well.

Eg: Maxwell's covariant field equation $$ \partial _\alpha G^{\alpha \beta}=0$$ takes the same form under lorentz transformation. That is $$ \partial _{\alpha} G^{\alpha \beta}= \partial _{\alpha '} G^{\alpha ' \beta '}=0$$

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Where as invariance means that an object remains same in all frames under some coordinate transformation.

Eg: The value of interval $$ds^2=dt^2-dx^2-dy^2-dz^2$$ is an invariant scalar under lorentz transform.