# In which sense equations of motion are covariant?

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.

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 $$\partial _{\alpha} G^{\alpha \beta}= \partial _{\alpha '} G^{\alpha ' \beta '}=0$$

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.

• I think that the form may change if the equation is covariant, for example: $$\partial_iB_i(x_1,...,x_n)=0$$ if we change the frame of reference become: $$\partial'_iB'_i(x'_1,...,x'_n)=0$$ that has another form since $B'$ is different form $B'$, but if $B$ has a symmetry respect the change of coordinates then the equation is form invariant: $$\partial'_iB'_i(x'_1,...,x'_n)=\partial_iB_i(x'_1,...,x'_n)=0$$ – SimoBartz Jun 21 at 13:31
• google.com/url?sa=t&source=web&rct=j&url=https://… . Form invariant means form of equation remains same. Refer part 5.3 in the above pdf. As an example, you can see that electromagnetic wave equation under galilean transform change its form. It takes some additional terms in new frame. Therefore the equation is not covariant under Galilean transform. To make it covariant we introduced lorentz transform. – walber97 Jun 22 at 4:28
• In this paper the examples are scalar equations. What if it's a vector equation? it should change in the form. – SimoBartz Jun 22 at 9:55
• Here $E$ is electric field, and it's a vector. Here we considered $x$ component of electric field only. If the electric field have y and z component, still the equation is invariant. The second partial derivative of electric field with respect to y and z remains same, but x component should have an additional term as you can see. – walber97 Jun 22 at 10:06
• Also in 3D it's an equality between scalar quantities since there is the divergence, my question is what if there is a vectorial equation? – SimoBartz Jun 22 at 10:12

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.

• Judging from the tags the OP did not ask about GRT – my2cts Jun 20 at 20:18