Implications of Einstein's Theories

[Drakkith]

I can see measuring the speed of light as a straightforward task, but my question is, how would they be able to tell that the observer will always experience the 'object' moving away at the speed of light? How do they know it's not the same as our everyday experience of (object A's velocity) - (object B's velocity)? Just because light always moves at c doesn't mean that an observer will always see it moving at c. Just like if someone were obsessed with driving 40 mph all the time, any observer standing still would observe him moving at 40 mph but if they were going 30 mph he would appear to be going 10 mph. What experiment did they do to conclude that light always appears to move at the speed of light regardless of the observer's velocity?

Plenty of examples in this article: http://en.wikipedia.org/wiki/Modern_searches_for_Lorentz_violation

I was under the impression that in the twin paradox there is no difference in aging which would imply their clocks manged to sync up again by the time the cosmonauts returned. I've heard this has something to do with the acceleration when the rocket turns around. Sorry for the vagueness.

That is incorrect. The returning twin will have aged less than the twin who stayed on Earth.

 

[JimiJams]

Drakkith, all those experiments came long after Einstein's theory. Were they able to prove that photons always appear to move at the speed of light regardless of the observer's velocity before Einstein began work on his theory? If not how did they come to that conclusion before experimentation?

 

[Simon Bridge]

Hey Simon, yes I fully understand that the length contraction and time dilation is only observed by the observer. My wording may sometimes be unclear to some,

It is really really important to avoid incautious wording when you are learning SR. There are established ways to talk about it that you should learn so people will understand you.

i.e. in this bit:

but I understand there are different experiences for both the observer and the observed, and the observer sees length contraction and his watch shows a slower time if it were at all possible to compare to the observed.

... it sounds like you just said that the observer's watch was slow. This is not the case - it is everybody else's (everybody with a non-zero relative velocity) watch that is slow.

I can see measuring the speed of light as a straightforward task, but my question is, how would they be able to tell that the observer will always experience the 'object' moving away at the speed of light? How do they know it's not the same as our everyday experience of (object A's velocity) - (object B's velocity)? Just because light always moves at c doesn't mean that an observer will always see it moving at c.

Measuring the speed of light is how you observe it moving at a particular speed. In this case the word "observe" is used in the sense of detecting or measuring something.

"Observing" light in any other way doesn't make sense.

Note: we cannot prove that all observers will always measure the same speed for light.
Nobody can - it is only possible to disprove it.

Just like if someone were obsessed with driving 40 mph all the time, any observer standing still would observe him moving at 40 mph but if they were going 30 mph he would appear to be going 10 mph. What experiment did they do to conclude that light always appears to move at the speed of light regardless of the observer's velocity?

The FAQ I liked to has many such experiments in it - go look.

I was under the impression that in the twin paradox there is no difference in aging which would imply their clocks manged to sync up again by the time the cosmonauts returned. I've heard this has something to do with the acceleration when the rocket turns around.

That impression is incorrect.
The earlier relativity and FTL FAQ I linked to has a detailed description of the Twin's Paradox.
tldr: the traveling twin returns younger.

the "paradox" of the title is that, from the POV of the traveling twin, it is his brother on Earth who is moving. Therefore, he argues, it is the Earthbound twin who should stay younger. What's wrong with the argument?

You are asking for a ot of information that requires a lot of writing. That is why you are directed to read some references ... which makes for a lot of reading for you but that cannot be helped.
The best way for you to get answers right now is to go read those references - I've tried to make them as accessible as I can while still answering the questions.

The relativity & FTL FAQ is very accessible and introduces you to the main tools you need to understand SR as well as a discussion about FTL that should be a nice break. The evidence FAQ itself is easy-ish to read, but the papers linked to are not so much which is tough. But at least you won't have to take my word for it that many experiments have been done.

