Speed of Light at near C speeds

[Lavar]

Hey guys this question is a bit basic but I always get caught up on the idea, I'm not in school for physics so I don't really have a prof to ask and most of the internet always skips ahead of this.

Say I am in an inertial frame traveling close to C speed, the only thing I see is a photon in-front of me. In this reference frame I couldn't distinguish between my velocity being close to C or me just being at rest, so how in this reference frame do I come to the conclusion that the speed of that photon is = c and not just a few m/s?

I'm assuming this has to do with time dilation but I don't know how to visualize that in this scenario.

Thanks for the help!

 

[PeterDonis]

Say I am in an inertial frame traveling close to C speed

There is no such thing in an absolute sense. There is only a frame traveling at close to C relative to some other frame.

how in this reference frame do I come to the conclusion that the speed of that photon is = c and not just a few m/s?

By understanding how velocities add in SR. See here:

https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

[Drakkith]

Say I am in an inertial frame traveling close to C speed, the only thing I see is a photon in-front of me.

For starters, you cannot see a photon in front of you. You have to interact with it, and thus destroy it, to detect it. But the crux of your question has to do with the idea that light travels at c for all inertial observers. If you set up an experiment to measure the velocity of a pulse of light (consisting of many photons), you will always find that it travels at c, no matter what your velocity is relative to any other object. It doesn't matter what the source's velocity was with respect to you, or how long you accelerated before prior to doing this experiment, it will always yield the same value for the speed of light: c.

Now, the frequency of the light may change depending on the relative velocity between you and the source, but the velocity will not.

 

[Lavar]

Dont get me wrong I understand that C is constant etc, but when people refer to light they say it will be constant no matter the reference frame just like you did. I'm really trying to ask in this specific reference frame if I could "see" the photon moving away at only a few m/s away from me and I have nothing else to reference to and hence in the formula s= (v+u)/(1 + (uv/c^2) I have no way of differentiating between my v = 0 or my v = nearly C. And hence S = a few m/s or the speed of light, or really anything in between as my speed could be anything in between..

 

[Drakkith]

I'm really trying to ask in this specific reference frame if I could "see" the photon moving away at only a few m/s away from me and I have nothing else to reference to and hence in the formula s= (v+u)/(1 + (uv/c^2) I have no differentiating between my v = 0 or my v = nearly C. And hence S = a few m/s or the speed of light, or really anything in between as my speed could be anything in between..

No. The light will move away from you at c.

 

[Lavar]

Yes the photon is moving at C, but in this frame of reference how can I prove that? It may be moving at C but its only gaining its (velocity - my velocity) of distance/time from me.

 

[PeterDonis]

It may be moving at C but its only gaining its (velocity - my velocity) of distance/time from me.

No, it isn't. It is gaining distance/time equal to C from you, in your frame. Look at the relativistic velocity addition formula I linked to before.

 

[Lavar]

Ugh I feel bad for continuing this thanks for trying to explain.

But say I use the formula under the assumption my v=0, then the speed of light=the velocity is gaining away from me. Or am I missing something?

(edit = assuming v=0 because I have no way to find what v is because I have no other reference.)

 

[PeterDonis]

say I use the formula under the assumption my v=0, then the speed of light=the velocity is gaining away from me.

I don't think you're understanding the formula correctly. $v$ is the speed of the thing you are observing--in this case the light--relative to the original frame. So for light we must have $v = c$. $u$ is the speed of your frame relative to the original frame, so in this case $u$ would be some speed close to $c$ but not quite at it. But you should be able to see that, if $v = c$, then $s$, the speed of the light relative to your frame, will also be $c$, regardless of what $u$ is.

 

[Lavar]

Thanks for your help, I got a little caught up and forgot that due to velocity being in my frame, hence time must pass and hence can compare initial to final.

Thanks a lot!

 

[DrGreg]

It may be moving at C but its only gaining its (velocity - my velocity) of distance/time from me.

To emphasize what's already been said, it's not gaining $velocity - myvelocity$ from you, it's gaining$$\frac{velocity - myvelocity}{1 - \frac{velocity \times myvelocity}{c^2}}$$where $velocity = c$.

