# Lorentz Transformation

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1. May 30, 2015

### bb1413

Hello,
I have some mathematics background but little to no physics background. I am very interested in physics and am beginning to learn about relativity. Upon exploring the derivation of the Lorentz Transformation equation I noticed something that confused me a little. Again, I don't have much experience or knowledge, so this is probably a rather naive question. My question is: when one takes the root of γ^2=1/(1-(v^2/c^2)) you get γ=±1/(√1-(v^2/c^2)). I understand the significance of the positive result, but what of the negative? Just curiosity, could not find a clear answer elsewhere. Thanks.

2. May 31, 2015

### Simon Bridge

Welcome to PF;
The negative result is not significant.

3. May 31, 2015

### ArmanCham

We define lorentz factor pozitive.So you take square of lorentz factor then you take the root.You gain again positive lorentz factor.I am trying to say lorentz factor defined $γ=1/√(1-(v/c)^2)$ pozitive.

4. May 31, 2015

### Fredrik

Staff Emeritus
As the other guys said, $\gamma$ is positive by definition. But you can consider Lorentz transformations with $-\gamma$ where $\gamma$ would normally appear. In addition to changing the velocity of the coordinate system, such a transformation also reverses the direction of both the time axis and the spatial axes.

5. May 31, 2015

### bb1413

Interesting. Thanks guys

6. Jun 4, 2015

### Stephanus

I don't know about physics, but in mathematic
If $x=\sqrt{16}$ then X is 4 not -4
But if $x^2=16$ then X is $±\sqrt{16}$
$\sqrt{something}$ is ALWAYS positive in math, even the imaginary is positive.
So the gamma factor in Lorentz transformation is ALWAYS positive.
$\sqrt{1-\frac{v^2}{c^2}}$ is always positive or positive imaginary number for v>c.

Wether we can apply gamma factor as negative, perhaps some physicist can step in

7. Jun 4, 2015

### Simon Bridge

I.e. if $x^2=a$ then $x=\pm\sqrt a$

So if $\gamma^2 =1/(1-(v/c)^2)$then $\gamma=\cdots$?

8. Jun 4, 2015

### Stephanus

Yes SimonBridge

If $\gamma^2 =1/(1-(v/c)^2)$then $\gamma=±\frac{1}{\sqrt{1-\beta^2}}$
But isn't the formula in physics is
$\gamma=\frac{1}{\sqrt{1-\beta^2}}$
not
$\gamma^2=\frac{1}{1-\beta^2}$
Or... we can put this in phisics? $\gamma^2=\frac{1}{1-\beta^2}$
If we can, no, no, no, don't explain to me how or the reverse direction as Fredik says. No use, I can't understand that. Hell, I can't even understand why the other participant experience time dilation, while the other don't. But I'm off topic here.

Last edited: Jun 5, 2015
9. Jun 4, 2015

### Staff: Mentor

That's easy.... No one ever experiences time dilation. The only way that we even know that time dilation is happening is that we compare what two clocks read at the same time ("same time" according to someone somewhere) twice. If they haven't changed by the same amount between the two readings, we say that one is time-dilated relative to other. Which if either is time-dilated will depend on who is deciding what "at the same time" means.

10. Jun 4, 2015

### Stephanus

Actually, I'm preparing a question right now about doppler effect, with graph.
But we are going off topic I think.

11. Jun 5, 2015

### Fredrik

Staff Emeritus
Yes. The convention is to take that first equation as the definition of $\gamma$.

12. Jun 5, 2015

### Stephanus

So actually there's only $\gamma = \frac{1}{\sqrt{1-\beta^2}}$
There's no
$\gamma^2 = \frac{1}{1-\beta^2}$, but we can change $\gamma$ sign at will, is that it?

13. Jun 5, 2015

### Fredrik

Staff Emeritus
That's like saying that there's only $2=\sqrt{4}$ and no $4=(\sqrt{4})^2$, but the former implies the latter. The definition $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ implies that $\gamma^2=\frac{1}{1-\beta^2}$.

