In particle physics we are going over 4-vector notation. However, my background on this is a little shaky, and I'm having difficulty differentiating the notation and visualizing what it actually means.

What is the difference between: $X^\mu,X^\nu$and $X^\sigma$? I'm not sure what the different superscripts mean, and I don't know how to visualize this. I know they are 4-vectors, but I don't know what they look like written out. Are each of them different? Or are the superscripts just placeholders to let you know that they have different components?

Another issues I have is the different between $\partial^\mu$ and $\partial^\nu$. $\partial^\mu=\big({\partial\over \partial t},-\vec \nabla\big) $ but what is $\partial ^\nu$?


$X^\mu,X^\nu$and $X^\sigma$ are all the same four-vector. The letter used for the superscript or subscript doesn't matter. If an index it isn't being contracted with another index to form a scalar, as in $X^\mu X_\mu$, then it is just a placeholder which can take the value 0, 1, 2, and 3 (or sometimes people use t, x, y, and z). For example, either

$$p^\mu=m u^\mu$$


$$p^\nu=m u^\nu$$

is just shorthand for four equations,

$$p^0=m u^0 \\ p^1=m u^1 \\ p^2=m u^2 \\ p^3=m u^3 \\ $$

When an index appears twice on the same side of the equation, once up and once down, this is called a contraction and you you have to sum over all four values of the index:

$$X^\mu X_\mu=X^0 X_0+X^1 X_1+X^2 X_2+X^3 X_3.$$

$\partial^\mu$ and $\partial^\nu$ both mean $\big({\partial\over \partial t},-\vec \nabla\big)$ (if you take the metric to be +---). But $\partial_\mu$ and $\partial_\nu$ both mean $\big({\partial\over \partial t},\vec \nabla\big)$.

  • $\begingroup$ Is it correct to say that the four-vectors are all 4x1 vectors? Then would $X_\mu$ be a 1x4 vector? $\endgroup$ – matryoshka Jan 31 at 19:26
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    $\begingroup$ The usual terminology in physics is that the $X^\mu$ are "contravariant components of a four-vector" and the $X_\mu$ are "covariant components of the same four-vector". You shouldn't think in terms of 1x4 vectors and 4x1 vectors, because if you do that you'll have problems when it comes to four-tensors with more than one or two indices, like, say, ${{{{R_\mu}^\nu}^\lambda}_\kappa}$. $\endgroup$ – G. Smith Jan 31 at 19:33
  • $\begingroup$ If I wanted to write out the four-vectors instead of using the short hand notation what would both of the types of vectors look like? $\endgroup$ – matryoshka Jan 31 at 19:37
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    $\begingroup$ $p^\mu=(p^0,p^1,p^2,p^3)$ and $p_\mu=(p_0,p_1,p_2,p_3)$ Or you could write them vertically instead. Gotta go now. $\endgroup$ – G. Smith Jan 31 at 19:40
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    $\begingroup$ side comment: common abuse of notation in physics uses $A^\mu$ to denote the 4-vector whereas strictly speaking this is the $\mu$’th component of a 4-vector. It’s usually clear from context if one refers to the whole 4-vector or any one of its components. $\endgroup$ – ZeroTheHero Jan 31 at 20:00

So there's a bunch of ways to understand this, but here is the "abstract index notation" view of those.

