How does the area of moving circle change?

Question

We know that the area of a circle at rest frame is

$$A=\pi x^2$$ from $\pi xy$ where y=x

If I move this circle with velocity $v$ in the x-direction, I would expect to see a contraction (I'm not talking about optical observation since in that case it will still look like a circle) in the x-direction and nothing on y-direction so my area will be

$$A=\pi x'y$$

Since length in y-direction hasn't changed and $x'=x{\sqrt{1-\frac{v^2}{c^2}}}$, area of a moving circle would be

$$A=\pi x^2 \sqrt{1-\frac{v^2}{c^2}}$$

I couldn't find anything on internet about contractions in area, this above reasoning would be how I would attack the problem, is that correct?

Answer 1

Your reasoning is correct. It is indeed possible and correct to calculate the are of some shape by using the contracted lengths.

For example, for a square with $A=x\cdot y$ where in the square's rest frame $x=y$, we would calculate the new area (assuming the square moves in $x$ direction with velocity $v$) by

$$A'=y \cdot x' = x^2{\sqrt{1-\frac{v^2}{c^2}}}=\frac{A}{\gamma}$$

(To avoid cunfusion: primed' indicates a measurement made on an object with relative velocity $v$ while non-primed is a measurement made in the object's rest frame)

In fact, it seems to me that the relationship $A'=\frac{A}{\gamma}$ also applies to volumes, for example for a cube with volume $V=x\cdot y\cdot z$ where in the cube's rest frame $x=y=z$, we have

$$V'=y\cdot z\cdot x'=yzx\sqrt{1-\frac{v^2}{c^2}}=\frac{x^3}{\gamma}=\frac{V}{\gamma}$$

The same would also apply to a sphere with $V=\frac 4 3\pi r^3$:

$$V'=\frac 4 3\pi r^2r'=\frac 4 3\pi r^2r\sqrt{1-\frac{v^2}{c^2}}=\frac{V}{\gamma}$$

It can thus be said, assuming constant velocity in one direction that

$$A'=\frac{A}{\gamma}\tag{1}$$

$$V'=\frac{V}{\gamma}\tag{2}$$

where $\gamma$ is the Lorentz Factor $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

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_{I am aware that the volume part isn't exactly related to the question, but I nevertheless wanted to expand a little in my answer, since the first part is basically just "yes"}

Answer 2

We know that the area of a circle at rest frame is

Accordingly, we are considering a subset of members of one specific inertial frame ("point particles sitting still in space relative to each other") whose distance relations among each other make them jointly having the shape of ... a circular disk (and its circular boundary).

$ A = \pi \, x^2 $

where $x$ apparently denotes the (length of the) radius of the circle to be considered; which, however, more usually is denoted by $r$.

If I move this circle with velocity $v$ in the x-direction,

Accordingly we're considering two distinct inertial frames (moving at constant speed $v$ straight along each other, in a direction which is in the plane of the circle):

• the inertial frame of which the constituents of the above-mentioned circle are a subset; let's call it "inertial frame $\mathsf D$" (as mnemonic for "domain"), and

• some specific other inertial frame ($\mathsf R$, for "range").

I would expect to see a contraction [...] in the x-direction and nothing on y-direction

For any two members of $\mathsf D$ (except the pairs oriented orthogonal to the direction of motion of $\mathsf R$'s members) their distance is larger than the length of their simultaneity projection in $\mathsf R$.

so [... the] area of a moving circle would be $A = \pi \, x^2 \, \sqrt{1 - \frac{v^2}{c^2}}$

But that's the area of each simultaneity projection in $\mathsf R$ of the circle!, which has the shape of an ellipse, with minor axis of length factor $\sqrt{1 - \frac{v^2}{c^2}}$ smaller than the diameter of the projected circle, oriented in the direction of motion of constituents of $\mathsf D$, and major axis of length equal to the diameter of the projected circle.

Instead, the area of the circle itself remains as before: $\pi \, x^2 $.

Answer 3

It might be better to rephrase your question to read 'How does the area of a circle change when it is considered from a reference frame moving relative to it?' The key point which @user12262 makes is that the circle itself does not change in its rest frame- in other frames it appears to be shortened in one direction, the direction and extent of the distortion varying according to the velocity of the frames. The apparent change in shape is caused by the fact that in any moving frame the leading edge of the circle is viewed at an earlier time than the trailing edge.