Lorentz Invariant Vectors and Grammatical Errors

[Phrak]

A vector in special relativity is the quantity:

$$V = V^\mu \hat{e_\mu}$$

On a change of coordinates, the basis vectors co-vary with the coordinate derivatives:

$$\hat{e_\mu'} = \frac{\partial x_\mu'}{\partial x_\mu} \hat{e_\mu}$$

The vector elements are the opposite. They are said to be contravariant.

$$V'_\mu = \frac{\partial x_\mu}{\partial x_\mu'} V_\mu$$

All basic stuff.Factually speaking, the vector itself is unchanged under a coordinate transformation; factually, the vector is invariant. The vector does not vary. A vector is unchanged under the Lorentz group. The coordinate system changes but the vector does not. That's why its called relativity! The immutable vector is unchanged by a change in coordinate system. It's elements and basis are not.

But the standard language is different. We say, "A vector is Lorentz Covariant". This is shorthand. It really means "The bases of a vector V is Lorentz Covariant." Through whatever route, this grammatical error has become standard. It is wrong-speak, but there it is.

However:-

From time to time I may slip and use the factual language rather than the standard shorthand without thinking. I would appreciate, if in the future this happens again, I am not again insulted over this and implied to be ignorant or stupid or pretentious, ad nauseum, in the subject matter of this relativity forum by either mentors, members or those having any other label.


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[dextercioby]

Your formulas are wrong. You should have used different summation indices.

As for the vectors, well, they are nothing but elements of an n-dimensional vector space tangent to a manifold at a specific point. Since the directional derivatives along curves passing through that point are linear independent, they can be taken as a basis in the vector space. A vector then has this decomposition

$$V|_{p} = V^{k} \partial_{k} |_{p}$$

 

[Fredrik]

This is related to an old-fashioned (and in my opinion really stupid) definition of "vector" as an association of an n-tuple of numbers with each coordinate system, such that the "contravariant" transformation law correctly describes how the n-tuple associated with one coordinate system differs from the n-tuple associated with another. All my physics teachers at the university used this retarded definition, and they didn't even get it right. They always left out that an n-tuple needs to be associated with each coordinate system, and just said that a vector is "anything that transforms like this:...", which makes no sense at all. It was a long time ago, but it still makes me angry when I think about it.

Edit: It was actually even worse than what the above suggests. They never defined what sort of thing the "anything" can be, or what "transforms" means. If you asked them, the answer would be something about experiments performed by different observers. They didn't even seem to understand that mathematical terms need mathematical definitions.

 

[bobc2]

...a vector is "anything that transforms like this:...", which makes no sense at all.

Hi, Fredrik. I can understand your frustration. But once you've established the transformation for displacement vector components between coordinate systems it actually useful to have that tool ("anything that transforms like this...) to test to see if some mathematical object qualifies as a vector. I actually think the definition your prof was providing is a cut above the old "something that has magnitude and direction." And you may not have been served well by your prof if that transformation concept was not adequately communicated to the class.

There are some pretty sharp physicists on the forum who could probably explain it in a way that would have you actually liking the concept.

 

[Phrak]

This is related to an old-fashioned (and in my opinion really stupid) definition of "vector" as an association of an n-tuple of numbers with each coordinate system, such that the "contravariant" transformation law correctly describes how the n-tuple associated with one coordinate system differs from the n-tuple associated with another. All my physics teachers at the university used this retarded definition, and they didn't even get it right. They always left out that an n-tuple needs to be associated with each coordinate system, and just said that a vector is "anything that transforms like this:...", which makes no sense at all. It was a long time ago, but it still makes me angry when I think about it.

Edit: It was actually even worse than what the above suggests. They never defined what sort of thing the "anything" can be, or what "transforms" means. If you asked them, the answer would be something about experiments performed by different observers. They didn't even seem to understand that mathematical terms need mathematical definitions.

It gets stood on it's head too. "A vector obeys the vector transformation law." becomes "If it obeys the vector transformation law, it must be a vector." Who wrote this rule?

 

[Phrak]

Your formulas are wrong. You should have used different summation indices.

Yes. The primes should go with the indices, not the variables. Other than that, and getting it backwards, it's probably fine.

 

[Fredrik]

Yes. The primes should go with the indices, not the variables. Other than that, and getting it backwards, it's probably fine.

No, he meant that you should have used two different symbols for the indices instead of using the same symbol all over the place.

 

[Phrak]

No, he meant that you should have used two different symbols for the indices instead of using the same symbol all over the place.