Here's those links again:
http://www.physicsguy.com/ftl/
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html

 

[Simon Bridge]

Drakkith, all those experiments came long after Einstein's theory. Were they able to prove that photons always appear to move at the speed of light regardless of the observer's velocity before Einstein began work on his theory? If not how did they come to that conclusion before experimentation?

I've already answered these questions.
Section 2 of the evidence faq I have repeatedly tried to get you to read, that I linked to in my last post, includes details of experiments before 1905. Please go read the references.

 

[JimiJams]

Simon, thanks for those links, I usually try to get an answer out of someone else though, but in this case I see I'll need to read more first.

Thanks for clarifying the clock situation haha. So the observer would see the the other's clock moving more slowly than his own. So for Einstein's formulas to make sense of light's constant speed of c regardless of observer's velocity, then the non-observer's time observed by the observer and the observed surrounding space outside the observer's reference need to be used for calculations. And length contraction will take place wherever the observer looks outside his reference frame, not just in the non-observer's reference frame, correct? Sorry for the lack of terminology, this is my last question before I read through those links tonight and perhaps the text again...

 

[Drakkith]

Simon, thanks for those links, I usually try to get an answer out of someone else though, but in this case I see I'll need to read more first.

Thanks for clarifying the clock situation haha. So the observer would see the the other's clock moving more slowly than his own. So for Einstein's formulas to make sense of light's constant speed of c regardless of observer's velocity, then the non-observer's time observed by the observer and the observed surrounding space outside the observer's reference need to be used for calculations. And length contraction will take place wherever the observer looks outside his reference frame, not just in the non-observer's reference frame, correct? Sorry for the lack of terminology, this is my last question before I read through those links tonight and perhaps the text again...

Let's use Observer A and Observer B instead of Observer and Non-Observer.

Observer A and B are in spaceships moving relative to one another. Observer A sees B's clock as moving slower than his own and also sees B as being length contracted. Observer B sees A's clock as moving slower than his own and also sees A as length contracted. So BOTH observers see the other person experiencing time dilation and length contraction but not themselves.

 

[JimiJams]

Does observer A see the space around observer B also length contracted?

 

[Drakkith]

Does observer A see the space around observer B also length contracted?

No, as you cannot see space.

 

[JimiJams]

Suppose space was a visible substance, a contracted B would appear like he was moving slower than the speed of light (if A intuitively knew what the speed of light looked like). But when he sees how slowly B's watch is ticking he is then able to validate that B is moving at the speed of light. So mathematically it seems like there is some type of proportionate relationship between the amount of length contraction and the amount of time dilation. My head hurts and I feel like I keep heading down the wrong path here. The link I went to just didn't clarify these questions of mine enough when it discussed length contraction and time dilation and they might not get cleared up until I get to Lorentz transformations I'm guessing.

 

[Drakkith]

Suppose space was a visible substance,

It is not, therefore anything said about visible space is pure speculation.

a contracted B would appear like he was moving slower than the speed of light (if A intuitively knew what the speed of light looked like). But when he sees how slowly B's watch is ticking he is then able to validate that B is moving at the speed of light.

This is wrong. No observer is ever able to move at the speed of light.

So mathematically it seems like there is some type of proportionate relationship between the amount of length contraction and the amount of time dilation. My head hurts and I feel like I keep heading down the wrong path here. The link I went to just didn't clarify these questions of mine enough when it discussed length contraction and time dilation and they might not get cleared up until I get to Lorentz transformations I'm guessing.

I believe the two are exactly proportional.

 

[JimiJams]

It is not, therefore anything said about visible space is pure speculation.

I'm just being reasonably hypothetical to gain a better understanding. Imagining a visible space won't skew any conclusion we come to from my specific question.

This is wrong. No observer is ever able to move at the speed of light.

Then call it a photon with a watch if it makes you happier.

 

[A.T.]

So mathematically it seems like there is some type of proportionate relationship between the amount of length contraction and the amount of time dilation.

