# B A Speed of Light Theoretical Question

1. Apr 4, 2016

### science_rules

Let's say a mother ship is traveling at the speed of light. If you shoot a pod out of the mothership at the speed of light(and that pod is already traveling at the speed of light), then wouldn't that pod be traveling at twice the speed of light? And if you have a second pod speeding away from that pod at the speed of light, wouldn't the second pod be traveling at three times the speed of light? If nothing can go faster than the speed of light, then how do you explain this theoretical possibility??

2. Apr 4, 2016

### Orodruin

Staff Emeritus
It cannot, mother ships are massive and therefore travels at sub-lightspeed.

Replacing all of your "speed of light" with "99.999% of the speed of light", the answer is no. You need to look up how relativistic addition of velocities works.

3. Apr 4, 2016

### Staff: Mentor

It is not a theoretical possibility. A ship cannot travel at the speed of light.

However, your question could be modified to make it feasible. Instead of having each one move at c relative to the last, have each move at 0.6 c relative to the last. Then the answer is relativistic velocity addition.

4. Apr 5, 2016

### newjerseyrunner

What you are asking is very similar to what Einstein first envision: a train traveling at the speed of lights turning on it's headlights.

5. Apr 5, 2016

### HallsofIvy

Staff Emeritus
The "relativistic addition of velocities" others referred to is $\frac{u+ v}{1+ \frac{uv}{c^2}}$. If, for example, a ship is moving at 99% the speed of light (relative to us) and fires a bullet ahead it at 99% the speed of light (relative to the ship), then the bullets speed, relative to us is NOT "0.99c+ 0.99c= 1.98c". It is $\frac{0.99c+ 0.99c}{1+ \frac{(0.99c)(0.99c)}{c^2}}= \frac{1.98c}{1+ 0.9801}= \frac{1.98c}{1.9801}= 0.99995c$ or 99.995% the speed of light.

6. Apr 8, 2016

### Battlemage!

Let's pretend your spaceship has zero mass, that way it can move at the speed of light.

If another massless space ship then shoots out at the speed of light relative to first ship at c, and a third and a fourth moving at c, ALL will see the others moving at c (if in opposite direction than -c).

Use the velocity addition formula Hallsoflvy gave, but replace u and v with c. What do you get?

(c+c)/(1+ c2/c2) = 2c/(1+1)
= 2c/2 = c.

This will work all the way through your chain of speeds, as long as in each step you apply THIS velocity addition formula rather than just u+v which, while intuitive, is wrong. (by the way, for speeds much slower than the speed of light, uv/c2 is approximately zero, and that causes the relativistic velocity addition formula to reduce to (u+v)/(1+0) = u+v, and that is why u+v seems to be correct).

If you want to know why u+v is wrong, you're going to have to do some research on the history/origins of special relativity, including the Michelson-Morley experiment and Lorentz transformations versus Galilean transformations. But in a nutshell, this all follows logically from the premises that in inertial reference frames the speed of light is independent of the motion of its source and that all inertial reference frames agree on the laws of physics. Both premises are based on observation.

Last edited: Apr 8, 2016
7. Apr 8, 2016

### Ibix

This is a rather dubious statement. It isn't possible to describe the perspective of anything moving at c - it contradicts Einstein's postulates, since light would have to be stationary with respect to you and it would have to be travelling at c at the same time.

You are correct that, using Halls' formula, any velocity added to c returns c - this is a manifestation of the frame invariance of the speed of light. However, that doesn't mean that adding any two velocities together is necessarily valid. Adding c and c isn't valid because you are implicitly boosting into a frame moving at the speed of light, which isn't possible (see previous paragraph).

8. Apr 8, 2016

### Staff: Mentor

This is not possible. You can't make a spaceship (or a person, for that matter) purely out of massless objects.

It is true that the relativistic velocity addition formula gives $c$ if either of the velocities being added is $c$. But you should just say that instead of trying to "explain" it by postulating something that violates the laws of physics.

9. Apr 11, 2016

### Battlemage!

I should point out the example was just a thought experiment to show the inherent flaw in his example, that his/her use of Newtonian/Galilean velocity addition is wrong. Of course it should have occurred to me that the Lorentz factor is division by zero at v=c but I was more interested in his notion that velocities add the old way.

However, now you have peaked some serious curiosity in me about the principle of relativity and photons.

