I How can the Universe know when you change directions?

[jnhrtmn]

If you and your frame are following a square path in space, and the Universe is the other frame, each leg of the square path instantly changes every photon in the Universe to suit your dimensional perspective of speed, length, frequency, and time. Is this new light situation actually different light from a completely different time dimension or are they altered somehow? If altered, how does the Universe instantly know throughout that you changed directions? If it's a different time dimension, then are there quantum time steps, or is it continuous somehow?

 

[FactChecker]

If you and your frame are following a square path in space, and the Universe is the other frame, each leg of the square path instantly changes every photon in the Universe to suit your dimensional perspective of speed, length, frequency, and time. Is this new light situation actually different light from a completely different time dimension or are they altered somehow? If altered, how does the Universe instantly know throughout that you changed directions? If it's a different time dimension, then are there quantum time steps, or is it continuous somehow?

Your time and space are related to how you measure them. Different observers traveling at different inertial speeds will measure the time and space between the same events differently. It is not something that the "Universe instantly knows". It is just something that your measurements in your inertial reference frame do automatically.

 

[jnhrtmn]

Your time and space are related to how you measure them. Different observers traveling at different inertial speeds will measure the time and space between the same events differently. It is not something that the "Universe instantly knows"

So, you just think the light is entering your eyes at one speed as a matter of opinion, and the rest of the light beam didn't actually change to get to you in the condition that it's in?

 

[phinds]

So, you just think the light is entering your eyes at one speed as a matter of opinion, and the rest of the light beam didn't actually change to get to you in the condition that it's in?

That is not even remotely what he said or implied.

 

[Ibix]

you and your frame are following a square path in space

What are you meaning by "your frame" here? Because in the usual SR meaning of "frame", "frame...following a square path" is nonsense.

Is this new light situation actually different light from a completely different time dimension or are they altered somehow?

Basically, the frequency you measure light to have depends on your state of motion. It isn't a property of light, but rather a property of the interaction between you and the light.

Similar statements can be made in Newtonian physics. For example, if I have a fixed speaker emitting a steady sine wave, microphones at rest and in motion will hear different tones.

 

[FactChecker]

So, you just think the light is entering your eyes at one speed as a matter of opinion, and the rest of the light beam didn't actually change to get to you in the condition that it's in?

Remember that when you measure the speed of something entering your eye, you must measure its time and position at two different points -- before your eye and at your eye. Since those measurements of time and distance are yours, others might legitimately (for them) disagree.
Also remember that one observer can measure the closure rate of two objects as being greater than the speed of light. He can not measure the speed of either object relative to him as being greater than the speed of light.

 

[phinds]

If you and your frame are following a square path in space, and the Universe is the other frame,

You would likely find it helpful to learn what a "frame of reference" really is and how they work. I think your misunderstanding about them is inhibiting your understanding.

EDIT: I see @Ibix beat me to it.

What are you meaning by "your frame" here? Because in the usual SR meaning of "frame", "frame...following a square path" is nonsense

 

[jnhrtmn]

The straight lined path of the square sets your inertial frame. When you turn left, new leg, new frame compared to the frame of the Universe. You see light at c, so new path, same c, my question was about the entire beam of light traveling to you. It travels to you, arrives at c only, and frequency information is in the beam somewhere, but the beam does not change to enter your eyes at c. Where does the Doppler shift come from, time?

 

[DaveC426913]

The straight lined path of the square sets your inertial frame.

This sentence is nonsensical. We don't know what this square is. Please stop using it.

When you turn left, new leg, new frame compared to the frame of the Universe. You see light at c, so new path, same c, my question was about the entire beam of light traveling to you. It travels to you, arrives at c only, and frequency information is in the beam somewhere, but the beam does not change to enter your eyes at c. Where does the Doppler shift come from, time?

OK, let's back up to some more basic concepts for a second.

Do you understand how Doppler shift in sound works?
If you are driving along parallel to a train that is blowing its horn, you will hear a certain frequency.
If you turn 90 degrees and drive toward the train the frequency will rise. This is not because the air is now moving toward you at a different speed; it is because you are gaining on the pressure waves of the air.

So it is with light, although the speed of light does not change, you encounter the pulses more frequently, which translates into a higher frequency of light.

