# Lorentz Factor

1. Apr 2, 2008

### A(s)

First off I apologize for lengthiness, but I want to be thorough... please read the entire post

I was just reading up on time dilation on wiki, and saw that the formula for determining time dilation in special relativity is essentially delta t times the Lawrentz factor.

in other words for time dilation,

t / [(1 - v^2 / c^2)^1/2]

The lawrentz factor describes things like time dilation, length contraction and relative mass, but I think there may be a problem with the application of it. The lawrentz factor is

c / [(c^2 - v^2)^1/2]

but it is commonly reduced to

1 / [(1 - B^2)^1/2]

where B = v / c

this reduction is done by simply pulling out a c^2 from under the square root, and then canceling it with the top c.

Here is the problem.

If we are dealing with a vector problem, as most time dilation theories do, (i.e. traveling out at a significant fraction of the speed of light, and coming back at the same speed therefore reducing the aging process) the reduced equation essentially assumes absolute values of the vectors. Here is an example.

Let us assume for the problem a reference point of (0,0,0) coordinates in space on an X,Y,Z axis.

The direction we will travel from point (0,0,0) to planet A will be in a positive Z direction at a rate of 2.0E8 meters per second with no X or Y components. Assume the distance is such that it takes 5 days on a spacecraft to get there.

Using the unreduced formula, the Lawrentz factor for the trip there will be

3.0E8 / [((9.0E16)-(4.0E16))^1/2] = 3.0E8 / 2.24E8 = 1.34

on the trip back we have

-3.0E8 / [((9.0E16)-(4.0E16))^1/2] = -3.0E8 / 2.24E8 = -1.34

the -3.08 on top must be negative because the speed of light and the velocity we determined are relative to a positive z vector, but now we are headed in a negative z direction... so a factor describing relative time and speed to a stationary point must maintain that relativity. The -3.0E8 and -2.0E8 on the bottom were both squared according to the equation, and thus lost the signs attributed to them, leaving them at 9.0E16 and 4.0E16 (i know this is basic, but i must point it out to avoid argument)

if we add the lawrentz factors, the total difference in relative time is 0

5days(1.34) +5days(-1.34) = 0

Finally, If we did this same equation with the reduced formula (where c had been canceled out), we would get 1.7 days fewer on both the trip there and the trip back, leaving us 3.4 days "younger" than our brothers at point (0,0,0)

How has nobody noticed this flaw...

To me, this is similar to the Doppler effect. If you hear an ambulance going away from you the pitch decreases. When it comes closer to you the pitch increases

If you had a telescope and you were able to watch the spaceship fly away from point (0,0,0) it would take you 1.7 days to observe 1 day of behavior on the ship... the light is stretched out... but if you watch the way back, it will only take you about .746 days to observe one day of behavior, because the light would be compacted and closer together... regardless, the entire trip could be observed in exactly 10 days. It would just appear to take 6.7 days for them to get there and 3.3 days for them to get back...

can anyone echo if I am on to something here? and why it hasn't been noticed before?

Last edited: Apr 2, 2008
2. Apr 2, 2008

### Antenna Guy

First-off, it's "Lorentz".

Careful - Doppler has to do with pitch, not amplitude. Amplitude does vary with distance, and pitch/frequency varies with relative velocity (i.e. approaching/receding).

Regards,

Bill

3. Apr 2, 2008

### A(s)

well pitch is frequency, which is cycles per second,

light waves more cycles per second = more densely perceived time

same as higher pitch = more densely perceived sound

In sound this is shown in the crests and troughs being closer together or further apart at the same speed

with light this would be shown with photons being closer together or further apart at the same speed

Last edited: Apr 2, 2008
4. Apr 2, 2008

### Antenna Guy

I realize what you meant to say, but "quieter"/"louder" means something else.

Regards,

Bill

5. Apr 2, 2008

### A(s)

thank you I will edit that

6. Apr 2, 2008

### Antenna Guy

Not exactly - what you said of sound would be true of light as well.

Regards,

Bill

7. Apr 2, 2008

### A(s)

so it might move off the visible spectrum because frequency increases or decreases, but the photons would still be released closer together or further apart depending on the direction of travel

8. Apr 2, 2008

### Mentz114

A(s):
Is it maybe just possible that you're wrong ? This has been scrutinised for over 100 years and you'd think any problems of the kind you point out would have been found.