  1. You have a vector space, call it $\mathcal V^\bullet$. In this case these are your four-vectors which are meant to transform like $(ct, x, y, z)$ do in relativity, but you can use this notation in other places too.
  2. You have a space of covectors, linear maps from $\mathcal V^\bullet$ to the real numbers $\mathbb R.$ It will turn out that these are in perfect correspondence to your vectors, so we call this space $\mathcal V_\bullet.$
  3. We create a bunch of copies of $\mathcal V^\bullet$ and $\mathcal V_\bullet,$ one for every symbol that you care about. So you might denote the copies with Greek letters replacing the $\bullet$, like the space $\mathcal V^\mu$ is a copy of $\mathcal V^\bullet.$ We also denote any inhabitant of a copy with a raised superscript index denoting which space it belongs to. So $v^\mu$ is a vector living in $\mathcal V^\mu.$ These are connected by the relabeling isomorphism $\delta^\bullet_\bullet$ which can take a vector in $\mathcal V^\mu$ and tell you what the same vector is in the space $\mathcal V^\nu.$
  4. Similarly we do this for the covectors, but when we see the same index repeated for a vector and a covector, we understand that it means to apply the covector to the vector to produce a scalar. So $v_\mu v^\nu$ is not a scalar but we call it an "outer product", which lives in a "tensor" space $\mathcal T^\nu_\mu$, the space of all multilinear maps from pairs $\mathcal V^\mu \times \mathcal V_\nu \to \mathbb R.$ But $v_\mu v^\mu,$ because the index is repeated, means to apply the covector to the vector to create a scalar, so it is a real number.
  5. We also have, as the previous point indicated, these tensor spaces, which are multilinear maps from the associated copies of vectors and covectors to our scalars. One key assumption that we have to require to make the above "contraction" axiom fully general, is that any tensor can be represented as a finite sum of products of vectors and covectors. So the relabeling isomorphism is a linear map from one space of vectors to another, but you can also look at it as a linear way to apply the covectors of that latter space to the former. $\delta^\alpha_\beta$ is in $\mathcal T^\alpha_\beta.$
  6. We extend the above notion of applying a covector to a vector by asserting, by fiat, that we only care about cases where every tensor can be written as a finite sum of outer products of vectors and covectors. So given some tensor $t^{\alpha\beta\gamma~~\epsilon}_{~~~~~~\delta}$ (usually the order of indexes matters a bunch because of the next point), living in $\mathcal T^{\alpha\beta\gamma\epsilon}_{\delta}$ (order doesn't matter for the tensor space though), we can contract, say, $\beta$ with $\delta$. We would write this as $\delta^\delta_\beta ~t^{\alpha\beta\gamma~~\epsilon}_{~~~~~~\delta}$ or just $t^{\alpha\beta\gamma~~\epsilon}_{~~~~~~\beta},$ and it is a tensor living in $\mathcal T^{\alpha\gamma\epsilon}$. So what we did here is that we expanded it out in a finite sum of vectors and covectors in $\mathcal V^\alpha \times \mathcal V^\beta \times \mathcal V^\gamma \times \mathcal V_\delta \times \mathcal V^\epsilon$ and then we applied all of the covectors living in $\mathcal V_\delta$ to all of the vectors living in $\mathcal V^\epsilon$ and each one of those gave us a real number times a tensor, which is another tensor, so we summed up all of the resulting tensors still in $\mathcal T^{\alpha\gamma\epsilon}.$
  7. We finally introduce one special tensor and its inverse, the "metric" tensor. You have heard this called a "dot product" before. Often it is written $g_{\bullet\bullet},$ it is the canonical way to take two vectors and produce a scalar. One can also view it as the canonical way to take a vector and produce a covector. It is symmetric: $g_{\alpha\beta} = g_{\beta\alpha}.$ As I told you above in point (2), I said the covectors are in perfect correspondence with the vectors: meaning that there is also an inverse tensor $g^{\bullet\bullet}$ that undoes it, so that $g^{\alpha\beta}~g_{\beta\gamma} = \delta^\alpha_\gamma.$ So there is a way from any vector $\vec v$ to get to the covector $(\vec v~\cdot~)$ and back.
  8. Using this tensor allows us to arbitrarily raise and lower any index, and therefore to contract any two indices, on any tensor. So remembering $t^{\alpha\beta\gamma~~\epsilon}_{~~~~~~\delta}$, we can form $$t^{\alpha\beta\gamma\delta\epsilon} = g^{\delta\phi}~{\alpha\beta\gamma~~\epsilon}_{~~~~~~\phi}$$and that is the canonical way to "raise" that $\delta$ index from lower to higher: we just implicitly understand that when you used the same base symbol $t$ to denote the tensor, you meant that you wanted us to consider these the same. That is also why I said above that the "order matters", it's because you want this to always be a common operation.

Now all of the above happen without any notion of coordinates or basis, but you may be interested to use those. So we might reserve the more usual alphabet of Roman/English letters for variables indicating numbers, and make them not part of the above abstract indexes. So we might invent a bunch of covectors that we could call "component extractors", $w_\alpha$ is a covector which takes a $v^\alpha$ four-vector and extracts its component in the $w=ct$-direction, $x_\alpha$ similarly might extract a component in the $x$-direction, $y_\alpha$, $z_\alpha$ likewise. And we might decide that it's easier to just number these as $b^{0,1,2,3}_\alpha$. And the key property is that there must also be some basis vectors in the space, $b^\alpha_{0,1,2,3},$ such that $$b^m_\alpha~b^\alpha _n = \{1 \text{ if } m = n \text{ else } 0\}.$$So notice that the conventional letters $m,n$ are just normal variables standing in for numbers, while the Greek indices are abstract indexes of the above sort.

  • $\begingroup$ It's praiseworthy that you put so much effort into this detailed answer, but this is wildly mismatched to the level at which the OP asked the question. $\endgroup$ – Ben Crowell Jan 31 at 22:31

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