The first postulate of relativity

metalrose

None of the above made any sense to me, thats because im still a junior in undergrad.
Any resources i could use at this point to understand the above ??

metalrose

Also,
I quoted my definition of the first postulate and inertial frames from the modern physics textbook by john r. Taylor.

I quote again,

Inertial frame: an inertial frame is any reference frame ( that is, system of coordinates x,y,z,t), where all the laws of physics hold in their simplest form.

I think what you describe is equivalent to above. I.e. spatial relations being euclidian and newton's first law , are some of the laws of physics in their simplest form.

metalrose

Also, my textbook, just after defining inertial frames in the above way, immediately says,

I think this is equivalent o the definition you gave.

Fredrik

In my opinion, you don't need to understand the mathematical things that strangerep mentioned to understand the issue you asked about. He mentioned them to me because we've been discussing ways to "discover" the Lorentz group in these threads: 1, 2.

One of the things he mentioned is however directly relevant to what you've been asking about: Isotropy. In the 1+1-dimensional case, we can view the statement that you're concerned about as a precise statement of the idea that "space is the same in all directions". I should probably have mentioned that myself.

To do this, you will at some point have to write down a mathematical statement that you view as the assumption that makes the idea of isotropy mathematically precise. So I'm curious about what that statement is. In my approach for the 1+1-dimensional case, ##\Lambda(v)^{-1}=\Lambda(-v)## is that statement. It can be viewed as the precise meaning of "isotropy" in 1+1 dimensions. In 3+1 dimensions, I think the mathematically precise statement of isotropy should be that the group of proper and orthochronous inertial coordinate transformations has a subgroup that consists of transformations of the form
\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix} where 0 is the 3×1 matrix with all components =0, and R is a 3×3 rotation matrix. This is at least one aspect of isotropy. I still haven't finished typing up my notes on the 1+1-dimensional case, so I haven't had time to think about what my complete list of assumptions will be in the 3+1-dimensional case. In particular, I haven't thought about what exactly should replace ##\Lambda(v)^{-1}=\Lambda(-v)##. The statement about rotations above will be a part of it, but I think I will probably also have to assume that boosts satisfy ##\Lambda(v)^{-1}=\Lambda(-v)##. I haven't thought this through enough to know if this rule can be derived from other mathematical assumptions.

DaleSpam

Sure it does. Think a little more about how you make any measurement. Any measurement is a comparison of your measured quantity against some standard quantity. So, how do you measure velocity? As with any measurement you compare it to a standard.

One way to do that is to produce standard distances and standard times and determine how many standard distances are traveled in one standard time. Another way is to produce a standard velocity. If your velocity standard and your distance/time standards agree in one reference frame and they disagree in another reference frame then the laws of physics are different in the different frames, which violates the principle of relativity.

So, if you equip two observers with a velocity standard calibrated to v and observer A sees the speed of B as v with the distance/time standards while observer B sees the speed of A as u with the distance/time standards, then in A's frame the velocity standard will match the time/distance standard and in B's frame the velocity standard will not match. This violates the principle of relativity.

Fredrik

I will elaborate a bit on DaleSpam's gun argument, to make the use of isotropy explicit.

Suppose that S and S' are both inertial coordinate systems such that S' can be obtained from S by a pure boost in the positive x direction. Denote the speed of S' in S by v, and the speed of S in S' by u. Suppose that a gun G is built at rest in S according to specifications that ensure that its bullets will come out with speed v relative to the gun. Suppose that a gun G' is built at rest in S' according to the same specifications. The principle of relativity strongly suggests that the bullets coming out of this gun will have speed v relative to the gun. Now suppose that the gun at rest in S' is rotated around the z axis by an angle of pi (so that it now points in the negative x' direction instead of the positive x' direction). If space is isotropic, the bullets coming out of the gun should still come out at speed v.

Now suppose that both guns are fired at the event where the fronts of the barrells pass each other. The bullet that comes out of G will be comoving with G'. Isotropy strongly suggests that the bullet that comes out of G' will be comoving with G. So the speed of G with respect to the bullet that came out of it (by assumption =u), is the same as the speed of the bullet that came out of G' with respect to G' (by isotropy and the principle of relativity =v). In other words, u=v.

metalrose

Could you expand a bit more on what you mean by isotropy of space and how the above statement should therefore be true?