9. Apr 2, 2008

### A(s)

Of course it is possible that I am wrong, but i would expect a better reason other than nobody else has realized it

if you have the equation of a line as y = (x^2 -1)/(x-1) even though it reduces to y = x+1 does not mean that they are the same line... one line has a hole or does not exist at x=1... the other exits at all values.

cancellation is a very dangerous thing if used incorrectly

10. Apr 2, 2008

### A(s)

to apply my previous statement directly to this problem... Y = b/ (b^2)^(1/2) does not equal 1 if b is negative... it equals -1

11. Apr 2, 2008

### Staff: Mentor

Come on: c stands for the speed of light; it's not a vector quantity.

The "Lorentz factor" is:

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

12. Apr 2, 2008

### JesseM

You're not dealing with a vector problem. The "c" and "v" in those formulas are only meant to be speeds, i.e. v is just the absolute value of the velocity vector, and c is just a constant speed. If someone changes speeds, then if you want to calculate the time elapsed on their clock between two coordinate times $$t_0$$ and $$t_1$$, and their speed as a function of time in your chosen coordinate system is v(t), then the total elapsed time is $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2 / c^2} \, dt$$

Last edited: Apr 2, 2008
13. Apr 2, 2008

### A(s)

ok, well even still, if the relative direction of the speed of light has no bearing, what if you pull out v^2 out of the square root. the negative velocity might apply.

gamma = plus or minus c / [v(c^2 / v^2) -1)]if v is negative, you still get a negative value

14. Apr 2, 2008

### A(s)

how are vectors irrelevant... with the doppler effect, you don't take absolute value and say that as a source of sound is approaching, the pitch decreases

15. Apr 2, 2008

### JesseM

Time dilation is not based on the doppler effect, so how is this relevant? Clocks moving at the same speed are slowed down by the same factor, regardless of their direction.

16. Apr 2, 2008

### A(s)

i suppose i am just wrong then... they seem really simmilar though

17. Apr 2, 2008

### JesseM

Just think of v as the absolute magnitude of the velocity vector, which can't be negative.

18. Apr 2, 2008

### Antenna Guy

The "visible spectrum" is a rather limited range of frequencies, but yeah - wavelengths could shift into/out of it via Doppler effects. Recall that one end of the spectrum is DC, while the other end is infinite frequency.

Whether the photons appear closer together or farther apart (relative to the frame in which they are emitted) depends upon the observer. If photons were produced at a certain rate within the frame of emission, they would be observed at a relative rate equivalent to the relative (proportional) change in wavelength (time dilation between frames).

Regards,

Bill

19. Apr 2, 2008

### Antenna Guy

Let $\frac{v}{c}=cos(\theta)$ and then $\theta=asin(\frac{1}{\gamma})$.

Regards,

Bill

20. Apr 2, 2008

### Staff: Mentor

Note that there is a relativistic version of the Doppler effect. It's derived similarly to the classical Doppler effect, but the effects of time dilation on the moving source are taken into account.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html

The frequency that is received by the observer increase if the source is approaching, and decreases if the source is receding, just as with the classical Doppler effect, but the details are different.

Last edited: Apr 2, 2008
21. Apr 2, 2008

### Antenna Guy

How does one get from the bi-directional classical version to the (Lorentz-consistent) unidirectional version without losing a sense of direction? Additionally, why does it appear that $v_{wave}$ changes to c in the process? .

Regards,

Bill

22. Apr 3, 2008

### JesseM

The relativistic doppler shift equation is bidirectional too (you see a clock ticking faster than your own as it approaches you, and slower than your own as it moves away, slowed down by an amount even larger than the time dilation factor), it's only time dilation that doesn't depend on direction.

23. Apr 3, 2008

### Staff: Mentor

On that page, for some reason they show the $\pm$ sign explicitly in the classical Doppler shift formula, but they hide it in the relativistic Doppler formula. In the classical formula, you substitute a positive number for the speed and choose either the + or - sign from the formula, but in the relativistic formula you substitute either a positive or negative number depending on the direction of motion.

Last edited: Apr 3, 2008
24. Apr 3, 2008

### pam

A simple derivation of the relativistic Doppler effect follows from the fact that
$$(\omega/c,\b{k})$$ is a four vector. Lorentz transformation gives
$$\omega'=\gamma\omega(1+v\cos\theta/c)$$, which includes a "transverse Doppler effect".

Last edited: Apr 3, 2008
25. Apr 3, 2008

### A(s)

that helps a bunch, thanks