Does the nabla operator has a unit?

[jwillie2000]

Hello Everyone,

I have a small question bothering me. I wan to know whether the nabla operator has a unit? I am thinking it does and it should be 1/m. I just want to make sure whether this is true. Thanks!

Jimmy

 

[HallsofIvy]

Mathematics operators in general don't have units. If, of course, you are applying them to specific physical quantities, the quantities carry their units with them. For example, if y is in meters and x is in seconds, then dy/dt would have units of "meters per second".

If f(x,y,z) gives the temperature, in degrees celsius, at point (x, y, z) where the coordinates are measured in meters from some fixed origin, then
$$\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$$
is a vector with each component in "degrees per meter". Or if f(x,y,z) is pressure in pounds per square foot and x, y, and z are in feet, then each component of $\nabla f$ is in "pounds per cubic foot". (Though it would probably be better to think "pounds per square foot per foot".)

 

[Rap]

Hello Everyone,

I have a small question bothering me. I wan to know whether the nabla operator has a unit? I am thinking it does and it should be 1/m. I just want to make sure whether this is true. Thanks!

Jimmy

HallsOfIvy is right. In physics, the nabla operator has units of 1/Length.

 

[jwillie2000]

Hi HallsOfIvy,

And thanks for replying to my post. If i got you well, it is better to leave it without unit when say defining it in the nomenclature of a thesis? In my thesis, length was measured mostly in metres. So do i indicate 1/m as the unit of the nabla operator or just leave it without unit?

Jimmy

 

[Pengwuino]

HallsOfIvy is right. In physics, the nabla operator has units of 1/Length.

One caveat to this is the fact that you can define gradients that aren't the typical vector calculus gradients you are use to seeing. I honestly can't think of specific examples, but I know there are gradients that are NOT the $${{\partial} \over {\partial x}} \hat x + {{\partial} \over {\partial y}} \hat y + {{\partial} \over {\partial z}} \hat z$$ typical kinda deal. If I'm not mistaken, you can define gradients for any space (not just the typical $$R^3$$ space). For example, if you want to work in some sort of momentum space, I suspect you could have gradients such as $${{\partial} \over {\partial P_x}} \hat P_x + {{\partial} \over {\partial P_y}} \hat P_y + {{\partial} \over {\partial P_z}} \hat P_z$$ that do not have units 1/length.

 

[Rap]

One caveat to this is the fact that you can define gradients that aren't the typical vector calculus gradients you are use to seeing. I honestly can't think of specific examples, but I know there are gradients that are NOT the $${{\partial} \over {\partial x}} \hat x + {{\partial} \over {\partial y}} \hat y + {{\partial} \over {\partial z}} \hat z$$ typical kinda deal. If I'm not mistaken, you can define gradients for any space (not just the typical $$R^3$$ space). For example, if you want to work in some sort of momentum space, I suspect you could have gradients such as $${{\partial} \over {\partial P_x}} \hat P_x + {{\partial} \over {\partial P_y}} \hat P_y + {{\partial} \over {\partial P_z}} \hat P_z$$ that do not have units 1/length.

Yes, I agree. The units depend on the units of the space you are dealing with.

 

[sweet springs]

Hi.
the nabla operator has a unit 1/m in MKSA.
Regards.