# Applying the lorentz transformation

1. Apr 9, 2004

### cephas

I am trying to understand how to use the lorentz transformation to calculate things and I have a few questions about it.

Suppose I have a circle of radius 1 in frame K which is at rest. Frame k' is moving at velocity v relative to frame k. What would the circle look like to someone in frame k' as they pass by it?

Can I just apply lorentz equation straight to the equation of the circle?

Rest frame k

x(u)=Cos(u)
y(u)=Sin(u)

moving frame k'

x'(u)=(cos(u)-v*t)*gamma
y'(u)=Sin(u)

then if I graph the above parametric equation for frame k' would that be what the circle would like to the person as they move by?

I really want to nail this stuff down. I don't understand why this reference frame stuff confuses me so much. It should be simple. Thanks for the help in advance

2. Apr 10, 2004

### chroot

Staff Emeritus
It will just look like an ellipse with major axis 1 and minor axis $1/\gamma$.

If you want to parameterize the ellipse, you'll need to relate u to t somehow to be able to finish the analysis.

- Warren

3. Apr 10, 2004

### cephas

Why would I need to relate u and t? x=Cos(u) just means I am transforming all x coordinates from 0<=u<=2Pi to x' I don't see why they need to be related.

Also, when I tried doing this, the circle stretched. I thought is was supposed to shrink. I have to be doing something wrong.

4. Apr 10, 2004

### cephas

Ok, now I got it so it contracts like it is supposed to. I don't understand why i have to take the reciprocal of gamma though. I will try studying my books some more.

5. Apr 10, 2004

### chroot

Staff Emeritus
All observed lengths are less than or equal to the proper length. Gamma is always greater than or equal to one. Ergo

$$L = L_0/\gamma$$

- Warren

6. Apr 11, 2004

### jdavel

That's not quite true is it?

Relativity says that if you MEASURE the length of a moving object by determining the locations of its ends simultaneously (in the rest frame) that the result will be less than that for the same object when its stationary. But what you SEE is not quite the same, because the light takes longer to get from the more distant parts of the object. This isn't an issue for a one dimensional object moving along its axis, but with a circle it comes into play.

Just thought I'd mention it.

7. Apr 11, 2004

### jdavel

It sounds to me like you're making a very common mistake, which is the result of using the length contraction equation instead of the compelte LT for length. Until you're VERY confident with this stuff, stick with the LTs, it's much harder to go wrong with them, because you can see which frame the simulataneous measurements have to be in.

I don't know if this helps or not.

8. Apr 13, 2004

### DW

Actually you didn't transform what a metal loop looks like from one frame to another. You transformed where a set of events occur from one frame to the other. If you apply the time transformation as well you will see that if they happen at the same time according to your initial frame that they occur at different times according to the second frame. If these events happen at the same time along a solid ring according to a frame at which it is at rest then they can not describe the shape of the ring according to any other frame because the ends of the ring have moved between the different times that the events occur. Most paradoxs or problems people seem to have in understanding relativity seem to be tracable to unaccounted for relative simultaneity.

Last edited: Apr 13, 2004