The relationship between the length contraction and time dilation can be visualized in a space-propertime diagram:
http://www.adamtoons.de/physics/relativity.swf

 

[Drakkith]

I'm just being reasonably hypothetical to gain a better understanding. Imagining a visible space won't skew any conclusion we come to from my specific question.

Attempting to imagine space as a visible substance is fundamentally wrong and will only lead to confusion.

Then call it a photon with a watch if it makes you happier.

Photons do not have inertial frames of reference, so we cannot use them as observers. An inertial frame of reference is one in which light will be observed to travel at c. Plugging in c as the velocity of an observer makes our math turn out nonsense, indicating that we cannot do it.

 

[mathskier]

Of course, if you consider two observers, one in frame S and the other in frame S' (moving at .99c relative to S), BOTH observers see the photon move at speed c... But the observer in S will see the person in S' move at nearly the same speed. That is, in S' the distance between the person and the photon after time t is ct (and the same can be said about the distance between the person in S and the photon, as measured in S), while in S the distance between the person in frame S' and the photon is only .01ct. Cool stuff!

As long as you stay in one inertial frame, your ordinary intuitions always work.

 

[Simon Bridge]

Suppose space was a visible substance,

You are persisting in imagining things that are unhelpful - things that you'd be steered away from if you had read the references. This leads us to suspect that you are not serious about learning and only want to contemplate artistic imagery.

For the kind of visualization you are contemplating, you need to pick a reference frame and them find a way to mark out a grid ... say: make the void dusty and have a grid of laser beams for objects to fly through.

a contracted B would appear like he was moving slower than the speed of light (if A intuitively knew what the speed of light looked like). But when he sees how slowly B's watch is ticking he is then able to validate that B is moving at the speed of light.

If B is moving at speed v<c in A's reference frame, then in what reference frame is B moving at c? There is no such frame. The observation of B's watch, by A, only confirms the relative speed of v.

This is what we and all the references are telling you about there being no preferred reference frame but you are still writing as if there is one. Get rid of that idea. It is nonsense. You need to start doing the math.

So mathematically it seems like there is some type of proportionate relationship between the amount of length contraction and the amount of time dilation.

Of curse, the amount of each depends on the relative speed.

You are having trouble with it because you failed to define you terms.
What do you mean by the amount of time dilation" and "the amount of length contraction"?

i.e. Alice and Bob both carry standard meter rulers with them.
When the rulers are at rest wrt to each other, the measure the same.
When Bob is moving at speed v in Alice's frame, Alice notices that Bob's ruler is shorter than her's.

It is shorter by $100(L_A-L_B)/L_A$ percent.

Alice could also use her watch to time how link a single tick (1s) on Bob's watch takes.

If you keep thinking in concrete terms like this you won't get quite so confused.
You should have noticed that the references are all full of those sorts of statements.
One of the things you learn with relativity is just how vague common ways of talking about lengths and time actually are. You have to get really specific about describing your ideas if you want to make sense. It's a pain, but, after a while, you get used to it and you stop needing to be quite so pedantic - but when you are learning it is essential.

 

[JimiJams]

I just wrote out an entire response, and just now deleted it. I would have to discuss this face to face with someone because I feel I'm failing to adequately state my question through this medium. I also think my hypothetical scenarios are being taken too literally. You need to imagine things when working with anything in science, just as most scientists have a clear image of what they believe may be taking place and then attempt to validate it through the mathematics. The spacetime grid is a great example of a visual aid that helps to understand a theory better. That's all I'm doing by saying visualize a measurable space. Thanks for any help with this, and I'm sorry if I failed to soak in anyone's attempt at clarifying things, as always I learn best through a book so I'll revisit my text regarding SR as it has been a while.

 

[Drakkith]

I just wrote out an entire response, and just now deleted it. I would have to discuss this face to face with someone because I feel I'm failing to adequately state my question through this medium. I also think my hypothetical scenarios are being taken too literally. You need to imagine things when working with anything in science, just as most scientists have a clear image of what they believe may be taking place and then attempt to validate it through the mathematics. The spacetime grid is a great example of a visual aid that helps to understand a theory better. That's all I'm doing by saying visualize a measurable space. Thanks for any help with this, and I'm sorry if I failed to soak in anyone's attempt at clarifying things, as always I learn best through a book so I'll revisit my text regarding SR as it has been a while.