 

[Dale]

The straight lined path of the square sets your inertial frame. When you turn left, new leg, new frame compared to the frame of the Universe.

If I understand your scenario, there is some inertial frame in which your spatial path forms a closed square. I believe that you intend that along this path you travel at constant speed with abrupt acceleration at each corner and with no proper acceleration otherwise.

If so, then your frame is not inertial. There are four different inertial frames in which your frame is temporarily at rest.

Are you asking to analyze things from your non-inertial frame or from one or more of the inertial frames?

Also, what is the frame of the universe? Please provide a reference for what you mean by that.

 

[Nugatory]

.... the frame of the Universe.

There is no such thing, and reasoning as if there is such a thing pretty much guarantees confusion and logical inconsistencies following from the flawed initial premise.

The non-existence of the "frame of the universe" isn't an Einstein relativity thing, it was first noted by Galileo and is present in classical physics as well (Google for "Galilean relativity").

 

[Ibix]

Here's a quick image of a wave and two observers:

The wave is moving to the right, as is the blue observer. The red observer is moving to the left. The image is drawn as they pass one another, which happens to be the same time as they pass the crest of a wave.

Now here's a picture drawn a moment later:

The wave has moved to the left by a long distance, marked by the grey arrow. The red and blue observers have moved the same distance as each other, but a shorter distance than the wave. This is marked by the red and blue arrows. But now look at where they are compared to the wave. The red observer is at the next crest, so has seen a full cycle of the wave go past. The blue observer isn't at the next crest, and has seen less than a full cycle of the wave go past. So they are seeing different numbers of cycles of the wave go past in the same time - they are seeing different frequencies.

Note that I haven't invoked relativity anywhere. This is simply a fact about waves and independent of the underlying model of the wave: the frequency depends on who is doing the observing. If two people observing the same wave and finding different frequencies bothers you then you have fundamental problems in your understanding of physics that you should probably fix before attempting to make sense of relativity.

 

[jbriggs444]

The straight lined path of the square sets your inertial frame. When you turn left, new leg, new frame compared to the frame of the Universe. You see light at c, so new path, same c, my question was about the entire beam of light traveling to you. It travels to you, arrives at c only, and frequency information is in the beam somewhere, but the beam does not change to enter your eyes at c. Where does the Doppler shift come from, time?

Have you ever used a paper map? If you have a map of the continental United States and rotate it by 90 degrees, do you expect that the people in New York City have to hang onto light posts to avoid being thrown off their feet?

When you rotate a frame of reference, it is like rotating a paper map. Nothing happens to the territory.

 

[jnhrtmn]

You are using the same wave train for two different observers here, each seeing c. I will have to come back with some accounting.

 

[Ibix]

View attachment 348739
You are using the same wave train for two different observers here, each seeing c.

Yes. A better diagram would be a Minkowski diagram, which incorporates the correct relativistic transforms. This is not incompatible with that - in fact the diagrams are essentially a pair of one-dimensional sections through a Minkowski diagram, but with added art.

 

[Ibix]

If you want to do the maths it's quite simple.

Start in the frame where the two observers are moving in opposite directions with velocities $\pm v$. At time zero, a wave crest and both observers meet at $x=0$.

The wave crests are generally at $x=ct + n\lambda$ where $n$ is an integer. One observer is at $x=vt$ and the other at $x=-vt$.

Now let's have a look at the light as measured using the frame where the observer moving at $+v$ is at rest. The path of the $n$th wave crest was $x=ct+n\lambda$, which Lorentz transforms to $$\begin{eqnarray*}
x'&=&\gamma(x-vt)\\
&=&\gamma(ct+n\lambda-vt)\\
&=&\gamma((c-v)t+n\lambda)
\end{eqnarray*}$$We don't want that dependence on unprimed quantities, so we can use the inverse Lorentz transform $t=\gamma(t'+vx'/c^2)$ to eliminate $t$, then solve for $x'$, which turns out to give$$x'=ct'+n\sqrt{\frac{c+v}{c-v}}\lambda$$We can compare this to the original expression for the positions of wavecrests in the unprimed frame. Clearly the speed is still $c$, but the spacing between successive crests has changed from $\lambda$ to $\lambda\sqrt{(c+v)/(c-v)}$ - the wavelength in this frame is longer. Clearly if the speed is the same in both frames but the wavelength isn't then the frequency is also different.

