# Lorentz transform low velocity limit

1. Jul 11, 2013

### geoduck

Why do we say that t'=t for Galilean transformation, when the low velocity limit of the Lorentz transformation is t'=t+vx/c2?

If x is really big, then doesn't time cease to be absolute, no matter how small v/c is?

2. Jul 11, 2013

### Simon Bridge

Consider: how big would x have to be before Galileo would notice?

3. Jul 11, 2013

### PAllen

If v is zero, the t'=t for any x. For any v > 0, as you note, there is some distance for which t and t' are are significantly different. This is the basis of the so called 'Andromeda Paradox' (which is not a paradox at all, just a feature of SR): if you are standing on a road, and I am riding a bicycle past you, then the events in the Andromeda galaxy that I consider simultaneous (using the same criteria as you) will differ by many years.

4. Jul 12, 2013

### geoduck

Thanks. It really bothered me that they say that t'=t in the nonrelativistic limit, but x'=x-vt. I thought to the same order of (v/c) [expanding the Lorentz transform in v/c], you had to add the extra term: t'=t-vx/c^2 to keep things to the same order. The equation for x' and t' also needs to be symmetrical, since this is a rotation in spacetime. So I'm relieved that technically you can't leave out the vx/c^2 term, although it would take large x for the effect to be noticeable.

5. Jul 12, 2013

### Staff: Mentor

I suspect that the limit that is really being contemplated in various texts when they talk about the Galilean transformation being a "non-relativistic limit" of the Lorentz transformation is not just v << c, but x << ct; in other words, the distances involved are short compared to the times (when both are put into the same units). The first limit does not necessarily imply the second, but the two will both be satisfied for a large range of phenomena. Certainly nothing that Galileo or Newton studied would have violated the second limit.

6. Jul 12, 2013

### PAllen

Another way to take the limit is to let c->∞. This simultaneously makes v/c -> 0 and vx/c^2 -> 0. Then, the Galilean transform is exactly recovered in this limit. Stating it practically, this means Galilean transform is good when light speed is effectively infinite for the problem under consideration: speeds are slow, and distances are such that light delay can be ignored.

Last edited: Jul 12, 2013
7. Jul 12, 2013

### geoduck

I suppose if a textbook claims that Newtonian mechanics occurs with the assumptions of small v/c and large non-quantum objects, then technically they need to add another assumption that all observers are close to what they're observing, x<ct. Or make the assumption that c=∞. I don't recall a classical mechanics book that has these as their assumptions. o well, it's nitpicking I suppose.

8. Jul 12, 2013

### Simon Bridge

When teaching introductory Newtonian/Galilean mechanics, it is unusual to use definitions that the student is not aware of needing: why bother? The texts say that Galileo's R only works for small speeds - which is correct. In pedantic-speak, that is "a necessary but not sufficient limit". But the books don't claim it is sufficient.

It is not surprising that Galilean relativity differs from the small-speed limit for special relativity: Galileo didn't know about SR. It seems reasonable to expect students to make the connection between discrepancies and the knowledge/equipment available to Galileo: for instance - he used to measure time by pouring water between containers and weighing the water. OTOH: it is understandable for students to forget the limitations our equipment imposes on the kinds of physics we can discover. It is also important for students to get used to the pronouncements in text books being incomplete or just plain wrong.

The question reminds me of a Greg Egan tale of a race who had to live with GR effects as an everyday reality: i.e. they were not small. They had to figure out geometry in a region where Euclid is not intuitive. This is the sort of question they'd come up with... so I don't think I want to mess with OPs intuitions too much here.

I suspect the key here is to realize what that vx/t2 bit is taking account of: covered in other replies.
Excuse me: I'm just intrigued when a question arrives the opposite way around from what I normally see.

9. Jul 13, 2013

### geoduck

Well, I got confused because my thinking went like this: an infinitesmal Lorentz transformation should equal an infinitesmal Galilleo transformation, because v/c is small for an infinitesmal transformation; however, a finite transformation can be built from a product of infinitesmal ones. Therefore how can the Galileo transformation be different from a Lorentz one if they are created from the same product of transformations?

But the distinction is the neglect of the vx/c^2 in the time component. This makes an infinitesmal Galileo transformation an additive group in the velocities for the space component, x'=x+(v_1+v_2+v_3+v_4+v_5+....)t, with time unchanging. Keeping the term vx/c^2 makes it the product of infinitesmal rotations, resulting in the Lorentz transformation.

I think one can say that the difference between Galileo transformation and Lorentz is the vx/c^2 in the time component, because knowing this:

x'=x+vt ===> x'=x+(v/c) ct
t'=t+vx/c^2 ===> ct'=ct+(v/c)x

allows you to pretend v/c is infinitismal, then take the product of infinitismal transformations, and out should come the Lorentz transformation as a finite rotation with finite parameter (v/c)*n=V/c where n is the number of infinitismal rotations v/c to get a finite parameter V/c.

Last edited: Jul 13, 2013
10. Jul 13, 2013

### Simon Bridge

I think that this way of thinking will stand you well in post-grad.
Just bear in mind that your courses are probably designed for people who think a bit differently to you.

11. Jul 13, 2013

### robphy

I think the proper procedure is to first define the limiting form of the metric (i.e. the dot product) then determine the transformations that preserve it.