# Special relativity and photons

1. Dec 18, 2013

### ViolentCorpse

Hello everyone!

There is a postulate of special relativity that says that the speed of light in vacuum is the same for all observers. This seems to imply (to me) that all the results of special relativity must only be used when dealing with observers who conform to the condition set by the postulate. If we don't keep this in mind, we run the risk of violating the postulate. So it confuses me when experts and non-experts alike use the equations of time dilation, length contraction etc. to assert that photons are "timeless" and have no space to travel through (it's contracted to 0).

Not only do I find it hard to imagine photons with such strange characteristics, I think (with every possibility of being wrong) that it is wrong to apply these results to something that doesn't even conform to the core postulate of relativity (for the speed of a photon in its own frame cannot be c and must be zero). But I'm only an unqualified amateur in physics and have no reason to trust my thinking enough to believe it, so I come here to ask, is it correct to apply these results to a photon? Why?

Thank you very much!

2. Dec 18, 2013

### DrGreg

Photons are "timeless" only in the sense that "time relative to a photon" is undefined (rather than zero).

See our FAQ: Rest frame of a photon.

3. Dec 18, 2013

### phinds

Violent, to restate DrGreg's correct response slightly differently, there IS no "its own frame of reference" for a photon so talking as though there is is pointless. In essence, that is exactly what you said. You came to the correct conclusion that there is no point in talking about a photon as though it had a frame of reference.

4. Dec 18, 2013

### ViolentCorpse

Thank you for the clarification, phinds!

There's one more thing I'd like to ask. We can use the relativistic energy equation to arrive at the result E=pc for a photon. Does deriving this result not require that the photon have an inertial frame of reference?

Thanks again!

5. Dec 18, 2013

### Naty1

E = pc is in your frame not the undefined photon frame.

There are some equations here which give additional perspectives:
http://en.wikipedia.org/wiki/Photon_energy#Physical_properties

This one, E = hc/λ, shows how the observed energy varies with observed wavelength ....as an observer [edit: moves] towards a light source it appears frequency shifted......

Last edited: Dec 18, 2013
6. Dec 18, 2013

### Staff: Mentor

No. The $E^2=(m_0c^2)^2 + (pc)^2$ relationship describes measurements by an observer, whether inertial or not, of an object that is moving (possibly with $v=p=0$, in which case we'd say that the object is at rest) relative to him.

7. Dec 18, 2013

### ViolentCorpse

Don't the time dilation and length contraction relationships too describe measurements of another frame by an observer's frame?

8. Dec 18, 2013

### Staff: Mentor

Not quite, because that bolded phrase above doesn't quite make sense - there's no such thing as "measurements of a frame". We measure actual physical quantities, and the time dilation and length contraction relationships tell us how one observer's measurements of these quantities are related to those of another observer who may be in motion relative to the first observer.

More generally: objects don't "have" frames, nor are they "in" frames. Frames are chosen by observers to make sense of their observations; observers generally choose to think in terms of the frames in which they are at rest.

9. Dec 18, 2013

### Naty1

I see Nugatory posted while I was composing....

I agree with his clarification. The Lorentz transform [with length contraction and time dilation] does convert the labels of all events [observations] as calculated in one frame to those as calculated in another frame. However, the transform applicability is limited to massive particles...those with speeds less than c and is also limited to flat spacetime...no curvature, no gravity.

10. Dec 18, 2013

### ViolentCorpse

Ah. And since no such transform is associated with the energy relationships, they're consistent with the postulates of SR. Right?

Nugatory: Thank you for clarifying it, but I happen to know that much. I apologize for my poorly-worded post. I should've been more exact. I was in fact trying to ask how is it that measurements involving Lorentz transforms made on a photon require a well-defined frame of reference of photons whereas the energy measurements don't, since neither class of measurements are being made in the undefined frame of the photon.

11. Dec 18, 2013

### Staff: Mentor

They do not. It's the observer who needs a frame of reference, not the observed - whether it's a photon or not.

Now I'm finding myself wondering... when you wrote "Lorentz transforms" above did you mean "the time dilation and length contraction formulas"?

12. Dec 18, 2013

### ViolentCorpse

Yes, that's what I meant.

13. Dec 18, 2013

### Staff: Mentor

14. Dec 18, 2013

### Staff: Mentor

What jtbell said.

Furthermore, if you want to understand relativity, you cannot start with the time dilation and length contraction formulas. You have to understand the Lorentz transformations first (and then you can use them to derive the time dilation and length contraction formulas if you want - it's a good exercise).

15. Dec 18, 2013

### Staff: Mentor

Even though you can derive length contraction and time dilation from the Lorentz transformation, they're not the "whole story." The LT also includes relativity of simultaneity, which is just as important as the other two phenomena.

Taken together, the set of three:

• Length contraction
• Time dilation
• Relativity of simultaneity

are equivalent to the Lorentz transformation, in physical content.

16. Dec 19, 2013

### ViolentCorpse

Thank you everyone!