You can do the equivalent computation for the frame where the observer moving at $-v$ is at rest. This time you'll get the light path to be $x''=\gamma((c+v)t+n\lambda)$, eliminate $t$ using $t=\gamma(t''-vx''/c^2)$, and get the equation for the positions of the light crests being$$
x''=ct''+n\sqrt{\frac{c-v}{c+v}}\lambda
$$which differs from the previous one only by where the signs are inside the square root. Again, the speed of light is the same but the wavelength has changed - this time it is shorter.

 

[PeterDonis]

each leg of the square path instantly changes every photon in the Universe

No, it doesn't. The rest of the Universe doesn't change when you change direction. All that changes is you. Light that happens to be passing you when you change direction will look different to you (its relativistic Doppler shift relative to you will change), but that's because of the change in you, not any change in the light.

Is this new light situation actually different light from a completely different time dimension

This is word salad and makes no sense.

how does the Universe instantly know throughout that you changed directions?

It doesn't. The spacetime of the Universe does not change. Only your state of motion through that spacetime changes.

 

[phinds]

And do you think your accounting will change the fact that both see the light coming at them a c but that they see different frequencies? [HINT: it won't]

 

[DaveC426913]

And do you think your accounting will change the fact that both see the light coming at them a c but that they see different frequencies? [HINT: it won't]

I suspect that is the very crux of the OP's confusion: how can the observed frequency of a given light beam change without the observed speed of it changing? Without the underpinnings of subject knowledge, this can seem like a paradox.

 

[phinds]

I suspect that is the very crux of the OP's confusion: how can the observed frequency of a given light beam change without the observed speed of it changing? Without the underpinnings of subject knowledge, this can seem like a paradox.

Based on his original post, I think his lack of understanding of what a FOR really is and how it works is also part of his problem.

 

[jnhrtmn]

I suspect that is the very crux of the OP's confusion: how can the observed frequency of a given light beam change without the observed speed of it changing? Without the underpinnings of subject knowledge, this can seem like a paradox.

Use 0.6c for simple numbers. Length cont. is 0.8x, time dilation is 1.25x, Doppler shifts are 0.5x and 2x. To me, these are not mechanisms, which is what I'm looking for in it, a mechanism. I can't relate the frequency change with the other changes yet, still looking at it. I would almost say that length and time transforms just cancel each other's effect. 1.25x0.8=1.0

 

[Ibix]

To me, these are not mechanisms, which is what I'm looking for in it, a mechanism. I

If I slice a sausage perpendicular to its length I get circular slices. If I slice it at an angle I get elliptical slices with the minor axis being the same as the circular slices' diameters and the major axis having been "length dilated".

Length contraction is the equivalent of that in Minkowski geometry. The thing you think of as "the object" is the 3d intersection between its 4d structure and whatever plane in spacetime you call "now". Different frames use different planes so the intersections are different shapes and "objects are length contracted".

What would you regard as "the mechanism" for why sausage slices might be circular or elliptical?

 

[PeterDonis]

To me, these are not mechanisms

Why not?

 

[PeterDonis]

I would almost say that length and time transforms just cancel each other's effect.

Length contraction and time dilation are not "transforms". The "transform" is the Lorentz transformation. Length contraction and time dilation are consequences of the Lorentz transformation, and you cannot consider either one of them in isolation, or even both of them in isolation, since a third consequence of the Lorentz transformation is relativity of simultaneity, and if you do not also consider relativity of simultaneity you will never be able to understand length contraction and time dilation. If you are going to consider any one of those three things, you have to consider all three. Or you could just ignore all three and look at the Lorentz transformations directly, which in the general case is much simpler.

 

[DaveC426913]

I would almost say that length and time transforms just cancel each other's effect. 1.25x0.8=1.0

No. They are the same effect examined two different ways.

There's a scenario where a relativistically-moving spaceship pilot shines a flashlight out his front window as he passes Earth. It demonstrates how length contraction and time dilation are two sides the same coin. But I won't go into it until you're ready for it, although you can always look it up on your own.

 

[Ibix]

I would almost say that length and time transforms just cancel each other's effect. 1.25x0.8=1.0

As others have noted, you would be wrong to say this.

