Constant Speed of Light: Low Speed Explained

[grounded]

I understand the relativistic effects at high speeds, what accounts for the constant speed of light at low speeds?
Example... If I travel towards a beam of light at 25 MPH, I will still measure the light as traveling towards me at the speed of light and not the speed of light plus 25 MPH.

 

[A.T.]

I understand the relativistic effects at high speeds, what accounts for the constant speed of light at low speeds?

It's the same for all speeds.

 

[Herman Trivilino]

I understand the relativistic effects at high speeds, what accounts for the constant speed of light at low speeds?

All relativistic effects occur at all speeds. It's just that the departure from Newtonian physics gets greater the higher the speed, but it is always there at all speeds, even if it's too small to measure with current technology. As our technology improves so does our ability to detect it at ever lower speeds.

 

[PeroK]

I understand the relativistic effects at high speeds, what accounts for the constant speed of light at low speeds?
Example... If I travel towards a beam of light at 25 MPH, I will still measure the light as traveling towards me at the speed of light and not the speed of light plus 25 MPH.

The velocity transformation formula is:$$u' = \frac{u + v}{1 + \frac{uv}{c^2}}$$Where $u$ is the speed being transformed from one reference frame to another with relative velocity $v$. If we take $u = c$, then:
$$u' = \frac{c + v}{1 + \frac{v}{c}} = c\frac{1 + \frac{v}{c}}{1 + \frac{v}{c}} = c$$And, we see that the speed $c$ is invariant for any relative velocity $v$ - even if $v$ is only $25 \ mph$.

 

[PeterDonis]

I understand the relativistic effects at high speeds, what accounts for the constant speed of light at low speeds?

There is no such thing as "high speeds" or "low speeds". There are only high or low speeds relative to something. Right now you are moving at "high speed" relative to cosmic ray particles that are bombarding the Earth or neutrinos coming from the Sun.

In short, no object has just one "speed" so it is meaningless to ask whether relativistic effects happen "at low speeds" as well as "high speeds".

 

[Ibix]

I understand the relativistic effects at high speeds, what accounts for the constant speed of light at low speeds?

Just to expand a little on what PeroK wrote, the formula he gave is correct at all speeds, but approximates to the familiar Galilean $u'=u+v$ when both $u$ and $v$ are much less than $c$. But you're talking about light speed, which is not less than light speed - i.e., one or other of $u$ and $v$ is not much less than $c$. So you can't use the Galilean approximation for light, ever, whatever the speed of the observer.

 

[grounded]

Thank you for the replies, however that’s not exactly what I was looking for, but I am aware of all of that, at least I think I am. My social skills are not the best, let me rephrase my question….

If I were to calculate the speed of light by measuring the frequency and wavelength of light coming from a source that is at rest relative to me, I will get the speed of light.

If I travel towards the source at a slow 25 MPH, the frequency and wavelength will have to change in a way that still equals the speed of light. Since the relativistic effects are non-linear, how does it account for the linear change in speed? If I travel at 1 MPH towards the source, the speed of light must change 1 MPH relative to me, and 2 MPH for 2 MPH and so on. What am I missing?

 

[Ibix]

If I travel towards the source at a slow 25 MPH, the frequency and wavelength will have to change in a way that still equals the speed of light.

The relativistic Doppler effect formula is easily Googled.

If I travel at 1 MPH towards the source, the speed of light must change 1 MPH relative to me, and 2 MPH for 2 MPH and so on.

No - this is exactly what @PeroK showed you. The speed of light does not change when you apply the correct velocity addition formula. Whatever speed you do, the speed of light remains invariant because velocities do not add as you are saying they do here.

 

[PeterDonis]

If I travel towards the source at a slow 25 MPH, the frequency and wavelength will have to change in a way that still equals the speed of light.

Yes. They do. Have you looked at the math?

Since the relativistic effects are non-linear, how does it account for the linear change in speed?

Instead of waving your hands, you should look at the math.

 

[PeroK]

What am I missing?

A knowledge of SR?

 

[vanhees71]

The relation between the frequency and wave vector of em. waves in a vacuum can be written in terms of a relatcistically covariant formula,
$$\omega^2/c^2-\vec{k}^2=0.$$
This means that in all inertial frames $\omega/|\vec{k}|=c$, i.e. the phase velocity is always the same, no matter what's the velocity of the light source.