You misunderstand. There is no reason to even visualize a spacetime grid because in SR spacetime never changes. So visualizing a spacetime grid and asking what happens to it is pointless, as the answer is nothing. The motions of objects has no effect on spacetime.

 

[TumblingDice]

That's all I'm doing by saying visualize a measurable space.

And everyone's trying to guide you away from "heading down the wrong path" as you've said yourself.

Can we just visualize a room or the old classic train car? Go ahead and give them the dimensions you want in their own inertial rest frames. You can even draw lines on the walls to define a ruler framework. We can do all of this to define a physical object with dimensions.

Just don't start visualizing space, because it's only math and geometry.

 

[Simon Bridge]

I would have to discuss this face to face with someone because I feel I'm failing to adequately state my question through this medium.

That may help - but they would still give you the same notes we have.

I also think my hypothetical scenarios are being taken too literally.

You should not be talking metaphorically about this stuff while you are still wrestling with the concepts.
You absolutely must must must be careful to state exactly what you mean.
It is difficult enough without using artistic language.

You need to imagine things when working with anything in science, just as most scientists have a clear image of what they believe may be taking place and then attempt to validate it through the mathematics. The spacetime grid is a great example of a visual aid that helps to understand a theory better. That's all I'm doing by saying visualize a measurable space. Thanks for any help with this, and I'm sorry if I failed to soak in anyone's attempt at clarifying things, as always I learn best through a book so I'll revisit my text regarding SR as it has been a while.

As already observed, movement through space does not affect space, in SR.
There is no movement in space-time.

But space-time grids are useful - scientists do use them - they are the lines in a space-time diagram that you find in the references we have given you and demonstrated in other posts.

You can also set up a grid of space points in some reference frame and then see what that looks like to observers moving wrt the grid.

But you have to set these things up carefully otherwise you will end up talking nonsense.

If you get nothing else from this discussion - take away the idea that you absolutely must use clear terms and clear language when talking about relativity. Intuitive concepts will just not work.

 

[JimiJams]

Drakkith, to put it simply and briefly, length contraction intuitively to me seems to conflict with calculations that show an observed object always moving at c regardless of observer's speed. Because the object would appear to contract or get smaller, so if the space (measurable space, imagine a uniform grid) it's traveling through does not contract it would appear to cover less distance in a given time.

I understand time dilation steps in and corrects this, but it would make more sense to me if the object appeared to grow larger as the observer's speed increased so it would appear to be covering more distance in a given time.

This is all I'll say for now because every time I try to expound on this I find it very difficult to get across on here and end up deleting it. What I typed above is purely from an intuitive standpoint. If length and time are compromised for something to always appear to be moving at c regardless of an outside observer's motion it would make more sense to me intuitively that time would slow and length would elongate. If something's traveling at c and the observrer at .9c then the object should appear to move at .1c. Length and time appear to alter in order for it to still appear to be moving at c, but you would think to make up the difference of .9c (mathematically) that length would not contract but elongate. But it contracts, so time dilation must really make up the bulk of that difference, mathematically.

I hope this is clear, and I'm sure it's wrought with errors, but without getting into mathematical detail this is the best I can do to describe what I would INTUITIVELY think would take place.

 

[TumblingDice]

JimiJams: You've mentioned objects moving at c several times now. For the purposes of this discussion, the only thing that travels at c is light. Nothing with mass ever travels at c.

Can we visualize a room or train car as I suggested in my post #53? The thought experiments you want to do right now should always involve rulers and/or clocks, depending on what you want to focus on in each 'experiment'. Train cars or rooms can be used if you need to establish a framework for your inertial reference frames (observers).

