# Low speed limit of Lorentz transformations

1. Feb 8, 2016

### ShayanJ

In the system of units where c=1, the Lorentz transformations are as follows:

$x'=\gamma (x-vt) \\ t'=\gamma (t-vx)$

In the limit $v \ll 1$, we have $\gamma \approx 1+\frac 1 2 v^2$, so we have, in this limit:

$x' \approx (1+\frac 1 2 v^2)(x-vt)=x-vt+\frac 1 2 v^2 x-\frac 1 2 v^3 t$
$t' \approx (1+\frac 1 2 v^2)(t-vx)=t-vx+\frac 1 2 v^2 t-\frac 1 2 v^3 x$

Now if we just keep terms first order in v, we'll get:

$x' \approx x-vt \\ t'\approx t-vx$

But this is a problem because we are promised to get Galilean transformations because of the correspondence principle!

What's wrong here?

Thanks

2. Feb 8, 2016

### Staff: Mentor

x is also small compared to t (more precisely, variations are small). vx is some sort of second order for relevant distances, while x and vt have the same order.

3. Feb 8, 2016

### ShayanJ

How do you justify that? What do you mean by relevant distances?

4. Feb 8, 2016

### Staff: Mentor

Distances traveled by objects with speeds of the order of v.

5. Feb 8, 2016

### Staff: Mentor

One way to see this is to put the $c$ back into the transformation. If we use the approximation $\gamma \approx 1$ (which is correct to first order), we have

$$x' = x - v t$$
$$t' = t - \frac{vx}{c^2}$$

So the $vx$ term is clearly of order $1 / c^2$ compared to all the others. That's why we can leave it out.

6. Feb 8, 2016

### ShayanJ

But that's not the interpretation of x! In case of coordinate transformations, its a label on a point and in case of interval transformations, its the coordinate distance between two points in a particular coordinate system. Neither have anything to do with motion!

7. Feb 8, 2016

### Staff: Mentor

Well, you cannot neglect relativistic effects to first order if you have objects at relativistic speeds.
Both coordinate transformation and motion of objects have to be slow to make the first order work in the intended way.

8. Feb 8, 2016

### Orodruin

Staff Emeritus
But it is! If you are comparing distances which are so large that light cannot traverse them in the time scales you are looking at, then light speed is relevant. The recovery of the Galilei transformations is when c goes to infinity.

9. Feb 8, 2016

### Staff: Mentor

What this "paradox" is telling you is that that is not the correct limit to take to get the Galilean transformation as a limiting case of the Lorentz transformation. More precisely, it's not a complete statement of the correct limit to take. Taking the limit $v / c \ll 1$ is part of it, but not all of it. You also have to take the limit $x \ll ct$, in other words, you have to restrict attention to spacetime intervals that satisfy that condition.

10. Feb 8, 2016

### Staff: Mentor

The Andromeda paradox is an example. This "paradox" appears when the two observers are walking in different directions on the surface of the earth so that v is vanishingly small compared with c but the distance involved (all the way to the Andromeda galaxy) is not small compared with ct.

11. Feb 8, 2016

### pervect

Staff Emeritus
An answer to a more complicated question regarding low speed limits was posted recently . Marc De Montigny, Germain Rousseaux, "On the electrodynamics of moving bodies at low velocities," http://arxiv.org/abs/physics/0512200

The answer to the simpler problem is more or less given in the introduction to the more complex problem, though it is unfortunately rather terse.

Basically, it turns out there are two possibile well-defined Gallilean limits.

The relevant section starts with

I believe the other limit would be
$$x' = x \quad t' = t - v \, x/c^2$$

So you don't get a Galilean limit until you make one of two additional assumptions:

The first limit should follow from the additional assumption x << ct. The second limit should follow from the assumption x >> ct. If you don't have either, i.e. if x is about the same value as ct, you don't have a Gallilean limit. If you have low velocities and timelike vectors, you get the first limit. With low velocities and space-like vectors, you get the second limit. If you have null or nearly-null vectors, you don't get a Gallilean limit - the rather sloppy way I'd put this is that the speed of light being constant for all observers means that a null worldline with x=ct just doesn't transform in a Gallilean manner.

