What does upper indices represent?
I saw that people were representing matrices in two ways.
It is representing a column matrix (vector actually) if we assume
What is it representing? At first, I thought it was a row matrix (vector) since it is the opposite of a column matrix (vector). But when I was writing the question I couldn't generate a row matrix using the 2nd equation. I became more confused when I saw
After searching a little bit I found that when we move components our vectors don't change. But, I can't get deeper into covariant and contravariant. I even saw some people use an equation like this:
I was reading https://physics.stackexchange.com/q/541822/, and those answers don't explain the covariant and contravariant for a beginner (those explanations are for those who have some knowledge of the covariant and contravariant).
I was watching the video, what he said that is, if we take some basis vectors and then find a vector using those basis vectors than if we decrease length of those vectors than that's contravariant vectors (I think he meant to say changing those components). But the explanation is not much more good to me. He might be correct also but I don't have any idea. If he is assuming that changes of basis vectors is contravariant then is "the original" basis vectors covariant? So how do we deal with covariant and contravariant altogether
2 answers
A textbook homework problem might ask, ‘Speedy the snail creeps along at a steady pace of 60 cm per minute. How far does Speedy travel each second?’ The correct answer is, of course, one centimeter. Hopefully your teacher would also accept 10 mm or 0.01 m as equally valid answers; after all, they all represent the same physical length. But how do we know that? And how do we get that answer from 60 cm/min? What even is a quantity like 60 cm/min, mathematically?
Well, I'll make the case that 60 cm/min is a tensor in
When we work with units such as meters, centimeters, minutes, seconds, etc., we bring a certain set of assumptions along. For starters, we assume that the quantities that we use units to represent can be added to each other if the units match: 1 meter + 1 meter = 2 meters. We also assume that these quantities can be multiplied by dimensionless quantities to get more dimensioned quantities in the same units: 5(2 m) = 10 m. We assume a bunch of intuitive facts that more or less boil down to not caring in what order any of these additions or multiplications are performed: 1 m + 2 m = 2 m + 1 m, 1 m + (2 m + 3 m) = (1 m + 2 m) + 3 m, (2 + 3)(10 m) = 2(10 m) + 3(10 m), 2(3(5 m)) = (2×3)(5 m), etc. Some of our units are compatible with each other, like meters and centimeters or minutes and seconds, and if we appropriately scale the quantities expressed in those units, they can also be added with each other. Finally, we assume that there's a ‘nothing’ associated with each of these families of compatible units, and that it doesn't matter which unit is used to express that nothing: 0 m = 0 cm. We expect it to behave like a proper nothing: 0 m + 5 m = 5 m, and 0(7 m) = 0 m.
Those assumptions are precisely what make each ‘family of compatible units’ into a vector space, for which the appropriate units are alternate choices of bases. I'll call the vector space for which meters are appropriate basis elements
Here's a question: is a hertz (Hz, or
However, there is a relationship between frequency and time: a frequency tells you how much of some dimensionless quantity you get for a given amount of time. If you type at 5 Hz for 1 s, you have pressed 5 keys. If you type at 5 Hz for 4 s, you have pressed 20 keys; the result scales linearly with the amount of time. So we can say that a frequency, in addition to being a vector in its own right, is also a linear function from a vector in
When I was talking about compatible units before, I was vague about how to scale values when converting between units. Let's hammer that out now. We know that one meter equals 100 centimeters. So if we want to express 2 m in centimeters, we know to multiply 2 by 100 to get 200 cm. In general, in a vector space, if you have expressed a vector in some basis, and then you want to change your basis by dividing it by some amount, you have to multiply the numbers in your vector representation by that same amount. That's all that's happening here: as we change basis from 1 m to 1 cm (dividing by 100), we have to multiply the number 2 by the same number, 100, to get 200 cm from 2 m. (And of course, since dividing by
The same logic applies for time: 2 min = 120 s because we divide a minute by 60 to get a second, and so we have to multiply 2 by 60 to get the equivalent representation in seconds. Similarly for frequency: 4 kHz = 4000 Hz, because 1 kHz = 1000 Hz. But now notice something interesting—frequency, remember, is a function from an amount of time to a quantity. If something is happening at 4 Hz for 60 seconds, it happens 240 times. What if we want to change units from seconds to minutes? We'd like to express 4 Hz, which is the same as 4
The words ‘contravariant’ and ‘covariant’ just distinguish between these two rules. When there is a particular vector space of interest
With all that out of the way, let's revisit Speedy's speed. When we say ‘60 cm/min’, we are expressing an exchange rate of a certain number of centimeters for every minute provided. Much like a frequency was a linear function from an amount of time to a dimensionless quantity, a speed is a linear function from an amount of time to an amount of one-dimensional space. If Speedy can travel 60 centimeters given one minute, Speedy can travel 120 centimeters given two minutes, and so on. Unlike frequency, the result of a speed when given an amount of time is not a dimensionless quantity, but another vector. So a speed is a linear function from vectors in one space to vectors in another.