 

[Battlemage!]

If the fact that light moves at c is an electromagnetic law, and the first postulate of special relativity says that the laws of physics are the same for all inertial reference frame, it would seem like a massive contradiction for you to be able to "move fast enough" so that you see light traveling at a few m/s. If you are in an inertial reference frame, then the same laws of physics must apply in your frame, which means light must travel at c according to you.

I don't know if the First Postulate is sufficient to get the results of SR (aside from implicit assumptions like the direction you move in space does not affect the laws of physics), but it seems if you look at it this way, that the Second Postulate is more like a resulting theorem than a postulate, which would mean it is sufficient (as long as you consider the electromagnetic laws as true).

And if that's so then simply assuming that the laws of physics apply to all inertial reference frames kind of short circuits the possibility of ever moving fast enough for light to move at anything other than c in your reference frame.Anyway, it seems to me that the only logical possibilities regarding whether or not "catching a beam of light" is possible are either (a) the speed of light will always be c in your inertial frame no matter "how fast" you are moving AND time/length/simultaneity will not be universal to all inertial frames OR (b) time/length/simultaneity are the same for everyone but the speed of light may vary depending on "how fast" you are moving (or it is infinite). It can't be both, however, which is what OP seems to be suggesting early on (agreeing that c is constant in all inertial frames and also entertaining the notion that you could "move fast enough" so that c is only a few m/s to you is completely contradictory, it seems to me).*"fast" phrases in parenthesis because how fast you move depends on a reference point (principle of relativity). Are you at rest? But Earth rotates at about 1000 mph. Moving at 1000 mph? But Earth orbits the sun. But the Sun orbits the galaxy. But the galaxy moves with respect to other galaxies. But the galaxy cluster moves. Etc.

 

[vanhees71]

Well, let's do a little calculation, but first of all forget photons for now. They are not little massless billard balls in any sense of the word. You cannot even define a position for them in the usual sense. Rather think about classical electromagnetic waves, e.g., light. A plane-wave packet in the vacuum moves with the speed of light with respect to any inertial reference frame. The wave vector of this plane wave $\vec{k}=\frac{2 \pi}{\lambda} \vec{e}_1$, where $\vec{e}_1$ is the direction of the wave propagation and $\lambda$ its wavelength (we assume that the wave packet has a spectrum narrowly peaked around this wave vector), builds together with the frequency $\omega=c |\vec{k}|$ a light-like four-vector
$$k^{\mu}=(k,k,0,0).$$
Now suppose you switch to the reference frame of another inertial observer running with speed $v$ in direction of $\vec{e}_1$. Such an observer describes the wave-four-vector via the Lorentz boost
$$k^{\prime \mu}={\Lambda^{\mu}}_{\nu} k^{\nu}.$$
Here it's sufficient to only look in the $k^0$-$k^1$ plane of Minkowski space time there you have with $\beta=v/c$
$$\begin{pmatrix} k^{\prime 0} \\ k^{\prime 1} \end{pmatrix}=\frac{1}{\sqrt{1-\beta^2}} \begin{pmatrix} 1 & -\beta \\ -\beta & 1 \end{pmatrix} \begin{pmatrix} k \\ k \end{pmatrix} = \frac{k(1-\beta)}{\sqrt{1-\beta^2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = k \sqrt{\frac{1-\beta}{1+\beta}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$
As you see, indeed the observer also sees a plane wave packet, but with a wave number $k'=k \sqrt{(1-\beta)/(1+\beta)}$, i.e., the wave is red or blue-shifted (depending on the sign of the relative velocity $v$ between the two reference frames) but still fulfills the dispersion relation $k'=c \omega'$ as it must be, and the speed of light stays the same also with respect to the new reference frame. That's, how the Lorentz transformation is derived after all! For more details on Special Relativity, see

http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Lavar]

Thanks a lot for all your responses! Got my head of the gutter and look at it all from a different point of view

You cannot even define a position for them in the usual sense.

Its just easier to say when looking at this question.

 

[Nugatory]

Its just easier to say when looking at this question.

You can always say "radio signal" or "flash of light" instead of "photon".