You could choose to denote the negative square root of $\frac{1}{1-\beta^2}$ by $\gamma$, if you want to confuse people. If that's not your goal, you should denote the negative square root by $-\gamma$.

14. Jun 5, 2015

### Stephanus

No, of course not.
In as I said in mathematic $\sqrt{4}$ is always 2, but if we want to use $-\gamma$ as reverse direction (or time?) we can add the minus sign at will.

15. Jun 6, 2015

### vanhees71

Well, if you have some background in mathematics, I'd just tell you that Lorentz transformations are those transformations leave the fundamental form of signature $(1,3)$ on $\mathbb{R}^4$ invariant. These transformations build a group, the $\mathrm{O}(1,3)$.

Now the question is, whether this full group is a symmetry group in nature. The answer is no! The weak interaction violate symmetry under spatial reflections, $t \mapsto t$, $\vec{x} \mapsto -\vec{x}$, which is in $\mathrm{O}(1,3)$. The weak interaction also violates time-reversal invarianc, $t \mapsto -t$, $\vec{x} \mapsto -\vec{x}$.

This tells you that the assumption that the entire $\mathrm{O}(1,3)$ is too large a group to serve as the symmetry group of Minkowski spacetime (without anything else than the space-time manifold itself you couldn't conclude this; for that you need more information, as the facts about the weak interaction envoked here). So we have to ask further, what's the minimal subgroup which must be fulfilled to make Minkowski space a consistent space-time model. The only thing you have to assume is that there is no physically realizable reference frame which is in any way distinguished from any other. Such a (inertial) reference frame is given by a bunch of synchronized clocks and a reference point in space and a basis of spacial vectors. The transformation from one such reference frame to another should be smoothly deformable from the group identity (i.e., doing nothing to the reference frame at all). So we have to look for the component of the $\mathrm{O}(1,3)$ which is smoothly connected to the identity matrix.

The first observation is that for ${\Lambda^{\mu}}_{\nu} \in \mathrm{O}(1,3)$ by definition you must have

\label{1}
\eta_{\mu \rho} {\Lambda^{\mu}}_{\nu} {\Lambda^{\rho}}_{\sigma}=\eta_{\rho \sigma}

with $(\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ (using natural units where $c=1$). This implies that
$$(\det \Lambda)^2=1 \; \Rightarrow \; \det \Lambda=\pm 1.$$
So in order to be deformable smoothly from the identity matrix, you must have $\det \Lambda=+1$. Thus the physical symmetry group must be a subgroup of $\mathrm{SO}(1,3)$, i.e., all Lorentz transformations with determinant 1.

The second observation is that (\ref{1}) implies that $({\Lambda^0}_0)^2 \geq 1$, i.e., ${\Lambda^0}_0 \geq 1$ or ${\Lambda^0}_0 \leq -1$. In order to be deformable smoothly from the unit matrix you must clearly have
$${\Lambda^0}_0 \geq 1.$$
It's easy to prove that the LTs fulfilling this constraint indeed define another subgroup, orthchronous Lorentz group $\mathrm{O}(1,3)^{\uparrow}$.

Now you take both observations together. Then you end up with the $\mathrm{SO}(1,3)^{\uparrow}$, the special orthchronous Lorentz group as the physically relevant symmetry group of Minkowski space.

Note that above I passed over the fact that I need to fix a point in space too, as well as a certain time at this point as the time $t=0$ of the synchronized clocks in that frame, but that choice shouldn't have a physical significance in empty space whatsoever. So in addition to the proper orthochronous Lorentz transformations also the space-time translations must be a symmetry, and thus also all compositions of such operations. This semidirect product of the $\mathrm{SO}(1,3)^{\uparrow}$ with the space-time translations defines the full physical symmetry group of Minkowski space-time, the proper orthochronous Poincare group.

That's the end of the story as far as classical (i.e., non-quantum) physics is concerned. Quantum theory leaves more possibilities to realize the symmetry group, and you have to extend the group to more general representations. For the Poincare group it boils down to substitute the $\mathrm{SO}(1,3)^{\uparrow}$ with it's covering group, the $\mathrm{SL}(2,\mathbb{C})$, which in addition to integer-spin representations of the rotation subgroup also admit half-integer-spin representations, which after all, build the fermionic fields describing matter!