 

[TumblingDice]

If length and time are compromised for something to always appear to be moving at c regardless of an outside observer's motion it would make more sense to me intuitively that time would slow and length would elongate.

You have the slower clock correct, because that allows more 'time' for light to travel at c relative to this observer. Making objects longer however would work against the slower clock, making light have to travel further. It needs to travel 'less' to maintain the locally observed speed of c.

Edit: What I wrote is a bit over-simplified. I hope it paints time dilation and length contraction a bit more intuitively to help you move forward.

 

[JimiJams]

Tumbling, you just clarified it for me. Measuring light inside a reference frame from another reference frame, it certainly would make sense that length contracts along with time dilation.

I misinterpreted what was meant by, "light always travels at c regardless of the observer's velocity."

So to go back to how I kept mistakenly visualizing this, if I was following behind a photon at .9c then it would actually appear to me to be moving away from me at .1c, correct? This seems to be unrelated to relativity because this example does not involve reference frames or any observers.

 

[Drakkith]

So to go back to how I kept mistakenly visualizing this, if I was following behind a photon at .9c then it would actually appear to me to be moving away from me at .1c, correct? This seems to be unrelated to relativity because this example does not involve reference frames or any observers.

If you are moving at 0.9c with respect to a stationary frame and traveling parallel to a light ray, then to the observer in the stationary frame the distance between you and the light will increase at 0.1c. However, in your frame the light moves away at 1.0 c.

 

[TumblingDice]

if I was following behind a photon at .9c then it would actually appear to me to be moving away from me at .1c, correct?

As Drakkith pointed out, light *always* is measured at the speed of c, to all observers.

This seems to be unrelated to relativity because this example does not involve reference frames or any observers.

Ah! Notice that Drakkith also mentioned you moving at .9c relative to a stationary reference frame. (Perhaps an observer on Earth, but could be anywhere relative to your .9c speed. You want to get into the habit of, whenever you mention moving at such-and-such speed, to always add, "relative to X". And in most cases you'll have at least two 'observers' moving relative to each other. Either one can validly take the perspective that their frame is 'at rest'.

 

[jtbell]

So to go back to how I kept mistakenly visualizing this, if I was following behind a photon at .9c then it would actually appear to me to be moving away from me at .1c, correct? This seems to be unrelated to relativity because this example does not involve reference frames or any observers.

To add onto Drakkith's answer, have you seen the "relativistic velocity addition" equation yet?

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html

Try the different examples, with the "projectile" being a photon, then see if you can set up the situation you describe above. You should find that observers A and B always both find the photon's velocity to be c.

 

[JimiJams]

I was trying to create a scenario where there is no other observer than myself. Imagine I'm on an empty train and I'm trailing behind a photon at .9c. You're saying that photon, in this situation where I am the only observer, will still appear to move away at c? If that's the case then there must be two relative reference frames for the math to work, how would we define them?

 

[jtbell]

I'm trailing behind a photon at .9c.

This implies an external observer, or more precisely, another reference frame besides the one in which you are at rest. You are moving at .9c with respect to whom or what?

With respect to yourself, you are stationary.

 

[PAllen]

I was trying to create a scenario where there is no other observer than myself. Imagine I'm on an empty train and I'm trailing behind a photon at .9c. You're saying that photon, in this situation where I am the only observer, will still appear to move away at c? If that's the case then there must be two relative reference frames for the math to work, how would we define them?

Traveling at .9c relative to what? It can't be relative to a photon, because a photon moves at c relative to all trains.

 

[JimiJams]

Let's say me and the photon are both moving above a stationary surface. I'm not in any flying object, my body is just moving. I'm traveling .9c relative to the stationary ground, and the photon is moving c relative to the ground. There are checkmarks on the ground indicating a certain distance, and I had two stopwatches. Every time the photon passes a checkmark I hit the button to record at what point past t=0 the photon passes that checkmark. I also hit my other stopwatch every time I pass a check mark.

