Is the Newtonian EOM Invariant under Galilei Boosts?

  • Thread starter haushofer
  • Start date
  • Tags
    Newton
In summary, the Galilei group does not hold the spatial metric invariant under boosts, unlike the Lorentz group. When considering the line element and Lagrangian in Newtonian physics, only time and space shifts and rotations leave the spatial metric invariant. To create an invertible metric, a time metric is added, but this does not give a metric invariant under the full Galilei group.
  • #1
haushofer
Science Advisor
Insights Author
2,944
1,474
Hi, I have some questions concerning Newtonian physics which confuse me, especially the Galilean boosts. First of all, the Galilei group is

[tex]
x^i \rightarrow R^i_{\ j}x^j + v^i t + d^i, \ \ \ \ \ \ \ \ x^0 \rightarrow x^0 + \xi^0
[/tex]
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)

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

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

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

[tex]
L = m\delta_{ij}\dot{x}^i \dot{x}^j
[/tex]

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

[tex]
\ddot{x}^i + \frac{\partial \phi}{\partial x^i} = 0
[/tex]
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,
[tex]
\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}
[/tex]
but how would one show this infinitesimally?

These may be confusing questions with simple answers, but I don't see it :)
 
Last edited:
Physics news on Phys.org
  • #2
haushofer said:
Hi, I have some questions concerning Newtonian physics which confuse me, especially the Galilean boosts. First of all, the Galilei group is

[tex]
x^i \rightarrow R^i_{\ j}x^j + v^i t + d^i, \ \ \ \ \ \ \ \ x^0 \rightarrow x^0 + \xi^0
[/tex]
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)

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

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

Under a Lorentz boost you would be getting:

[tex]
ds'^2 = \delta_{ij}dx'^i dx'^j
[/tex]

The presence of extraneous terms under a Galilei/Newton boost makes the metric frame-variant. Same story with the lagrangian.
 
  • #3
For the Galilei structure you have frame invariant contravariant "metric tensor". I use quotation marks because it is degenerate: [tex]g^{00}=0.[/tex]. So it does not have the inverse.
Another invariant object is the "time form" [tex]dx^0[/tex].

But you have invariant 3d metric in the fibres [tex]x^0=\mbox{const}[/tex].
 
Last edited:
  • #4
starthaus said:
Under a Lorentz boost you would be getting:

[tex]
ds'^2 = \delta_{ij}dx'^i dx'^j
[/tex]

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

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

[tex]

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

[/tex]

with

[tex]
dx^{'i} = \frac{\partial x^{'i}}{\partial x^j}dx^j + \frac{\partial x^{'i}}{\partial t}dt

[/tex]

For time translations I have trivially

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

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

[tex]
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
[/tex]

For spatial translations I have trivially

[tex]
dx^{'i} = dx^i
[/tex]

But for boosts I get extra terms and certainly then

[tex]
ds^2 \neq ds^{'2}
[/tex]Why?
 
  • #5
arkajad said:
For the Galilei structure you have frame invariant contravariant "metric tensor". I use quotation marks because it is degenerate: [tex]g^{00}=0.[/tex]. So it does not have the inverse.
Yes, but I'm trying to show that explicitly. :)
 
  • #6
starthaus said:
Under a Lorentz boost you would be getting:

[tex]
ds'^2 = \delta_{ij}dx'^i dx'^j
[/tex]

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

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

[tex]
dt = 0
[/tex]

If this is not the case, then certainly

[tex]
ds^2 \neq ds^{'2}
[/tex]

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:
  • #7
haushofer said:
Yes, but I'm trying to show that explicitly. :)

You can easily show explicitly that the contravariant degenerate metric is invariant by applying the transformation laws of a contravariant tensor.
 
  • #8
arkajad said:
You can easily show explicitly that the contravariant degenerate metric is invariant by applying the transformation laws of a contravariant tensor.

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

[tex]
x^{'\mu} = \Lambda^{\mu}_{\ \nu}x^{\nu} + a^{\mu}
[/tex]

We then have

[tex]
\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
[/tex]

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

[tex]
\Lambda^{'\alpha}_{\ \mu} \Lambda^{'\beta}_{\ \nu} h^{\mu\nu} = h^{'\alpha '\beta}
[/tex]

The 00 component equation for the RHS gives

[tex]
h^{00} = h^{'0'0}
[/tex]

The ij component gives

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

and the 0j component gives

[tex]
R^j_{\ k} h^{0k} + v^j h^{00} = h^{'0'j}
[/tex]

This tells me that for h to be invariant,

[tex]
h^{00} = h^{0j} = 0
[/tex]
 
  • #9
haushofer said:
Hi, I have some questions concerning Newtonian physics which confuse me, especially the Galilean boosts. First of all, the Galilei group is

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
haushofer said:
This tells me that for h to be invariant,

[tex]
h^{00} = h^{0j} = 0
[/tex]

That's it.
 
  • #11
Petr Mugver said:
Rather, the Galilean group leaves TWO metrics invariant: the (0+++) and the (+000).

This is correct provided you think of (0+++) as contravariant, and not as a covariant one.
 
  • #12
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
arkajad said:
This is correct provided you think of (0+++) as contravariant, and not as a covariant one.
Why do you rule out the covariant possibility?
 
  • #14
Petr Mugver said:
Why do you rule out the covariant possibility?

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 [tex]g^{\mu\nu}[/tex] and the closed absolute time form [tex]\theta=dx^0[/tex].
 
Last edited:

1. What is a Newton boost and a Galilei boost?

A Newton boost is a transformation of coordinates in Newtonian mechanics that describes the motion of objects in an inertial frame. A Galilei boost is a similar transformation used in Galilean relativity, which is a simplified version of special relativity that applies to non-accelerating frames. Both boosts involve changing the reference frame to observe the motion of an object.

2. What is the difference between a Newton boost and a Galilei boost?

The main difference between the two boosts is that a Newton boost takes into account the time dilation and length contraction effects predicted by special relativity, while a Galilei boost does not. This means that a Newton boost is more accurate in describing the motion of objects at high speeds, while a Galilei boost is only accurate for low speeds.

3. How are Newton boosts and Galilei boosts used in physics?

Both boosts are used to describe the motion of objects in different reference frames. They are essential in classical mechanics and are also used in special relativity. They allow scientists to compare observations made in one reference frame to those made in another, which is crucial in understanding the laws of motion.

4. Are there any real-life applications of Newton boosts and Galilei boosts?

Yes, there are many real-life applications of these boosts. For example, they are used in navigation systems, such as GPS, to accurately track the motion of objects. They are also used in engineering to design and analyze the motion of machines and vehicles. In addition, these boosts are used in physics research to study the behavior of particles at high speeds.

5. Are Newton boosts and Galilei boosts still relevant in modern physics?

While special relativity has replaced Galilean relativity as the most accurate theory of motion, Galilei boosts are still used in certain situations where the effects of special relativity are negligible. Newton boosts are also still relevant in many areas of physics, particularly in classical mechanics. However, for high-speed objects, the predictions of special relativity must be taken into account.

Similar threads

  • Special and General Relativity
Replies
3
Views
984
Replies
47
Views
5K
  • Special and General Relativity
Replies
3
Views
2K
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
15
Views
1K
  • Special and General Relativity
Replies
2
Views
785
  • Special and General Relativity
Replies
18
Views
2K
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
Replies
4
Views
1K
  • Special and General Relativity
Replies
11
Views
2K
Back
Top