# Confused about tangent to a circle (polar/Cartesian convert)

• TomServo
In summary, the conversation is discussing problem 5.6 from D'Inverno's book on GR, which involves using the Jacobian/transformation matrix to convert the tangent to a circle centered at the origin of radius A from Cartesian to polar coordinates. The person is able to solve the problem using the matrix, but gets an extra factor of A when trying to do it without using the matrix. The conversation also discusses issues with rendering equations and a possible error in the book's answer. The expert suggests that the problem may be due to normalizing the polar basis vectors, and explains why this could affect the solution.
TomServo

## Homework Statement

The problem is from D'Inverno's book on GR, problem 5.6. We're using the Jacobian/transformation matrix to convert the tangent to a circle centered at the origin of radius A from Cartesian to polar coordinates. I can do the problem and get the book answer, that's okay. What confuses me is when I try to do the conversion without using the matrix and just do it the old-fashioned calculus class way I get an extra factor of A.

## Homework Equations

Polar coordinate system is primed, Cartesian is unprimed.

$$x=a cos(\phi)\\ y=a sin(\phi)\\ \vec{x}=x\hat{x}+y\hat{y}\\ \vec{x'}=a\hat{r}\\ \vec{T}=\dot{x}\hat{x}+\dot{y}\hat{y}=-y\dot{\phi}\hat{x}+x\dot{\phi}\hat{y}$$
$$\vec{T'}=\frac{d}{dt}a\hat{r}=a\dot{\phi}\hat{\phi}\\ \hat{r}=cos(\phi)\hat{x}+sin(\phi)\hat{y}\\$$
$$\hat{\phi}=-sin(\phi)\hat{x}+cos(\phi)\hat{y}$$

$$\dot{\hat{r}}=\dot{\phi}\hat{\phi}$$
I don't know why this last one won't render, I've tried everything I could think of.
Mod note: It's fixed now. Instead of \dot\hat{r}, use \dot{\hat{r}}

## The Attempt at a Solution

When I try to convert $$\vec{T}$$ to polar coordinates by substituting x and y in terms of $$r, \phi$$ I get:$$\vec{T}=-(a sin(\phi))\dot{\phi}\hat{x}+(a cos(\phi))\dot{\phi}\hat{y}=a\dot{\phi}\hat{\phi}=\vec{T'}\\$$

The problem is that using the transformation matrix I get $$\vec{T'}=\dot{\phi}\hat{\phi}$$, no factor of a, and this is what the book has as an answer too. Where am I going wrong? Thanks.

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I do not have d'Inverno, but probably your problem is that you have normalised your polar basis vectors.
TomServo said:
I don't know why this last one won't render, I've tried everything I could think of
\dot\hat{x} is not a good TeX command. You must do \dot{\hat{x}} or \dot{\hat x}

Orodruin said:
\dot\hat{x} is not a good TeX command. You must do \dot{\hat{x}} or \dot{\hat x}
I edited the OP before noticing that you had addressed this problem.

Orodruin said:
I do not have d'Inverno, but probably your problem is that you have normalised your polar basis vectors.

\dot\hat{x} is not a good TeX command. You must do \dot{\hat{x}} or \dot{\hat x}

Yes but my Cartesian basis vectors (xhat and yhat) are also normalized. I get the feeling that the book answer is wrong, just by considering units, and that there should be a factor of a in the transformed tangent vector.

TomServo said:
Yes but my Cartesian basis vectors (xhat and yhat) are also normalized. I get the feeling that the book answer is wrong, just by considering units, and that there should be a factor of a in the transformed tangent vector.
Not the Cartesian basis, the polar basis. The Cartesian basis is normalised by default.

Orodruin said:
Not the Cartesian basis, the polar basis. The Cartesian basis is normalised by default.
Aren't the polar vectors always normalized? Where would the normalization enter into the derivation of the transformation matrix?

TomServo said:
Aren't the polar vectors always normalized? Where would the normalization enter into the derivation of the transformation matrix?
No. Why would they be?

Edit: A normalised basis would mean a displacement rate of one when changing the corresponding coordinate. The displacement rate when changing the polar angle (and therefore the tangential basis vector) depends on the radius. Of course, you can then normalise the basis, but then you must take this into account and your transfer coefficients will not be given by the partial derivatives.

## 1. What is a tangent to a circle?

A tangent to a circle is a line that touches the circle at exactly one point, called the point of tangency. The tangent line is perpendicular to the radius of the circle at the point of tangency.

## 2. How do you find the equation of a tangent to a circle in polar form?

To find the equation of a tangent to a circle in polar form, you first need to know the radius of the circle and the angle at which the tangent line intersects the circle. Then, you can use the formula r = psecθ to find the length of the tangent line, where r is the radius of the circle, p is the distance from the origin to the point of tangency, and θ is the angle at which the tangent line intersects the circle. Finally, you can use the equation y = mx + b to find the equation of the tangent line, where m is the slope of the tangent line and b is the y-intercept.

## 3. How do you convert a tangent to a circle from polar form to Cartesian form?

To convert a tangent to a circle from polar form to Cartesian form, you can use the formula x = rcosθ and y = rsinθ, where r is the length of the tangent line and θ is the angle at which the tangent line intersects the circle. These equations will give you the coordinates of the point of tangency, which you can then use to find the equation of the tangent line in Cartesian form.

## 4. What is the difference between a tangent line and a secant line?

A tangent line is a line that touches a circle at exactly one point, while a secant line is a line that intersects a circle at two points. In other words, a tangent line is external to the circle, while a secant line is internal to the circle.

## 5. Can there be more than one tangent to a circle at a given point?

No, there can only be one tangent to a circle at a given point. This is because the tangent line is perpendicular to the radius of the circle at the point of tangency, and a circle can only have one perpendicular line at a given point.

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