The original math problem asked us to evaluate the function ‘60 cm/min’ on the vector ‘1 s’. We could do this by translating 1 s to
All this without having to talk about tensors yet! Let's fix that.
Speedy's cousin Sticky is treasure hunting and has mounted a compass-like metal detector to her shell. At Sticky's current location, for every centimeter Sticky creeps north, the needle of the detector dips down 0.017 mm and to the left 0.003 mm; for every centimeter Sticky creeps east, the needle moves to the left 0.058 mm. Assuming the detector has a linear response in Sticky's vicinity, how far does the needle move in total if she slides a total of 1.5 cm north and 0.5 cm west?
The intended solution to this problem is quite straightforward: the needle moves
Just as a speed expresses an exchange rate between one-dimensional space and time, the numbers in this problem express exchange rates between two different two-dimensional spaces—namely, Sticky's displacement and the needle's displacement. The problem tells us to assume this relationship is linear, so it's natural to try to look at these numbers as a linear function from
Instead, let's look at each element individually, and without dropping the units. The first element we're given is that the needle dips down 0.017 mm for every centimeter Sticky travels north—which is to say, each vector
This motivates the definition of a tensor product of vector spaces (over a common field of scalars): a new vector space generated by tuples of one basis element from each of the factor vector spaces. We write
So Sticky's metal detector response can also be represented as a tensor in
Finally, let's look at what happens when we solve the Sticky problem. Just like Speedy's speed was a function that we applied to an input vector to get an output vector, Sticky's detector response is also such a function. We simply have to apply the above tensor as a function to the vector
Now we can save quite a bit of writing by abstracting over the basis vectors with indices. Let the index
This operation of matching a covariant index with a contravariant index and summing is called contraction. It is extremely common, and also quite intuitive once you're comfortable with the underlying concepts—you're just feeding an output vector into a linear form which expects an input vector! Contraction is such a common operation in tensor algebra that there is a convention, Einstein notation, which enables the
So in summary:
- Tensors are combinations of vectors from multiple vector spaces.
- Basis vectors are like physical units; tensors are like multidimensional quantities involving several units.
- A space of tensors can be covariant or contravariant in an underlying vector space, depending on whether it combines vectors or dual vectors from that space. The terminology just describes how the coefficients of tensors need to be adjusted if the basis of that vector space is scaled.
- But tensors covariant in a given vector space also represent linear functions consuming vectors from that space, while tensors contravariant in a vector space represent producing vectors in that space.
- A variable with upper and/or lower indices represents a coefficient from a tensor given some set of bases for the vector spaces being tensored.
- Upper indices correspond to contravariant basis vectors and lower indices correspond to covariant basis vectors.
- Matching upper and lower indices are meant to be summed over, which corresponds to feeding contravariant vectors to linear forms, a.k.a. covariant vectors.
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In general sense, convariant and contravariant isn't a interesting thing. But what confuses here that is the meaning of those words.
Actually, if we think of two different basic vectors than a basis vector will be covariant another will be contravariant. But a basis vector must be smaller than another basis vector. Let we have a vector which looks like this :
We usually use the method to define spacetime from different parts. There's possible way to transform (It's not the one which I had seen but it describes little bit) them also.
When using contravariant and covariant in a single term that represent that, we are trying to represent a vector using different basis. Like as, we are going to define plain line respectively through x,y and z axis
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