Why isn't temperature a vector quantity?

[nouveau_riche]

why isn't temperature a vector quantity?

 

[Pengwuino]

Why should it be?

 

[I like Serena]

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.

I'm not sure if that's what you mean though.

 

[iRaid]

Temperature doesn't have a direction does it?

 

[ZealScience]

Vectors have components while temperature can't be represented by components.

 

[I like Serena]

Temperature doesn't have a direction does it?

Not a spatial direction, but in the vector I described it has an other type of generalized direction.

 

[ZapperZ]

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.

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.

 

[I like Serena]

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.

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.

 

[Hootenanny]

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.

 

[chrisbaird]

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 T_z and non-zero transverse temperature T_xy. 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.

 

[nouveau_riche]

Why should it be?

bcoz i need a direction to specify it's magnitude

Temperature doesn't have a direction does it?

according to me it should have,rest of the world goes against me

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 T_z and non-zero transverse temperature T_xy. 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.

i think the temperature is taken to be constant because what they call as temperature is the average k.E energy of the system,which neglects the random behavior of particles
that is why the temperature of room seems to be constant over it's domain,but it may vary outside that

 

[Dale]

according to me it should have,rest of the world goes against me

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]

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.

the same discrete divisions can be done with vectors,say a velocity can be superposition of many
so what makes them a vector?

 

[HallsofIvy]

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.

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"?

 

[Hootenanny]

the same discrete divisions can be done with vectors,say a velocity can be superposition of many
so what makes them a vector?

Vectors obey a specific set of rules, as do Matrices. They have to have certain properties, otherwise they are simply arrays, as HOI says.

 

[Dale]

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:

http://en.wikipedia.org/wiki/Vector_space#Definition

Note that advanced physics, like general relativity, makes heavy use of tensors and Riemannian geometry where vectors are defined a little differently, but I think the above link is probably more relevant for your current questions.

 

[I like Serena]

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"?

Errr... yes it makes sense to add or multiply these vectors, or as you may call them, arrays.
Think averages, variances, physical attributes with which you may want to multiply entry by entry, convolutions, ...

 

[Dale]

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.

 

[I like Serena]

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.

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.

 

[Hootenanny]

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.

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.

There are occasions where defining temperature vectors are useful, but not in the sense you propose. For example, suppose we are looking at heat condition in a simple 2D - biatomic lattice. Then we could define a vector $\boldsymbol{\theta} = [\theta_1,\theta_2]^\text{T}$, where $\theta_i$ is the temperature of the $i^\text{th}$ lattice node in the elementary cell. One could then define the heat flux as a matrix product $A\boldsymbol{\theta}$ where the matrix $A$ described the heat conduction through the lattice links. Such an application could then make use of the properties of vector spaces and matrix algebra to simplify the problem and express the heat condition problem in the whole lattice in terms of matrix products.

In other words, defining the discretised temperature (or other scalar) field as a vector is useful for quantifying the relationship (or flow of the field) between different points or mesh vertices. So, although adding the temperature at two different points doesn't make sense, summing the flux from neighbouring points does.

As I said earlier, I fear that we are veering wildly off course here and away from the OP's intended question.

 

[Dale]

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.

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 $\overline{\overline{a+b}+c}\neq \overline{a+\overline{b+c}}$

I just don't see a natural way to fit the physical concept of temperature into the mathematical formalism of vectors and vector spaces. Therefore, temperature is not a vector. You can make some mathematical object which satisfies all of the properties of vectors and you can even give it units of Kelvin, but that object wouldn't correspond very well to the physical concept of temperature.

 

[I like Serena]

As I said earlier, I fear that we are veering wildly off course here and away from the OP's intended question.

It baffles me that 4 (5?) established members tell me I'm wrong and as yet none to support my view.
As far as I can tell I can counter any of the arguments posed, although we may quibble about what the word "vector" exactly means and when that word should be used.

But I agree that this is off-topic and as far as I'm concerned we can let this lie.

 

[Dale]

As far as I can tell I can counter any of the arguments posed

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]

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.

All right.

First let me adjust the term I will be trying to use.
We're not talking about a temperature vector, it's a vector of temperatures.
But I'll use the term array of temperatures to avoid word-definition-discussions.
Note that this array is well-defined as a vector in a vector space.

I'm not really sure what you're referring to with the additive inverse problem or with a vector add problem though.
You did say the mean-operation is not associative.
But I never said so and indeed it isn't.

Your argument is about the mathematical validity to treat this as a vector space, so I'll zoom in on that.

You can find the definition of a vector space for instance on wiki:
http://en.wikipedia.org/wiki/Vector_space
If you check, you will see that all axioms are satisfied.

More specifically.
Regular addition is defined for all elements and is associative.
The additive inverse is defined and contains negative temperatures (physically these are temperatures to be subtracted from some other temperature).
Multiplication (or division) by a constant is defined for all elements and is distributive to regular vector addition.

So no, the mean is not associative, but the regular vector addition is, and the division by a non-zero constant is properly defined.

 

[Dale]

Regular addition is defined for all elements and is associative.

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?

The additive inverse is defined and contains negative temperatures (physically these are temperatures to be subtracted from some other temperature).

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.

Multiplication (or division) by a constant is defined for all elements and is distributive to regular vector addition.

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]

@Dalespam:

So we're not talking any more about the mathematical validity, but about the physical validity?

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.
Are you saying temperature as a scalar is not properly defined?
Or that you can't add temperatures to obtain a mean?
A vector of scalars behaves exactly the same.

Anyway, in general intermediate results do not have to make physical sense.
It's only the initial set up and the final conclusion that have to make physical sense.
To illustrate, what's the physical sense in multiplying 2 masses when applying Newton's law of gravity?

 

[Dale]

So we're not talking any more about the mathematical validity, but about the physical validity?

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.

Contrast this to displacement. The addition of two displacements results in another displacement and has a clear and useful physical meaning. Every displacement has an equal and opposite displacement, the additive inverse. Scalar multiplication of a displacement also results in another displacement and has a clear physical meaning. So the mathematical structure of vectors is useful in describing the physical concept of displacement, and therefore we say displacement is a vector.

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.

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?

 

[WannabeNewton]

Yes, in order for temperature to be a vector it would have to obey the vector transformation laws when defining new coordinate functions but temperature can remain invariant after coordinate transformations as DaleSpam stated. Plus, if you give temperature in terms of a vector field instead of a scalar field then how would you take the gradient.

 

[nouveau_riche]

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?

depends upon the K.E distribution of particles around

 

[Hootenanny]

depends upon the K.E distribution of particles around

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?

 

[nouveau_riche]

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.

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.

the key idea in that temperature does not follow an addition of 300k + 400k=700k is the fact that temperature is seen as averages,when you say 300k ,this is the state of equilibrium,in that context the energy has been distributed uniformly giving it a scalar touch
and that is what @i like serena is trying to highlight
i suppose

 

[nouveau_riche]

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?

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

 

[Hootenanny]

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

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.

 

[Dale]

depends upon the K.E distribution of particles around

KE is not a vector either.

Please see the above discussion, temperature does not form a vector space.

 

[HallsofIvy]

Since we are talking about vectors in space, the crucial point is how they convert to a different coordinate system. If your "vector" has temperatures at different locations as components, how do they change if you rotate the coordinate system?