Here is a Minkowski diagram illustrating what's going on. If you haven't come across Minkowski diagrams before, they are basically the same as displacement-versus-time graphs that you probably came across in school physics. We draw the time axis up the page and a position axis across the page, and we record the position of an object at each time and plot a line. The interesting thing about these diagrams in relativity is that we can take them fairly literally as maps of spacetime.

This diagram shows the circumstance I described above; a red observer is moving to the left (so his line gets further to the left as you go up the page) and a blue observer to the right (so the blue line slopes the other way). A ray of light is propagating to the right; I've marked the positions of seven wave crests in orange. Again, these lines slope to the right and, as is conventional, I've picked units so that the speed of light is 1 and hence the gradient of the light rays is 1.

Immediately we can see from the diagram that the blue line takes a lot longer to meet the wave crest after the one it meets at the origin than the red line does, just as I said before.

I said above that we take the diagram fairly literally as a map of spacetime. And just like a geographic map, we can draw more than one set of axes. A glance at the Lorentz transforms will show you that lines of constant $t'$ have slope $1/v$ and lines of constant $x'$ have slope of $v$ in the unprimed frame - so we can add the axes of the primed frame (blue's rest frame) to this diagram:

The primed frame axes are drawn in blue. Because the blue object is at rest at the origin in this frame the time axis is hidden behind it and you can only see the $t'$ numbers, but you can see the $x'$ axis as a fine blue line. You can also see the orange lines crossing the black axis at intervals of one unit and crossing the blue axis at intervals of two units, consistent with $\sqrt{(c+v)/(c-v)}$ for $v=0.6c$.

The absolutely critical factor that you must understand is that the blue axis is not parallel to the black one. That means that "all of space at one time" means a completely different thing to the two frames - some parts of what the blue frame calls "now" are in the past of any given moment according to the black one and some parts in the future, and vice versa.

Failure to comprehend this fact, that "all of space at one time" for each frame is a completely different and non-parallel "slice" of spacetime, is the single biggest cause of failure to understand relativity.

We can add a few things to the diagram to illustrate time dilation and length contraction:

I've dropped a line from $t'=2$ on the blue axis to the black axis. It strikes at $2\gamma=2.5$. The sloped blue dashed line is the path of an object moving at the same speed as the blue observer but some distance ahead of him. You can see that it passes through the blue axis' $x'=2$ position, so the blue frame would say it is 2 units ahead of the blue observer, and it passes through the black frame's $x=1.6$ position, so the black frame says it is only 1.6 units ahead of the blue observer. This is length contraction.

Note that there's no "mechanism" as such. It's just that the two frames are measuring different parts of spacetime and calling them both "length".

Finally, we can draw the same diagram using the blue frame's axes as the Cartesian ones:

Now the black frame axes aren't represented as orthogonal (they are still orthogonal in the Minkowski geometry sense, but that cannot be represented wholly accurately on a Euclidean plane) but the blue ones are. The blue dashed line is clearly at rest at $x'=2$, and you can see that the wave crests are separated by 2 units - the wave is redshifted.

But the key point is that nothing has happened to the wave. Just like length contraction and time dilation, the redshift is entirely a product of the blue frame using axes with a different slant to the ones the black frame uses. The measured frequency isn't a property of the wave - it's a property of the interaction between the wave and the frame we've chosen to analyse it in.

 

[Dale]

To me, these are not mechanisms, which is what I'm looking for in it, a mechanism.

The proper distance between the emitter and receiver are changing. What more mechanism do you need?

 

[jnhrtmn]

As others have noted, you would be wrong to say this.

Here is a Minkowski diagram illustrating what's going on. If you haven't come across Minkowski diagrams before, they are basically the same as displacement-versus-time graphs that you probably came across in school physics. We draw the time axis up the page and a position axis across the page, and we record the position of an object at each time and plot a line. The interesting thing about these diagrams in relativity is that we can take them fairly literally as maps of spacetime.

