# Understanding Special Relativity Through Polar Coordinates

• arestes
In summary, Dalespam found that changing coordinates from Cartesian to polar coords results in the same path being found as using the standard Minkowski metric.

#### arestes

Hi guys!
I was reviewing some basic stuff in Special Relativity, specifically the part where it can be proven that a straight line connecting two events is the path that maximizes the interval between these two events. The proof is easy using the metric with cartesian coordinates

$$ds^2 = -c^2dt^2 + dx^2$$

but I should get the same using polar coordinates (I'm working with two dimensions for simplicity). In that case I should work with the interval element
$$ds^2 = -\theta_0^2dt^2 + t^2d\theta^2$$
where t stands for a "radial" coordinate and $$\theta$$ is the angle in Minkowski spacetime and $$\theta_0$$ is a real constant that is there for dimensionality reasons . This way the interval is

$$\Delta s = \int_{p_1}^{p_2}d\theta \sqrt{-\theta_0^2 t'^2 + t^2}$$
where $$t'$$ is the derivative of t wrt $$\theta$$. Now, I will use the Euler-Lagrange equations to get the stationary path

$$\frac{d}{d\theta} \frac{\partial \sqrt{-\theta_0^2 t'^2 + t^2}}{\partial t'} = \frac{\partial \sqrt{-\theta_0^2 t'^2 + t^2}}{\partial t}$$

Operating I get (I've checked this many times but please re-check it)

$$-\frac{t''}{t} + 2\frac{t'^2}{t^2} = \frac{1}{\theta_0^2}$$
which, by changing variables $$t \rightarrow 1/y$$
$$y'' - \frac{1}{\theta_0^2} y = 0$$
which has as solutions hyperbolic sines and cosines or, equivalently, a multiple of a hyperbolic cosine with a phase. This should be wrong because what I expect is to get a straight line in Minkowski spacetime as solution. A straight line in polar coordinates is something of the form
$$t = t_0 sec[\frac{1}{\theta_0} \theta + \phi]$$
which means that the differential equation for y should have a PLUS sign in front of $$1/\theta_0^2$$.
What is wrong here?

Also, I would like to know as to what the best way to see whether the stationary function obtained by the E-L equations is a maximum or a minimum (or just a "saddle point") is.

Going from Cartesian to polar coords, you could work in the (t,r) plane instead of (t,x), in which case the metric is
$$ds^2=-c^2dt^2+dr^2$$

The surface of a sphere of radius R is

$$ds^2=-c^2dt^2+R^2d\theta^2+R^2\sin(\theta)^2d\phi^2$$

and you need two spatial dimensions. The 'straightest' path is a segment of a great circle connecting two points.

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When people say polar coordinates in flat space-time, they usually mean the following metric:

$$ds^2 = c^2dt^2 - dr^2 - r^2d\theta^2 - r^2sin^2(\theta)d\phi^2$$

I've never seen anyone try and include time in polar coordinates. There is a hyper-spherical metric where there is a boost hyper-angle corresponding to Lorentz boost, but it still looks very different from what you're defining.

arestes said:
Hi guys!
I was reviewing some basic stuff in Special Relativity, specifically the part where it can be proven that a straight line connecting two events is the path that maximizes the interval between these two events. The proof is easy using the metric with cartesian coordinates

$$ds^2 = -c^2dt^2 + dx^2$$

but I should get the same using polar coordinates (I'm working with two dimensions for simplicity). In that case I should work with the interval element
$$ds^2 = -\theta_0^2dt^2 + t^2d\theta^2$$
where t stands for a "radial" coordinate and $$\theta$$ is the angle in Minkowski spacetime and $$\theta_0$$ is a real constant that is there for dimensionality reasons .
What is your coordinate transformation to your "polar coordinates"? Using the standard polar coordinate transformation:
$$x=r \; cos(\theta)$$
$$t=r \; sin(\theta)$$
I am getting a metric which is much more complicated than the one you posted, even setting c=1. It involves cross terms that do not cancel out:
$$d\theta^2 r^2 \cos (2 \theta )+dr^2 (-\cos (2 \theta ))+2 d\theta dr r \sin (2 \theta )$$

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Hi guys! I just figured out my mistake... It was in the very metric you all said was wrong. To make it clear, I didn't want to work on the t-r plane. Also, I shouldn't have used "t" as the radial coordinate because it is not time as in cartesian coordinates. I didn't mean to work with time in polar coordinates, it gets mixed up with the spatial coordinate.

In any case, I get the same metric as DaleSpam with a minus sign. Using
$$x=r \; cos(\theta)$$

$$ct=r \; sin(\theta)$$
I get
$$ds^2 = -d\theta^2 r^2 \cos (2 \theta )+dr^2 (\cos (2 \theta ))- 2d\theta dr r \sin (2 \theta )$$
and from here I get the desired result, namely-

$$r = r_0 sec\left[\theta + \phi_0\right]$$
which describes a straight line in Minkowski spacetime.
I just did this for fun, since I knew that changing coordinates should give me the same result as using the standard Minkowski metric.

thanks!

## 1. What is special relativity?

Special relativity is a theory in physics that was developed by Albert Einstein in the early 20th century. It describes how objects move and behave at high speeds, close to the speed of light.

## 2. How does polar coordinates relate to special relativity?

Polar coordinates are a way of describing the position of an object in space using two variables: distance and angle. In special relativity, these coordinates are used to define the space-time interval between two events, which is a fundamental concept in the theory.

## 3. What is the significance of the speed of light in special relativity?

The speed of light, denoted by "c", is a fundamental constant in the theory of special relativity. It is the maximum speed at which any object can move in the universe, and it plays a crucial role in determining the behaviors and limitations of objects at high speeds.

## 4. How does special relativity affect our understanding of time and space?

Special relativity shows that time and space are not absolute, but rather relative to the observer's frame of reference. This means that measurements of time and space can vary depending on an observer's relative speed and position, which has significant implications for our understanding of the universe.

## 5. What are some real-world applications of special relativity?

Special relativity has many practical applications, including GPS technology, nuclear power, and particle accelerators. It also helps us understand phenomena such as time dilation and length contraction, which have been confirmed through various experiments and observations.