# A few questions about the Lorentz factor

kotreny

1)

The consequences of faster-than-light speed. I noticed that $$\gamma$$ would then involve complex numbers, and seeing as complex analysis is definitely more than imaginary, I was wondering what it could mean, both mathematically and physically (wormholes?). Unfortunately, I haven't gone very far into math higher than calculus yet. Hopefully soon to change.

2)

$$\gamma$$ bears more than a passing resemblance to the derivative of arcsin. Details?

Bob S
$$\gamma$$ bears more than a passing resemblance to the derivative of arcsin. Details?
You are nearly right. When I started high energy physics ~50 years ago, we did not have pocket calculators. We had only slide rules and trig tables. If you look up β in the sin(θ) column of the trig tables, then cos(θ) = 1/γ.

Consider β = 1/2. Look up sin(θ) = 1/2, and find cos(θ) = 0.86603, so γ = 1.155

Also if you set one side of a right angle triangle to the rest mass and the other side to the relativistic momentum (in energy units) then the hypotenous is the total energy.

So who needs pocket calculators?

Bob S

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kotreny
If you integrate $$\gamma$$=c/(c^2-v^2)^.5 wrt v, then you get c*arcsin(v/c), since c is constant.

Geometrically, this means if you have a point on a circle of radius c, with vertical component v, then the angle will be arcsin(v/c) in radians. Multiplying this by c will yield the arc length. $$\gamma$$ is therefore dS/dv, S being arc traversed, measured in units of velocity.

That's about as far as I've gone. No idea what it means though; still learning the ropes. Don't forget to address 1)--suggesting some online resource or book would be helpful.

Bob S
hello kotreny-

Please look at my thumbnail. I have drawn 3 triangles, all with the same included angle θ.
One triangle is your description, one shows the relationship between β γ and θ. the third shows the relation between total energy, momentum (in energy units), and rest mass. See also the trig relations.

Bob S

#### Attachments

• Relativistic_betagamma.jpg
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Gold Member
It turns out that there are good reasons for using hyperbolic functions (sinh, cosh, tanh) instead of trig functions (sin, cos, tan). The relevant equations then become

$$\begin{array}{rl} \sinh \phi & = \gamma \beta = \gamma v / c\\ \cosh \phi & = \gamma \\ \tanh \phi & = \beta = v/c \\ \\ p & = m_0 c \, \sinh \phi \\ E & = m_0 c^2 \cosh \phi \\ \end{array}$$​

The quantity $\phi$ is called rapidity, and is another way to measure motion instead of velocity. At non-relativistic speeds, the rapidity approximates to v/c. The rapidity of light is infinite, relative to any inertial observer.

qraal
This is so cool. It makes more sense than ever.

Here's another one which I thought worth noting. The travel time, as seen by a stationary observer, for a continuously accelerated motion (from v=0 to v=0 again)...

t2 = 4.s/a + (s/c)2

...notice how it compares to the Newtonian version

t2 = 4.s/a

...thus differing only by the light-travel time component.

Really does make it all space-time rather than just space and time.

Staff Emeritus
The quantity $\phi$ is called rapidity, and is another way to measure motion instead of velocity. At non-relativistic speeds, the rapidity approximates to v/c. The rapidity of light is infinite, relative to any inertial observer.

You didn't mention the best thing of all about rapidity - it, unlike velocity, is additive.

Gold Member
You didn't mention the best thing of all about rapidity - it, unlike velocity, is additive.
True.

Also the proper acceleration of an object is $c \, d\phi / d\tau$, where $\tau$ is proper time.

The Lorentz transform is

$$\begin{array}{rcrcr} ct' & = & ct \cosh \phi & - & x \sinh \phi \\ x' &= & -ct \sinh \phi & + & x \cosh \phi \end{array}$$​

remarkably similar to a rotation in Euclidean space.

Doppler blue shift is $e^{\phi}$. Doppler red shift is $e^{-\phi}$.

One drawback, though. Rapidity is a scalar, not a vector. Think of it as the angle between two worldlines in 4D spacetime.

kotreny
Thank you for the reply, Dr Greg, but I have yet to thoroughly familiarize myself with the hyperbolic functions! I look forward to reviewing this thread in the near future.

Hello Bob, interesting thumbnail. I will definitely explore this connection more deeply once I am comfortable with the formulas involved. I can see why the triangles are similar.

This is digressing, but I just noticed yesterday that if space and time were expressed in the same units, then c would be a dimensionless ratio, and E=mc^2 would officially allow energy and mass to be expressed in the same units as well. Imagine: a unit for both space and time!

Again, could someone please give me more info about the mathematical results of faster-than-light speed? At least let me know if there's a proof that the complex numbers involved cannot have meaning. Thanks.

kotreny
I found another triangle that should be added to Bob's three. It puts a new perspective on length contraction.
The hypotenuse is L (the rest length), the opposite side is $$\beta$$L, and, interestingly, the adjacent is L*sqrt(1-$$\beta$$^2). It's easy to see that the triangle is similar.

My interpretation is this: Visualize a circle, tilted at an angle to you. The circle will look like an ellipse in your view, and by my calculation the minor axis will be the diameter times the cosine of the tilt angle. The length contraction would be calculated in exactly the same manner if the circle were moving relative to you. Also note that when an object is tilted, there is no change along the axis of revolution.
Could this mean that, somehow, space "tilts away" when there is motion? It also makes it easy to see that contraction of the other is observed in both frames of reference, because the "tilt" is the same in both perspectives, just in opposite directions.

Bob S
I found another triangle that should be added to Bob's three. It puts a new perspective on length contraction.
The hypotenuse is L (the rest length), the opposite side is $$\beta$$L, and, interestingly, the adjacent is L*sqrt(1-$$\beta$$^2). It's easy to see that the triangle is similar.

My interpretation is this: Visualize a circle, tilted at an angle to you. The circle will look like an ellipse in your view, and by my calculation the minor axis will be the diameter times the cosine of the tilt angle. The length contraction would be calculated in exactly the same manner if the circle were moving relative to you. Also note that when an object is tilted, there is no change along the axis of revolution.
Could this mean that, somehow, space "tilts away" when there is motion? It also makes it easy to see that contraction of the other is observed in both frames of reference, because the "tilt" is the same in both perspectives, just in opposite directions.
Hi Kotreny-
You need to read this paper in Phys Rev
http://prola.aps.org/abstract/PR/v116/i4/p1041_1
Sorry, it is pay per view.
Bob S