Why can coefficient "a" between spacetime intervals depend on velocity between systems?

[Mike_bb]

1. The absolute value of relative velocities is in both cases not $4m/s$. If the individual velocities of your 4 frames refer to a certain unnamed inertial reference frame, then you must use the relativistic velocity addition formula.
2. From equation $(2.6)$ in the scan in the OP (posting #1) follows $ds_2^2=ds_1^2$ and $ds_3^2=ds_4^2$.

At first, I consider two frames:

Frame 1: 1m/s Frame 2: 5m/s; (1-5)m/s and (5-1)m/s => absolute value of relative velocity is 4m/s

Then, I consider another two frames:

Frame 3: 2m/s Frame 4: 6m/s; (2-6)m/s and (6-2)m/s =>absolute value of relative velocity is 4m/s

What's wrong?

 

[Ibix]

What's wrong?

Relative velocity means the velocity one thing measures the other to have. Here, you've calculated the velocity difference between two things as measured by a third party (someone at rest in the unnamed frame in which you specified the velocities of frames 1-4), which I usually call the separation rate. In Newtonian physics the separation rate and relative velocity are the same but not in relativity, and you need to use the relativistic velocity addition formula.

For such low velocities as 4m/s the error will be on the order of one part in $10^{16}$, but it does mean that your two pairs of frames don't have the same relative velocities. I didn't mention this because it makes no difference to the discussion of your $K$, which is entirely about choosing consistent units. But @Sagittarius A-Star is correct to note that frames 1 and 2 will not measure the other to have the same velocity that frames 3 and 4 will.

(Edit: e.g. imagine I measure object A to be doing +0.9c and object B to be doing -0.9c. The separation rate I measure is 1.8c, but this cannot be the velocity that A measures B to have since it exceeds $c$. To get the speeds they measure I would use the relativistic velocity addition formula$$u'=\frac{u-v}{1-uv/c^2}$$which predicts that A and B would measure the other to be doing about 0.994c. But at the velocities you specified for your frames the $uv/c^2$ term is tiny so the relativistic formula becomes $u'\approx u-v$, which is the Newtonian result you implicitly used.)

 

[Mike_bb]

On page 5 according to your screenshot you can conclude from the first sentence, that $a$ must be constant. L&L doesn't explain this, but in the 1960 book "Special Relativity" of Wolfgang Rindler, he did it in §8, page 16:

View attachment 353291​

Hello!

I read about irreducible polynomials and constant factor between them but it was written that polynomials were depended of the same variables. For example, P(x) and Q(x), but in the case of the example we have two polynomial with different variables: P(dt, dx,dy,dz) and Q(dt',dx',dy',dz'). Why does it work in this case?

Thanks.

 

[Mike_bb]

Perhaps there is mistake in the book:

Maybe: $P=aT'^2+bX'^2+cY'^2+dZ'^2 +....$

In his sold-out 1st edition (1982) of the book "Introduction to Special Relativity", Rindler described this. Unfortunately, this was removed from the 2nd edition of the book, in which the LT is derived differently.

https://www.amazon.com/Introduction-to-Special-Relativity/dp/0198531818?tag=pfamazon01-20

View attachment 353298


​

 

[Sagittarius A-Star]

Hello!

I read about irreducible polynomials and constant factor between them but it was written that polynomials were depended of the same variables. For example, P(x) and Q(x), but in the case of the example we have two polynomial with different variables: P(dt, dx,dy,dz) and Q(dt',dx',dy',dz'). Why does it work in this case?

Thanks.

As written in the screenshot, this works because the transformation is linear. Therefore, you can re-write all primed variable in terms of the unprimed variables in the following way:
$t' = Ax+By+Cz+Dt$ and so on. This means you can write both polynomials so, that they depend on the same variables.

 

[Sagittarius A-Star]

Perhaps there is mistake in the book:

Maybe: $P=aT'^2+bX'^2+cY'^2+dZ'^2 +....$

No, see above in posting #35.