# Moving at c

1. Aug 3, 2011

### DonB

I'm a relative newbie to relativity (no pun intended -- I know you've heard that one too many times), as well as to this forum, so forgive me if this is a dumb question....

As I understand it, there is a significant percentage of those that believe that people, spaceships, etc. will never travel at the speed of light (c), right?

If that is so, where is my logic (below) faulty:

1. For any two given things (A and B), the speed of A relative to B will always be the same as the speed of B relative to A.

2. Relativity insists that light always travels at c relative to all things.

Thus, if light is moving at c relative to me, then am I not moving at c relative to that particular beam of light (...and, in fact, all beams of light at any given moment in time)? And thus, the speed of light is not only something that is not unattainable, but has rather never been unattained?

What am I missing?

2. Aug 3, 2011

### Pengwuino

It does not make sense to speak of "what does a photon see?" and thus, "what velocity does a photon see". The idea of something observing a velocity depends on two things; the ability to measure distances and the ability to measure times. Neither of these things can be done for photons.

3. Aug 3, 2011

### I like Serena

Welcome to PF, DrDon!

This very reasoning had been confusing physicists.
Then Einstein and others came with the revolutionary idea to say, what if c is constant for any observer?
Through what hoops do we have to wring the mathematics of it (Lorentz transformations), to make it work?

There! Relativity theory!

Last edited: Aug 3, 2011
4. Aug 3, 2011

### Staff: Mentor

When we say that A is moving at some velocity relative to B, we mean that A is moving at that velocity in a reference frame in which B is at rest. However, if B is a light beam or photon, there is no such (inertial) reference frame! Light travels at speed c in any inertial reference frame. So the general consensus is that it's meaningless to talk about the velocity of something relative to a photon or light beam.

Note also the following FAQ in the FAQ section at the top of this forum:

Rest frame of a photon

5. Aug 3, 2011

### DonB

If I may interact, without intending (or appearing) to argue....

Your response seems to make relativity equal to perception -- what does something "see". But Einstein's arguments seem to indicate other he didn't feel this way, for he often referenced movement relative to the ground, a train, etc. If his references are valid, then why is it not equally valid to ref. to another inanimate object like light?

6. Aug 3, 2011

### DonB

Thank you for the welcome.

That's what I'm saying..., if c is constant to any observer, then isn't each observer moving at that constant c relative to the light?

7. Aug 3, 2011

### ZapperZ

Staff Emeritus
Please start by reading the FAQ subforum in the Relativity forum.

Zz.

8. Aug 3, 2011

### WannabeNewton

You cannot lorentz boost to a photon's reference frame so you cannot say you are moving at 'x' relative to the photon because for that you would need to have lorentz boosted to the photon's rest frame which doesn't exist and the lorentz transformations fail in the process.

9. Aug 3, 2011

### DonB

Thanks for the response. If I may interact with your ideas, without intending (or appearing) to argue that I'm right (just trying to understand)....

This doesn't seem to match what I've read by Einstein. He gives examples of referencing movement of men who are walking or moving on trains (e.g., walking 3 MPH on a train moving 60 MPH).

Furthermore (I am asking in ignorance), do we really know that light is not at rest relative to its own frame of ref.?

Interesting. How do we know that? Is it just that it's a "theoretical" impossibility (which assumes all the thought supporting the theory is absolutely sound), or is there experimental evidence from which this is drawn?

Interesting thought (just thinking out loud).... If one accepts this, and yet one has yet to prove by experimentation that light has no ref. frame, then arguably he has to allow the possibility that relativity requires that light must travel at c relative to itself. Interesting indeed!

I thought I had read all the FAQ's before, but I may have missed this one. I'll check it out.

Thanks!

10. Aug 3, 2011

### DonB

Thanks Zz. As I explained above, I thought I had. Apparently I missed a section of it. My apologies.

11. Aug 3, 2011

### nitsuj

Who would agree that you are moving at c? It is c that is constant, your velocity varies. But again, even though your velocity changes you will always measure c as constant.

If that is your point, what is the consequence of saying as far as EM waves/photons go, you have always moved at c and they (photons) NEVER move? How long does that thought hold up for you? What movement do you perform when lights shine on the left and right of you, from the photons perspective ofcourse

12. Aug 3, 2011

### DonB

Thanks for pointing this link out. DH does a great job explaining this fascinating concept.

I know this will give me major points on the crackpot scale, but that's not where I'm coming from. I'm nothing more than a newbie genuinely trying to understand (and who can't find anyone locally who knows beans about this to help me out). DH concludes that trying to grasp the concept of a rest ref. frame for a photon is "undefined" and "indeterminate" (which I understand and do not question). But is it legit to resign ourselves to the conclusion that undefined and indeterminate means non-existent? Can I legitimately say that because I do not understand X=3/0 -- which is undefined -- that there is no expression "X=3/0"? In my naivety I understand "undefined" and "indeterminate" to speak of my personal inability (specifically, to wrap my mind around something), but not necessarily the inability of the item in question to exist. Am I missing something?

Thanks again for the link.

13. Aug 3, 2011

### DonB

I don't know; that's why I'm asking in the first place.

But that first requires that we determine what the velocity is referenced to, right? Does my velocity vary in ref. to a beam of light, which as you say, is always constant?

No, it isn't my point.

I have moved, and they have not moved...., relative to what? I probably misunderstand, but your question seems to argue for some absolute point of ref. that (in my understanding) relativity argues against. I thought that relativity argues that for two entities that are moving in ref. to each other, it is impossible to determine which is moving and which is at rest -- or, if in fact both are moving.

None; but then again, there is no movement as I pass a friend as we both float through space. Am I moving? Or is he moving? Relativity seems to argue that answering that question is impossible without first determining the point of ref. I'm simply apply the same relativity principles to me passing a beam of light out in space...., and asking if there is some reason that that is different.

