# Photons are clocks and SR says they shouldn't tick

1. Apr 1, 2004

### StarThrower

Suppose the time dilation formula is true:

$$\Delta t = \frac{\Delta t^\prime}{\sqrt{1-v^2/c^2}}$$

if v = c then the denominator is zero.

If the denominator is zero and there is division by zero error then the numerator is nonzero.

If the denominator is zero and there isn't division by zero error then the numerator is zero.

Assume the time dilation formula doesn't lead to division by zero error.

Thus, if the denominator in the formula is zero then the numerator is zero.

Consider a photon moving through an inertial reference frame.

By the fundamental postulate of the theory of special relativity, the speed of this photon is c.

It follows that the denominator is zero. For the theory of special relativity to avoid self-contradiction, it follows that the numerator must be zero.

Thus, if there were a clock attached to a photon that moves along with a photon, that clock cannot be ticking (since delta t` is an amount of time measured by a clock moving at speed v in an inertial reference frame).

Thus, any clock moving along with a photon, cannot be ticking.

A photon has an intrinsic frequency. Thus, something within the photon is cycling at regular intervals. Thus, a photon constitutes a clock which is ticking.

This contradiction indicates that there is an error in the special theory of relativity.

Regards,

The Star

2. Apr 1, 2004

### Hurkyl

Staff Emeritus
No it doesn't. Different reference frames give different measurements for a photon's frequency. We call this the Doppler effect.

And, in the limit, the Doppler effect predicts that the frequency of a photon would be redshifted to zero in the photonic "frame", so there's no problem here even when assuming the limit is valid.

Last edited: Apr 1, 2004
3. Apr 1, 2004

### DrChinese

Why does this argument sound oh so familiar? Is it because... StarThrower gave us a substantially similar argument in a thread StarThrower started YESTERDAY?

And why is it that SR predicts that all observers will measure c to have an identical value, while not making the same prediction about photon frequency (which is subject to redshift and blueshift)? Is it because SR makes correct predictions? Why, yes, that is the case!

On the other hand, why is StarThrower seemingly incapable of putting forth an alternative predictive theory as a part of his apparent obsession with dethroning SR? He has dismissed the idea that SR does not apply when the velocity of the reference frame v=c. Yet he offers no substitute for use when v<c other than SR itself.

Hmmm. Maybe I will start a thread critical of SR because it does not account for leprechauns.

4. Apr 1, 2004

### Tom Mattson

Staff Emeritus

StarThrower, perhaps it has escaped your attention, but pet theories that attempt to disprove relativity (especially those based on gross misunderstandings of the theory, such as your pet theories) are to be posted in the Theory Development Forum.

*kick*

Off you go.

5. Apr 1, 2004

### StarThrower

Yes, there is a problem here, you are practically standing on top of it.

Kind regards,

The Star

6. Apr 1, 2004

### StarThrower

Why would I offer an alternative theory to SR, Newton gave it almost 400 years before SR.

Kind regards,

The Star

7. Apr 1, 2004

### StarThrower

I'm not propounding a theory, I'm just doing algebra.

Kind regards,

The Star

P.S. You know, it really is scary how many Einstein worshippers are here. You guys are like mystical about a theory which is so obviously erroneous as to make me want to vomit.

A reference frame in which a photon is at rest happens to be an inertial reference frame, so the whole theory is self-contradictory.

Last edited: Apr 1, 2004
8. Apr 1, 2004

### Hurkyl

Staff Emeritus
Then where did you get

and
?

9. Apr 1, 2004

### Tom Mattson

Staff Emeritus
Yes, you are. You have in a single sentence propounded a theory of the photon (that "intrinsic frequency" nonsense, remember?)

And you aren't even doing that correctly.

Pffft.

Since there is no reference frame in which a photon is at rest, this is moot.

10. Apr 2, 2004

### Severian596

Hey Tom, could I bother you for some sources on this discussion? I'm a novice with SR and the topic of a photon's reference frame has me puzzled. I've determined on my own that a photon's spacetime diagram would be very unorthodox (that the length of each unit on the ct axis would be $$\infty$$ and the length of each unit on the x axis would be 0). But I wondered if you have some sources that discuss this in great detail.

I have a feeling that in the end it will turn up like dividing by zero...I'll just have to trust that it's undefined. Don't get me wrong, I want that to be true!

Thanks!