Understand "Manifestly Covariant" in Relativity Notes

[Kalidor]

In my relativity notes, I have several remarks like the following one:
"The Lorentz condition on the potentials can be written in manifestly covariant form in this way: $$\partial_i A^i = 0$$ , where the A^i are the components of the 4-potential."
This made me realize I probably have not understood what covariant means. Why would that be manifestly covariant?

 

[Mentz114]

Any expression written in tensor form is covariant. The transformation properties of tensors define covariance. Physical quantities are contractions of 4-vectors, 4-tensors and remain invariant when these objects are transformed.

Lorentz covariance just means transforming under Lorentz transformations so that physical quantities are invariant.

I haven't expressed this very well, but I hope it helps some.

 

[Kalidor]

Any expression written in tensor form is covariant. The transformation properties of tensors define covariance. Physical quantities are contractions of 4-vectors, 4-tensors and remain invariant when these objects are transformed.

Lorentz covariance just means transforming under Lorentz transformations so that physical quantities are invariant.

I haven't expressed this very well, but I hope it helps some.

Hmm this all seems a bit above my current relativity knowledge.

 

[superg33k]

In my relativity notes, I have several remarks like the following one:
"The Lorentz condition on the potentials can be written in manifestly covariant form in this way: $$\partial_i A^i = 0$$ , where the A^i are the components of the 4-potential."
This made me realize I probably have not understood what covariant means. Why would that be manifestly covariant?

Manifestly covarint and covariant are (annoyingly) different ideas. Covariant refers to the tensors transform under the Lorentz transform. Manifestly covarient refers to an equation of tensors that holds under transform, even if the individual components of the equation do not.

 

[Mentz114]

Hmm this all seems a bit above my current relativity knowledge.

An example of a covariant object is the 4-velocity. The contraction ( i.e. square of the proper length) of the vector is invariant and has the value -c².

Manifestly covarient refers to an equation of tensors that holds under transform, even if the individual components of the equation do not.

Thanks. I was never sure about that.

 

[Ben Niehoff]

No, manifestly covariant just means that the fact that the equation/expression is covariant is obvious; i.e. manifest.

To summarize:

Invariant means that something transforms as a scalar under Lorentz transformations (i.e., that it doesn't transform).

Covariant means that something transforms like a tensor under Lorentz transformations.

Manifestly covariant means that something has either of the above properties in an obvious way. For example, in index notation, if all the quantities in an expression are tensors, and all their indices are properly matched or contracted, then the expression is manifestly covariant.

 

[Sam Gralla]

It doesn't pay to read too much (or anything) into words like "covariant" unless they have been precisely defined by the author. There are just too many things they can mean. Lots of reputable people use them in lots of inequivalent ways. Usually when people say "manifestly covariant" they are the type that thinks the word "covariant" has no content (since anything that makes any sense at all can be made covariant by introducing special tensor fields representing the preferred directions or coordinate systems or whatever).

In other words, don't worry about it. Just understand the physics.

 

[superg33k]

Manifestly covariant means that something has either of the above properties in an obvious way. For example, in index notation, if all the quantities in an expression are tensors, and all their indices are properly matched or contracted, then the expression is manifestly covariant.

This is not correct. Manifestly covarient refers to an equation that holds under lorentz transform. E.g. an equality of energy-momentum tensor in one frame therefore be true it in any other frame, it is manifestly covarient equation

Consider 4 objects in a collision. Any 2 observers could disagree about the energy-momentum of all of the particular objects. However they will both agree that energy-momenum was conservered in the collision.

Even further, a manifestly covarient equation does not have to be made up of tensors. Covarient differentiation is a manifestly covarient equation, although neither of its components (the reimann curvature tensor, or a derivitate) generally transforms like a tensor.

 

[Kalidor]

But then, for instance, why is the lorenz condition covariant in the way I wrote it in the first post?

 

[Ben Niehoff]

This is not correct. Manifestly covarient refers to an equation that holds under lorentz transform. E.g. an equality of energy-momentum tensor in one frame therefore be true it in any other frame, it is manifestly covarient equation

No, I'm afraid you are wrong. "Covariant" means that something transforms as a tensor under Lorentz transformations (or equivalently, that a given equation holds in any Lorentz frame).

"Manifest" is being used in its ordinary English sense. It means "obvious", "explicit", or "clear from the notation itself". Hence my reference to index notation, which is designed specifically to make covariance manifest.

Even further, a manifestly covarient equation does not have to be made up of tensors. Covarient differentiation is a manifestly covarient equation, although neither of its components (the reimann curvature tensor, or a derivitate) generally transforms like a tensor.

This does not contradict what I said. Yes, covariant derivatives are another example of an object that is manifestly covariant.

To the OP:

To see the difference between "covariant" and "manifestly covariant", consider the temporal gauge

$$A_0 = 0$$

This expression obviously breaks Lorentz covariance, because one component of A is being singled out for special treatment. The temporal gauge condition is not preserved under Lorentz transformations.

This would seem to break the Lorentz covariance of the physics, but the neat fact is that it doesn't! The vector potential A is not a directly physical quantity; only the electromagnetic 2-form F is. And it turns out that even with the above gauge choice, the actual physics remains Lorentz covariant. So here is an example of covariance which is not manifest.

 

[superg33k]

So your saying the equation in the OP is covarient but not manifestly covarient?

 

[Ben Niehoff]

So your saying the equation in the OP is covarient but not manifestly covarient?

No, and I don't see why you would think that.

 

[Kalidor]

To the OP:

To see the difference between "covariant" and "manifestly covariant", consider the temporal gauge

$$A_0 = 0$$

This expression obviously breaks Lorentz covariance, because one component of A is being singled out for special treatment. The temporal gauge condition is not preserved under Lorentz transformations.

This would seem to break the Lorentz covariance of the physics, but the neat fact is that it doesn't! The vector potential A is not a directly physical quantity; only the electromagnetic 2-form F is. And it turns out that even with the above gauge choice, the actual physics remains Lorentz covariant. So here is an example of covariance which is not manifest.

Ok I hope I have understood this difference, but why is the equation I wrote in the first post even just covariant?

 

[Mentz114]

Ok I hope I have understood this difference, but why is the equation I wrote in the first post even just covariant?

It's a scalar isn't it ?

 

[superg33k]

No, and I don't see why you would think that.

I have skimmed my textbooks and you could be right. I realize the only reason I have for beleiving I am correct is I had an assignment marked correct where I stated opposite of stated below:

But just to confirm.

The equation for covariant differention:
$$\nabla_\lambda T^\mu=\frac{\partial{T^\mu}}{\partial{x^{\lambda}}}+{\sum}_{\rho}{\Gamma}^{\mu}_{\rho \lambda}T^{\rho}$$
Its manifestly coverint if I write an equation as:
$$\nabla_\lambda T^\mu = 0$$
Since its all covarient tensors. But if I write it as:
$$\frac{\partial{T^\mu}}{\partial{x^\lambda}}+{\sum}_{\rho}{\Gamma}^{\mu}_{\rho \lambda}T^{\rho} = 0$$
it is still a covarient equation, however it is not manifestly covarient becuase it isn't obvious because neither the reimann tensor nor partial derivatives are covarient.

 

[Kalidor]

It's a scalar isn't it ?

Then shouldn't I say it's invariant?

 

[Mentz114]

Then shouldn't I say it's invariant?

Yes. I would.

...it is still a covariant equation, however it is not manifestly covariant becuase it isn't obvious because neither the reimann tensor nor partial derivatives are covariant.

I think that's it. It is possible for individual terms to be not-covariant when their sum is. Like
your example and also differences between Christoffel symbols.