12. Feb 9, 2016

### PAllen

Going back to the OP, the simplest way to state the flaw (IMO) is to just look at units. vx has the wroing units for time in conventional systems. To get the right units, you need the vx/c2. The x transform equation comes out right directly for conventional units. Then you can say that for earth scales you recover the Galilean transform.

You can get the same result using units consistent with c=1. Then the v is light seconds/second, and x is light seconds. You can then see in these units the multiplication of very small quantities for vx, that you then ignore for the stated purpose. Meanwhile, in the x transform equation, even though the speed is small, t can be as large as you want, and the vt term has (anyway) the same order of magnitude as the x term.

13. Feb 11, 2016

### haushofer

To be honest, I don't see the problem. The limit consists of v/c<<1, a dimensionless statement.That's as far as I can see enough to obtain the Galilei group from the Poincaré group.

The "Andromeda-paradox" was also unfamiliar to me, but I'll check that one out.

14. Feb 11, 2016

### haushofer

I see the point now, which is kind of straightforward: an arbitrarily small v/c can be compensated by an arbitrarily large distance x, such that the dilation 'builds up'. So indeed, v/c on itself doesn't erase time dilation.

I like to think of these kind of limits in terms of contractions of the underlying Lie algebra (Poincare), and there the contraction leading to the Galilean algebra involves (among others) a rescaling of the spatial translation generator. I guess this rescaling takes into account both 'small velocities' and 'small distances'.

I'v never realized this. So thanks! :)

15. Feb 11, 2016

### ShayanJ

Is this something completely group theoretical? Can you clarify it a bit?

16. Feb 11, 2016

### haushofer

Yes. It is called an Inonu-Wigner contraction. It consists of rescaling part of the generators in the Lie algebra, calculate the brackets and take a limit on the rescaling. See e.g.

http://arxiv.org/abs/1011.1145

17. Feb 11, 2016

### PAllen

One more thought on this ...

The correct way to take a single limit from Lorentz to Galilean coordinates is to let c->∞. This makes both velocity dependent effects and simultaneity effects go to zero - it restores the notion of absolute global simultaneity. Once you let c=1, you lose the ability for this effect of the correct limit to occur.

In terms of the Lorentz transform with c present, you have two separate ratios: v/c and x/c (the product of these being present in the time transform equataion). C going to infinity makes both vanish. Letting c=1, and making v/c small, fails to make the second one vanish. This again shows that Galilean limit is one of low relative speed, and 'normal' distances.

Last edited: Feb 12, 2016
18. Feb 12, 2016

### haushofer

Yes. The contraction parameter in the Lie algebra can be regarded as the speed of light, as my reference also shows. You can also take the opposite limit, resulting in the so-called Carroll-algebra, describing ultrarelativistic spacetimes. See e.g. http://arxiv.org/pdf/1505.05011.pdf.

But I guess that the usual textbook "Newtonian limit" of GR then also needs to be supplemented by this restriction on distances, right? I've never seen in any textbook this restriction being mentioned. Amazing how you can learn new stuff about something you thought you knew thoroughly.

19. Feb 12, 2016

### haushofer

Also, in Dautcourt's paper

https://inis.iaea.org/search/search.aspx?orig_q=RN:23072570

it is shown that even (a slight generalization of) Newton-Cartan theory can be obtained from GR by a c --> oo limit procedure. But it is far more complicated than the usual textbook limit, which is not on the level of geometry but on the level of equations of motion.

20. Feb 13, 2016

### haushofer

Something else which pops up in my mind is whether cosmology based on a Newtonian approximation where v<<c instead of c --> oo makes sense. Something to think about this weekend :P