You didn't understand. That is what I said in so many words.

$$V^{\mu '} = \frac{\partial x^{\mu '}}{\partial x^\mu}V^{\mu}$$

$$\partial_{\mu '} = \frac{\partial x^\mu}{\partial x^{\mu '}}\partial_{\mu}$$

This is how Sean Carroll does it. Do you have any problem with this notation?

And as a matter of fact, I am fairly certain I placed the primes on the indices rather than the coordinates in post#1, but Latex thought otherwise. Try it yourself. x_\mu' will place the prime and on x rather than mu if \mu' is not enclosed in curly brackets.

And here it is. $x_\mu'$

 

[Fredrik]

Dextercioby's point was just that

$$A^\mu = \Lambda^\mu_\nu B^\nu$$

makes more sense than

$$A^\mu = \Lambda^\mu_\mu B^\mu.$$

Do I have a problem with the notation in post #8? Not really, but I'm not thrilled about it either. It avoids the problem that dextercioby pointed out, assuming that $\mu'$ is to be interpreted as a different symbol than $\mu$, but it also (in my opinion) makes it look like the purpose of the prime is to just give us a larger alphabet, and it hides the fact that we're dealing with two different coordinate systems (by not priming the x in the numerator). So to me, your

$$V^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^\mu}V^{\mu}$$

looks like

$$V^{\nu} = \frac{\partial x^{\nu}}{\partial x^\mu}V^{\mu}.$$

Of course, if Carroll explains what he really means (and I assume that he does), people who have read the explanation won't interpret his notation this way.

Unfortunately, I don't think there's a notation that's self-explanatory, so we will always have to explain what we mean. I prefer to say something like this: If v is a tangent vector at p, let V be its matrix of components in the coordinate system x, and V' its matrix of components in the coordinate system x'. Then we have

$$V'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}V^\nu.$$

In my notation, the primes in this equality must be exactly where they are.

 

[HallsofIvy]

This is related to an old-fashioned (and in my opinion really stupid) definition of "vector" as an association of an n-tuple of numbers with each coordinate system, such that the "contravariant" transformation law correctly describes how the n-tuple associated with one coordinate system differs from the n-tuple associated with another. All my physics teachers at the university used this retarded definition, and they didn't even get it right. They always left out that an n-tuple needs to be associated with each coordinate system, and just said that a vector is "anything that transforms like this:...", which makes no sense at all. It was a long time ago, but it still makes me angry when I think about it.

Edit: It was actually even worse than what the above suggests. They never defined what sort of thing the "anything" can be, or what "transforms" means. If you asked them, the answer would be something about experiments performed by different observers. They didn't even seem to understand that mathematical terms need mathematical definitions.

And it all goes back to using "position vectors" in flat space as our first introduction to vectors. When I was in high school, I spent a lot of time worrying about what "position vectors" on a sphere looked like! Did they curve around the sphere or did they go through the sphere?

I eventually learned to drop the whole concept of "postition vector". All vectors are derivatives and they live in the tangent space to the manifold, not the manifold itself. Those "transformations" are really the chain rule.

 

[Phrak]

Dextercioby's point was just that

$$A^\mu = \Lambda^\mu_\nu B^\nu$$

makes more sense than

$$A^\mu = \Lambda^\mu_\mu B^\mu.$$

Do I have a problem with the notation in post #8? Not really, but I'm not thrilled about it either. It avoids the problem that dextercioby pointed out, assuming that $\mu'$ is to be interpreted as a different symbol than $\mu$, but it also (in my opinion) makes it look like the purpose of the prime is to just give us a larger alphabet, and it hides the fact that we're dealing with two different coordinate systems (by not priming the x in the numerator). So to me, your

$$V^{\mu'} = \frac{\partial x^{\mu'}}{\partial x^\mu}V^{\mu}$$

looks like

$$V^{\nu} = \frac{\partial x^{\nu}}{\partial x^\mu}V^{\mu}.$$

Of course, if Carroll explains what he really means (and I assume that he does), people who have read the explanation won't interpret his notation this way.

Unfortunately, I don't think there's a notation that's self-explanatory, so we will always have to explain what we mean. I prefer to say something like this: If v is a tangent vector at p, let V be its matrix of components in the coordinate system x, and V' its matrix of components in the coordinate system x'. Then we have

$$V'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}V^\nu.$$

In my notation, the primes in this equality must be exactly where they are.

I agree.

I do recall being baffled by Carroll's placement of primes, and his non-explanation, "Notice that we put the prime on the index, not on the x." This is not a harsh criticism of Carroll. He taught me what general relativity I know, where others were far too dry to pursue.

In retrospect, it really makes no sense (consider a 90 degree rotation in space) unless the prime is meant to apply to both the index and the coordinate.