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

If this is not the case, then where would I assign the two reference frames?

 

[WannabeNewton]

No because you're using the Galilean conception of relative velocity. In Galilean relativity, if a particle $O$ has velocity $\vec{v}$ relative to an inertial frame and another particle $O'$ has velocity $\vec{v}'$ relative to the same inertial frame then $O'$ has a velocity $\vec{v}' - \vec{v}$ relative to $O$. This is the formula you're trying to apply in your scenario but you cannot do this because Galilean addition of velocities is only accurate to 1st order in $\frac{v^2}{c^2}$ when $v << c$.

When you have $v\sim c$ you need to use the special relativistic velocity addition formula: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel.html

 

[Drakkith]

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

No, as you aren't measuring the photons velocity relative to yourself, but to the ground. You would be measuring the rate that the distance between yourself and the photon increases as seen from the stationary ground.

Imagine you had a very long set of distance markers stretched out in front of you, with 1 meter between each marker. An observer on the ground also has markers set up, with 1 meters in between each of them. You are moving at 0.866c relative to the observer on the ground. In one second, as measured by you, the photon will pass 300 million of your markers. To the observer on the ground, the photon passes 300 million of his markers per second according to his watch.

 

[ghwellsjr]

I was trying to create a scenario where there is no other observer than myself. Imagine I'm on an empty train and I'm trailing behind a photon at .9c. You're saying that photon, in this situation where I am the only observer, will still appear to move away at c? If that's the case then there must be two relative reference frames for the math to work, how would we define them?

Here's what I think you want. I'm going to show you stationary in your own rest frame and you emit a photon. I'm defining light to travel at 1000 feet per microsecond (usec). You are shown as the thick black line with dots every microsecond. The photon is defined (not measured, not observed, not seen) to travel at 1000 feet per microsecond in any Inertial Reference Frame (IRF) and is shown as the thin black line. After 5 microseconds we (not you) see that the photon is 5000 feet in front of you:

Next we transform (using the Lorentz Transformation process as I described in post #25) this scenario to one in which you are traveling at 0.9c (by using β=-0.9c), chasing the photon:

Now we (not you) see that the photon is just 1000 feet in front of you after 10 microseconds.

Do you understand why I repeatedly said that we and not you can see the propagation of the photon?

[NOTE: I made a mistake in post #25 where I said that "it takes 3 nanoseconds for each flash to make the round trip". I should have said "it takes 6 nanoseconds".]

 

[ghwellsjr]

Let's say me and the photon are both moving above a stationary surface. I'm not in any flying object, my body is just moving. I'm traveling .9c relative to the stationary ground, and the photon is moving c relative to the ground. There are checkmarks on the ground indicating a certain distance, and I had two stopwatches. Every time the photon passes a checkmark I hit the button to record at what point past t=0 the photon passes that checkmark. I also hit my other stopwatch every time I pass a checkmark.

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

If this is not the case, then where would I assign the two reference frames?

First, you should realize, as I have repeatedly said, you cannot see the photons. You cannot tell when they hit the checkmarks. All you can do is wait for the light to reflect back from the checkmarks and then you can measure the roundtrip time it took for the light to leave you (back at time zero) propagate to a checkmark and reflect back to you. You have no idea where during that roundtrip it hit the checkmark.

But what you do if you want to follow the precepts of Special Relativity, is you assume that the light took exactly the same amount of time to propagate to the checkmark as it took for the reflected light to propagate back to you. And you also assume that the light propagated at c in both directions. This allows you to construct your own reference frame in which you are at rest.

Let's see how this works starting with the second diagram that I drew from my previous post except this time I've added in two checkmarks every 5000 feet. I've shown the light that reflects back to you and you can determine when your stopwatches would measure both the roundtrip light signals and when you pass the checkmarks:

It looks to me like you see the light from the first checkmark at your time of 2.3 usecs and the second one at 4.6 usecs. You divide these times in half to assign the times that light reflected off the checkmarks and you calculate how far light travels in that time to determine how far away they were from you at that time. So you would determine that the first checkmark was 1150 feet away from you at your time of 1.15 usecs and the second checkmark was at 2300 feet away at your time of 2.3 usecs.

