# A few questions about fields

Tags:
1. Jul 25, 2015

### Isaac0427

Hi all! I am a 13-year-old that is very interested in physics. I am currently studying fields, and I have gotten into some advanced parts of field theory. With that, I have 3 big questions:
1. In vector and higher order tensor fields, what does it mean to "observe the field from point (x,y,z)?" I fully understand how to use the equations to get a column (or row) vector, and how to draw a vector from it, but the vector seems to change depending on what point I plug in (with most equations). I understand how this works mathematically, but I would love to know how it works applied to a real-life vector field so I can understand it better.
2. What just doesn't make sense to me is a 3rd-order tensor. Again, I understand it as a mathematical concept, but not at all as a concept applied to real-life physics.
3. This one is fairly straight foreword: Does the curl of a vector field change depending on the point it is measured at? From what I understand the answer is yes, but I just want to check.

2. Jul 25, 2015

### cpsinkule

1)observing a field is just a physicist saying "the value of the field at point such and such". the vectors are supposed to change from point to point. the vector fields you are studying are "real-life" vector fields.
2)What tensor are you speaking of? Most tensors don't have a "picture this in your head" view, but there are some, the energy momentum tensor represents the flow of momentum\energy through surfaces of constant coordinate values same goes for strain-stress tensors.
3)The curl, in general, changes from point to point

3. Jul 25, 2015

### Isaac0427

So if I measure vector a at point (3,2,1), and the row vector I get is (1,2,3), then would vector a go through (3,2,1) and (1,2,3) when measured at point (3,2,1)?

4. Jul 25, 2015

### cpsinkule

I'm having trouble understanding your question. The components of a vector field change from point to point unless the vector field is constant.

5. Jul 25, 2015

### Isaac0427

Why does the field change depending on where I view it from. I'm looking for an answer that applies it to an actual vector field, not a mathematical answer.

6. Jul 25, 2015

### jbriggs444

Take the field defined by the gravitational force from the earth that would be felt by a hypothetical 1 kg test mass at each point in space. The field value measured at one point will not, in general, be the same as that measured at another point. First, because the magnitude will vary with distance from the center of the earth. The two points may not be the same distance from the Earth's center. Second, because the direction of the force will point toward the center of the earth. The two points may not be in the same direction from the center of the Earth.

7. Jul 25, 2015

### Staff: Mentor

Here's an easy example: the Earth's gravitational field. You're already familiar with idea that the force on an object of mass $m$ from the Earth's gravity is given by $GmM_E/r^2$ where $r$ is the distance to the center of the earth and the force vector points towards the center of the earth.

We can restate this in the language of fields: the force on an object of mass $m$ from the earth's gravitational field is given by $\vec{F}=m\vec{G}$ where $\vec{G}$ is the gravitational field at the point where the object is - it is a vector pointing towards the center of the earth with magnitude $GM_E/r^2$. That vector is the value of the gravitational field vector at that point, and you can see that it has different values at different points.

8. Jul 25, 2015

### Staff: Mentor

To carry the gravitational field example from the previous two posts a bit further, the gravitational field at the top of the tower at One World Trade Center in New York doesn't depend on whether you view the tower from Manhattan, or from New Jersey, or from the International Space Station.

(just in case that's what you were really asking)

9. Jul 25, 2015

### Isaac0427

Thanks guys. I'm just a little confused on this: say you measure gravity from point (3,3,3) and when you go through the function, you get the vector from (5,6,7) to the origin (ie the function is F=(xy-4)i+(z-y+2x)j+(x^2-2)k). The point (3,3,3) is not even on that vector, although it is where you measure the field from.

10. Jul 25, 2015

### jbriggs444

There is no reason that it should be. If the field is arbitrary then its value (which is a vector) at any point is arbitrary. In particular, the direction of its value there is arbitrary and need not point at the origin.

Edit: In addition, it is more standard to say that the vector value is (5,6,7) rather than trying to say that it is some kind of arrow pointing from (5,6,7) to (0,0,0).

11. Jul 25, 2015

### Isaac0427

But isn't the vector (5,6,7) a vector from (5,6,7) to (0,0,0)? And if I'm getting this right, the length of the vector just represents the strength of the force. Lastly, if the vector (5,6,7) doesn't point to the origin, where does it point to and how would someone know? Thank you for all the help.

12. Jul 25, 2015

### cpsinkule

ohhh I see your difficulty here. In high school your teachers teach you that a vector is something that originates at the origin and terminates at some point. This is not how it works in general. What is really happening is at every point in space you define a linear vector space at that point (called the tangent space). The reason you don't notice this is because the tangent space at a point to Rnis itself, Rn. To make this clearer, imagine a sphere and imagine at every point on that sphere there is a tangent plane. Those tangent planes are the tangent spaces and the vectors defined on the sphere live in those tangent planes, and the set of all of the tangent spaces is called the tangent bundle. Now just imagine the sphere as being R2, you would not be able to visually identify the tangent space from the space itself. I think you are imagining the vector given by the function and then "dragging" it to the origin, which is not what you do. At least, that's what I got from your reply.

13. Jul 25, 2015

### jbriggs444

You can model vectors as directed displacements. That much is good. One would conventionally regard the vector (5,6,7) to correspond with a directed line segment (depicted as an arrow) from the origin to the point with coordinates (5,6,7).

[The representation you have in mind has the arrow pointing the opposite way. But that's not a problem. It's merely a sign convention]

If this is done, one would often regard the displacement from (0,0,0) to (5,6,7) as being equivalent to the displacement from (1,1,1) to (6,7,8). So that the origin is irrelevant and all that matters is the direction and length of the arrow.

Given that, you could imagine a "field" as an array of arrows. Each arrow starting at a point with a length and direction corresponding to the field value at that point. (Kind of like a field full of weather vanes or wind socks)

14. Jul 25, 2015

### Isaac0427

Ok so please tell me if this is correct: in my example, the strength of the vector field would be √110 of whatever unit r-hat is. The direction would be the direction of the vector (5,6,7).

15. Jul 25, 2015

### jbriggs444

Assuming that "strength" is another word for "magnitude" then yes, that is the strength of the field at the point (3,3,3). The strength and direction of the field at other points would be different, of course.

16. Jul 25, 2015

### Isaac0427

Yes, my apologies. I meant magnitude.