Lorentz Transformation negative result

[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.

 

[Simon Bridge]

Welcome to PF;
The negative result is not significant.

 

[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.

 

[Fredrik]

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.

 

[bb1413]

Interesting. Thanks guys

 

[Stephanus]

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.

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

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.

 

[Simon Bridge]

I don't know about physics, but in mathematic
If $x=16$−−$\sqrt x=\sqrt{16}$ then x is 4 not -4
But if $x^2=16$ then X is $\pm\sqrt{16}$

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

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

 

[Stephanus]

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

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

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.

 

[Nugatory]

Hell, I can't even understand why the other participant experience time dilation,

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.

 

[Stephanus]

For that definition "see", their respective experiences will be described by the Doppler Shift analysis section of http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

...That is because all clocks show their own spacetime interval. Forget time dilation/length contraction. Clocks and Doppler is where the physics is.

Hi Nugatory, I still remember your answer to me.
Actually, I'm preparing a question right now about doppler effect, with graph.
But we are going off topic I think.

 

[Fredrik]

But isn't the formula in physics is
$\gamma=\frac{1}{\sqrt{1-\beta^2}}$
not
$\gamma^2=\frac{1}{1-\beta^2}$

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

 

[Stephanus]

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

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?

 

[Fredrik]

So actually there's only $\gamma = \frac{1}{\sqrt{1-\beta^2}}$
There's no
$\gamma^2 = \frac{1}{1-\beta^2}$

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 $\gamma^2=\frac{1}{1-\beta^2}$.

but we can change $\gamma$ sign at will, is that it?

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 $-\gamma$.

 

[Stephanus]

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 $-\gamma$.

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.

 

[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 spatial 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
\begin{equation}
\label{1}
\eta_{\mu \rho} {\Lambda^{\mu}}_{\nu} {\Lambda^{\rho}}_{\sigma}=\eta_{\rho \sigma}
\end{equation}
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!

 

[Stephanus]

Well, if you have some background in mathematics...

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 high school. 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.
What about Newton gravitational formula
$F = \frac{G.m1.m2}{r^2}$ 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"

 

[PeterDonis]

I remember decades ago in high school. My math teacher once said
"The square root of 9 is not -3. It's always 3. But if $x^2 = 9$ then $x = ±\sqrt{9}$

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.

I don't know how this applies to physics.

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

What if v≥1? Would physics allow that?

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

What about Newton gravitational formula
$F = \frac{G.m1.m2}{r^2}$ What if R <0?

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

 

[Stephanus]

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.

Ahhh, all these years!.

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

$\gamma = \frac{1}{\sqrt{1-v^2}}$

Is v≥1 allowed?

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

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.

 

[PeterDonis]

While mathematic allows $\gamma$ to be negative, physyic somehow doesn't allow that.

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.

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

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.

 

[Stephanus]

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

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.

double d;
d = sqrt(9);

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

What about Newton gravitational formula
$F = \frac{G.m1.m2}{r^2}$ What if R <0? Math allow that...

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.

 

[vanhees71]

Ok, let me try again without group theory. Let's go to 1+1-dimensional space, i.e., one time and one space dimension. The Lorentz transformations are linear mappings from one coordinate system to another, i.e.,
$$\begin{pmatrix} t' \\ x' \end{pmatrix} =
\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix}t \\ x \end{pmatrix}$$
which leave the Minkowski-bilinear form invariant, i.e.,
$$t'^2-x'^2=t^2-x^2.$$
It's easy to see that for the transformation matrix this means that
\begin{equation}
\label{1}
a^2-c^2=1, \quad b^2-d^2=-1, \quad a c-b d=0.
\end{equation}
Now from the first and second equation it follows that you can find real numbers $\eta$ and $\eta'$ such that
\begin{equation}
\label{2}
a=\cosh \eta, \quad c=- \sinh \eta, \quad b=- \sinh \eta', \quad d=\cosh \eta'.
\end{equation}
From the third equation in (\ref{1}) you have
$$\cosh \eta \sinh \eta = \cosh \eta' \sinh \eta' \; \Rightarrow \sinh(2 \eta)=\sinh(2 \eta') \; \Rightarrow \eta=\eta'.$$
There we have already used that we must choose $a \geq 1$ in order to have orthochronos Lorentz transformations. So our Lorentz matrix takes the form
$$\Lambda=
\begin{pmatrix}
\cosh \eta & -\sinh \eta \\ -\sinh \eta & \cosh \eta
\end{pmatrix}.$$
Now what's the physical meaning of this? We write down the transformation explicitly
$$t'=\cosh \eta t-\sinh \eta x, \quad x'=\cosh \eta x-\sinh \eta t.$$
Now the origin $x'=0$ of the new frame fulfills the equation of motion
$$\cosh \eta x-\sinh \eta t=0 \; \Rightarrow \; x=t \tanh \eta.$$
That means that the constant velocity of the new frame relative to the old one is
$$v=\tanh \eta \; \Rightarrow\; -1<v<1.$$
This means that the relative velocity between two physically realizable reference frames must be always less than the speed of light.
Now we have
$$\cosh \eta=\frac{1}{\sqrt{1-\tanh^2 \eta}}=\frac{1}{\sqrt{1-v^2}}, \quad \sinh \eta=\frac{\tanh \eta}{\sqrt{1-\tanh^2 \eta}}=\frac{v}{\sqrt{1-v^2}}.$$
Usually one introduces the Lorentz factor
$$\gamma=\frac{1}{\sqrt{1-v^2}},$$
so that the Lorentz transformation in physical terms reads
$$t'=\gamma (t-v x), \quad x'=\gamma (x-v t).$$
As you can easily see, there are no sign ambiguities, and of course the square root used above is defined to be positive, as usual in real analysis.