Since we cannot boost in to a reference frame in which a photon is at rest, wouldn't that mean that photons in a way have a special or preferred reference frame? That the principle of relativity does not apply to photons? If the principle of relativity DOES apply to photons, then the laws of physics are equally valid in the frame of reference of a photon, which means light travels at the speed of light from the perspective of a photon (in the least it is true that the velocity addition formula holds here). But since photons must travel at the speed of light in all inertial reference frames, there cannot be an inertial reference frame in which a photon is at rest and has a proper time (this is what you were saying to me, right?) This seems to be a paradox, which to me suggests that the principle of relativity is not actually universal: it only applies to objects that move slower than light in inertial reference frames.

Since it appears that the frame of reference of a photon cannot be inertial (since there can't be a frame in which the photon is at rest), does general relativity address this issue, since it deals with all kinds of reference frames rather than just inertial ones? I have pretty much zero knowledge of it. I'm reading a book about it and much of the discussion so far is about arbitrary coordinate systems and tensors (which are killing me), but I'm suspecting that locally the same rules have to apply, which makes me nervous about even GR being able to describe such a frame.

You are right. However I think that the impossibility was assumed in his original example.
What I should have done is started with "suppose you have something moving at c with respect to some inertial observer," but still I think that would pose a logical problem now that I think about it.

10. Apr 11, 2016

### Staff: Mentor

The principle of relativity does not apply to objects at all. The principle of relativity states that the laws of physics are the same in all inertial reference frames. So the principle of relativity applies to inertial reference frames and laws of physics.

The laws of physics that govern light are Maxwells equations. They are equally valid in all inertial frames. So the principle of relativity applies fully to light.

11. Apr 12, 2016

### Battlemage!

Thanks for the replies.

But since light itself doesn't appear to have its own inertial frame (that is, photons can't be at rest in an inertial reference frame), would those laws theoretically speaking apply in the reference frame of a photon? Or is it just a meaningless topic to consider?

If photons don't seem to experience proper time (how can they if c2dt2 - dxi2 = 0 ?) would any motion at all from their perspective make sense? Or is there some other way of defining "time" from the perspective of a photon?

12. Apr 12, 2016

### jbriggs444

It is meaningless to consider.

13. Apr 12, 2016

### Staff: Mentor

We can string words together to say "reference frame of a photon", just as we can string words together to say "the even factors of an prime number"... But both strings of words are meaningless, for about the same reason.

14. Apr 12, 2016

### Mister T

The Principle of Relativity asserts that the laws of physics are equally valid in all inertial reference frames. Since you cannot have a reference frame moving at speed $c$, the rest frame of a photon doesn't exist. Such nonexistence doesn't speak to the validity of an assertion that deals with reference frames that do exist.

15. Apr 12, 2016

### Staff: Mentor

I don't know how you can expect to get even a marginally useful response to a self-contradictory question.

16. Apr 13, 2016

### Battlemage!

Well, one thing I asked earlier is whether or not something that deals with non-inertial reference frames has an answer for it. I was doubtful of course.

In any case, no matter what the result of the supposedly useless question in terms of theory, one thing that HAS happened is my own personal understanding has increased. In my opinion, that makes considering such a self-contradictory question quite worth it. You see, prior to these last few posts, I had merely looked at the velocity addition math and noticed that c as an input for both u and v leads to c as an output, and following that math I had an incorrect understanding. However after having this discussion my understanding has been helped. How is this a bad thing?

17. Apr 13, 2016

### Staff: Mentor

I am glad that your understanding has improved. Just try to be cognizant of other people on the forum too.

At that point there is nothing left to say. Once the questioner is aware that the question is self contradictory nothing more can be said. It is frustrating to the respondents to be placed in such a position.

18. Apr 13, 2016

### Staff: Mentor

OK, so the principle of relativity for SR refers specifically to inertial reference frames. So my further comments have no bearing on the principle of relativity as discussed above.

It is certainly possible to make coordinate systems in which the axes are lightlike worldlines. One example is called light cone coordinates. You can write Maxwell's equations in those coordinates if you wish, but they will be decidedly different.

Usually in SR there is usually little distinction made between coordinates and reference frames because the coordinate basis of inertial coordinates is an inertial frame. However, here is an example where the distinction does need to be made. The coordinate basis of light cone coordinates is not orthonormal, so it is not a valid reference frame (inertial or not)

19. Apr 15, 2016

### Battlemage!

I now realize that what I was asking about could be fodder for quacks. Shame on me for that, haha. The direction my mind was going though was related to the notion for momentum of light, despit the notation for massive particles also being infinity at c=v (obviosly there is no problem, but when I first learned of relativistic momentum I was similarly confused about how light could have momentum)

In any case I looked up light cone coordinates and I am simply not ready to investigate that. I can barely follow tensor notation for a metric at this point, and it looks like understanding a metric and notation for it is the very basics here. I do appreciate you showing me the direction to look on this topic, however.

20. Apr 16, 2016