Calculating Triangle Area in Relativity Theory

[LagrangeEuler]

Area of triangle from picture
https://en.wikipedia.org/wiki/Special_right_triangle#/media/File:45-45-triangle.svg
is ##A_0=\frac{1}{2}##. If that triangle staying still in system S' and S' moving across one of the sides of length ##1## in respect to system ##S## area of the triangle in the system S will be ##A=A_0\sqrt{1-\frac{u^2}{c^2}}##. I am confused what will happen if triangle moving across side of length ##\sqrt{2}##. What area of triangle in the system S will be?

 

[Dale]

If that triangle staying still in system S' and S' moving across one of the sides of length ##1## in respect to system ##S## area of the triangle in the system S will be ##A=A_0\sqrt{1-\frac{u^2}{c^2}}##. I am confused what will happen if triangle moving across side of length ##\sqrt{2}##. What area of triangle in the system S will be?

It will be ##A=A_0\sqrt{1-\frac{u^2}{c^2}}##

 

[LagrangeEuler]

And could you explain me why?

 

[Ibix]

What's the formula for the area of a triangle?

 

[Dale]

And could you explain me why?

If you take any figure and change the scale in one direction then the area will change by that same amount. The details of the figure don’t matter.

Consider breaking an arbitrary figure into a bunch of square pixels in the rest frame. When you transform it, you have the same number of squares, but now they are rectangles.

 

[LagrangeEuler]

If you take any figure and change the scale in one direction then the area will change by that same amount. The details of the figure don’t matter.

Consider breaking an arbitrary figure into a bunch of square pixels in the rest frame. When you transform it, you have the same number of squares, but now they are rectangles.

I am confused with that. Because only one direction is shorten.

 

[LagrangeEuler]

So would someone who sits in S see triangle anymore in that case?

 

[Dale]

I am confused with that. Because only one direction is shorten.

Can you explain your confusion? If you shorten one direction of a square, is it not obviously a rectangle?

 

[LagrangeEuler]

End what about this case? Will observer in S will see triangle?

 

[Nugatory]

And what about this case? Will observer in S will see triangle?

Yes, but...

We need to be careful about using the word "see" here. When we say that we "see" an object, we're actually seeing the image on the retina of our eyes (or the pixel array of a digital camera, or a sheet of photographic film if we're using an old-fashioned camera) formed by light reflected from the object. Different parts of the object are different distances from the eye so the light that forms the image left different parts of the object at different times; for a fast moving object the image we see won't be an accurate representation of the entire object at anyone moment.

Thus, to observe length contraction, we have to calculate where the various parts of the object are at the same time. In principle we would fill all of space with tiny observers, all at rest relative to us and holding synchronized clocks and pieces of paper. At exactly the stroke of midnight they all write down whether a piece of the triangle is passing by them just then; afterwards, we gather up the pieces of paper and use them to reconstruct exactly where the triangle was at the stroke of midnight. When people talk about "seeing" that the triange is length-contracted, that's usually what they mean, and that's what I meant when I answered "yes" above.

The triangle will still be an isoceles triangle, but because it is length-contracted along the direction of travel, the angle at the apex will be greater and the two base angles will be correspondingly smaller.

 

[New Simplicio]

In Special Relativity, the y and z directions are not contracted

 

[Nugatory]

I am confused with that. Because only one direction is shorten.

Divide the arbitrary shape into squares all aligned with the direction of motion, and length contraction will turn them into rectangles of smaller area. The total area is the sum of the areas of each individual rectangle, so will also be less.

Yes, there’s a bit of error at the edges of the object where the squats don’t exactly fit... but that error can be made arbitrarily small by using more smaller squares.

 

[Dale]

Divide the arbitrary shape into squares

I tried that approach earlier in post 5. @LagrangeEuler doesn’t seem to recognize that a square with one direction shortened is a rectangle, but he wouldn’t explain why. Maybe your more detailed version will work for him.

 

[Ibix]

My approach was going to be to observe that if the triangle moves parallel to one of its sides, we can declare that side to be the base, then observe that the area is half base times height. Base is Lorentz contracted; height is not; straight lines remain straight. The conclusion is obvious.