Newton and Galilei boosts

1. Sep 28, 2010

haushofer

Hi, I have some questions concerning Newtonian physics which confuse me, especially the Galilean boosts. First of all, the Galilei group is

$$x^i \rightarrow R^i_{\ j}x^j + v^i t + d^i, \ \ \ \ \ \ \ \ x^0 \rightarrow x^0 + \xi^0$$
where R is an element of SO(3), v is a constant velocity, d is a constant vector and xi^0 is a constant shift in time. The boost is explicitly (R=I, d=0)

$$x^i \rightarrow x^{'i} = x^i + v^i t$$
We know that in Newtonian physics we have a spatial metric which is Euclidean (space is flat!). So I have the line element

$$ds^2 = \delta_{ij}dx^i dx^j$$
Under a boost however, I get via
$$dx^{'i} = \frac{\partial x^{'i}}{\partial x^j}dx^j + \frac{\partial x^{'i}}{\partial t}dt = dx^i + v^i dt$$
that
$$ds^2 \rightarrow \delta_{ij} [dx^i + v^i dt][dx^j + v^j dt] = \delta_{ij}dx^idx^j + 2 \delta_{ij} dx^i v^j dt + \delta_{ij}v^i v^j dtdt$$

What does it mean for the Euclidean line element NOT to be invariant under Galilei boosts? A same situation I ofcourse have if I calculate the variation of the Lagrangian

$$L = m\delta_{ij}\dot{x}^i \dot{x}^j$$

under boosts; that becomes a total derivative, but the action is invariant, so that's OK (the Noether charge belonging to boosts has to be adjusted) and I do understand that.

Second, how would I proof that the Newtonian EOM

$$\ddot{x}^i + \frac{\partial \phi}{\partial x^i} = 0$$
is invariant under boosts infinitesimally? Physically I understand that it should be invariant, but mathematically I have some problems showing this explicitly. For finite transformations it's quite obvious,
$$\ddot{x}^i \rightarrow \ddot{x}^i, \ \ \ \ \frac{\partial \phi}{\partial x^i} \rightarrow \frac{\partial x^j}{\partial x^i}\frac{\partial \phi}{\partial x^j} = \frac{\partial \phi}{\partial x^i}$$
but how would one show this infinitesimally?

These may be confusing questions with simple answers, but I don't see it :)

Last edited: Sep 28, 2010
2. Sep 28, 2010

starthaus

Under a Lorentz boost you would be getting:

$$ds'^2 = \delta_{ij}dx'^i dx'^j$$

The presence of extraneous terms under a Galilei/Newton boost makes the metric frame-variant. Same story with the lagrangian.

3. Sep 28, 2010

For the Galilei structure you have frame invariant contravariant "metric tensor". I use quotation marks because it is degenerate: $$g^{00}=0.$$. So it does not have the inverse.
Another invariant object is the "time form" $$dx^0$$.

But you have invariant 3d metric in the fibres $$x^0=\mbox{const}$$.

Last edited: Sep 28, 2010
4. Sep 29, 2010

haushofer

I don't see that. Let's do all the calculations explicitly:

$$ds^2 = \delta_{ij}dx^i dx^j$$

with

$$dx^{'i} = \frac{\partial x^{'i}}{\partial x^j}dx^j + \frac{\partial x^{'i}}{\partial t}dt$$

For time translations I have trivially

$$dx^{'i} = \frac{\partial x^{'i}}{\partial x^j}dx^j = dx^i$$
because here R=I, the identity, and v=d=0.

For rotations I have
$$dx^{'i} = \frac{\partial x^{'i}}{\partial x^j}dx^j = R^{'i}_{\ j}dx^j$$
but because the rotations are orthogonal I get

$$ds^{'2} = \delta_{'i'j}dx^{'i} dx^{'j} = \delta_{ij}R^{'i}_{\ k}R^{'j}_{\ l}dx^k dx^l = \delta_{kl}dx^k dx^l$$

For spatial translations I have trivially

$$dx^{'i} = dx^i$$

But for boosts I get extra terms and certainly then

$$ds^2 \neq ds^{'2}$$

Why?

