Galilean vs Lorentz Transformations: Correct Understanding?

[David Lewis]

TL;DR

What are Galilean and Lorentz Transformations' respective domains of applicability?

In the frame of Observer C standing by the side of the road, the speed of Car A with respect to Car B = v1 + v2. (Galilean Transformation).

In the frame of Car A, the speed of Car B < v1 + v2 (Lorentz Transformation).

Please tell me if this understanding is correct.

 

[Nugatory]

[What are Galilean and Lorentz Transformations' respective domains of applicability?

The Galilean transformations are never exactly correct, but they are applicable when the speeds are small compared to $c$ so that the difference between the two is negligible.

So your understanding is correct: using a frame in which car A is at rest (much better than saying "in the frame of car A") we can use the exact relativistic velocity addition rule to calculate that the speed of the other car will be slightly less than $v_1+v_2$, and likewise using a frame in which car B is at rest.

Nonetheless, we can save ourselves some work and use the Galilean velocity addition rule instead and claim that the speed of the other car is $v_1+v_2$. This works because with cars neither $v_1$ nor $v_2$ is going to be much greater than 100 meters/second (fastest I've been in a car is about 65 meters/sec relative to the road) and at those speeds the difference between the two calculations is too small to affect the applicability of either.

 

[Ibix]

I think you are missing the point slightly, yes. The Galilean and Lorentz transforms link measurements made in different frames, and the example you give of "Galilean velocity addition" is not a frame change.

There are two distinct concepts related to relative speed. One is the rate of change of distance (aka the "separation rate") between two objects measured using some frame (not necessarily the rest frame of either). In this case you can just add speeds relative to this frame. But this is a separate concept from relative velocity, the velocity A measures B to have. To get that, you need to use the velocity transform to calculate B's velocity in A's frame.

That distinction can be made in either Newtonian or Einsteinian physics, but is a distinction without a difference in the Newtonian case because the Galilean velocity transform is so simple.

So in your example, if the pedestrian wants to know the time he has until the cars collide then he uses the separation rate. However, if he wants to know the time the drivers will measure until collision he needs to Lorentz transform the positions and velocities into the rest frame of one car and then apply the separation rate in that frame (which should be easy as one car's velocity will be zero). You aren't using the Galilean transforms at all.

"Domain of applicability" is a slightly different concept. It usually refers to the range of velocities/energies/whatever over which an approximate theory (e.g. Newtonian physics) is indistinguishable from a more precise one (e.g. relativity). To work that out you usually Taylor expand something, find that the first term is the simpler theory, and look at when the second term is measurable to your degree of precision. So Lorentz velocity addition says$$\begin{eqnarray*}
u'&=&\frac{u+v}{1+uv/c^2}\\
&=&(u+v)\left(1-\frac{uv}{c^2}+\ldots\right)
\end{eqnarray*}$$You can see that ignoring all terms in the series after 1 gives you Galileo, so Galilean transforms are fine as long as speeds are low enough that a fractional error of $uv/c^2$ in your velocity measurements is tolerable.

 

[Sagittarius A-Star]

View attachment 299724
In the frame of Observer C standing by the side of the road, the speed of Car A with respect to Car B = v1 + v2. (Galilean Transformation).

No, that is not a Galilean Transformation. You did not transform from one reference-frame to another.

What you calculate, is the closing speed.

 

[David Lewis]

Thank you, all. So if a car turns on its headlights then (in Frame C) the separation rate of the light with respect to the car is c-v?

 

[Ibix]

I'd say that the separation rate of the light and the car is $c-v$ to avoid privileging the car, as your phrasing kind of does. But that's nitpicky - yes, you are basically correct.

 

[cianfa72]

There are two distinct concepts related to relative speed. One is the rate of change of distance (aka the "separation rate") between two objects measured using some frame (not necessarily the rest frame of either). In this case you can just add speeds relative to this frame. But this is a separate concept from relative velocity, the velocity A measures B to have.

Indeed the separation rate is always given by adding the velocities of the involved objects w.r.t. the given reference frame, regardless of the reference frame employed.

That distinction can be made in either Newtonian or Einsteinian physics, but is a distinction without a difference in the Newtonian case because the Galilean velocity transform is so simple.

Yes, since using Galilean transformations to evaluate the positions and velocities of both objects from the pedestrian rest frame to the rest frame of car A for example, we get that the velocity of A is of course null while the velocity of B is exactly the same as the separation rate evaluated in the pedestrian rest frame.

 

[David Lewis]

...to evaluate the positions and velocities of both objects from the pedestrian rest frame to the rest frame of car A... we get that the velocity of A is... null while the velocity of B is exactly the same as the separation rate evaluated in the pedestrian rest frame.

Are you saying in this case the Galilean Transformation is exact?

 

[cianfa72]

Are you saying in this case the Galilean Transformation is exact?

No, I was just saying that if you decided to apply the Galilean transformation from the pedestrian inertial rest frame to the rest inertial frame of car A then you would get that result.

 

[Ibix]

Are you saying in this case the Galilean Transformation is exact?

No - I think the point being made is a reiteration of the fact that if the Galilean transforms were exact then there would be no meaningful distinction between separation rate and relative velocity.

 

[cianfa72]

the Galilean transforms were exact then there would be no meaningful distinction between separation rate and relative velocity.

Yes, exactly.