 

[jartsa]

Clocks and rulers account for the measurement result being c, when measuring the speed of light.

At low speeds clocks are much more important than rulers. I mean clocks and rulers bolted on a slowly moving lab. Motion of the lab has a small effect on the length of the ruler, large effect on the synchronization of the two clock at the ends of the ruler.

 

[PeterDonis]

Motion of the lab has a small effect on the length of the ruler, large effect on the synchronization of the two clock at the ends of the ruler.

Motion of the lab relative to something else has no effect at all on measurements made by the lab itself.

 

[jartsa]

Motion of the lab relative to something else has no effect at all on measurements made by the lab itself.

If we want to answer the question that was asked, we need a moving measuring equipment. And moving rods are contracted.

Well I am just guessing here that it's the measuring system that is moving at the slow speed that OP mentions.

Oh yes, OP say that OP moves and measures. So there must be an observer that observes OP moving and measuring, and the question is why according to this observer the result of OP's measurement is c.

 

[PeterDonis]

OP say that OP moves and measures. So there must be an observer that observes OP moving and measuring, and the question is why according to this observer the result of OP's measurement is c.

And the answer is that the OP's measurement result is an invariant, because, as I said, motion of the OP and his measuring devices relative to something else has no effect at all on the OP's own measurements. It has nothing whatever to do with any length contraction or time dilation that someone else sees the OP having.

The OP later, in post #7, clarified his question, saying his concern is motion relative to the light source. That is a different question and has a different answer than the one I gave above (and several posters have given the answer in response to post #7), but the answer still has nothing whatever to do with how the OP is moving relative to some other observer.

 

[Sagittarius A-Star]

If I were to calculate the speed of light by measuring the frequency and wavelength of light coming from a source that is at rest relative to me, I will get the speed of light.

If I travel towards the source at a slow 25 MPH, the frequency and wavelength will have to change in a way that still equals the speed of light. Since the relativistic effects are non-linear, how does it account for the linear change in speed? If I travel at 1 MPH towards the source, the speed of light must change 1 MPH relative to me, and 2 MPH for 2 MPH and so on. What am I missing?

You are missing to choose the longitudinal relativistic Doppler formula over the classical Doppler formula.

The Lorentz transformation for energy is

$E' = \gamma (E - \frac{v}{c} * cp_x)$

With $cp_x=c \frac{h}{\lambda} =hf = E$ for a photon moving in x-direction follows:

$f' = f \gamma (1 - v/c) = f \frac{1}{\gamma (1 + v/c)}$

$\lambda ' = \lambda \frac{1}{\gamma (1 - v/c)}= \lambda \gamma (1 + v/c)= c / f'$

The time-dilation factor $\gamma = 1/\sqrt{1-v^2/c^2}$ makes it possible, that the speed of light is the same in both frames without a contradiction to the principle of relativity. You are wrongly assuming Newton's absolute time.

In your example, it the source is at rest in the primed frame and you in the unprimed frame, and $v= +1\text{MPH}$, you would measure:

$f = f' \frac{1}{\gamma (1 - v/c)}$
$\lambda = \lambda' \gamma (1 - v/c)$

 

[vanhees71]

For an elementary treatment of the Doppler effect for light (em. waves) in a vacuum, see Sect. 4.2.1 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

For the more general case for em. waves in dispersive media or other waves (like sound waves) as far as normal dispersion is concerned, see

https://itp.uni-frankfurt.de/~hees/pf-faq/rela-waves.pdf

 

[Herman Trivilino]

If I were to calculate the speed of light by measuring the frequency and wavelength of light coming from a source that is at rest relative to me, I will get the speed of light.

If I travel towards the source at a slow 25 MPH, the frequency and wavelength will have to change in a way that still equals the speed of light.

That is correct.

If I travel at 1 MPH towards the source, the speed of light must change 1 MPH relative to me, and 2 MPH for 2 MPH and so on. What am I missing?

You're contradicting yourself. First you say the speed of light doesn't change, then you say it does!

What you are missing is the foundation and the logic used to derive the consequences. The speed of light is the same for all observers. That's part of the foundation. You seem to accept the consequences, such as length contraction and time dilation, but you don't seem to want to accept the foundation.

 

[grounded]

Thanks guys for all the replies and help, I got what I was looking for.
You can close this thread.