5. Sep 29, 2010

haushofer

Yes, but I'm trying to show that explicitly. :)

6. Sep 29, 2010

haushofer

Yes, you're right; Galilei boosts do NOT hold the spatial metric invariant! Imagine I have a rod. I measure one end of it, and then the other, and from that I can construct the length of the rod.

Now imagine someone traveling with constant speed. If he wants to measure the length of the rod, he has to measure both ends simultaneously, so

$$dt = 0$$

If this is not the case, then certainly

$$ds^2 \neq ds^{'2}$$

which the calculation also implies.

Measuring the other endpoint at some later time wouldn't make sense. So we can conclude that the spatial metric delta is only invariant under constant time and space shifts and rotations, but NOT under boosts.

To make up an invertible metric, one adds a time metric, which in certain coordinates would only have a 1 at the 00 entry and for the rest is zero. This metric is trivially invariant under all the Galilei transformations, but adding the spatial metric delta and this time metric doesn't give me a metric in spacetime which is invariant under the full Galilei group.

It's funny how deceiving these things can be. Thanks for the answers anyway, and if people have more to say about this I'm happy to hear! :)

Last edited: Sep 29, 2010
7. Sep 29, 2010

You can easily show explicitly that the contravariant degenerate metric is invariant by applying the transformation laws of a contravariant tensor.

8. Sep 29, 2010

haushofer

Ok, let's do that explicitly. Let's write the Galilei transformation as

$$x^{'\mu} = \Lambda^{\mu}_{\ \nu}x^{\nu} + a^{\mu}$$

We then have

$$\Lambda^{i}_{\ j} = R^i_{\ j}, \ \Lambda^{i}_{\ 0} = v^i, \ \Lambda^{0}_{\ 0} = 1, \ \ \Lambda^0_{\ j} = 0, \ \ a^i = d^i, \ \ a^0 = \xi^0$$

Under rotations and boosts (I think the spatial and temporal shifts are trivial) I have

$$\Lambda^{'\alpha}_{\ \mu} \Lambda^{'\beta}_{\ \nu} h^{\mu\nu} = h^{'\alpha '\beta}$$

The 00 component equation for the RHS gives

$$h^{00} = h^{'0'0}$$

The ij component gives

$$v^i v^j h^{00} + 2v^i R^j_{\ k}h^{0k} + R^{i}_{\ k} R^j_{\ l}h^{kl} = h^{'i'j}$$

and the 0j component gives

$$R^j_{\ k} h^{0k} + v^j h^{00} = h^{'0'j}$$

This tells me that for h to be invariant,

$$h^{00} = h^{0j} = 0$$

9. Sep 29, 2010

Petr Mugver

The galilean group, unlike the Lorentz group, is not a set of transformations that keep a metric invariant. So it's kind of unuseful to introduce a space-time metric ds2 as you did. Rather, the Galilean group leaves TWO metrics invariant: the (0+++) and the (+000). There's no need to introduce the (+---), or whatever treates space and time on the same footing, as there's no speed of light to preserve here.

10. Sep 29, 2010

That's it.

11. Sep 29, 2010

This is correct provided you think of (0+++) as contravariant, and not as a covariant one.

12. Sep 29, 2010

BTW: this kind of structure comes naturally as the result of a dimensional reduction from 5-dimensional (-++++) Kaluza-Klein space, if you consider the fibers to be light-like instead of space-like, as it is done usually.

13. Sep 29, 2010

Petr Mugver

Why do you rule out the covariant possibility?

14. Sep 29, 2010

Because covariant (0+++) metric is not invariant under Galilei boost - which was the reason for starting this thread. Contravariant is. This can be generalized to include space curvature and a torsion-free affine connection compatible with the degenerate contravariant metric $$g^{\mu\nu}$$ and the closed absolute time form $$\theta=dx^0$$.