Fredrik

The assumption that space is isotropic means roughly that space is the same in every direction. Every theory that assumes that space is isotropic or proves it as a theorem does it in a different way.

Since we're not working within the framework of a specific theory, there's no exact definition that we can use. We're just talking about how to apply some loosely stated ideas. I'm just saying that if we build a gun and it turns out that the speed of the bullet that comes out of it depends on the direction we aim, then I would interpret that as a violation of isotropy.

metalrose

I get that. That's what i thought you meant.

How can you say this?
All that isotropy of space implies is that the bullet fired from G' will move with speed v relative to the gun G', irrespective of wether the gun G' points in the positive x' or negative x' direction.

I don't understand how isotropy of space implies that the bullet coming out of G' is comoving with G.

I think your argument for this is something as follows ?? (Correct me if i'm wrong)
The bullet coming out of G' moves with speed v w.r.t. G' to the left. "Therefore G' moves with v relative to this bullet towards the right."

Also since G' moves with v towards right (+ve x), w.r.t. G, therefore the bullet from G' anf G are comoving.

I think the flaw in such an argument(if this i indeed what you are saying), would be to presume what i stated in the double quotes above, because that is exactly what we are trying to prove.

Fredrik

No, that's not the argument. G and G' are identical guns at the same event in spacetime. The only difference between them is that they're aimed in opposite directions. If the bullet from G ends up comoving with G', and the bullet from G' doesn't end up comoving with G, then what can possibly be the reason? This scenario doesn't involve anything other than space, time, the guns and the bullets. The guns are identical. The bullets are identical. The events where the bullets are fired are as close together as is physically possible. If we want them to be closer, we can consider smaller guns. So it's not a matter of their separation in space, time, or spacetime. There are no factors left to consider other than directions in space.

strangerep

OK, sorry. My message was more intended for Fredrik since we've been discussing closely related stuff in other threads. Perhaps I should have sent it as a PM instead of putting it in your thread here.

[Edit: I notice that Fredrik is continuing part of that conversation in subsequent posts and I'm about to reply. But if you feel this sub-discussion is too off-topic for your thread just say so and we'll stop.]

Rindler's textbook gives a treatment of alternate group-theoretic derivations of Lorentz transformations without assuming the light postulate up front. Fredrik has also mentioned other threads where we've been discussing this.

Yes, they're close enough. I previously only wanted to point out that when you said:you weren't saying the same thing as either Rindler or Taylor.

strangerep

I'm uneasy about writing out a fully rigorous statement of it in this thread, so I'll just summarize...

A local observer can choose his spatial directions to be orthogonal. If (e.g.,) a transformation squeezed 2 axes closer together (like a scissor action), this is detectable locally. Hence, preservation of spatial angles at the origin must be a feature of transformations between inertial frames. This leads quickly to SO(3) as a subgroup of the group of IMTs (inertial motion transformations).

Proceeding to the case of velocity boosts, we first use up the space- and time-translation freedoms to make the old and new origins coincide, along with the spatial axis directions. (I'm still uncertain about whether one must also invoke parity invariance here to make the chiralities coincide, but I'll get back to that.)

Then, we choose an arbitrary (spatial) direction and consider IMTs for which that direction remains fixed. Denote a velocity ##\mathbf v## along that direction as ##v \, \hat v## and regard ##\hat v## (a unit vector) as fixed. Then we try to find a 1-parameter Lie group with ##v\in\mathbb{R}## as its parameter. First we perform orthogonal decomposition of an arbitrary position vector (in both the old and new frames) with respect to ##\hat v##, which I'll express as ##\mathbf{r} = \mathbf r_\| + \mathbf r_\perp##. Spatial isotropy then means the transformation equations must be invariant under rotations around ##\mathbf r_\perp## which reverse ##\hat v## but leave ##\mathbf r_\perp## invariant. IOW, if we reverse both ##\mathbf r_\|## and ##\mathbf r'_\|##, the transformation equations should be unchanged. This is sufficient to prove that the unknown functions of ##v## in the transformation equations must be even in ##v##. Later, when investigating the inverse transformation, this evenness property can be exploited to obtain the result relating ##v## and ##-v##.

metalrose

That clears it up. Thanks a ton.