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nouveau_riche
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why isn't temperature a vector quantity?
iRaid said:Temperature doesn't have a direction does it?
I like Serena said:You might consider putting temperature in a vector to represent for instance what the temperature is at different locations in a pipe.
That's what you would do if you want to calculate the gradient of the temperature in the pipe.
ZapperZ said:But that is no different than having a potential field. It still doesn't turn it into a vector. What you are describing isn't a temperature vector, but rather a position vector.
Zz.
Pengwuino said:Why should it be?
according to me it should have,rest of the world goes against meiRaid said:Temperature doesn't have a direction does it?
chrisbaird said:This is actually a deeper question than first appears. Temperature can be a vector. Temperature is a measure of the average kinetic energy of an ensemble of particles. If we were to get the particles to behave slower in one dimension on average than in the other dimension, then we would need separate temperatures to describe the distributions in the different directions. This is exactly what happens in quantum nanostructures. For instance, in a quantum well state, electrons will have a zero axial temperature Tz and non-zero transverse temperature Txy. In fact, there are separate states in the quantum well, each with its own electron temperature. Conventionally, the different temperature components are just treated separately and not formed into vector mathematics, but there is no reason why they cannot be.
So according to you, if I am facing due north at the equator, should 300 K be towards the left or towards the right of 100 K, and why?nouveau_riche said:according to me it should have,rest of the world goes against me
Hootenanny said:I think that we're drifting a little here.
It is clear that the OP didn't mean a discretised potential field. Temperature is a scalar field and like any scalar field can be discretised if we so wish, but that doesn't mean we would refer to it as a vector.
Well, an array. But an array is not either a "vector" or a "matrix" unless it has algebraic properties. Does it make sense to add your "vector" or "matrix" of temperatures to another? What about multiplication of your "matrices"?I like Serena said:Yes, it's a potential field.
But I think we're splitting words here.
Note that the word "vector" has many meanings.
Certainly temperature itself is not a vector, but you can have a vector, or perhaps I should call it a matrix, of temperature values.
Certainly this is not a position vector, although each entry in the matrix corresponds to a position vector in my example.
Vectors obey a specific set of rules, as do Matrices. They have to have certain properties, otherwise they are simply arrays, as HOI says.nouveau_riche said:the same discrete divisions can be done with vectors,say a velocity can be superposition of many
so what makes them a vector?
The primary thing that makes something a vector is that they must be elements of a vector space. Vector spaces have two operations, the addition of two vectors and the multiplication of a vector by a scalar, and those two operations have to satisfy several axioms:nouveau_riche said:the same discrete divisions can be done with vectors,say a velocity can be superposition of many
so what makes them a vector?
HallsofIvy said:Well, an array. But an array is not either a "vector" or a "matrix" unless it has algebraic properties. Does it make sense to add your "vector" or "matrix" of temperatures to another? What about multiplication of your "matrices"?
DaleSpam said:More importantly, for every element of a vector space there is an additive inverse which is also an element of the vector space. In most systems a temperature of 300 K makes sense, but a temperature of -300 K does not. So for most systems temperatures wouldn't be elements of a vector space.
I am also not certain that addition of temperature makes sense physically. I mean, if you add a system of 300 K to a system of 400 K you don't usually get a system of 700 K. Contrast this to momentum where if you add a system of 300 kg m/s to a system of 400 kg m/s you do get a system of 700 kg m/s.
If the defining operations of a vector space do not make physical sense, then you have to question the usefulness of defining a "temperature vector". I was with you with the discretisation, but calling something a vector for the sake of it isn't useful.I like Serena said:The operations are well defined mathematically.
Left on their own they may make no physical sense, but as an intermediate step to a result, they do make sense.
So you add 2 temperature vectors. The result makes no physical sense.
Then you divide it by 2.
There! You have the average of the temperatures, which does make physical sense.
Same thing for calculating a variance, where you would add the additive inverse of the vector to the vector with the mean temperatures.
Next you would multiply the vector with itself to find the vector with squared errors.
That is a clever approach. There is certainly no reason that you have to restrict the addition operation in a vector space to be the obvious or usual addition operation. The problem is that you then violate associativity with this definition. I.e. the mean is not associative [itex]\overline{\overline{a+b}+c}\neq \overline{a+\overline{b+c}}[/itex]I like Serena said:Left on their own they may make no physical sense, but as an intermediate step to a result, they do make sense.