It also looks to me like you passed the first checkmark at your time of 2.4 usecs and the second one at 4.8 usecs.

Now you have all the information you need to calculate the speeds of the checkmarks relative to you.

For the first checkmark, you determined that at your time of 1.15 usecs, it was 1150 feet away and then at 2.4 usecs, it was 0 feet away. Therefore, it traveled 1150 feet in (2.4-1.15) = 1.25 usecs for a speed of 1150/1.25 or 920 feet per usec. The actual speed is 900 feet per usec, so this is pretty good for eyeballing.

For the second checkmark, you determined that at your time of 2.3 usecs, it was 2300 feet away and then at 4.8 usecs, it was 0 feet away. Therefore, it traveled 2300 feet in (4.8-2.3) = 2.5 usecs for a speed of 2300/2.5 or 920 feet per usec, same as for the first checkmark.

Finally, for the speed of the photons, as I said before, you simply assume that they travel at c. That's what you did to determine the speed of the checkmarks. There is no way that you can actually measure their speed apart from assuming their speed to begin with.

So here is the frame that you construct from the measurements you took and the assumption you made about the speed of light and the calculations that you made:

This is exactly the same diagram that you get by transforming the above frame to a speed of 0.9c (in fact, that's just how I got it).

 

[Simon Bridge]

Let's say me and the photon are both moving above a stationary surface. I'm not in any flying object, my body is just moving. I'm traveling .9c relative to the stationary ground, and the photon is moving c relative to the ground. There are checkmarks on the ground indicating a certain distance, and I had two stopwatches. Every time the photon passes a checkmark I hit the button to record at what point past t=0 the photon passes that checkmark. I also hit my other stopwatch every time I pass a checkmark.

You have to say how you know where the photon is.

Lets modify this experiment a bit.
Instead of following a single photon, you are following a well-located pulse of light.
The pulse will spread out as it travels so we'll keep the experiment short enough in duration that this is not a problem.

We will define two reference frames -
A: the rest-frame of the ground. Observer - "Alice"
B: the rest-frame of you, "Bob", which has relative velocity of v wrt tA "Alice" and the ground.

The light travels at speed c in both frames.

At regular (say: 1 light-second, as measured by Alice) intervals there are light detectors stationary wrt Alice. When the pulse maximum intensity is detected, they emit a flash of light. In this way Bob and Alice can track the progress of the same pulse as it speeds away from them.

This should avoid a bunch of objections.
There are still a few technical considerations, but they are mainly engineering problems.

At the end wouldn't I be able to calculate my speed at .9c and the photon's at c, showing that the photon is really only moving .1c relative to me?

You will always calculate your own speed as zero.
This is where you have to be careful in your language - when you want to work out a speed, you have to say who is doing the observing: what is the speed relative to? If you don't say, then the convention is that it is relative to yourself (or the last observer). Your speed relative to yourself is always zero.

You could work out that Alice would calculate that you were doing 0.9c and that the light pulse is moving at 0.1c relative to you. That is correct.
But so what?

There is a habit you get into with Newtonian mechanics, where you pretend that something is "really" stationary and everything else moves. We end up thinking that there really is such a thing as absolute rest. In Galilean relativity that idea looks a bit shaky but Special relativity kills it stone dead.

If you want to say that you are "really" traveling at 0.9c, then you must be saying that Alice's POV is somehow special. But why pick hers? Why not someone elses? The choice is arbitrary.


Take a more extreme example:
If you and I were to head at 0.9c to A, in opposite directions, then Alice would calculate our relative velocity to each other as 1.8c. Either of us could calculate that Alice would calculate that but, again, so what?
What doe that mean to either of us?