This diagram shows the circumstance I described above; a red observer is moving to the left (so his line gets further to the left as you go up the page) and a blue observer to the right (so the blue line slopes the other way). A ray of light is propagating to the right; I've marked the positions of seven wave crests in orange. Again, these lines slope to the right and, as is conventional, I've picked units so that the speed of light is 1 and hence the gradient of the light rays is 1.
View attachment 348746

I appreciate this greatly. I have to work some things out myself. Everything here deals with the consequences of light-constant-per-observer with time being a real physical thing and not just in your head as a function of thought. If time exists, I can see it. Thanks

 

[Dale]

I appreciate this greatly. I have to work some things out myself. Everything here deals with the consequences of light-constant-per-observer with time being a real physical thing and not just in your head as a function of thought. If time exists, I can see it. Thanks

The fact that relativity is a real-physical thing is conclusively established by one of the largest and most precise and varied bodies of experimental evidence produced by science. See this brief summary: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[Nugatory]

with time being a real physical thing

"real physical thing" is not a very precisely defined term - what sort of experiment could possibly distinguish between real physical things and not so real physical things?

The best way to think about time in physics is from Einstein: “Time is what a clock measures”. That sounds a bit cryptic at first, but it will make sense when you compare it with “Distance is what a ruler measures”. These are definitions of time and distance, not of clocks and rulers.

 

[jnhrtmn]

The best way to think about time in physics is from Einstein: “Time is what a clock measures”.

Clocks are always some form of oscillator. Math is using time as a sequence marker to describe the oscillation or periodic effect. Send a pendulum into space and the clock stops measuring what it was measuring. Remember and anticipate are in the brain. How can I see time as more than numbers on a page?

 

[Ibix]

How can I see time as more than numbers on a page?

The point of Einstein's "time is what clocks measure" is to suggest that you stop philosophising and follow the evidence. Clocks are apparently measuring something because we get repeatable, predictable and consistent measurement from them.

So go back and look at the last diagram I drew. Notice how the spatial axis of the original frame isn't horizontal? That means that a ruler at rest in that frame is not measuring distance in what this frame calls space, but some mixture of space and time. Likewise, the time axis is not vertical, so clocks at rest in that frame do not measure just what this frame calls time, but some mix of space and time. The point is that it's a mistake to think of "time" as a thing in relativity. There is only spacetime. You may split it into space and time, and it's often convenient to do so, but it's an arbitrary distinction that you choose to make.

That flexibility means that there are actually several distinct concepts in relativity that more or less correspond to the Newtonian notion of time. "Proper time" (proper in the Latin sense where "property" comes from) is what clocks measure. There is a concept in spacetime closely analogous to "distance" in space, and proper time is a measure of that analogue of distance that the clock has travelled. As with ordinary distance, the distance along different paths can be different - this is the resolution of the so-called Twin Paradox, where two clocks take paths of different "length" through spacetime and have different readings at the end.

But proper time is personal - two clocks may start at the same place synchronised, separate and return, and be showing different times (not due to mechanical malfunction) when they meet again because they took routes of different "length". It is possible to define a "global" time, which we call coordinate time by having a flock of clocks at rest with respect to each other and synchronised. But it turns out that the synchronisation process is arbitrary, so that global time comes at the cost of being something you can choose differently.

Fundamentally, spacetime is a 4d manifold. Its equivalent of Pythagoras' Theorem is $\Delta s^2=\Delta x^2+\Delta y^2+\Delta z^2-(c\Delta t)^2$, and that one term having an opposite sign is why one "direction" is different from all the others. You can't freely reverse that direction and you can't freely exchange it with the others, and that is why time is different from space.

 

[Sagittarius A-Star]

Send a pendulum into space and the clock stops measuring what it was measuring.

A "pendulum clock" is not a clock. The system of "pendulum clock" and a gravitational field is a clock, according to your own definition:

Clocks are always some form of oscillator.

 

[Nugatory]

Send a pendulum into space and the clock stops measuring what it was measuring

And when we send a wooden ruler into a sufficiently hot fire the ruler stops measuring what it was measuring.

Whether we’re measuring time or distance, we need measuring devices that won’t fail while we’re measuring. That tells us something about how we should design our measuring devices, but has nothing to do with the definitions of time as what a clock measures or distance as what a ruler measures.

 

[cianfa72]

But proper time is personal - two clocks may start at the same place synchronized, separate and return, and be showing different times (not due to mechanical malfunction) when they meet again because they took routes of different "length". It is possible to define a "global" time, which we call coordinate time by having a flock of clocks at rest with respect to each other and synchronized. But it turns out that the synchronization process is arbitrary, so that global time comes at the cost of being something you can choose differently.