Now you can figure out, which different sign patterns in (\ref{2}) are possible, because from (\ref{1}) you could have chosen other relative signs for the matrix elements. Here I argue in another way.

The above constructed matrices are the $\mathrm{SO}(1,1)^{\uparrow}$ transformations, i.e., they have $a \geq 1$ and $\mathrm{det} \Lambda=1$. Now, if you give up the demand that the physical transformations are deformable from unity smoothly, you can have $a<-1$ and $\mathrm{det} \Lambda = \pm 1$. Now we need just one special Lorentz transformation for each sign combination of these two quantities. Except the special orthochronous Lorentz transformations derived above we then have the cases
(1) $a \geq 1, \quad \mathrm{det} \Lambda=-1,$
(2)$a <-1, \quad \mathrm{det} \Lambda=-1,$
(3)$a<-1, \quad \mathrm{det} \Lambda=1.$
Case (1): Here the special $\mathrm{O}(1,1)$ matrix is given as $P=\mathrm{diag}(1,-1)$. That's a spatial reflection ("parity transformation"). For large parts of physics this is a good symmetry. Only the weak interaction breaks this symmetry.
Case (2): The special matrix is $T=\mathrm{diag}(-1,1)$. This is the time-reversal transformation. Also here only the weak interaction breaks the corresponding symmetry in nature (which is a bit subtle to interpret physically, because you cannot somehow run time really backwards).
Case (3): This is the "grand reflection", which can be written as $PT=\mathrm{diag}(-1,-1)$.
Now any Lorentz transformation can be obviously written as the product of a proper orthochronous Lorentz transformation, derirved above, and one of the special transformations discussed as Cases (1)-(3). So you have the complete set of Lorentz transformations in 1+1 dimensions.

 

[Fredrik]

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

There is only one result. 9 has two square roots, but $\sqrt{9}$ denotes the unique non-negative square root of 9, and the sqrt function only finds the non-negative square root of the input.

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.

Right. Newton's theory of gravity says that if two masses m and M are a distance r apart, the force that M exerts on m has the magnitude $G\frac{mM}{r^2}$ and is in the direction towards M. Distance is by definition non-negative, so it makes no sense to plug in a non-negative number instead of r.

There's no mystery here. The difference between math and physics is that a theory of physics is a piece of mathematics plus a set of statements that tells you how to use the math to make predictions about results of experiments. That's really all there is to it.

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?

No, but position coordinates can. The negative numbers represent locations on your left, while the positive numbers represent locations on your right. Similarly, negative time coordinates represent times before the moment in time that you have chosen to label 0, and positive time coordinates represent times after that moment.

 

[Stephanus]

Math allows d<0, while radius <0 is meaningless in physics.
Right. ...so it makes no sense to plug in a non-negative number instead of r.

At last, one of my opinion is right

There's no mystery here. The difference between math and physics is that a theory of physics is a piece of mathematics plus a set of statements that tells you how to use the math to make predictions about results of experiments. That's really all there is to it.

Thanks!

 

[1977ub]

One example is that velocity is a vector and any of its components can be negative, depending upon one's coordinate system. "Speed" on the other hand, is the absolute value of the velocity - is a magnitude , and only makes sense as a positive number (or zero).

 

[Stephanus]

One example is that velocity is a vector and any of its components can be negative, depending upon one's coordinate system. "Speed" on the other hand, is the absolute value of the velocity - is a magnitude , and only makes sense as a positive number (or zero).

Good point. Thanks.