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I was experimenting with the formula for length contraction, when I realized that anything traveling at the speed of light shrinks out of existence. This is the formula for length contraction: $$T=T'\sqrt{1-\frac{v^2}{c^2}}$$ Where $T$ is the observed length and $T'$ is the proper length. When the speed of light, $c$, is plugged in for $v$, then the formula simplifies to this: $$T=T'\sqrt{1-1}$$ Which further simplifies to this: $$T=T'(0)$$ Therefore, anything traveling at the speed of light will not be seen, regardless of actual length. So, wouldn't we not be able to see any particle traveling at the speed of light?

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    $\begingroup$ Does a photon actually have a length. ...I am not asking, just wondering. $\endgroup$
    – user108787
    Jul 15, 2016 at 0:56
  • $\begingroup$ Afaik you've hit the limits of the classical (non-QM) mechanics. Photons are handled the QED (quantumelectrodynamics) correctly, which is relativistc quantum mechanics. $\endgroup$
    – peterh
    Jul 15, 2016 at 1:21
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    $\begingroup$ More to the point, your SR formula applies to objects with mass. Photons have no mass, so your equation does not apply to them. $\endgroup$
    – Jon Custer
    Jul 15, 2016 at 2:03
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    $\begingroup$ The derivation of length contraction assumes the existence of an inertial reference frame in which the object is at rest. But there is no inertial reference frame in which a photon is at rest and, thus, there is no rest length to speak of; no $T'$ exists for a photon to place on the RHS of your equations. $\endgroup$ Jul 15, 2016 at 2:07
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    $\begingroup$ Photons are not things, they are states, i.e. properties of a quantum field. There is a quantum of energy exchanged where the source is and another one at the detector, but between the two talking about a photon as some "thing" with properties like position and size doesn't make much sense. $\endgroup$
    – CuriousOne
    Jul 15, 2016 at 2:12

2 Answers 2

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I am only a layman, so don't take this answer seriously.

This length contraction formula, and the whole Special Relativity in its original form, is for macro-sized, non-quantummechanical objects. Thus, the formulas work if you want to calculate the size of a spaceship nearing the speed of the light. And not if you want to calculate the size of photons with exactly the speed of the light.

Photons are elementary particles. Or waves. Or both of them. They are quantummechanical objects, described by QED (quantumelectrodynamics), which is a field theory.

So, the theories seem so:

      1. Classical (newtonian, non-QM) mechanics
             /                   \
2. Quantummechanics            3. Special relativity
               \                /
      4. Relativistic Quantummechanics (QED, QFT)

You are now thinking in (3), but your problem is answered by (4).

Essentially, photons are the waves of the electromagnetic field, and these waves propagate with c.

Unfortunately, QFT hasn't so simple and beautiful formulas as the SR, but it is tremendous fun even trying to understand them.

Extension: There is also a possibility, that you are calculating only with classical electrodynamics, without any QM. But there are no photons in classical electrodynamics, there is only the electromagnetic field described by the Maxwell Equations. Here it is possible to calculate light as a wave packet. In this case, you don't have any particles, there is only a propagating wave of the EM field.

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  • $\begingroup$ I think the OP's question would apply equally to a macroscopic, classical EM wave though (that is a wave packet with some finite extent). And the special theory of relativity certainly applies in that case- indeed, it was designed to. $\endgroup$
    – Rococo
    Jul 15, 2016 at 2:08
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    $\begingroup$ In addition, Lorentz transformations are present in QED with the same form as classical special relativity. $\endgroup$
    – Rococo
    Jul 15, 2016 at 2:10
  • $\begingroup$ @Rococo "And the special theory of relativity certainly applies in that case- indeed, it was designed to." If it would be so, then there would be a "length" of a photon and it would be zero. In addition, the length contraction formula wasn't designed for v=c cases, SR has special formulas for them everywhere. $\endgroup$
    – peterh
    Jul 15, 2016 at 2:16
  • $\begingroup$ @Rococo In addition, afaik there are wave packets and quantum fields in the QED and these are working, not the SR formulas. In the case of photons, for example, there is $\delta_\nu\delta^\nu A = e\overline{\phi}\gamma^\nu\phi$. It seems to me a little more complex as the length contraction formula et el. But write a good answer fast, and I will be happy to upvote it or edit this and I will be happy to accept your edit. The OP asked very cleary from photon particles, and now your critic mentions very clearly wave packets and other QM terminology. $\endgroup$
    – peterh
    Jul 15, 2016 at 2:25
  • $\begingroup$ I cannot improve on the answer given by Alfred in the question I have linked to, which I why I have marked this question as a duplicate. $\endgroup$
    – Rococo
    Jul 15, 2016 at 2:43
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Anything moving at speed of light loses its reference system. This is the reason why your calculation yields strange results. Everything is multiplied by zero so that not only photons (which indeed have no length) but also long distances are reduced to zero.

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  • $\begingroup$ Does this mean that from the theoretical point of view of the photon (inertia frame of the photon) it gets everywhere "instantly"? I mean, any distance the photon travels contracts to zero? $\endgroup$
    – Steeve
    Dec 5, 2018 at 10:45
  • $\begingroup$ @steeve, the spacetime interval of lightlike phenomena (photons in vacuum, fields) is zero. That means that the event of emission and the event of absorption are adjacent in spacetime, even if there are billions of miles between them from our point of view. $\endgroup$
    – Moonraker
    Dec 5, 2018 at 11:41
  • $\begingroup$ This makes me wonder how or where the c "speed limit" arises from. $\endgroup$
    – Steeve
    Dec 5, 2018 at 11:45
  • $\begingroup$ @steeve, you have to distinguish two aspects: on one hand the point of view of observers which is relative, on the other hand the spacetime interval which is observer-independent. Perhaps you noticed that the second postulate of SR does not say that light is moving at c, but only that light is observed as moving at c. This is the whole difference. The absolute spacetime interval is zero. $\endgroup$
    – Moonraker
    Dec 5, 2018 at 12:57

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