Ok, therefore the flocks of clocks entirely filling the space you were talking about, are actually "at rest w.r.t. each other" as measured for instance by constant round-trip travel time of bouncing light/radar pulses exchanged between them. Otherwise which could be a different physical criterion to define "at rest w.r.t. each other" ?

 

[Dale]

Clocks are always some form of oscillator. Math is using time as a sequence marker to describe the oscillation or periodic effect. ... How can I see time as more than numbers on a page?

Is the oscillator a number on a page? The math describes the oscillator, but the oscillator is not numbers on a page.

 

[Dale]

The point of Einstein's "time is what clocks measure" is to suggest that you stop philosophising and follow the evidence. Clocks are apparently measuring something because we get repeatable, predictable and consistent measurement from them.

Personally, I think that it is even more than that. The thing that clocks measure is a useful quantity. It deserves a name. That name is "time". If we didn't call it "time" then we would have to call it something else, and as a practical matter that would be the word that we use for talking about when things happen. Time.

 

[cianfa72]

The math describes the oscillator, but the oscillator is not numbers on a page.

You mean that the math provides "a map" that maps what clocks read into a real number.

 

[Nugatory]

Ok, therefore the flocks of clocks entirely filling the space you were talking about, are actually "at rest w.r.t. each other" as measured for instance by constant round-trip travel time of bouncing light/radar pulses exchanged between them. Otherwise which could be a different physical criterion to define "at rest w.r.t. each other" ?

This is drifting into a thread hijack, but be aware that @Ibix is just providing an abridged description of Taylor and Wheeler's ("Spacetime Physics") model of an inertial frame with associated simultaneity convention.

To avoid further drift, you might want to review that section of Taylor and Wheeler and then if you have followup questions, start a new thread.

 

[jnhrtmn]

Back to Doppler, it seems overcomplicated. Using 0.6c added to c gives 1.6c, (contracted length) / (speed multiplier), 0.8/1.6=0.5. The other is 0.8/0.4=2 This works throughout. Is it because1.6c can't be real, but even sqrt 1.6 is still 1.265. Using 1 instead of contracted length is the classical version. I don't understand the use of source and observer in the math if the relative velocity between the frames is the only factor.

 

[Sagittarius A-Star]

Back to Doppler, it seems overcomplicated. Using 0.6c added to c gives 1.6c, (contracted length) / (speed multiplier), 0.8/1.6=0.5. The other is 0.8/0.4=2 This works throughout.

Yes.
$\sqrt{1+v/c \over 1-v/c} = \sqrt{(1+v/c)(1-v/c) \over (1-v/c)(1-v/c)} = {1 \over \gamma (1-v/c)} [= \gamma (1+v/c)]$

Usually, the $\gamma$ factor in the relativistic Doppler formula is explained via time-dilation, but because of $c=\lambda f$ also an argumentation via length contraction is possible.

Is it because1.6c can't be real

The closing speed between 2 objects can be greater than $c$. Only the speeds of each object relativ to the choosen inertial reference frame cannot be greater than $c$.

I don't understand the use of source and observer in the math if the relative velocity between the frames is the only factor.

See
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

 

[Dale]

it seems overcomplicated.

Sometime things are complicated. Nature has no obligation to be convenient or simple.

The theoretically predicted and experimentally validated formula is derived here: https://www.mathpages.com/home/kmath587/kmath587.htm

 

[Ibix]

I don't understand the use of source and observer in the math if the relative velocity between the frames is the only factor.

You need to be careful here. If you and I stand stationary near each other, we share a rest frame. I throw a ball to you. The ball has a different rest frame with a single velocity with respect to our frame. But to you, light from the ball is blue shifted while it is red shifted to me. So the $v$ in the Doppler shift formula cannot be a frame velocity since we couldn't get different Doppler shifts if it were. And we could get opposite Doppler shifts by reversing our $x$ axes so the velocity switched from $+v$ to $-v$.

The $v$ in the Doppler shift formula is the closing rate (or separation rate, depending on your sign convention) of the source with respect to a specific observer, not a frame velocity. It will typically be equal to plus-or-minus the velocity of the source's rest frame in the observer's frame, but it is a distinct concept.