16. Jun 6, 2015

### Stephanus

Oh no, no. I don't have math background, much less physics background and I can't even barely grasp what you write
It's just that I remember decades ago in highschool. My math teacher once said
"The square root of 9 is not -3. It's always 3. But if $x^2 = 9$ then $x = ±\sqrt{9}$
That's why I keep telling myself that $x≠\sqrt{x^2}$ if x is less (or equal?) then zero.
I don't know how this applies to physics.
Clearly if I see Lorentz equation.
$\gamma = \frac{1}{\sqrt{1-v^2}}$ What if v≥1? Would physics allow that? Because math wouldn't allow that, or at least the result is imaginary number.
$F = \frac{G.m1.m2}{r^2}$ What if R <0? Math allow that, would physics allow that? Or somehow we have to change the formula?
After all Neil Degrasse Tyson once (only once?) said that "Mathematic is the language of the universe"

17. Jun 6, 2015

### Staff: Mentor

Your math teacher was being a pedant, insisting on a particular definition of the term "the square root" which is not really very reasonable. The key error is the definite article "the". 9 does not have one square root: it has two, +3 and -3. There is no mathematical reason to prefer one over the other, or to insist that +3 is "the" square root of 9.

Physically, negative numbers make sense in some contexts but not in others. For $\gamma$, negative values don't make sense.

It depends on whether you think tachyons are physically possible. Most physicists don't think so.

What would that mean, physically? R = 0 at the center, and takes positive values everywhere else. So where could R be negative?

18. Jun 6, 2015

### Stephanus

Ahhh, all these years!.

While mathematic allows $\gamma$ to be negative, physyic somehow doesn't allow that.
While mathematic doesn't allow v≥0, of course the result would be imaginary number, physica somehow allows that.

Last edited: Jun 6, 2015
19. Jun 6, 2015

### Staff: Mentor

Because a negative $\gamma$ would mean a negative energy (since energy is $\gamma$ times rest mass, which is positive), which doesn't make sense physically. Note that we are assuming here that $v \lt 1$, so that $\gamma$ and the rest mass are both real; for the case $v \gt 1$, see below.

I assume you mean $v \gt 1$; I should have pointed that out before. (Note that $v = 1$ is not allowed, since that would make $\gamma$ undefined.) You can make a consistent theory of tachyons (at least, for some values of "consistent"), but only if the tachyons have imaginary rest mass. That way, the energy and momentum of the tachyon are still real--they are the product of an imaginary $\gamma$ and an imaginary rest mass.

20. Jun 7, 2015

### Stephanus

I just remember something.
There's mathematic
there's physic
and there is computer math.

double d;
d = sqrt(9);

Even if math gives two result for $\sqrt{9}$, the computer only gives 1 result.

Suppose if we change all those variable into meaningless ones.

$A = \frac{B.c1.c2}{d^2}$
Math allows d<0, while radius <0 is meaningless in physics.

I I remember something I calculated years ago.
Two cars A and V
A's speed is 300 m/s and is accelerating 1 m/s2, to the east
V's speed is 100m/s heading to the east, and 1950 meters behind A, in the west of A.
When will they meet?
Looking this problem at a glance suggest that it's a qudratic equation. Can be solve by ABC formula, so...
A distance = V distance
$\frac{1}{2} 1 t^2 + 300 t = 100 t - 1950$
$0.5t^2 + 200 t + 1950 = 0$
t1 = -10
t2 = -390
That gives A will meet V at coordinate: -2950 m and -40950.
Can time really be smaller than 0?
It seems that A WILL never meet V, but A MET V 10 seconds ago and 390 seconds ago.
Distance can be negative?
It seems that A met V twice. 390 seconds ago, A actually headed west and cross with V and 10 seconds ago A headed east catching up B.
Okay..., so far Math and Physic somehow agree.