14. Aug 3, 2011

### nitsuj

1.) it is for you to think about, if no one would agree you are moving at c, then you are not moving at c

2.) Yes in "reference" to the beam of light. You are suggesting that is what determind you move at c all the time. I am suggesting that isn't true, since you can move at different velocities and c doesn't.

3.)okay

4.) Relative to each other, you suggested a photon can be used as a frame of reference, which can't be done. I offered the thought of having a light shine on the left and right of you means you travelled towards the photons at c, while the light source travelled away at c, and that this happened on both sides of you at the same time.

5.)"I'm simply apply the same relativity principles to me passing a beam of light out in space...., and asking if there is some reason that that is different." Ah okay, so you now know you can't. That photon reference frame FAQ explains pretty clearly why it doesn't makes sense to use a photon as a rest frame.

I recently learned the difference between coordinate time and proper time. That helps me understand why a photon is not remotely avaible in coordinate time as a frame of reference for something that is subject to proper time. Turining the idea on it's head means nothing and despite that being a great "tool" for conceptualizing SR/GR, it can't be used for the laws of physics itself.

Laslty. EM is massless and HAS TO move at c, you are not massless and have to not move at c.

Lastly lastly, I'm a relative, relativity "newbie" too. It's fun stuff for sure!

Last edited: Aug 3, 2011
15. Aug 3, 2011

### DonB

1) So, you're saying that the reality of my speed is totally dependent upon whether others will agree that I am moving? Sounds to me like reasoning that I can't know how fast I'm traveling in my Jeep without getting someone else to tell me the speed.

2) No, that is not what I'm suggesting. Two rocks traveling through space are going a given speed relative to each other, even though neither one "determined" the speed of the other. As long as the two rocks exists, they are going a given speed relative to each other, and that speed is the same for A relative to B, or B relative to A. So, if we can validly claim that a beam of light is going at c relative to me, then my question is why can we not similarly say that I am moving at c relative to the light? As the photon passes by me, does it not "see" me passing by it at c?

4.a) I'm not arguing that your statement is wrong; but I am wanting to know how we know that it is right -- the 'why'.

4.b) Sorry, I just don't follow you.

5) Begging to differ in my ignorance, but as I spelled out in an earlier post, I'm missing how we can legitimately jump from something being "undefined" to conclude that it is "non-existent."

16. Aug 3, 2011

### WannabeNewton

The reason you can say that is because you can freely lorentz boost from B to A and A to B so that you can evaluate relative speeds from the rest frames of A and B. You cannot lorentz boost to the frame of a photon so how do you suppose the same statement carries over?

17. Aug 3, 2011

### pervect

Staff Emeritus
You can assume that a photon has a reference frame like you're used to, or you can assume that Maxwell's equations work.

You can't do both.

Maxwell's equations are not consistent with having an unchanging electric field and an unchanging magnetic field just propagating through space. Which is what you'd get if you assumed a photon, somehow, had a reference frame.

People looked for along time at how to fix up Maxwell's equations without success, but there didn't appear to be anything wrong with them.

So, Einstein realized that the problem was ssuming (based on no particular necessity to do so) that a photon had a reference frame in the first place. The problem was solved.

Once it's been pointed out that it leads to a contradiction to assume that a photon has a reference frame, it seems to me to be not particularly productive to ask "why". It's a classic reducto ad absurdum proof - assume A, show that you get a contradiction, then you realize that A must be false.

It's only slightly more complicated here, because the proof that Maxwell's equations do in fact work is experimental.

18. Aug 3, 2011

### ghwellsjr

If you are A and a particular beam of light is moving at c relative to you and I am B and that same beam of light is moving at c relative to me and you and I are moving relative to each other, then how can we both be moving at c relative to that beam of light?

19. Aug 3, 2011

### DonB

Good question. But the same thing can be equally said about turning the question around and asking how can light be moving at a constant c relative to both of us when we are moving relative to each other..., yet it is accepted nonetheless. So my question becomes, why accept the latter as a given while thinking the former impossible and unworthy of consideration?

20. Aug 3, 2011

### DonB

Thank Pervect. This is the kind of "why" I was looking for. Now off to research what all Maxwell says.

21. Aug 3, 2011

### ghwellsjr

Are you aware that in the Theory of Special Relativity, the speed of light is only c for observers at rest within a Frame of Reference and that for moving observers the speed of light is not c?

22. Aug 3, 2011

### PAllen

What? The speed of light is c for all inertial observers (in vaccuum, etc). For accelerating observers (e.g. inside a rapidly accelerating rocket), it is still c as a local measurement. Nonlocal measurements for an accelerated observer run into the same issues of coordinate choice and conventions as GR.

What is 'at rest in a frame of reference'?

23. Aug 3, 2011

### nitsuj

What!??

ghwellsjr You have a unique ability to point out these blantant details in SR/GR that I miss.

I have always thought c is constant, and signed off on that as being...well...constant when perceiving things sr/gr.

What is a scenario that a moving observer wouldn't measure c to be constant?

24. Aug 3, 2011

### nitsuj

I won't be able to even remotely define for you why this is the way physics explains it i.e. conclude it's undefined/non-existant ect. But I will say from my understanding there is little difference between c and proper time. Given that, you suggest it is you who exists purely in the "time dimension" and that EM is in 3D.

see how physics conlcudes one of those two perspectives you offer is right.

25. Aug 3, 2011

### WannabeNewton

Well the speed of light is measured as c in inertial frames, or locally flat space i.e. where the general metric reduces to the minkowski metric, so global measurements wouldn't give c because you can't even define velocity properly globally due to the path dependence of parallel transport.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook