# Can we do other than Lorentz (or Poincare) Transformation in SR?

1. Apr 11, 2013

### altsci2

I want to discuss this because I afraid that the answer is no. In SR we stuck with the transformations from Poincare Group because this transformations leave invariant the exact form of the Lorentz Metric tensor. Any other transformation will change the components of the Lorentz Metric Tensor and screw up all the calculations that we used to do. Because of that we can not introduce an accelerated or rotated coordinate system by some strange transformation of coordinates. For the same reason we can not do even Galilran Transformation (see attached file).
Suppose we have a coordinate system with arbitrary metric tensor: gmn where the indexes indicate coordinates and can take values: m,n = 0,1,2,3. If we take 2 points: P1(0,0,0,0) and P2(x0,x1,x2,x3) and connect them with a straight line then the invariant length of this line will be s2=gmnxmxn where we have 2 contractions. In t-x plane it will be: s2=g00(x0)2+2g01x0x1+g11(x1)2 (this shows how contractions work). For a resting clock (x1=0) we have: s=x0√g00. The reading of a clock will reflect the invariant length s and will be not equal to the reading of time coordinate x0. The square root is a “time factor” (see attached article). The same way there exist “x-factor”, “y-factor”, and “z-factor”. The simple measurement is possible only if the factors on all axes equal to 1. That is: g00=|g11|=|g22|=|g33|=1. This is true for Lorentz Metrics. Einstein’s “Frames” can represent the corresponding coordinate systems only if all the axes factors equal to 1.

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• ###### Metric Tensor.pdf
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2. Apr 11, 2013

### Bill_K

Altsci2, By this same line of reasoning, Newtonian physics can only be done in Cartesian coordinates!

Of course you can do SR in any coordinate system you like: spherical, cylindrical, etc. You do need to include the components of the metric tensor explicitly. Lorentz frames are nice to work with, but SR is much more general than that.

Accelerating and rotating coordinate systems are limited in range. You can't cover the entire spacetime with them because you'll reach a place where the velocity of light is exceeded. So what you need to use instead is a bit more general - a set of basis vectors at each point, known as a tetrad or vierbein field.

3. Apr 11, 2013

### ghwellsjr

Or use the radar method to establish a consistent coordinate system for an accelerating observer. It covers the entire spacetime and has a constant velocity of light (since it uses that to define the coordinate system). But I don't know if there is any transformation process that will automatically get you to/from such a coordinate system and an IRF.

4. Apr 11, 2013

### Staff: Mentor

Including regions behind a (for example) Rindler horizon?

5. Apr 11, 2013

### ghwellsjr

I was addressing the issue that Bill brought up, not a Rindler horizon. If an observer cannot see something, then he cannot assign coordinates to it because he can have no knowledge of it, correct? But it's not an issue of the speed of light, is it?

6. Apr 11, 2013

### Passionflower

Not correct.

7. Apr 11, 2013

### altsci2

Agree that in any coordinate system. But: where comes the metric tensor from? People never calculate the metric tensor explicitly. Take for example GT. People just assume that the time factor is 1 which is incorrect, it is √(1-v^2)

8. Apr 11, 2013

### Staff: Mentor

Sure, you can easily do a Galilean transform. Suppose that (t,x,y,z) is a standard inertial coordinate system (c=1). Then the metric is $ds^2=-dt^2+dx^2+dy^2+dz^2$. If we perform the Galilean transform as follows:
$T=t$
$X=x-vt$
$Y=y$
$Z=z$

This gives:
$dt^2=dT^2$
$dx^2=v^2 dT^2+2v \, dT dX+dX^2$
$dy^2=dY^2$
$dz^2=dZ^2$

So the new metric is $ds^2=-(1-v^2)dT^2+ 2v \, dTdX + dX^2+dY^2+dZ^2$.

The GT is a legitimate transform, it is just not inertial because the form of the metric is not the same when expressed in the new coordinates.

9. Apr 11, 2013

### WannabeNewton

Just because there are a preferred class of transformations on $(\mathbb{R}^{4},\eta_{ab})$ i.e. the elements of the Poincare group, doesn't mean we cannot define other transformations on Minkowski space-time or some open subset of it (usually an open subset). That's like saying, as Bill elucidated, just because Newtonian space-time has a preferred class of transformations coming from the Galilean group, we cannot transform to coordinates of rotating reference frames or uniformly accelerating reference frames. There is nothing that says you can only use transformations that preserve the metric tensor.

10. Apr 12, 2013

### altsci2

Yes, I agree, arbitrary transformation of coordinates in Minkowski space (and everywhere) is quite legitimate. What I mean is that: in SR we have a huge accent on “frames” rather than on coordinate systems. A metric tensor in a frame is not available for measurement. Therefore, by definition “frame” correspond to such a coordinate system where all coordinate factors are equal to 1 (|g00|=g11=g22=g33=1). This requirement is necessary to be able to do “measurements”. Recall the Einstein’s idea of equivalent inertially moving frames with observers. Every observer says: I can measure my axes and time by physical clocks and physical meter sticks and will mark on the axes correspondingly. This already sets all coordinate factors to unity. That means that between frames we can do only transformations that preserve Lorentz Metrics. Example: after GT we have a coordinate system that does not correspond to any frame. Also any accelerated coordinate system is just a coordinate system and not a frame.

11. Apr 12, 2013

### Bill_K

Repeating what I said earlier, altsci2, the concept you need to become familiar with is the tetrad field, which gives you the ability to describe more general situations such as acceleration and rotation, by means of an independent Lorentz frame at each point.

12. Apr 13, 2013

### altsci2

Quote from Bill-K:

I am talking about "frame" that imitate a whole coordinate system. The ability to describe acceleration and rotation is not at question in any coordinate system. I do not know what kind of independent Lorentz frame at each point you are talking about, but independent coordinate system in each point is impossible.

13. Apr 13, 2013

### Staff: Mentor

Nonsense. If you want coordinate systems then I provided that answer. If you want a reference frame then Bill_K has pointed you to the correct concept. There is no such thing as an imitation coordinate system.