So you add 2 temperature vectors. The result makes no physical sense.
Then you divide it by 2.
There! You have the average of the temperatures, which does make physical sense.
Hootenanny said:As I said earlier, I fear that we are veering wildly off course here and away from the OP's intended question.
Really? As far as I can tell you didn't even try to counter additive inverse problem and the counter you proposed for vector addition didn't work since it violated some of the other axioms.I like Serena said:As far as I can tell I can counter any of the arguments posed
DaleSpam said:Really? As far as I can tell you didn't even try to counter additive inverse problem and the counter you proposed for vector addition didn't work at all.
I thought we agreed earlier that temperature addition doesn't make sense physically. I certainly don't agree with this. Is there any physics formula you can think of that involves the addition of two temperatures?I like Serena said:Regular addition is defined for all elements and is associative.
The additive inverse of a vector is itself a vector. So a negative temperature needs to exist as a temperature in its own right, not simply as an argument to some operation. For most systems this is not the case.I like Serena said:The additive inverse is defined and contains negative temperatures (physically these are temperatures to be subtracted from some other temperature).
I don't have a problem with scalar multiplication of temperatures, although I can't think of a physics formula that does that either.I like Serena said:Multiplication (or division) by a constant is defined for all elements and is distributive to regular vector addition.
Both. We are talking about whether or not a certain mathematical structure (vector) is a useful formalism in describing a given physical concept (temperature). It seems to me that the mathematical formalism of vectors is very ill-suited to the physical concept of temperature. Addition of temperatures has no physical meaning of which I am aware, and most systems do not have negative temperatures. So the mathematical structure of vectors is not useful in describing the physical concept of temperature, and therefore we say temperature is not a vector.I like Serena said:So we're not talking any more about the mathematical validity, but about the physical validity?
As far as I know the term scalar refers to a quantity that is unchanged under coordinate transformations. Temperature doesn't change under rotation or translation, but it is possible that it changes under boosts (in which case it wouldn't be a scalar). I don't think that scalars form a space with an addition operation and an additive inverse. Do you have a reference that supports that?I like Serena said:Your arguments would be the same for temperature as a scalar.
A scalar is a real number, meaning it can be added and has an additive inverse.
DaleSpam said:So according to you, if I am facing due north at the equator, should 300 K be towards the left or towards the right of 100 K, and why?
No it doesn't! You are standing in the same spot. Just because you are looking in a different direction doesn't mean that the temperature will be different.nouveau_riche said:depends upon the K.E distribution of particles around
I like Serena said:The operations are well defined mathematically.
Left on their own they may make no physical sense, but as an intermediate step to a result, they do make sense.
So you add 2 temperature vectors. The result makes no physical sense.
Then you divide it by 2.
There! You have the average of the temperatures, which does make physical sense.
Same thing for calculating a variance, where you would add the additive inverse of the vector to the vector with the mean temperatures.
Next you would multiply the vector with itself to find the vector with squared errors.
DaleSpam said:More importantly, for every element of a vector space there is an additive inverse which is also an element of the vector space. In most systems a temperature of 300 K makes sense, but a temperature of -300 K does not. So for most systems temperatures wouldn't be elements of a vector space.
I am also not certain that addition of temperature makes sense physically. I mean, if you add a system of 300 K to a system of 400 K you don't usually get a system of 700 K. Contrast this to momentum where if you add a system of 300 kg m/s to a system of 400 kg m/s you do get a system of 700 kg m/s.
Hootenanny said:No it doesn't! You are standing in the same spot. Just because you are looking in a different direction doesn't mean that the temperature will be different.
The temperature only depends on where you are, which is why it is a scalar [field] and not a vector!
Try it: Take a thermometer and stand in your house/school/gym/park facing east (or any other direction you choose). Then, turn by 90 degrees. Does the temperature change?
That still wouldn't change anything. The temperature only depends on your position, not your orientation. It doesn't matter whether your environment is in equilibrium or not. Provided your thermometer is in the same position, it will register the same temperature regardless of its orientation.nouveau_riche said:as i said what you might call as temperature is the equilibrium state,where energy is distributed,do the same experiment as u suggest but with a condition that in that room there is a heater which is just being switched on
KE is not a vector either.nouveau_riche said:depends upon the K.E distribution of particles around